Nonequilibrium Thermodynamics, 2nd Edition Transport and Rate Processes in Physical, Chemical and Biological Systems
by Yasar Demirel
• ISBN: 0444530797 • Publisher: Elsevier Science & Technology Books • Pub. Date: September 2007
PREFACE* Natural phenomena consist of simultaneously occurring transport processes and chemical reactions. These processes may interact with each other and lead to instabilities, fluctuations, and evolutionary systems. The objective of this book is to explore the unifying role of thermodynamics in natural phenomena. Nonequilibrium thermodynamics is based on the entropy production character of nonequilibrium and irreversible processes and provides a link between classical thermodynamics and transport and rate processes. In 1850, Clausius introduced the concept of noncompensated heat as a measure of irreversibility. In 1911, Jaumann formulated the rate of entropy production as the product of flows and thermodynamic forces reflecting the second law of thermodynamics, stated as, "A finite amount of organization may be purchased at the expense of a greater amount of disorganization in a series of interrelated spontaneous actions." Every process in nonequilibrium conditions operates with thermodynamic forces, such as pressure, temperature, or chemical potential gradients, which cause flows. After the publication of Onsager's reciprocal relations in 1931, Casimir, Meixner, Prigogine, and De Groot made early attempts to formulate a general theory for irreversible processes. Nonequilibrium thermodynamics mainly formulates the rate of entropy production and the relations between the conjugate flows and the forces in a system. The rate of entropy production can be derived using the general balance equations of mass, momentum, energy, entropy, and the Gibbs relation based on local equilibrium. The rate of entropy production equation identifies the independent forces and flows that are related by phenomenological equations containing the proportionality constants called phenomenological coefficients. A matrix of these coefficients is symmetric, according to Onsager's reciprocal relations. The theory treating near global equilibrium phenomena is called linear nonequilibrium thermodynamics, and phenomenological equations linearly relate forces and flows. This edition updates and expands most of the chapters of the first edition by focusing on the balance equations of mass, momentum, energy, and entropy together with the Gibbs equation for coupled processes of physical, chemical, and biological systems. Every chapter contains example problems and practice problems to be solved. Chapter 1 briefly describes basic elements of classical thermodynamics, such as irreversibility, equilibrium state, thermodynamic principles, the Gibbs equation, and the phase equilibria. Chapter 2 briefly introduces transport processes and chemical reactions. Here, nonequilibrium systems (with thermodynamic branch), transport coefficients, and some well-known examples of coupled phenomena are briefly introduced. Chapter 3 discusses the general balance equations, which are used in the Gibbs equation to derive the rate of entropy production based on local equilibrium. Chapter 4 focuses on thermodynamic analysis of transport processes, chemical reactions, and power-generating systems. It also introduces the concept of exergy, the equipartition principle, and pinch analysis with various example problems to underline the contribution of thermodynamic analysis toward creating optimum designs and assessing the performance of existing designs. Chapter 5 introduces the thermodynamic optimum approach in the design and optimization of various processes. Here, the concept of thermoeconomics is emphasized. Under proper financial, normative, environmental, and technical constraints, thermoeconomics can minimize the overall cost by identifying the thermodynamically optimum design and operation. Chapter 6 summarizes the diffusion in nonelectrolyte and electrolyte systems and explores some related applications. Chapter 7 describes coupled heat and mass transfer and the level of coupling without chemical reaction. Chapter 8 briefly summarizes chemical reactions and coupled phenomena. Chapter 9 combines transport processes and chemical reactions; it focuses on the dynamic balance equations consisting of coupled flows as well as the coupling between chemical reactions of scalar character with the heat and mass flows of vectorial character. Chapter 10 briefly describes passive, facilitated, and active transport through membranes. Chapter 11 introduces various applications of thermodynamics in biological systems as well as energy conversion and coupling phenomena in bioenergetics. Based on the nonequilibrium thermodynamic approach, the stability aspects of various transport processes and chemical reactions are covered in Chapter 12. Stability analysis cannot predict the behavior of evolutionary processes. However, it can predict when a system will become unstable. Chapter 13 briefly describes some organized structures maintained with the outside supply of energy and matter. Some biological systems are good examples of maintaining dissipative organized structures. Chapter 14 summarizes some of the other thermodynamic approaches, such as extended nonequilibrium thermodynamics. The appendix supplies some data needed in the example and practice problems. *An instructor resource containing the Solution Manual can be obtained from the author -
[email protected] http ://books.elsevier.com/Demirel/thermodynamics/
or from
xviii
Preface
This book introduces the theory of nonequilibrium thermodynamics and its use in simultaneously occurring transport processes and chemical reactions of physical, chemical, and biological systems. Thus, it provides a unified approach in describing natural phenomena and would be effective in senior and graduate education in chemical, mechanical, system, biomedical, tissue, biological, and biological systems engineering, as well as physical, biophysical, biological, chemical, and biochemical sciences. All through the first and second edition, the work of many people who contributed to both the theory and applications of thermodynamics, transport processes, and chemical reactions has been visited and revisited. I acknowledge and greatly appreciate all these people. I am also thankful for the help I received from Dr. Yelizaveta P. Renfro and the production team of Macmillan India Ltd. for editing and reviewing the chapters. Y. Demirel, 2007
PREFACE TO FIRST EDITION Classical thermodynamics is based on a limited number of natural laws, which have led to a vast number of equations describing macroscopic behavior of various types of systems. However, classical thermodynamics is mainly limited to energy conversion in equilibrium, and particularly applied to reversible and closed systems. Beside the equilibrium, there are instabilities, fluctuations, and evolutionary processes. The objective of this book is to bring out and emphasize the unifying role of thermodynamics in transport phenomena, chemical reactions, and coupled processes in physical and biological systems by using the nonequilibrium thermodynamic approach. The development of nonequilibrium thermodynamics is based on the entropy-generating character of irreversible processes to provide a link between the classical thermodynamics, and the transport and rate processes. In 1850, Clausius introduced the concept of noncompensated heat as a measure of irreversibility. In 1911, Jaumann introduced the concepts of the entropy production and the entropy flux. Donnan and Guggenheim in 1934 related the coupled natural processes to the second law of thermodynamics stated as "A finite amount of organization may be purchased at the expense of a greater amount of disorganization in a series of interrelated spontaneous action." After the publication of Onsager's reciprocal relations in 1931, Casimir, Meixner, Prigogine and De Groot made early attempts of a macroscopic and general theory for irreversible processes. Irreversible processes cause entropy generation because of net thermodynamic forces and flows within the system. Nonequilibrium thermodynamics is mainly concerned with the analysis of entropy generation and the study of the relations between the conjugate flows and the forces. Using the general balance equations of mass, momentum, energy, entropy, and the Gibbs relation, entropy generation or the dissipation function can be derived. The rate of entropy generation or the dissipation function identifies the flows and forces that are related by the phenomenological equations. These equations contain the proportionality constants called the phenomenological coefficients. This coefficient matrix is symmetric according to the Onsager's reciprocal relations. Such an analysis is necessary to understand the complex, coupled transport and the global behavior of physical and biological systems. The theory treating near-equilibrium phenomena is called the linear nonequilibrium thermodynamics. It is based on the local equilibrium assumption in the system and phenomenological equations that linearly relate forces and flows of the processes of interest. Application of classical thermodynamics to nonequilibrium systems is valid for systems not too far from equilibrium. This condition does not prove excessively restrictive as many systems and phenomena can be found within the vicinity of equilibrium. Therefore equations for property changes between equilibrium states, such as the Gibbs relationship, can be utilized to express the entropy generation in nonequilibrium systems in terms of variables that are used in the transport and rate processes. The second law analysis determines the thermodynamic optimality of a physical process by determining the rate of entropy generation due to the irreversible process in the system for a required task. Some processes may have forces operating far away from equilibrium where the linear phenomenological equations are no longer applicable. Such a domain of irreversible phenomena, such as some chemical reactions, periodic oscillations, and bifurcation, is examined by extended nonequilibrium thermodynamics. Extending the methods of thermodynamics to treat the linear and nonlinear phenomena, and such dissipative structures are attracting scientists from various disciplines. This book introduces the theory of nonequilibrium thermodynamics and its use in transport and rate processes of physical and biological systems. The first chapter briefly presents the equilibrium thermodynamics. In the second chapter, the transport and rate processes have been summarized. The rest of the book covers the theory of nonequilibrium thermodynamics, dissipation function, and various applications based on linear nonequilibrium thermodynamics. Extended nonequilibrium thermodynamics is briefly covered. All the parts of the book can be used for senior- and graduate-level teaching in engineering and science. All through this book, the work of many people who contributed to both theory and the applications of nonequilibrium thermodynamics has been visited and revisited. All those whose work has contributed in preparing this book are acknowledged, and greatly appreciated. Y. Demirel, 2002
LIST OF SYMBOLS a
A A* B,C Be Br
C,c
Ct,,Cv CSs D Da Ds,e DD, e DT, e e eX
E Ex
f fi
F g G zXG~ zXI-tr h H I J
4 4 k
kB K
ke L Le Lik Lqr Lsr
Jr m
M n
N
Nr Nu P Pe
activity, interfacial area Helmholtz free energy, chemical affinity, area nondimensional affinity virial coefficients, Bejan number Brinkman number concentration, cost heat capacities at constant-pressure and constant-volume, respectively reactant concentration at surface diffusion coefficient, diameter Damk6hler number effective diffusion coefficient for the substrate S coupling coefficient related to the Dufour effect coupling coefficient related to the Soret effect energy specific exergy activation energy of the chemical reaction, energy exergy fugacity of pure component fugacity of species in a mixture force acceleration of gravity Gibbs free energy Gibbs free energy change on reaction reaction enthalpy enthalpy (specific) enthalpy, Henry's law constant electric current diffusive mass flux conduction heat flux W m 2 reaction velocity thermal conductivity Boltzmann constant chemical equilibrium constant resistance coefficients effective thermal conductivity first-order reaction rate constant characteristic half thickness Lewis number phenomenological coefficients (conductance) coupling coefficient between chemical reaction and heat flow coupling coefficient between chemical reaction and mass flow volumetric reaction rate mass molecular weight number of moles, number of components number of moles number of independent reactions Nusselt number pressure, volumetric rate of entropy production Peclet number
xxii Pr
Pi
er qo q
Q R Re S
S Sc prod
t 7"
rr U V
V W
W
Ws X
X Y Z
Z
List of symbols
Prandtl number partial pressure of species i reduced pressure degree of coupling between processes i and j heat flow vector volumetric flow gas constant Reynolds number specific entropy entropy Schmidt number entropy production time temperature reduced temperature internal energy velocity vector volume mass fraction work shaft work liquid mole fraction, ratio of forces thermodynamic force vapor-phase mole fraction dimensionless distance compressibility factor
Greek letters
~e
#
3' "Yi
S S~j E:
Y q~ KT
/x P
0 A P II O" T 60
eigenvalues effective thermal diffusivity coefficient of thermal expansion, thermicity group specific heat ratio activity coefficient small deviations Kronecker delta function dimensionless coefficient related to Soret effect, extent of a chemical reaction, elasticity coefficient generalized Thiele modulus, dimensionless local entropy generation rate Arrhenius group, dimensionless efficiency, shear viscosity dimensionless temperature isothermal compressibility viscosity, chemical potential stoichiometric coefficient, kinematic viscosity dimensionless concentration controlling parameters density osmotic pressure dissipation function, dissipated available energy dimensionless parameter dimensionless time accentric factor, dimensionless parameter related to Dufour effect
===
List of symbols
Subscripts av az b c
d D e eq EOS f L liq m rain mix opt P prod q r
rev R s
sat S T th V vap
average azeotrope backward, bulk critical, charging discharging Dufour effective equilibrium state equation of state forward, formation liquid liquid melting minimum mixing or mixture optimum packed production heat reaction, reduced property reversible residual properties surface, isentropic property along a two-phase coexistence line Soret thermal diffusivity thermodynamic vapor vapor
Superscripts E R
sat
0 o
excess reduced saturation rate pseudo reference (initial) conditions standard conditions
XXlll
Table of Contents
Chapter 1 - Fundamentals of Equilibrium Thermodynamics Chapter 2 - Transport and Rate Processes Chapter 3- Fundamentals of Nonequilibrium Thermodynamics Chapter 4 - Using the Second Law: Thermodynamic Analysis Chapter 5 - Thermoeconomics Chapter 6 - Diffusion Chapter 7 - Heat and Mass Transfer Chapter 8 - Chemical Reactions Chapter 9 - Coupled Systems of Chemical Reactions And Transport Processes Chapter 10 - Membrane Transport Chapter 11 - Thermodynamics and Biological Systems Chapter 12 - Stability Analysis Chapter 13 - Organized Structures Chapter 14 - Nonequilibrium Thermodynamics Approaches
1 FUNDAMENTALS OF EQUILIBRIUM THERMODYNAMICS 1.1
INTRODUCTION
The name thermodynamics stems from the Greek words therme (heat) and dynamis (power). Thermodynamics is a science of energy in which, temperature as it is related to the average of molecular motion is an important concept. Guggenheim defines thermodynamics as that part of physics concerned with any equilibrium property's dependence on temperature. Thermodynamics also formulates the average changes taking place among large numbers of molecules; therefore, it is a macroscopic science. Thermodynamics first emerged as a science after the construction and operation of steam engines in 1697 by Thomas Savery and in 1712 by Thomas Newcomen in England. Later, Carnot, Rankine, Clausius, Kelvin, Gibbs, and many others developed formulations of thermodynamic principles for describing the conservation and conversion of energy. The theoretical formulation of classical thermodynamics is a set of natural laws governing the behavior of macroscopic systems; these laws lead to a large number of equations and axioms that are exact, based entirely on logic, and attached to well-defined constraints. As natural phenomena are far from reversible, adiabatic, isothermal, equilibrium, or ideal, the engineer must exercise a pragmatic approach in applying the principles of thermodynamics to real systems. Some new texts have attempted to present thermodynamic principles and formulations in ways applicable to students and engineers, which would allow the formulations to be applied in modeling, designing, and describing some natural and complex phenomena. Principles of thermodynamics find applications in all branches of engineering and the sciences. Besides that, thermodynamics may present methods and "generalized correlations" for the estimation of physical and chemical properties when there are no experimental data available. Such estimations are often necessary in the simulation and design of various processes. This chapter briefly covers some of the basic definitions, principles of thermodynamics, entropy production, the Gibbs equation, phase equilibria, equations of state, and thermodynamic potentials.
1.2 1.2.1
BASIC DEFINITIONS Systems
A thermodynamic system is a part of the physical universe with a specified boundary for observation. A system contains a substance with a large amount of molecules or atoms, and is formed by a geometrical volume of macroscopic dimensions subjected to controlled experimental conditions. An ideal thermodynamic system is a model system with simplifications to represent a real system that can be described by the theoretical thermodynamics approach. A simple system is a single state system with no internal boundaries, and is not subject to external force fields or inertial forces. A composite system, however, has at least two simple systems separated by a barrier restrictive to one form of energy or matter. The boundary of the volume separates the system from its surroundings. A system may be taken through a complete cycle of states, in which its final state is the same as its original state. In a closed system, material content is fixed and internal mass changes only due to a chemical reaction. Closed systems exchange energy only in the form of heat or work with their surroundings. In an open system, material and energy content are variable, and the systems freely exchange mass and energy with their surroundings. Isolated systems cannot exchange energy and matter. A system surrounded by an insulating boundary is called a thermally insulated system. A system and its surroundings are considered the universe.
2
1. Fundamentals of equilibrium thermodynamics
The properties of a system based on the behavior of molecules are related to the microscopic state, which is the main concern of statistical thermodynamics. In contrast, classical thermodynamics formulate the macroscopic state, which is related to the average behavior of large groups of molecules leading to the definitions of macroscopic properties such as temperature and pressure.
1.2.2 Processes Energy through its conversion and degradation can cause physical and chemical processes to occur. A process takes place in a system. Any process within an adiabatic system is known as an adiabatic process. A process that takes place with only an infinitesimal change in the macroscopic properties of a system is called an infinitesimalprocess. The classification of processes according to Planck considers three independent infinitesimal processes. They are natural processes, unnatural processes, and reversible processes. Natural processes actually occur and always proceed in a direction toward equilibrium. Unnatural processes are those that proceed in a direction away from equilibrium that never occurs. A reversible process is a case between natural and unnatural processes and proceeds in either direction through a continuous series of equilibrium states. Guggenheim provides the following simple example: consider the evaporation of a liquid at an equilibrium pressure Peq. If P < Peq, a natural evaporation takes place. However, when P > Peq, the evaporation is unnatural. I f P - P e q - 6 , where 6 > 0, evaporation takes place, and in the limit ~ ~ 0 the process becomes reversible. Variables in a system that remain constant with time are in a steady-state process, while those that change with time are in an unsteady-state process or in a transient process. At steady state, a system exchanges energy or matter at a constant rate.
1.2.3 Thermodynamic Properties Thermodynamic properties or coordinates are derived from the statistical averaging of the observable microscopic coordinates of motion. If a thermodynamic property is a state function, its change is independent of the path between the initial and final states, and depends on only the properties of the initial and final states of the system. The infinitesimal change of a state function is an exact differential. Properties like mass m and volume V are defined by the system as a whole. Such properties are additive, and are called extensive properties. Separation of the total change for a species into the external and internal parts may be generalized to any extensive property All extensive properties are homogeneous functions of the first order in the mass of the system. For example, doubling the mass of a system at constant composition doubles the internal energy. The pressure P and temperature T define the values at each point of the system and are therefore called intensive properties, some of which can be expressed as derivatives of extensive properties, such as the temperature T -- ( O U/OS)v,N i. IfXdenotes any extensive property (not necessarily a thermodynamic property) of a phase, we may derive intensive properties denoted by X/and called as partial properties as follows:
, (i 4: j)
(1.1)
For any partial property, we have dX = Ei (OX/Oni)dni = "ElXidl'li at constant T and P. The Euler theorem shows that X = ]~i Yini.
1.2.4 Energy Energy may be transferred in the form of heat or work through the boundary of a system. However, the conversion of work to heat or heat to work should be used with caution, as they are not numerically equal to each other. In a complete cycle of steady-state process, internal energy change is zero and hence the work done on the system is converted to heat ([work[ = Iheatl) by the system. The mechanical work of expansion or compression proceeds with the observable motion of the coordinates of the particles of matter. Chemical work, on the other hand, proceeds with changes in internal energy due to changes in the chemical composition (mass action). Potential energy is the capacity for mechanical work related to the position of a body, while Idnetic energy is the capacity for mechanical work related to the motion of a body. Potential and kinetic energies are external, while sensible heat and latent heat are internal energies.
1.2 Basicdefinitions
3
The conservation o f mass in an open system states that the change in the total mass is equal to the mass exchanged with the surroundings. In a system, we may consider two changes to the mass of species j: the internal change dimj and the external change with the surrounding demi (1.2)
dmj = dem j + dim j
During a process, energy can be transferred and converted from one form to another, while the total energy remains constant. This is known as the conservation o f energy principle.
1.2.5
Entropy
Entropy change is determined by the following equation: dS-
6qrev
T
(1.3)
where 6qrev is the reversible heat flow. When a fluid system changes from state A to state B by an irreversible process, then the change of its entropy is to be AS = SB - SA. Some important properties of entropy are: • • • •
• • • • •
Entropy is a state function and an extensive property. The determination of entropy requires the measured enthalpy and the use of relation T(OS/OT)p = (OH/OT)p = Cp. For a single phase, dS >- q/T, the inequality is for a natural change, while the equality is for a reversible change. The change of entropy is expressed as, dS = deS + diS where deS (des = q/T) is the change due to the interaction of a system with its surroundings, and dis is the increase due to a natural change, such as a chemical reaction, within the system and is always positive for irreversible changes (dis > 0) and zero at equilibrium (diS = 0). The entropy of a system is the sum of the entropies of all changes within the system. The entropy change of ice melting at 273.15 K is A S m - - ~ m / T = 21.99 J/(mol K). The entropy change of water vaporization at 373.15 K is 2~Sv = zXttv/T = 108.95 J/(mol K). The entropy of an insulated closed system remains constant in any reversible change, increases in any natural change, and reaches a maximum at equilibrium. Entropy remains constant for any reversible adiabatic change so that dS = O. For any complete cycle, the change of entropy is zero.
Some research topics in entropy are the determination of entropy changes in the mixing of very similar species (e.g., isotopes), in very cold systems (i.e., T ~ 0), and in highly dispersed systems.
1.2.6
Changes in Enthalpy, Entropy, and Volume in Terms of Temperature and Pressure
Some general property relations in terms of temperature and pressure are
clH :
-~
dr+
-~
dP : Cp d r + V - T --~
P
dS _
T
//
_0_~ S
dV =
dT +
OV
/
dP
(1.4)
T
--~ r dP - C P - - ~ -
dT +
-Of p
dP = ~ Vd T - K VdP
P
(1.6)
T
We can determine some of the partial derivatives above by using the following Maxwell relations, thermal expansion/3, and isothermal compressibility K:
/
0--fi)r - -
/
-~-
p
--/3V
After dividing d H = TdS + VdP by dP at constant temperature and using the Maxwell relations, we have
(1.7)
4
1. Fundamentals of equilibrium thermodynamics Table 1.1 Molar heat capacities for some gas compounds at T = 298.15 K and P = 1 atm
Species
Cp (J/mol)
Cv (J/mol)
y - Cp/Cv
(5/2)R (7/2)R 20.79 37.11 28.82 29.36 29.12
(3/2)R (5/2)R 12.47 28.46 20.44 20.95 20.74
5/3 7/5 1.6672 1.3039 1.4099 1.4014 1.4040
Ideal monoatomic gas Ideal diatomic gas Noble gases CO2 H2 O2 N2
Source: Kondepudi and Prigogine (1999).
= (1- fiT) V
_p(a~p) =(KP-flT) V T
(1.8)
(1.9)
T
P
and
--v
,,
v
(1.10)
For ideal gases, we have dH = Cp dT
(1.11)
dP ds:c --di-T _ R 7-
(1 12)
Table 1.1 shows the heat capacities at constant pressure and volume for some gas compounds at 298.15 K and 1 atm. Tables B3-B5 list the heat capacities of various compounds as function of temperatures. With small changes in T and P, and constant values of/3 and K, integration of Eq. (1.6) yields
In V2= ~(T2-T1)-K(P2-Pi) Vm
(1.13)
Equation (1.13) is a practical representation of a compressible fluid. For liquids, we have dH = Cp d r + (1 - fl T) VdP dT dS = Cp - 7 -
~ VdP
(1.14)
(1.15)
For incompressible fluids, thermal expansion fl and isothermal compressibility K are practically zero, and Eqs. (1.14) and (1.15) become dH = Cp dT + VdP
dS = Cp
dT
T
Table 1.2 shows expansivity and isothermal compressibility for some compounds at 298.15 K at 1 atm.
(1.16)
(1.17)
1.2 Basic definitions
5
Table 1.2 Expansivityand isothermalcompressibilityof some liquids and solids Species
/3 x 104 ( l / K )
Water Benzene Ethanol Tetrachloromethane Mercury Copper Lead Iron
KX
2.1 12.4 11.2 12.4 1.8 0.501 0.861 0.354
106 (1/atm) 49.6 92.1 76.8 90.5 38.7 0.735 2.21 0.597
Soume: Kondepudi and Prigogine (1999).
1.2.7
Change of Internal Energy and Entropy in Terms of Temperature and Volume
Some general property relations in terms of temperature and volume are dU-
OU d r + - ~ ~"
-~
dS-
-~
--~
dr+
r
V
dV
(1.18)
dV
(1.19)
T
From the following Maxwell relations, thermal expansion/3, isothermal compressibility K, and other relations, we have -~,
-T
~
~,
T
v
(1.20)
(1.211
After dividing d U - T d S - PdV by dV at constant temperature and using the Maxwell relations, we have
/, - P : r
7
77
(1.22)
Therefore, Eqs. (1.18) and (1.19) become (1.23)
dT dS-C,.--~+
(OP)dV=C - ~ ,"
dT+~dV --T- K
(1.24)
"~(~ dT dT [3 --2- - ~ VdP = C v ~ + -- d V 1 T K
(1.25)
where
f-
K
From Eqs. (1.15) and (1.24), we have dS
6
1. Fundamentals of equilibrium thermodynamics
Equation (1.25) yields a relation between Cp and Cv for real gases (1.26) For incompressible fluids, Eq. (1.26) reduces to Cp = Cv.
1.2.8
Chemical Potential
For ideal systems, the chemical potential is expressed by [&j = ]j,jo ( T , P ) + R T lnxj
(1.27)
where xj is the mole fraction of species j. The chemical potential can also be defined in terms of the concentration of species j, cj = N / V [&j -- ~j~jo ( T , V ) + R T lncj
(1.28)
For nonideal systems, we use activity in place of concentration, and Eq. (1.27) becomes [Ubj = [lbjo ( T , P ) + R T In ~ j X j
(1.29)
In the presence of external fields, the potential energy is included in the chemical potential. When the external field is an electric field, we get electrochemical potential/29 - = ]&jo ( T , P ) + R T ]&j
In "~jXj + FzjqJ
(1.30)
where F is the Faraday, or electric charge per mole (F = 96,500 C/mol), zj the valence of the species j, and ~ the electric potential.
1.2.9
Absolute Activity
The absolute activity is related to the molar chemical potential by
a=exp( ) The absolute activity is often used in phase equilibrium. For example, the equilibrium condition for the distribution of species i between vapor and liquid phases is a yap = a] iq
1.3
(1.32)
REVERSIBLE AND IRREVERSIBLE PROCESSES
Consider the equations that describe time-dependent physical processes; if these equations are invariant with regard to the algebraic sign of the time, the process is called a reversible process; otherwise it is called an irreversible process. Reversible processes are macroscopic processes that occur in the vicinity of global equilibrium. We can reverse the reversible process at any stage by a slight change in an external parameter. Guggenheim describes the reversible process and reversible change as follows: a process in a system interacting with its surroundings is a reversible process if the system and its surroundings are in equilibrium throughout the process. If equilibrium is not established, the result is a reversible change. For example, if heat flows from one system in equilibrium to another system in equilibrium, then a reversible change occurs; it is not a reversible process unless these two systems are at the same temperature. In a reversible process, it would be possible to perform a second process in at least one way to restore the system and its environment to their respective original states, except in case of differential changes higher than the first order. A reversible process proceeds with infinitesimal driving forces (i.e., gradients) within the system. Hence, for a linear transport system, reversible change occurs slowly on the scale of macroscopic relaxation
1.3
Reversible and irreversible processes
7
times, and dissipative effects cannot be present. Time appears only through its arithmetic value in the equations for reversible processes. For example, the equation describing the propagation of waves in a nonabsorbing medium is 1
02bl --
(1.3 3)
V2U
2 c o Ot 2
where Co is the velocity of propagation and u is the amplitude of the wave. Equation (1.33) is invariant in the substitution of t for (-t); hence, the propagation of waves is a reversible process. For a simple reversible chemical reaction, if one path is preferred for the backward reaction, the same path must also be preferred for the reverse reaction. This is called the principle of microscopic reversibility. Time can be measured by reversible, periodic phenomena, such as the oscillations of a pendulum. However, the direction of time cannot be determined by such phenomena; it is related to the unidirectional increase of entropy in all natural processes. Some ideal processes may be reversible and proceed in forward and backward directions. The Fourier equation 10T
(1.34)
-- V2T
c~ Ot
is not invariant with respect to time, and it describes an irreversible process. The term a is the thermal diffusivity. Irreversibility is a consequence of the dynamics of collisions in which the transfer of mass, energy, and momentum takes place. Hydrodynamics specifies a number of nonequilibrium states by the mass density, velocity, and energy density of the fluid. Hydrodynamic equations thus comprise a wide range of relaxation processes, such as heat flow, diffusion, or viscous dissipation, which are all irreversible.
1.3.1
Arrow of Time
Entropy in an isolated system increases dS/dt > 0 until it reaches equilibrium dS/dt = 0, and displays a direction of change leading to the thermodynamic arrow of time. The phenomenological approach favoring the retarded potential over the solution to the Maxwell field equation is called the time arrow of radiation. These two arrows of time lead to the Einstein-Ritz controversy: Einstein believed that irreversibility is based on probability considerations, while Ritz believed that an initial condition and thus causality is the basis of irreversibility. Causality and probability may be two aspects of the same principle since the arrow of time has a global nature. The term irreversibility has two different uses and has been applied to different "arrows of time." Although these arrows are not related, they seem to be connected to the intuitive notion of causality. Mostly, the word irreversibility refers to the direction of the time evolution of a system. Irreversibility is also used to describe noninvariance of the changes with respect to the nonlinear time reversal transformation. For changes that generate space-time symmetry transformations, irreversibility implies the impossibility to create a state that evolves backward in time. Therefore, irreversibility is time asymmetry due to a preferred direction of time evolution.
1.3.2
Dissipative Processes
All natural processes proceed toward an equilibrium state and dissipate their driving power; phenomenological relations, such as Fourier's law of heat conduction, can express them. Real physical processes progress with dissipative phenomena, such as mechanical or electrical friction, viscosity, and turbulence. These dissipative phenomena internally generate heat, and decrease the amount of energy available for work. In an isolated composite system, the change in the internal energy of a subsystem equals the change in heat d U - 6q -
OU
dS
(1.35)
OS
This phenomenon is associated with the level of entropy production due to the irreversibility of the process. Entropy is not conserved; it is the extensive parameter of heat.
1.3.3
Some Properties of Reversible Processes
• A reversible process can be reversed at any point by external conditions, retrace its path, and restores the original state of a system and its surroundings. • Equations that describe time-dependent processes are invariant with regard to the algebraic sign of the time.
8
1. Fundamentalsof equilibrium thermodynamics
• • • • • •
Processes are differentially removed from equilibrium. Reversible processes traverse a succession of equilibrium states. Reversible processes are characterized by frictionless flow. Flow (heat or mass) occurs when the net driving force is only differential in size. A reversible process represents a limit to the performance of actual processes. For cyclic processes, clS = 0
1.3.4 • • • • • • • • • • •
(1.36)
Some Properties of Irreversible Processes
Irreversible processes are actual processes carried out in finite time with real substances. Equations that describe time-dependent processes are not invariant with regard to the algebraic sign of the time. No infinitesimal change in external conditions can reverse process direction. In irreversible processes, heat transfer occurs through a finite temperature difference. In irreversible processes, mass transfer occurs through a finite chemical potential difference. An example of an irreversible process is a spontaneous chemical reaction or electrochemical reaction. Irreversible processes are characterized by the flow of fluids with friction, and sliding friction between any two matters. An example of an irreversible process is electric current flow through a conductor with a resistance. An example of an irreversible process is magnetization or polarization with hysteresis. An example of an irreversible process is inelastic deformation. In Agtota 1 > 0, no single process is possible for which the total entropy decreases. d s > 6Cl ~ dS = O T'
1.4
(1.37)
EQUILIBRIUM
If a physical system is isolated, its state changes irreversibly to a time-invariant state in which no physical or chemical change occurs, and a state of equilibrium is reached in a finite time. Some conditions of equilibrium are: (i) for a system thermally insulated with an infinitesimal change at constant volume: dS = O, d V = O, d U = 0, (ii) for a system thermally insulated with an infinitesimal change at constant pressure: dS - O, dP = O, d H - 0, (iii) for a system thermally insulated with an infinitesimal change at constant volume and temperature: dA = O, d V = O, d T = 0, and (iv) for a system thermally insulated with an infinitesimal change at constant pressure and temperature: d G = O, d T = O, dP = O. At equilibrium, all the irreversible processes vanish, and temperature, pressure, and chemical potentials become uniform; this means that no thermodynamic force exists in the system. No perturbation will cause a change in a neutral equilibrium. Any two phases in hydrostatic equilibrium must have the same pressure; in thermal equilibrium, any two phases must have the same temperature. If two phases are in equilibrium with respect to any species, then the chemical potential of that species must have the same value in these phases. Consider an elementary general chemical reaction vsS = vpP
(1.38)
where ~'s and Vp are the stoichiometric coefficients. The condition for chemical equilibrium is Vs/Xs = Vp~p
(1.39)
where/tz i is the chemical potential of a species i. This chemical equilibrium condition is equivalent to the vanishing affinity A defined by A=-~_vi~
i -0
(1.40)
1.4 Equilibrium
9
A system may be in a stable, metastable, unstable, or neutral equilibrium state. In a stable system, a perturbation causes small departures from the original conditions, which are restorable. In an unstable equilibrium, even a small perturbation causes large irreversible changes. A metastable system may be stable or unstable according to the level and direction of perturbation. All thermodynamic equilibria are stable or metastable, but not unstable. This means that all natural processes evolve toward an equilibrium state, which is a global attractor. The emergence of macroscopic reversibility from microscopic irreversibilities is referred to as dynamic equilibrium with the mechanisms of cancellation of the opposite molecular processes. An extremum principle minimizes or maximizes a fundamental equation subject to certain constraints. For example, the principle of maximum entropy (dS)u- 0 and, (d2S)u < 0, and the principle of minimum internal energy (dU)s = 0 and (d2U)s> 0, are the fundamental principles of equilibrium, and can be associated with thermodynamic stability. The conditions of equilibrium can be established in terms of extensive parameters U and S, or in terms of intensive parameters. Consider a composite system with two simple subsystems of A and B having a single species. Then the condition of equilibrium is (1.41)
d U -- (T A - TB )dS A - ( P A -- PB)dVA + (/d'A --/XB)dNA = 0
Hence the thermal, mechanical, and chemical equilibrium conditions in terms of the intensive properties are
(1.42)
rA - rB, PA - PB, ~A -- "B
since dSA, dVA, and d~A are the infinitesimal changes in independent variables. Similarly, the equilibrium conditions are expressed in terms of entropy
dS_( 1
~ rA
TB
(
PA rA
PB =0, rB
/XA rA
1]dUA_(PA
TA
TB )
PB
)
/d'A
]d'B
TA
TB
dNA= 0
(1.43)
and the equilibrium conditions become
1
1
ra
TB
=0,
/XB = 0 rB
(1.44)
Example 1.1 Equilibrium in subsystems Consider a closed isolated cylinder with two subsystems of 1 and 2 containing air with an equal volume of 1 L and equal temperatures of 298.15 K. There is a fixed piston at the boundary of the subsystems, which have different pressures of P1 = 2 atm and P2 = 1 atm. Estimate the temperature, volume, and pressure of the subsystems when the piston is released. Assume that the piston is impermeable to air, freely movable, and heat conducting. Solution: Assume that the air is an ideal gas with constant heat capacity. The initial states (i) of the subsystems are PliVli - nlRTli ' f 2iV2i - n2RT2i
(1.45)
8~ v~ = nlR r~r, Pay Vat - n2Rr2s
(1.46)
In the final state (f), we have
where the total volume is constant, and we have
Vt~ +V2;-2V1,-2L According to the first law and constant heat capacity Cv, we have (UI / - U l i ) + ( U 2 t , - L i
2, )-- 0, n i t v (Tlf - T l i ) +
n2C v ( T 2 f - T 2 i )
= 0
(1.47)
10
1. Fundamentals of equilibrium thermodynamics
Since Tlf-- T2f, Tli = T2i, and nli ::fi n2i , we have T V = Tli This is a natural result as the internal energy of an ideal gas depends on the temperature only, and the system is isothermal at the initial and final conditions. From Eqs. (1.45) and (1.46), we have
eli - nli - Vlf - 2 P2i n2i V2f
(1.48)
With a total volume of 2 L, the final volumes and pressures become 4
vv . 7L,.
1.5
.
2
3
. 7L, s.
2 atm
THE FUNDAMENTAL EQUATIONS
The f u n d a m e n t a l equations relate all extensive properties of a thermodynamic system, and hence contain all the thermodynamic information on the system. For example, the fundamental equation in terms of entropy is (1.49)
S = S ( U , ... , X j , . . . )
The extensive properties of U and X are the canonical variables. The fundamental equation in terms of internal energy U is U = U (S, ... , Xj , . . .)
(1.50)
For the entropy and internal energy, the canonical variables consist of extensive parameters. For a simple system, the extensive properties are S, U, and V, and the fundamental equations define a fundamental surface of entropy S = S(U,V) in the Gibbs space of S, U, and V. Differential forms of the fundamental equations contain the intensive thermodynamic properties. For example, dS and d U are
dS-~ (O~g) X
d U - + ' ~i (O~Xi) dXi " U,Xj=/=Xi
d U = --~ x dS
"3I-
"
-~i
S , X j @X i
(1.51)
(1.52)
Here, the first-order partial derivatives are the intensive properties T,/, and Y. In terms of the intensive properties, Eqs. ( 1.51) and ( 1.52) become d s = l d u + ~_ Ii dXi T i
(1.53)
dU
(1.54)
= TdS +
~_, Yi dXi
i The first term on the right side of Eq. (1.53) or Eq. (1.54) represents heat associated with the thermodynamic temperature T, and the remaining terms are the work terms. The pairs of intensive and extensive properties, such as 1/T and U, or Ii and X,., are the conjugate properties.
1.6
1.6
The thermodynamiclaws
11
THE THERMODYNAMIC LAWS
A set of thermodynamic laws governing the behavior of macroscopic systems lead to a large amount of equations and axioms that are exact, based entirely on logic, and attached to well-defined constraints. These laws are summarized in the following sections.
1.6.1
The Zeroth Law of Thermodynamics
Two systems in thermal contact eventually arrive at a state of thermal equilibrium. Temperature, as a universal function of the state and the internal energy, uniquely defines the thermal equilibrium. If system 1 is in equilibrium with system 2, and if system 2 is in equilibrium with system 3, then system 1 is in equilibrium with system 3. This is called the zeroth law of thermodynamics and implies the construction of a universal temperature scale (stated first by Joseph Black in the eighteenth century, and named much later by Guggenheim). If a system is in thermal equilibrium, it is assumed that the energy is distributed uniquely over the volume. Once the energy of the system increases, the temperature of the system also increases (dU/dT> 0).
1.6.2
The First Law of Thermodynamics
A change in a state function accompanying the transition of a system from one state to another depends only on the initial and final states and not on the path between these states. If the system returns to its original state, the integral of the change is zero au
- o
(~.55)
Such systems are called cyclic processes. The Poincare statement of the first law states that in a cyclic process, the work done by the system equals the heat received by it. According to the first law of thermodynamics, the state function of internal energy Uin a closed system is equal to the sum of the heat received by the system 6q and the mechanical work 6 Wperformed on the system by the surroundings dU = 6q + 6 W
(1.56)
Heat and work always refer to the system, and the sign convention for q and W chosen specifies which direction of energy transfer to the system is positive. The sign convention adapted here assumes that heat transferred into the system from the surroundings is positive, while work transferred into the system (work done on the system) at which energy is transferred into the system from the surroundings is positive. For compression work we have 6 W = - P d V as the compression leads to - d V and positive work. The signs of heat and work referring to the surroundings would then be opposite qsur = --q and W~ur= - W . Changes of heat and work depend on the path of a change and are not state functions. Therefore, it is not possible to define a function q or W that depends only on the initial and final states. When we consider an open system, we have a flow of energy due to heat transfer and to exchange of matter. The conservation of energy states that the total energy is conserved in any change of state. The total energy is the sum of the internal, kinetic, and potential energy of the system, heat, and type of work (such as electrical, mechanical, or chemical work). In general, the term 6 W represents all different forms of work. Work is the product of an intensive variable and a differential of an extensive variable. For example, if the system is displaced by a distance dl under a force F, it performs the work o f - F d l . I f - d N i moles of substance i with the chemical potential/x i flow from the system to its surroundings, the chemical work o f - t z d N , occurs. Thus the total work becomes
6 W = - P d V + Fdl + Ode + ~ tx i dN i + ' "
(1.57)
i=1
Here, - P d V refers to the sign convention recommending that work done on the system is positive as the compression work leads to - d V and positive work. For an open system, an additional contribution to the energy due to the exchange of matter dUm occurs dU - 6q + 6 W
+ dU m
(1.58)
12
1. Fundamentals of equilibrium thermodynamics
For systems with chemical reactions, the total energy may be considered a function of T, V, and N/: U = U(T, V, Ni). The total differential of U is
dU=(O~T) dT+(O-~V ) dV+~(O~Ni) dNi=6q+6W+dUm V, Ni
T ,Ni
i
(1.59)
V ,T,Ni, ~
The exact form of the function U(T, V,N/) for a certain system is obtained empirically.
Example 1.2 Relationships between the molar heat capacities Cp and
Cv
The first law of thermodynamics leads to a relation between the molar heat capacities. The change in internal energy expressed in volume and temperature U = U(T, V) is
dU=(O-~T ) dT+(O-~V ) d V = 6 q - p d V V
(1.60)
~.
The heat effects are
OU dT + P+ - ~ 6q= - ~ v
r
dV
(1.61)
At constant volume, the heat capacity is (1.62) V
V
and at constant pressure, the heat capacity is
G=U
+P+
(1.63)
The difference between Eqs. (1.63) and (1.62) is
C p - C v = P+ --~ T
--O-T p
(1.64)
The fight side of Eq. (1.64) shows the energy effect due to the expansion of volume at a constant pressure process. For a mixture of ideal gases, the internal energy is a function of temperature only, and hence Eq. (1.64) and PV = RT yields
Cp - Cv = R
(1.65)
The gas constant R is common to all gases and determined by the product of the Boltzmann constant and the Avogadro number R=(1.3805×10 -23) J/K (6.0225×1023) mo1-1 =8.3143 J/(mol K) In contrast, for a fluid, the enthalpy and entropy may be expressed as a function of temperature and pressure and in terms of thermal expansion/3 = (0 V/OT)p/V and isothermal compressibility K = -- (0 V/OP)r/V as follows:
dH = Cp d T + VdP = Cp d T + (1- [3T) VdP
(1.66)
dT dS = Cp --T- - ~ VdP
(1.67)
1.6
The thermodynamiclaws
13
For the internal energy and entropy, from Eqs. (1.23) and (1.24) we also have dU - CvdT +[-~ T-
P]dV
d S - Q ~dT + --[3d V T K
By comparing Eqs. (1.67) and (1.25), we have dT dT dS = Ci, ---2-- ~ VdP = Q ~ + - - d V 1
T
(1.68)
K
or
Cp-Cv=T/3V
~-f v
K
(1.69)
-~ p
At constant volume, Eq. (1.69) yields (1.70) V
For an incompressible fluid/3 = 0, and Cp = Cv = C. The second partial derivatives of the state functions at constant volume and entropy are V
,
=asV,
-
=KsV
(1.71)
I"
where Cv is the heat capacity at constant volume, as is the adiabatic thermal expansion, and Ks is the compressibility. The second partial derivatives of the state functions at constant pressure and temperature are OH
= Cp , P
-~
= ~T ' P
= apV P
-
= KrV
(1.72)
T
Here, Cp is the heat capacity at constant pressure, at, is the isobaric thermal expansion, and Kp is the isothermal compressibility. Table 1.1 shows the molar heat capacities of some gas compounds.
1.6.3
The Second Law of Thermodynamics
The work of Carnot, published in 1824, and later the work of Clausius (1850) and Kelvin (1851), advanced the formulation of the properties of entropy and temperature and the second law. Clausius introduced the word entropy in 1865. The first law expresses the qualitative equivalence of heat and work as well as the conservation of energy. The second law is a qualitative statement on the accessibility of energy and the direction of progress of real processes. For example, the efficiency of a reversible engine is a function of temperature only, and efficiency cannot exceed unity. These statements are the results of the first and second laws, and can be used to define an absolute scale of temperature that is independent of any material properties used to measure it. A quantitative description of the second law emerges by determining entropy and entropy production in irreversible processes. If a system is in equilibrium, then all the forces X~ are fully known from external parameters ai, so that the first law is (Sq - d U - (5 W = d U - ~ X i (a i )da i
(1.73)
Equation (1.73) is a Pfaffian equation and a; is an independent variable. Caratheodory's theory states that starting from a known original state, there may be other states that cannot be reached by an adiabatic process along the path 6q = 0. This shows the existence of an integrating factor for 6q; hence we have
14
1. Fundamentalsof equilibrium thermodynamics A d S = dr/
(1.74)
where dr/is a total differential of the variables a i. Therefore, r/must be a state function known as entropy S, and the integrating factor is the reciprocal of absolute temperature T, so that d S - ~qrev
(1.75)
T Equation (1.75) is a mathematical statement of the second law of thermodynamics for reversible processes. The introduction of the integrating factor for 6q causes the thermal energy to be split into an extensive factor S and an intensive factor T. Introducing Eq. (1.75) into Eq. (1.56) yields the combined first and second laws dU = TdS + ~ W
(1.76)
Every system is associated with an energy and entropy. When a system changes from one state to another, the total energy remains constant. However, the total entropy is not conserved, and increases in irreversible processes while remaining unchanged in reversible processes. The notion of entropy is not a directly intuitive concept. We can relate the entropy of an irreversible process to the external and internal properties, regardless of the energy content of the system. We can attain the same distribution of internal parameters imposed both reversibly and irreversibly by a set of external parameters. These different paths result in different work and energy changes in the system. However, we assume that a set of local parameters determines the entropy, and we can devise an ideal process that would reversibly bring the system to any configuration of the irreversible process. For example, diffusion of a substance is a nonequilibrium process, and the local concentration profile is necessary to define the system. We may apply reversibly a centrifugal field to the system to maintain the same concentration profile in a state of equilibrium. The energy applied reversibly to the centrifugal field is different from an irreversible diffusion process. Thus, the thermodynamic states of an irreversible diffusion process and the corresponding equilibrium system are different. Entropy may be computed as the corresponding entropy of the real system.
Example 1.3 Entropy and distribution of probability Entropy is a state function. Its foundation is macroscopic and directly related to macroscopic changes. Such changes are mostly irreversible and time asymmetric. Contrary to this, the laws of classical and quantum mechanics are time symmetric, so that a change between states 1 and 2 is reversible. On the other hand, macroscopic and microscopic changes are related in a way that, for example, an irreversible change of heat flow is a direct consequence of the collision of particles that is described by the laws of mechanics. Boltzmann showed that the entropy of a macroscopic state is proportional to the number of configurations 1~ of microscopic states a system can have S = k In f~
(1.77)
where k is the Boltzmann constant (k = 1.3805 × 10 -23 J/K). Consider a system with two chambers containing a total number of particles n. The total number of possibilities for distributing the particles between the two chambers is f~, which is the total number of distinct microstates with nl number of particles in chamber 1 and n 2 number of particles in chamber 2 f~ = (nl + n2)! nl! n2!
(1.78)
Equation (1.77) shows that disorganization and randomness increase entropy, while organization and ordering decrease it, and equilibrium states have the maximum value of D. In the above system, 1~ reaches its maximum value when n I = n 2. In parallel, the increase in entropy corresponds to the increase in the number of microscopic states or states with higher probability. The concept of entropy as a measure of organized structures is attracting scientists from diverse fields such as physics, biology, and communication and information systems.
1.7
BALANCE EQUATIONS
Balance equations consist of conserved, such as mass and energy, and nonconserved, such as entropy, properties. The following sections summarize the general balance equations of mass, energy, and entropy.
1.7 Balance equations
1.7.1
15
Mass Balance
In an open system, mass flow rate for the flowing streams through the boundary is (1.79)
rh = yAp
where v is the average velocity, A the cross-sectional area, and p the density. Assuming that the flow is positive when it enters into the control volume and using Eq. (1.79), the mass balance is dm
)in -- - -
d m + E (rhi )out -- Z ( g h i dt i i
+- Z
dt
i
(vAP)i, out - Z (vAP)i, in = 0 i
(1.80)
This form of mass balance is also called the continuity equation. At steady state, the accumulation term (dm/dt) becomes zero, and we have
E (/~i)out -- E i
1.7.2
(vAP)i,out - Z (vAP)i,in
(ghi)in - - E
i
i
0
=
(1.81)
i
Energy Balance
Each unit of mass flow transports energy e at a rate, O = [U + (1/2)v 2 + zg]rh, where z is the elevation above a datum level, and g is the acceleration of gravity. However, considering the flow work of all the entering and leaving streams in terms of the product of the pressure and volumes of each stream, we have a total energy associated with a stream i defined by
I/ ,v2 / U+-
+zg
J E{
rh+(PV)rh
-
2
H+-
i
+zg 2
/] rh
=O i
(1.82)
i
Using Eq. (1.82), we have a compact form of energy balance
d(mU__~_~)+ E (ei)out -- ~ dt
i
( e i ) i n -- ~/-F- m
(1.83)
i
where q and W are heat transfer rate and shaft work. Equation (1.83) assumes that the center of mass of the control volume is stationary. If the kinetic and potential energy changes are small enough, then Eq. (1.831) reduces to
d(mU)) + ~_~(I;t~)out i
dt
~ (/;/~)in
-
-
0 + f¢l
(1.84)
i
Equation (1.84) is widely applicable to many thermal engineering systems. If a system is at steady state, then the accumulation term vanishes
Z (~'i)out-- Z (el)in -i
(1.85)
4 Jr- ~/1~
i
or
•
1.7.3
(( ,v2 !/
(/ ,v2 i)
/,out
2
(1.86) /,in
Entropy Balance
The rate form of entropy balance is ±(rhS) +
d(mS) dt
+
dSsulT dt
0 Xpr°d > --
(1.87)
16
1. Fundamentals of equilibrium thermodynamics
where A(rhS) is the net entropy change of the flowing streams, and Sprod is the rate of entropy production. The term dSsurr/dl is the entropy changes within the surroundings defined by
dSsurr ~ Oi dt
-.
~
(1.88)
With Eq. (1.88), and for a steady-state condition, Eq. (1.87) reduces to A ( g h S ) - t~.. -~/ ~0 qi __ Sprod "
(1.89)
Equation (1.89) determines the rate of entropy production due to irreversibility within a control volume. The concept of entropy production is elaborated further in the next section. 1.8
ENTROPY AND ENTROPY PRODUCTION
By using Eq. (1.76) with a pressure-volume work, we have
dS =
dU + PdV T
(1.90)
The entropy of a system is an extensive property, and it changes through the exchange of mass and energy• If a system consists of several processes, the total entropy change is equal to the sum of the entropy changes in each process. The total change of the entropy dS results from the flow of entropy due to exchanges with surroundings (deS) and from the changes inside the system (diS)
dS=deS+diS
(1.91)
The value of diS is zero when the change inside the system is reversible, and it is positive when the change is irreversible
diS = 0
(Reversible change)
(1.92)
diS > 0
(Irreversible change)
(1.93)
For an isolated system, there is no interaction with the surroundings so that
dS = diS > 0
(1.94)
The rate of entropy production Sprod is expressed by Sprod = d i S ~ 0
dt
(1.95)
Irreversible processes produce entropy in any isolated, open, or closed system and Eq. (1.94) holds. In every macroscopic region of the system, the entropy production of irreversible processes is positive. A macroscopic region contains enough molecules for microscopic fluctuations to be negligible. The second law of thermodynamics states that the sum of the entropy production of all processes for any system and its environment is positive. When interfacial phenomena are considered, the entropy production is based per unit of surface area. The entropy source strength • is the rate of entropy production per unit volume (I) -- Spr°d ~ 0
(1.96)
dV The product of the entropy source strength and the absolute temperature is called the dissipation function = T~ >_ 0
(1.97)
1.8 Entropyand entropy production
17
When the T is the environmental temperature, the dissipation function represents the energy dissipated to the environment. The entropy source strength and the dissipation function are not state functions, and they depend on the path between the given states.
Example 1.4 Entropy production and subsystems Equation (1.91) can be applied to various irreversible processes. Let us consider a system consisting of two closed subsystems of I and II, and maintained at uniform temperatures of T I and Tn, respectively. The total entropy dS is expressed as dS -
6I q
d S I + d S II
6n q
= U + ri-----f-
(1.98)
The interactions of heat in each subsystem are given by 6Iq = 6~q + 6~q,
(3IIq = 6~Iq + 6~Iq
(1.99)
Using Eq. (1.99) and the conservation of energy 6~Iq + 6~q = 0, Eq. (1.98) yields
dS=---fif-+-~
,
q
T II = d e S + d i S
TI
(1.100)
The entropy production per unit time is
dt
dt
TI
T II > 0
(1.101)
Equation (1.101) shows that the rate of entropy production is the product of flow (heat flux) (6~q/dt), and the thermodynamic force, (1/T I - 1/Tn). Equation (1.90) is the total differential of the entropy as a function of the variables U and V only. To generalize this relation, we also consider the changes in the amounts of species. Using the mole amounts for the species, we have a general expression for the change of entropy from the Gibbs relation ~j
dS = d U + PT d v - E ---T-
(1.102)
J
Equation (1.102) is one of the main relations for calculating entropy production.
Example 1.5 Entropy production in a chemical reaction in a closed system Derive an expression for the entropy production for a chemical reaction in a closed system. Solution: The reaction is a single elementary reaction and homogeneous. Entropy production due to a chemical reaction in a closed system is given by dS = -6q - + - - dAe
T
T
(1.103)
where A is the affinity of the chemical reaction, A = - E vjix2, s the extent of the reaction, and v/the stoichiometric coefficient of species j. Equation (1.103) shows that the entropy change contains two contributions: one is due to interactions with the surroundings deS = 6q/T, and the other is due to a change within the system diS = Ade/T. Therefore, the rate of entropy production can be expressed in terms of the rate of reaction Jr diS dt
1 - --AJ r > 0 T
(1.104)
18
1. Fundamentals of equilibrium thermodynamics
where
Jr--
d8
dt
Equation (1.104) is similar to Eq. (1.101) in relating the rate of entropy production to the product of the flows (here the rate of reaction) and the scalar thermodynamic force that is A/T. Equation (1.104) can be readily extended to several chemical reactions taking place inside the system
diS_ 1 t dt T E AkJrk >- 0
(1.105)
k=l
When the chemical reaction reaches equilibrium, affinity vanishes A = -Eufl~j = 0. The entropy generated per unit time and unit volume is called the rate of volumetric entropy production or the entropy source of density = A Jr ~> 0
(1.106)
VT
Example 1.6 Entropy production in mixing Consider the following two mixing processes: Mixing process 1: In a steady mixing process, 14.636 kg/s of saturated steam (stream 1) at 133.9 kPa is mixed with 15.0 kg/s of saturated steam (stream 2) at 476 kPa. Mixing process 2: In a steady mixing process, 11.949 kg/s of super heated steam (stream 1) at 773.15 K and 6000.0 kPa is mixed with 60.60 kg/s of saturated steam (stream 2) at 2319.8 kPa. Assume that the mixing processes are adiabatic. Determine and compare the rate of entropy production and the lost work for these two mixing processes producing a product of stream 3. Solution: Assume that the surroundings are at 298.15 K and the kinetic and potential energy changes are negligible. Mixing process 1: Available data from the steam tables in Appendix D are th1 = 14.636 kg/s,
/'h 2 --
15.0 kg/s
H 1 = 2688.3 kJ/kg, H 2 = 2745.4 kJ/kg S 1 = 7.2615kJ/(kgK), S2 = 6.8358 kJ/(kg K) To = 298.15K Assuming negligible kinetic and potential energy changes, mass, energy, and entropy balances yield 0 = rh3 - rh2 - rh1 ~ rh3 = 29.636 kg/s
0 = ( t h H ) 3 - ( t h H ) 2 - ( r n H ) l --~ H 3 =
[(thH)2 + (thH)l ]
= 2717.2 kJ/kg
th3
From the steam table, we read T3 = 401.15 K and $3 = 7.0462 kJ/(kg K). The entropy balance yields the rate of entropy production, Sprod Sprod -- (thS)3 - (thS)2 - ( t h S ) l = thl (83 - S1 ) +/J/2 ($3 - $2 ) = 0.004724 kW/K
1.8 Entropyand entropy production
19
The rate of work loss Elos~,becomes /~loss = ToSprod - 1.4083 k W
The dissipated work potential is small as we mix two saturated steams at relatively low temperature levels. Mixing process 2: Available data from the steam tables are rnI - 11.949 kg/s, rn2 = 60.60 kg/s P1 - 6000 kPa, P2 = 2319.8 kPa (saturated) H l = 3422.2 kJ/kg, H 2 = 2799.9 kJ/kg S, - 6.8818 kJ/(kg K), S 2 - 6.2817 kJ/(kg K) From the mass and energy balances, we have 0 = /~/3 -- #12 --/~1 ~ th3 = 7 2 . 5 4 9 kg/s
0 = (thH)3 - (thH)2 - (rhH)l
--, H 3 =
[(,n/-/)2 + (rh/4)~ ]
= 2902.4 kJ/kg
rh3 From the steam table we read T3 - 523.15 K, S 3 - 6.5454 kJ/(kg K), and P3 = 2000 kPa. The entropy balance yields the rate of entropy production, Sprod Sprod -- (?T/S)3 - (tI/8)2 - ( t~/S)l -- #/l (83 - S1 ) + 1~/2(33 - 32 ) = 11.9603 kW/K
The rate of work l o s s /~loss becomes /~loss -- ToSprod -- 3565.96 k W
These two simple examples show that mixing the saturated steam with the superheated steam in the second mixing process causes much greater entropy production and lost work potential than mixing two saturated steams in mixing process 1.
1.8.1 ThermodynamicCoupling When a system exchanges mass and energy with its surroundings, it also exchanges entropy, and moves away from the equilibrium state. As the system produces entropy due to irreversible processes taking place, the entropy flowing out of the system is greater than the entropy flowing into the system. As the system keeps exchanging entropy with the environment, it may reorganize itself and transform into a higher order state maintained by matter and energy being exchanged with the environment. Various irreversible processes inside the system may continue and interact with each other as the mass and energy exchange continue. These interactions are called thermodynamic couplings, which may allow a process to progress without its primary driving force or in a direction opposite to the one imposed by its own driving force. For example, in thermodiffusion, a species diffuses not because of a concentration gradient but because of a temperature gradient. Sometimes, a species may flow from a low to a high concentration region, which must be coupled with a compensating spontaneous process with positive and larger entropy production. The principles of thermodynamics allow the progress of a process without or against its primary driving force only if it is coupled with another process. This is consistent with the second law, which states that a finite amount of organization may be purchased at the expense of a greater amount of disorganization in a series of coupled spontaneous processes. Such coupled processes are of great importance in physical, chemical, and biological systems, such as the Bdnard instability and biological pumps of sodium and potassium ions.
20 1.9
1.
Fundamentals of equilibrium thermodynamics
THE GIBBS EQUATION
By introducing Eq. (1.57) into Eq. (1.76), we have
dU = T d S - PdV + Fdl + ~, de + ~ [LidN i -+-".
(1.107)
i=1
Equation (1.107) relates the total change in internal energy to the sum of the products of intensive variables T, P, F, /zi, ~, and the changes in extensive properties (capacities) of dS, dV, dl, dN,., and de. The Bronsted work principle states that the overall work A W performed by a system is the sum of the contributions due to the difference of extensive properties AK across a difference of conjugated potentials X,.,1-X,.,2
(1.108)
AW = ~ (Xi,I - Xi,2)AKi i=1
Equation (1.107) is more useful if it is integrated with the Pfaffian form; however, this is not a straightforward step, since intensive properties are functions of all the independent variables of the system. The Euler relation for U(S, V, l, e, Ni) is
+ ~ N. V,l,e,Ni
S,I,e,Ni
V,S,e,Ni
V,I,S ,Ni
i=1
OU
+""
(1.109)
S,V,I,e,Nj
Comparing Eq. (1.109) with Eq. (1.107) yields the definitions of intensive properties for the partial differentials
-~
V,/,e,/Vi : T,
- . ~ S,',e,Ni---P,
-~
V,S,e,Ni
(1.110) V,I,S,Ni
=
°2v i S,V,I,e,Nj
= I~i
The chemical potential/z indicates that the internal energy is a potential for the chemical work (or mass action) txidNi, and it is the driving force for a chemical reaction. The chemical potential cannot be measured directly, and its absolute values are related to a reference state. However, the change of chemical potential is of common interest. By introducing the definitions given in Eq. (1.110) into Eq. (1.107), we obtain the integrated form of the Gibbs equation
U = T S - P V + FI + ~te+ ~ I~iNi
(1.111)
i=l Differentiation of Eq. (1.111) yields
dV
=
rdS + S d r
-
P d V - VdP + Fdl + ldF + qJde + e&b + ~
P'i d N i
+ ~ Ni d]Dl,i
(1.112)
Comparison of Eq. (1.112) with Eq. (1.107) indicates that the following relation must be satisfied:
SclT- VdP + ldF + ed~p + ~_~N i d lz i = 0
(1.113)
Equation (1.113) is called the Gibbs-Duhem relation, which becomes particularly useful at isobaric and isothermal conditions, and when the force and electrical work are neglected, we have
~Nidld, i : ~Ni i=l j=l
( OIdfi I : 0 , tONj),
j:l,2,...,n,
i=/=j
(1.114)
1.9
21
The Gibbs equation
Equation (1.114) determines the changes in chemical potential with the addition of any substance into the system. From the Gibbs fundamental equationf(U,S,V,N), we have the three functions of S, V, and N, the respective differential relations, and the Euler equations given by dU + p ~_dV dN T iX-T-,
S = S(U, V, N ) - ~ dS = r
= s
1 V u-+P--iX-T T
V = V ( U , S , N ) - - , dV = - d U + T ~dS+ ix dN V=U--1 + s T#+ N_ P P ---fi-, P P
N
T _ P
(1.115)
(1.116)
and N = N(U,S,V)-,dN
- dU tx
T dS + p dV tx tx
N =U 1 IX
s T +v P tx ix
(1.117)
Using the molar-specific volume (v-- V/N) and molar-specific entropy (s = S/N), a simplified version of the Gibbs-Duhem relation results dix = - s d T + vdP
(1.118)
By partial differentiation, Eq. (1.118) can be transformed to a form called the thermal equation
( O~pp) = v = v ( T,P )
(1.119)
T
and the corresponding caloric equation is ( O~T)
= s = s(T,P)
(1.120)
P In Eq. (1.73), ifXk are the external variables to maintain the nonequilibrium distribution of the internal parameters of ~:kin a state of equilibrium, we have a potential energy o f - E~kXk.This energy is the additional work of the external parameters to maintain the distribution of internal parameters. The internal energy of equilibrium system Ueq is related to the internal energy of the nonequilibrium system U by
Ueq
=U--Z~kX
k
(1.121)
The irreversible work 6 Weq is related to the work necessary in reaching the same conditions reversibly
meq
- ~ m nt- ~ ~kdYk
(1.122)
In contrast, the entropy change in the corresponding reversible process is TdS = d geq
- • Weq
( 1.123)
By inserting Eqs. ( 1.121 ) and (1.122) into Eq. (1.123), we have rdS = dU - ~ X k d ~ . - ~ ~kdYk + 8 W + Z ~kdXk - dU - ~ W - ~ X k d ~
(1.124)
The entropy term TdS in Eq. (1.124) is the same for the irreversible process and the corresponding reversible process. Therefore, Eq. (1.124) represents the Gibbs equation for irreversible process. With the first law of thermodynamics, dU - 6 W = 6q, Eq. (1.124) becomes TdS = 6 q - ~_, XkdsCk
(1.125)
22
1. Fundamentals of equilibrium thermodynamics
For an adiabatic process 6q = 0, we have (1. 126)
TdS = m~__~X k d ~ k
Equation (1.126) represents the change of entropy for an irreversible process in an adiabatic system as a function of the internal and external parameters. This may be an important property to quantify the level of irreversibility of a change, and hence yields (i) a starting point to relate the economic implications of irreversibility in real processes, and (ii) an insight into the interference between two processes in a system.
1.10
EQUATIONS OF STATE
Equations of state relate intensive properties to extensive properties, and are obtained from the Euler equation as partial derivatives. In the entropy representation, we have the following equations of state:
-
,
V,l,e,Ni
-
,
T
(1.127)
-
V,l,e,Ni
T
V,l,e,Ni
In contrast to fundamental equations, equations of state do not contain all the information on a system, since the intensive properties are partial derivatives of the extensive ones. To recover all the information, all the equations of state are inserted into the Euler equation.
1.7 Heat c a p a c i t i e s for real gases For real gases, the internal energy U is not a function of temperature only because of molecular interactions, such as the collision of molecules, which depends on the distance between the molecules. Therefore, the change in volume affects the energy. The molecular forces have a short range. At low densities, the molecules are far apart from each other; hence the effect of interactions is small and negligible, and it vanishes as the volume approaches infinity. Perfect gas has low enough pressure or density so that all virial coefficients are ignored. A gas is slightly imperfect when all virial coefficients are ignored except the second virial coefficient B. Integration of the Helmholtz equation (OU/O V)T = T2[O(P/T/OT)]v yields Example
(1.128)
real'
A similar expression is also obtained from the Gibbs free energy equation when pressure approaches zero. If the partial differentiation inside the integral is determined from an equation of state (EOS), then Urealcan be calculated. For example, (P/T) from the van der Waals EOS is P m nR T V-nb
1 an 2
(1.129)
T V2
Therefore, from Eqs. (1.128) and (1.129), we get
rea
)
(1.130)
The second term on the fight side of Eq. (1.130) represents the energy of molecular interactions per unit volume. As the volume increases, the interactions get smaller, as is the case for gases at low density. If we use the Berthelot equation
P -
nRT
1 an 2
V - nb
T V2
(1.131)
23
1.10 Equationsof state
in Eq. (1.128), we obtain (1.132)
Ureal- Uidea1-- a 7 Equation (1.130) also yields a relation for the heat capacity of a real gas at constant volume
Cv,rea1 =
(0ereal)
=
(0eideal)
+
0 [IVT2 [ 0 (P)])
dV
(1.133)
After reorganizing Eq. (1.133), we have
Cv,real: Cv,idealq-Ivr~OT2)v dV
(1.134)
If the second partial derivative inside the integral is determined from an EOS, then the heat capacity of a real gas at constant volume can be calculated. For example, the integral in Eq. (1.134) vanishes for the van der Waals equation, and as Eq. (1.129) shows, pressure is a linear function of temperature. However, by using the Berthelot EOS, (Eq. 131), the heat capacity is obtained from Eq. (1.134).
Example 1.8 van der Waals isotherms Use the critical parameters for ammonia and estimate the van der Waals constants a and b. Plot the van der Waals isotherms for ammonia at T1 = 200 K, T2 = 406 K, and T3 = 550 K, when the volume changes from 0.04 to 0.25 L.
Solution: Use the van der Waals equation (Eq. 1.129) p-
nRT V-
nb
an
2
V2
with the parameters
(1.135)
Figure 1.1 shows the isotherms obtained from the MATHEMATICA code below. (*Ammonia*) Tc = 405,7; Pc = 11fi.8"0,9860 (*atm*); R = 0,082 l; (% atm/Mol K*) a = (27/64)*(R"2)*(Tc"2)/Pc
b = (1/8)*(R*Tc/Pc) p[V_,T_]: = (R*T/(V-b)-(a/VA2)); Plot[{p[V200],p[V406], p[-V,550]}, {V,0.04,0.2}, Frame ~ True, OridLines ~ Automatic, PlotStyle -, {Thickness[0,0085]}, FrameStyle --, Thickness[0.007], FrameLabel ~ {"V, L", "P, atm "}, RotateLabel ~ True, DefaultFont --. {"Times-Roman", 12}] a = 4.2043 b = 0.0374004
1. Fundamentals of equilibrium thermodynamics
24
3000
/
2000
a2 lOOO
0
~ .
.
.
0.05
.
0.075
, .
O. 1
.
O. 125
O. 15
O. 175
0.2
V,L
Figure 1.1. Isotherms obtained from van der Waals equation.
Example 1.9 Estimation of molar volume of a gas at high pressure (a) Estimate the molar volume of 100 mol of methane from the van der Waals equation at 310 K and 15 atm. (b) Plot pressure versus temperature when the temperature changes from T = 250 to 450 K and V = 75 L. Solution: Use the van der Waals equation nRT
an 2
V - nb
V2
p-
At the critical point, the vapor, liquid, and critical volumes are the same, ( V v - 3v v + 3 v g v - v? = 0
Vc)3 =
0,
and expansion gives
(1.136)
or in polynomial form, we have
V3-
b+
RT¢ ) V2 + - a- V - ~ =ab O Pc
(a) Tc = 190.6; Pc = 45.99*0.9869; (*Tc, in K, Pc in atm*)
R = 0.0821; (*L atm/Mol K*) a = (27/64)*R^2*(Tc^2)/Pc;(*L^2 atm mol^-2 *) b = (1/8)*R*Tc/Pc; (*L mol^-l*) n = 1 0 0 ; P - 15;T= 310; (*Solve for V*) Solve[{P-(n*R*T/(V-n*b)-(a*n^2)/V^2) = = 0},V] {{V~4.4772- 4.42488i},{V~4.4772 + 4.42488i},{V~ 165.029}} V ~ 165.029 L/mol
(b) (*methane*) n = 100;V= 75; PIT_]: = n*R*T/(V-n*b)- (a*n^2)/V^2 Plot[P[T], {T, 250,400}, Frame ~True, GridLines ~Automatic, FrameLabel~ {" T, K", "Patm"},
Pc
(1.137)
1.10 Equations of state
25
42.5 40 37.5
E
35 32.5
30 27.5 25 f - : . . . . ' . , . 260 280
~ ' 300
' ' ' 320 340 T,K
i... 360
, .... 380 400
Figure 1.2. Change of pressure with temperature for methane.
RotateLabel --,True, DefaultFont--, {"Times-Roman", 12}] Clear Figure 1.2 shows the change of pressure with temperature.
Example 1.10 Estimation of volume of a gas at high pressure using generic cubic equation of state Estimate the volume of n-butane at 15 bar and 400 K. Solution: A generic cubic equation of state to calculate the molar volume is
RT +b P
V=
a(r) V-b P (v + eb)(V + o-b)
(1.138)
The solution can be obtained by iterative methods or using a software package with an initial estimate from the ideal gas law. Using V = ZRT/P, equations for vapor and vapor-like root Zvap and for liquid and liquid-like root Zliq are obtained Zvap -- 13
Zva p ---1 + / ~ - q/3
(Zva p + e/3)(Zva p + o'/~) ' ( 1 -- Zliq + ~ )
Zliq =/3 + (Zli q + el3)(Zli q + o'~)
q13
(1.139)
where
/3-
~Pr Tr
,
q-
¢"~(Tr) , ~-~rr
Tr -
r ,/Dr Tc
=-
P Pc
Iterative calculations using the implicit Eqs. (1.139) start with an initial value of Zvap close to one and an initial value of Zli q close to zero. Once the value of compressibility is obtained, then volume is obtained from V = ZRT/P. Table 1.3 lists the parameters for the generic equation of state. (*n-butane Redlich-Kwong equation*) R = 83,14; (*bar cm3/Mol K*) T = 400 (*K*) P = 15; (*bar*)
26
1.
Fundamentals of equilibrium thermodynamics
Table 1.3 Parameters for the generic equation of state
z~
Equation of state
a(Tr)
o-
van der Waals Redlich-Kwong Soave-Redlich-Kwong Peng-Robinson
1 Tr-1/2 al a2
0 1 1 1 + ]2
0 0 0
1/8 0.08664 0.08664 0.07779
1 - ]2
27/64 0.42748 0.42748 0.45724
3/8 1/3 1/3 0.3074
al = [1 + (0.480 + 1.574o) -0.176o)2)(1 - Tr-1/2)] 2. a2 = [ 1 + (0.37464 + 1.54226o) - 0.26992o)2)(1 - Tr- 1/2)]2. o) (Omega) is the accentric factor. Smith et al. (2005)
Source:
Vo = R*T/P (*cm3*) Zo = 0.9; Tc = 425.1; Pc -- 37.96; T1 = T/Tc; Pr = P/Pc; bb = 0.08664*Pr/T1; q = 0.42748"(T1"(- 1/2))/(0.08664"T1); FindRoot[Z -- = 1+bb-q*bb*(Z-bb)/(Z*(Z+bb)), {Z, Zo}] Clear 2217.07 initial volume {Z~0.815911} V = R T Z / P = 1808.90 c m 3
1.10.1
Joule Thomson Coefficient
The Joule Thomson coefficient is the ratio of the temperature decrease to the pressure drop, and is expressed in terms of the thermal expansion coefficient and the heat capacity
OT)
(OH/OP)T _ _ V ( 1 - a T ) /, :-
(0H/0r)
(1.140)
-
Example 1.11 Entropy of a real gas Determine the entropy of a real gas. Solution: Using
(OAIOV)r = - P or (OG/OP)r= V, we have A (T,V,n)= A (T, Vo,n) -
pdV
(1.141)
The difference between real and ideal systems of the Helmholtz energies yields v
Areal
(T,V,n)-Aideal (T,V,n)=-I (Preal-Pideal)dV
(1.142)
By using the van der Waals EOS in Eq. (1.142), we obtain Areal--A i d e a l - a ( ~ ) - n R T l n (
V-nb
(1.143)
where Aidea 1 = gidea 1 -- TSidea 1 = gidea 1 - n T
s O + C v InT + R
In
Using Eqs. (1.143) and (1.128) in A r e a l - - g r e a l - TSreal, we can calculate the entropy of a real gas. For example, using the van der Waals EOS, we get
27
1.10 Equations of state
Sreal =
n[s o +C v l n T + R ln(V-nb)]n
(1.144)
The ideal gas form of S is
[
Sidea1 = n s o + Cv In T + R In
(1.145)
Comparison of Eq. (1.144) with Eq. (1.145) shows that the van der Waals gas entropy has ( V - nb) instead of V.
Example 1.12 Chemical potential of a real gas Similar to Eq. (1.142), the Gibbs free energy for a real gas is
Greal(T,P,n)-Gideal(T,P,n ) --
(1.146)
.foP (Vreal -- Videal)dP
Using the definition for chemical potential/.t = (OG/On)p,Tand Eq. (1.146), we get ~real
(T,P)-/~ideal (T,P) = So (/Areal --
(1.147)
Videal)dP
Using the compressibility factor Z, the volume of a real gas is Vreal= ZRT/P. Therefore, the chemical potential in terms of Z in Eq. (1.147) is
~real(T,P)-Id, ideal(T,P)= R T f ~ ( Z - 1
dP
(1.148)
The chemical potential can also be expressed in terms of fugacityf (1.149)
/Zreal (T,P)-/Zideal (T,P) = R T l n ( f ) where ]Zidea 1(T, P) = / z 0 (T, P0 ) 4- R TIn P
and
In
-
P
From various EOSs, it is possible to determine the compressibility and hence the chemical potential of a real gas. For example, using the virial equation, which is valid for gases at low density, we have
Z-
PV - 1+ B'(T)P +C'(T)P 2 +... RT
(1.151)
By using Eq. (1.151) in Eq. (1.148) and ignoring the term with p2, we have #rea] (T, P) = #ideal(T, P) + RTB'(T)P +...
(1.152)
In terms of the virial coefficient B(=B'RT), the approximation in Eq. (1.152) becomes /Zreal (T, P) = ~ideal
(T, P) 4- BP +...
(1.153)
Similarly, by using the van der Waals EOS and the chemical potential from the Helmholtz energy/z(V,P) = (OA/On)v,v, we obtain (C
1 ]RT-T I.t(c,T) = (U o - 2ac)+| -__~v+ 1-bc ~R where c = n/V
[
so + C v l n T - R l n
c
(1.154)
28
1.
1.10.2
Chemical
Affinity
Fundamentals of equilibrium thermodynamics
of Real
Gases
The affinity of a chemical reaction is A = -~pi[.£i, where b,i is the stoichiometric coefficient, which is negative for reactants and positive for products. For a real gas mixture, the affinity is determined by P
Areal
-~ Aidea' -- Z
VifO (V/'real -
Vi'ideal )dP
(1.15 5)
i
Using the relation of RT/P for an ideal gas and an appropriate EOS for the volume of a real gas, the affinity for a real gas may be determined. 1.10.3
The
Clapeyron
Equation
In the case of a pure compound, vaporization and melting processes need latent heat to be supplied at certain temperatures. Still, under isothermal and isobaric conditions, various phases of a compound can coexist at phase equilibrium. A compound can be at various phases under suitable pressure and temperature. Phase diagrams (see Figures 1.3-1.5) show these different phases. For example, using a P versus T diagram (Figure 1.3), one can find the melting and vaporization temperatures at certain pressures. The saturation curve shows the temperature and pressure at which two phases are in equilibrium with each other. A Txy diagram for a binary mixture (Figure 1.6) can display the boiling temperatures for changing mixture compositions at a specified pressure. Chemical potentials of a species in vapor and liquid phases are equal at phase equilibrium. Entropy production due to the irreversible processes at equilibrium must be zero, and hence the affinity of liquid-vapor conversion must vanish for each compound
(1.156)
A=/Xva p ( T , P ) - t Z l i q ( T , P ) = 0, and /Zvap(T,P ) = ~liq(T,P)
The Clapeyron equation relates pressure to temperature, and hence boiling or melting points can be calculated with changing pressure. By using Eq. (1.156), we can equate the Gibbs-Duhem equation for two phases Carbon Dioxide: Temperature - Pressure Diagram 10000.0 ............................... i
t ................................
......................... i
........................... t
1000.0
........................................ ~ .................................
! .................................................... I
i
I........
......
i Liquid L
100.0
¢¢t
L_ ............................................~........................................
el
i
lO.O iii il .............................................; .................................. !
ii
iiiiiiiiiii iiii illI
....................................
............... I............................... t......................i............................... I~
i
i
i Vapori
1.0 ....~......
......i ..... i ....... i
0.1
.....
.................................... ~..............................
i ........................................ I .........................I ..........................I .................................... I,,
CO~Ta~ V1.0
Drawn
......
i
q.~
-100
-90
-60
-70
-6O
-50
-40
-30
-20
-10
0
10
20
30
40
Temperature. °C Figure 1.3. Temperature pressure diagram for carbon dioxide (with permission from Chemicalogic Corporation).
1.10
29
Equations of state
Carbon Dioxide: Pressure - Entlmlpy Diagram I .000
i
#
i ~'~i 1
/ ~'tl 1. Table 2.3 shows the units that are commonly used for thermal conductivity and heat transport.
Example 2.4 Estimation of heat flow through a composite wall with constant thermal conductivities A pipe with an outside diameter of 10 cm and a length of 110 m is carrying hot fluid. The pipe is insulated with 0.5 cm thick silica foam and 10 cm thick fiberglass. The pipe wall is at 120°C and the outside surface of the fiberglass is at 30°C. Estimate the heat flow in the radial direction of the pipe. The thermal conductivities of silica foam and fiberglass are 0.055 and 0.0485 W/(m K), respectively. Solution: Assume radial heat flow only, and the thermal conductivities of the insulation layers are constant. For heat flow in the radial direction of a pipe only, Eq. (2.26) is
dT qr = -kA ~ dr
(2.36)
2.4
Transportphenomena
65
// q
q
q
r4
Temperature profile
Temperature profile (a)
(b)
Figure 2.5. Insulation layers around (a) a pipe and (b) a slab.
where A (=27rrL) is the surface area of the pipe with a length L normal to the heat flow qr in the radial direction. Inserting the area relation and integrating Eq. (2.36) between an inner radius, ri, and outer radius, r o, we find (2.37)
qr = 27rkL(Ti - T° )
ln(ro/~) If we have two layers of insulation with thermal conductivities of ka and kb around a pipe (Figure 2.5), then Eq. (2.37) becomes qr =
2 7rL(T i - T o ) overall
(2.38)
ln((r2/t] )/k a ) -~ ln((r3/r2)/k b )
Heat flow for a composite slab (Figure 2.5) is
qr
(Ti -- To ) overall ( ~ ( a / k a M) + (~¥'b/kb N) -t- (z~kXc/kcA)
(2.39)
Equation (2.38) is used in the solution of the problem qr =
2rr(110)(120- 30) l n ( ( 0 . 0 5 5 / 0 . 0 5 ) / 0 . 0 5 5 ) + ln((O. 105/0.055)/0.0485)
= 4126.8W
Another example of one-dimensional heat flow is the radial heat flow through the wall of a hollow sphere. Starting with Eq. (2.36) dT
qr = - k A
dr
and the area normal to the radial heat flow ofA = 47rr 2, we have qr -- - k 4 7 r r
2 dT
(2.40)
dr
By integrating this equation with the boundary conditions: r-
ri,
r-ro,
T-
Ti
T-To
we get qr = - 4 rrk
~-Vo (1/ri ) + (1/ro)
(2.41)
66
2.
Transport and rate processes
For one-dimensional steady-state heat conduction in the x-direction, we have a general relation for the temperature profile obtained from (V2T = 0) -- xs dx
= 0
(2.42)
where s is the shape factor: s = 0 for a rectangular shape, 1 for a cylindrical shape, and 2 for a spherical shape. Using the boundary conditions in Eq. (2.42), the temperature profiles are Rectangular shape T(x)-T2-Tlx+T L
1,
a t x = 0 , T = T 1 and x = L , T = T
(2.43)
2
Cylindrical shape Ti . - .T° l n r ,
T ( r ) . T i.
ln(ro/r~)
rl
a t r = r i , T = T i and r = r o , T = T O
(2.44)
Spherical shape
T ( r ) = Ti -
(2.45)
a t r = r i , T = T i and r = r o , T = T o
1/ri _ 1/ro
The temperature profiles are linear for a rectangular shape, logarithmic for a cylindrical shape, and hyperbolic for a spherical shape. The thermal resistances for various shapes in estimating the one-dimensional heat flow with constant thermal conductivity are Geometry Thermal resistance
Rectangular
Cylindrical
Spherical
L/Ak
ln( (ro/ri)/27rLk)
(ro/ri)/47rrirok
Example 2.5 Estimation of heat flow with temperature-dependent thermal conductivity The temperatures at the surfaces of a 0.2 ft thick rectangular box are 40°F and 120°E The box is filled with air. The thermal conductivity of the air is a linear ftmction of temperature: k = k0(1 + aT) with k = 0.0140 Btu/(ft h °F) at T = 32°F and k = 0.0183 Btu/(ft h °F) at T = 212°E Estimate the heat flow and the temperature profile in the air when the resistances at walls are negligible. Solution: Assume that the resistances at the walls are negligible, and the system is at steady state and has one-dimensional heat flow. k = 0.0140Btu/(fth°F)
atT = 32°F
k = 0.0183Btu/(fth°F)
atT = 212°F
Using the k values above, we find the parameters: k0 = 0.0134 and a = 0.001719, which are valid between 32 and 212°E The heat flow is q A
-k -
dT
dx
---,
q ix=2 A x=O d x
:
-k
f T2=120°F o a T~=32°F (1 + a T ) d T :
-k o
I
a T2
T +-~
ll20Of
:-6.669Btu/(ft 132°F
2 h)
2.4
Transportphenomena
67
The temperature profile is obtained from
q - - k dT ~ - -qx = - k o ( T + a2T) A dx A 2 D
q _ _ k ( T 1-T2) A L where k is an average value for the thermal conductivity estimated as 120°F
T2 - T 1
r 2 -1"1
2
= 0.0151Btu/(ft h °F) 132°F
Using the value of k in the heat flow equation above, we determine the temperature profile
aT2 + T + 0.0151(T1 - T2) 7x = 0 2 This quadratic equation will have two solutions and the solution with the positive sign must be chosen to satisfy the boundary conditions - 1 + x/1 - 4(a/2)[0.015 I(T1 - T2)x/L] T
2.4.9
m
Estimation of Thermal Conductivity
Thermal conductivity can vary from - 0 . 0 1 W/(m K) for gases to --~1000 W/(mK) for pure metals. Tables 2.4-2.6 show some experimental values of thermal conductivities. When available, experimental values should be used in calculations; otherwise, several empirical relations may provide satisfactory predictions. Table 2.4 Thermal conductivities, heat capacities, and Prandtl numbers of some gases and liquids Substance Gases at 1 atm pressure Hydrogen, H2 Oxygen, 02
Carbon dioxide, CO~_ Methane, CH 4 NO
r (K)
k (W/(m K))
Cp (kJ/(kg K))
Pr
100 300 100 300 200 300 200 300 200 300
0.06799 0.1779 0.00904 0.02657 0.0095 0.01665 0.02184 0.03427 0.01778 0.02590
11.192 14.316 0.910 0.920 0.734 0.846 2.087 2.227 1.015 0.997
0.682 0.720 0.764 0.716 0.783 0.758 0.721 0.701 0.781 0.742
300 350 400 250 300 350 250 350 250 300 350 200 300
0.6089 0.6622 0.6848 0.1808 0.1676 0.1544 0.1092 0.0893 0.1478 0.1274 0.1071 0.1461 0.1153
Liquids at their saturation pressures Water, H20
Ethanol, C2HsOH
Carbon tetrachloride, CC14 Diethyl ether, (C2Hs)20
1-Pentene, CsH~0
4.183 4.193 4.262 2.120 2.454 2.984 0.8617 0.9518 2.197 2.379 2.721 1.948 1.907
6.02 2.35 1.35 35.8 15.2 8.67 16.0 5.13 5.68 4.13 3.53 8.26 3.72
Source: Bird et al. (2002); Data Compilation of Pure Compound Properties, Design Institute for Physical Property Data, AIChE, New York, NY (2000).
68
2.
Transport and rate processes
Table 2.5 Thermal conductivities, heat capacities, and Prandtl numbers of some liquid metals at atmospheric pressure
Metal
T (K)
k (W/(m K))
Cp (kJ/(kg K))
Pr
Mercury, Hg
273.2 372.2 977.2 366.2 644.2 422.2 700.2
8.20 10.50 15.1 86.2 72.8 45.2 39.3
0.140 0.137 0.0146 0.0138 0.0130 0.795 0.753
0.0288 0.0162 0.013 0.011 0.0051 0.0066 0.0034
Lead, Pb
Sodium, Na Potassium, K
Source: Bird et al. (2002); Liquid Metals Handbook, 2nd ed., U.S. Government Printing Office, Washington, DC (1952); E.R.G. Eckert and R.M. Drake, Jr., Heat and Mass Transfer, 2nd ed., McGraw-Hill, New York (1959).
Table 2.6 Thermal conductivities of some solids
Solid
T (K)
k (W/(m K))
Aluminum, A1
273.2 373.2 273.2 291.2 373.2 273.2 373.2 291.2 373.2 273.2 373.2
202.5 205.9 387.7 384.1 379.9 55.4 51.9 46.9 44.8 418.8 411.9
Copper, Cu
Cast iron, Fe Steel Silver
Source: Bird et al. (2002); Griskey (2002); Reactor Handbook, Vol. 2, Atomic Energy Commission AECD-3646, U.S. Government Printing Office, Washington, DC (1955).
2.4.10
Effect of Temperature and Pressure onThermal Conductivity
In the corresponding states approach, the reduced thermal conductivity, kr = k/kc, is plotted as a function of the reduced temperature and the reduced pressure, as shown in Figure 2.6, which is based on a limited amount of experimental data for monatomic substances, and may be used for rough estimates for polyatomic substances. Figure 2.6 shows that the thermal conductivity of a gas approaches a limiting function of T at --~1 atm pressure. Thermal conductivities of gases at low density increase with increasing temperature, while they decrease with increasing temperature for most liquids. This correlation may change in polar and associated liquids. For example, water exhibits a maximum in the curve of k versus T. The corresponding states provide a global view of the behavior of the thermal conductivity of fluids.
Example 2.6 Estimation of thermal conductivity at specified temperature and pressure Estimate the thermal conductivity of methane at T = 380 K and P = 30 atm. At 300 K and 1 atm, the thermal conductivity is
0.03427 W/(m K). Solution: Assume Figure 2.6 can be used for polyatomic gases. Use the critical parameters and estimate the reduced pressure and temperature to use from Figure 2.6 for reading the reduced thermal conductivity approximately. For methane Tc = 190.7 K and Pc = 45.8 atm. The reduced conditions are Tr = 300/190.7 = 1.57 and Pr = 1/45.8 = 0.022. Using these reduced properties in Figure 2.6, we read the approximate reduced thermal conductivity as kr = 0.5. Estimate the critical thermal conductivity kc = 0.03427/0.5 = 0.0684 W/(m K).
2.4
Transportphenomena
69
10 .
.
.
.
8
7
-40~.,,
3o b-
I
10~
3
~
......
¢o
!1 I \ \ , ' ~ ' - - - ~ ~ "
\ l.s rf
07 0.6
i '1 0 ~ ~ v
o.,
o.3
0.2
o.1 S 0.3
~~d~/q ~"
0.4
0.6
0.8 1.0
2
3
4
5
6 7 8 910
Reducedtemperature,T r = T I T c Figure 2.6. Change of reduced thermal conductivity with reduced temperature and reduced pressure for monatomic substances [O.A. Hougen, K.M. Watson and R.A. Ragatz, Chemical Process Principles Charts, 2nd ed., Wiley, New York (1960)].
At the specified temperature and pressure the reduced conditions are Tr _ 380 _ 2.0 and Pr - 30 - 0 . 6 5 190.7 45.8 With the reduced parameters, Figure 2.6 yields kr = 0.66 (approximately). Therefore, the predicted value is k = kckr = 0.0684(0.66) = 0.0451W/(m K). The predicted values for polyatomic gases may not be satisfactory as Figure 2.6 is based on a limited set of data from monatomic gases only.
2.4.11
Thermal Conductivity of Gases at Low Density
The thermal conductivities of dilute monatomic gases are well understood. The thermal conductivity of a dilute gas composed of rigid spheres of diameter d is expressed as k -
2 x/vTmnT 3vr
-
-
7rd2
C~
( m o n a t o m i c gas)
(2.46)
where m is the mass of a molecule and K the Boltzmann constant. Equation (2.46) predicts that k is independent of pressure, and the prediction is satisfactory up to --- 10 atm for most gases. The predicted temperature dependence is weak.
70
2.
Transport and rate processes
The Chapman-Enskog formula for monatomic gases at low density and temperature T produces better predictions, and is given by k = 1.9891 × 10 . 4 x/-T/M
O-2~-~k
(2.47)
where k is in cal/(cm s K), T in K, o- in A~, and the collision integral for thermal conductivity, Ok, is identical to that for viscosity, 1~,. The values of collision integrals are given for the Lennard-Jones intermolecular potential as a function of the dimensionless temperature KT/s in Table B2. A simple semiempirical equation for polyatomic gases at low densities is given by k=
+-~R
(2.48)
when Cp is in cal/(mol K), R in cal/(mol K), and ~ is the viscosity in g/(cm s). This equation is called the Euckenformula, and it can provide a simple method of estimating the Prandtl number for nonpolar polyatomic gases at low density P r - Cptx_
k
Cp Cp + (5/4)R
(2.49)
The thermal conductivities for gas mixtures kmix at low densities are
kmix --- ~ i:1
xiki ZjXj~ij
(2.50)
where x i is the mole fraction of species i and ki the thermal conductivities of the pure gases. The values of ~/j are identical to those appearing in Eq. (2.24) in the viscosity equation.
Example 2.7 Estimation of thermal conductivity of monatomic gases Estimate the thermal conductivity of helium at T = 400 K and P = 1 atm and compare with the experimental value of 4.41 cal/(cm s K) (W.K. Saxena, S.C. Saxena, J. Phys. DAppl. Phys., 1 (1968) 1341). Solution: Assume that the pressure is low. Use Eq. (2.47): k = 1.9891 × 10 .4 ,~-f/M or2[~k where k is in cal/(cm s K) and T in K. Read the Lennard-Jones parameters for helium from Table B 1 and B2" o" = 2.576, -e = 10.2 K, and M = 4.003 k At T = 400 K, we have TK/e = 400/10.2 = 39.21. From Table B2, we have the approximate value for collision integral, 12k = 0.673. From Eq. (2.47)" k = 1.9891 × 10 . 4
x/400/4.003 = 4.45×10 .4 cal/(cms K) 2.5762(0.673)
Comparing this result with the experimental value of 4.41 × 10-4 cal/(cm s K) yields a deviation of 0.9%, which is low.
2.4
71
Transportphenomena
Example 2.8 Estimation of thermal conductivity of polyatomic gases Estimate the thermal conductivity of NO at 200 and 300 K at low pressure. Use the data in the following table" T (K)
Cp (cal/(mol K))
200 300
7.283 7.154
Solution:
Assumptions" The pressure is low. Read the Lennard-Jones parameters for NO from Table B 1 and B2: o- = 3.47, -e __ 119 K, M = 30.01 k With these parameters, we calculate the values of TK/e and the collision integrals, and use them in the following equation (Eq. 2.20) for estimating the viscosity" /z = 2.6693x10 -5 ~ / M T _ 2.6693×10_ 5 ~/30.01(200) = 1"363X10- 4 g/(cms) O"2~-~ 3.472 (1.260) A semiempirical Eucken equation is used to estimate the thermal conductivity of polyatomic gas with R = 1.987 cal/(mol K). k = Cp +-~R
=(7.283+2.484) 1"363×10-4 = 0.0185 W/(m K) 30.01
The results are shown in the following table" T (K)
200 300
Cp,ex p (Table 2.4) (cal/(mol K))
TK/e(K)
7.283 7.154
1.68 2.52
~ - ~ 1.260 1.093
/~ X 104 (g/(cm s))
kest (W/(m K))
kexp(Table 2.4) (W/(m K))
Deviation (%)
1.363 1.924
0.0185 0.0258
0.01778 0.0259
4.0 0.0
Example 2.9 Estimation of thermal conductivity of gas mixtures at low density Estimate the thermal conductivity of the following gas mixture at 293 K and 1 atm using the data given in the following table: Species
Yi
M
k (W/(m K))
/x (mPa s)
Air Carbon dioxide
0.5 0.5
28.97 44.01
0.026 0.01601
0.0181 0.0146
Solution: Assume that the system is at low density. Thermal conductivities for gas mixtures kmi×at low densities may be estimated from Eq. (2.50)" kmi x -- ~ i=1
yiki EjYj*ij
where x i is the mole fraction of species i and k i the thermal conductivities of the pure gases. The coefficients ~ij are identical to those appearing in Eq. (2.24) in the viscosity equation
,
(,+ i 'J21 / /TM ]2
72
2.
Transport and rate processes
i
j
Mi/Mj
Mj/M i
t.zi/[Lj
1
1 2 1 2
1 0.658 1.519 1
1 1.519 0.658 1
1 1.2397 0.8066 1
2
~1 £jYj% Yiki
k m i x = "=
0.5(0.026) + 0.5(0.0160) 0.806
0.701
f~iJ" 1 0.613 0.403 1
22j=lYj~ij 0.806 0.701
= 0.0161+0.0114 = 0.0275 W/(mK)
No result is available for comparison at these conditions.
2.4.12
Estimation of Thermal Conductivity of Pure Liquids
Based on the theory of Bridgrnan, the following expression can be used to estimate the thermal conductivity of a pure liquid: N)
k - 2.80 7-
/3
KVs
(2.51)
where N is Avogadro's number, K the isothermal compressibility [K =-1/V(O V/OP)T], Vthe molar volume, and vs the sonic velocity defined by
/ Cp
OP
lj2 (2.52)
U V where p is the density and the quantity (OP/Op)Tmay be obtained from isothermal compressibility data.
Example 2.10 Estimation of thermal conductivity of pure liquids Estimate the thermal conductivity of water at 300 K and 1 atm. The density of liquid water at 300 K is 995.7kg/m 3, and its isothermal compressibility u = -1/p(Op/OP)T is 0.4477 × 10-9 m2/N (= pa-1). Solution: Assume that Cp = Cu for water. Estimate
OP _ 1 _1 = 2.24 × 106 m 2/S 2 Op p(1/p)(Op/OP) 995.7(0.4477 × 10-9) The speed of sound is
Cp OP Vs=-~v-~p r
= 1497.7 m/s
Using the Avogadro number, the Boltzmann constant, and the molar volume of water (M/p = 0.01807 m3/kg), we have k = 2.80(6"023 × 1026 kg mol 2/3 1.3805 × 10.23 J/K (1497.7 m/s) = 0.599 W/(m K) 0.01807 m3/kg The experimental value of k at 300 K is 0.608 W/(m K) (Table 2.4), and the deviation is --~1.4%.
2.4 Transportphenomena 2.4.13
73
Mass Transfer
We consider a thin, horizontal, fused-silica plate of area A and thickness H. Initially, both horizontal surfaces of the plate are in contact with air (Figure 2.3c). We assume that the air is completely soluble in silica. At time t - 0, the air below the plate is replaced by pure helium, which is appreciably soluble in silica. The helium slowly penetrates into the plate by molecular motion, and eventually appears in the air above the plate. This molecular transport of one substance relative to another is known as d![jFusion. The air above the plate is removed rapidly, so that there is no measurable helium concentration there. In this system, the index i shows helium andj shows silica, and the concentrations are given by the mass fractions w; and w/, respectively. Eventually, the concentration profile tends toward a straight line with increasing t, and we have w ; - w~0 at the bottom surface, and w i = 0 at the top surface of the plate. At steady-state diffusion, the molar flow vector that is the flow rate of helium per unit area j~ is proportional to the concentration gradient in terms of the mass fraction V w i (2.53)
Ji = - P D ! j V w i
where p and D~9 are the density and the diffusivity coefficient of the silica-helium system, respectively. Similarly, we relate the molar flow vector to the concentration gradient by (2.54)
Ji = - c D ! j V x i
Equations (2.53) and (2.54) are called Fick's first law of diffusion, and indicate that mass flows from a high to a low concentration region. It is valid for any binary fluid or solid solution, provided that j; is defined as the mass flow relative to the mixture mass average velocity v defined by v-
1 P ~_~piv i = ~ . w , v i i
(2.55)
i
and in general we have
(2.56)
Ji - - P W i (Vi -- V)
Here the term ( v i - v) is called the diffusion velocity. The mass flow ]j is defined analogously. As the two chemical species interdiffuse, there is a shifting of the center of mass in the y-direction if the molecular weights of components i andj differ. The flows Ji and jj are measured with respect to the motion of the center of mass, and Ji + J j - 0. Molar average velocity is 1
VM
--'--ECiVi --EXiVi C
t•
(2.57)
i
and in general we have Ji
-- Ci(Vi
(2.58)
--VM)
We can use the molecular mass flow vector j; and the convective mass flow vector pv to define the combined mass flow vector ni, and combined molar flow vector Ni n; - ji + piv
(2.59)
Ni = Ji + civ
(2.60)
Equation (2.59) is the mass flow in terms of mass average velocity v, and using the diffusion velocity (v;- v), we obtain Ji - - P i ( V i
- - V) - - n i - pi v - - - p D ( / V w
Ji = ci(vi - VM) = Ni - civ = - c D i j V x i
For pair i-j there is just one diffusivity coefficient D!i = D/;.
i
(2.61)
(2.62)
74
2.
Transportand rateprocesses
Other type of mass flow is the one that is relative to the plane of no net volume flow. When JV,A and JV,B are the vectorial molar mass flows of species A and B relative to the plane of no net volume flow, we have
JV,AVA + Jv,BVB = 0
(2.63)
where VAand VB are the partial molar volumes of species A and B in the mixture. The mass flows are
vB
JV,A----~-JA
and
VA
JV,B=--V--JB=0
(2.64)
where V is the molar volume of the mixture. When the partial molar volumes are equal to each other, then JV, A = JA"
Example 2.11 Mass flow across a stagnant film Consider that a liquid A in a tube with a large diameter is evaporating into a stagnant gas B. Derive the relations for mass flow and the concentration profile. Assume that the liquid level is maintained at y = Yl. Solution: Assume that at the liquid-gas interface, the concentration of A is XA1, which is the gas-phase concentration of component A corresponding to equilibrium with the liquid at the interface. The mole fraction XA1 is the vapor pressure of A divided by the total pressure provided that A and B form an ideal gas mixture and that the solubility of gas B in liquid A is negligible. A stream of gas mixture A-B of concentration XA2 flows slowly past the top of the tube, to maintain the mole fraction of A at XA2.The entire system is kept at constant temperature and pressure. There is a net flow of gas upward from the gas-liquid interface. The transport process is in the y-direction and at steady state with no convective mass transfer, and the reaction source is
-- VNAy =
0
or
dNAy - 0 dy
(2.65)
For a molecular mass transport of component A, we have the mass flow in the y-direction
NAy=JAy+XANAy or NAy--
Jay
(2.66)
1 m XA
Substituting Eq. (2.66) in Eq. (2.65) and using Fick's law of diffusion, we find
d(CDAB dXA) -- 0 dy 1 - x A dy
(2.67)
For an ideal gas mixture c = P/RT, so that at constant temperature and pressure c must be constant. Furthermore, for a binary gas system, DAB is very nearly independent of the composition. Therefore, the product cDAB is constant, and Eq. (2.67) reduces to
d( 1
dXA) = 0 dy 1--x A dy
(2.68)
This is a second-order differential equation for the concentration profile expressed in terms of mole fraction of component A. With the boundary conditions Y = Y l , XA -- XA1 Y = Y2, XA = XA2 integration of Eq. (2.68) yields
--ln(1--XA)=IlY+I 2
(2.69)
2.4
75
Transportphenomena
After determining the integration constants Il and I2, the final solution becomes
2.4.14
1--XA
1-- XA2 exp( Y -- Y] )
I--XA1
1--XA1
(2.70)
Y2 --Yl
Estimation of Diffusivities
In general, diffusivity depends on pressure, temperature, and composition. With respect to the mobility of molecules, the diffusion coefficients are generally higher for gases and lower for solids. The diffusivities of gases at low densities are almost independent of concentration, increase with temperature, and vary inversely with pressure. Liquid and solid diffusivities are strongly concentration dependent and generally increase with temperature. Tables 2.7 and 2.8 show some of the experimental binary diffusivities for gas and liquid systems.
Table 2.7 Molecular diffusivities of some gases at atmospheric pressure
Gas pair
(i-j)
Air-ammonia Air-carbon dioxide Air-ethanol Air-n-octane Air-water Air-chlorine Air-benzene Air-naphthalene Air-hydrogen Carbon dioxide-benzene Carbon dioxide-ethanol Carbon dioxide-methanol Carbon dioxide-water
Carbon dioxide-carbon monoxide Hydrogen-ammonia
Hydrogen-benzene Hydrogen-ethane Hydrogen-water Oxygen-ammonia Oxygen-benzene Oxygen-ethylene Ammonia-hydrogen
Propane-iso-butane Propane-n-butane
T (K)
D~/(cm2/s)
273 273 317.3 298 313 298 298 313 273 298 298 298 318 273 298.6 298 307.2 328.6 273.2 273 293 358 273 311.3 273 293 293 296 311.3 293 263 298 298 378.2 378.2 437.7
0.198 0.136 0.177 0.132 0.145 0.0602 0.260 0.288 0.124 0.0962 0.0611 0.410 0.0715 0.0693 0.105 0.164 0.198 0.257 0.139 0.745 0.849 1.093 0.317 0.404 0.439 0.850 0.253 0.094 0.101 0.182 0.58 0.233 0.0439 0.0823 0.0768 0.107
Source: Bird et al. (2002); Griskey (2002); J.O. Hirschfelder, C.E Curtiss, and R.B. Bird, Molecular Theo~ of Gases and Liquids, 2rid ed., Wiley, New York (1964); Reid et al. (1987); S. Gotoh, M. Manner, J.P. Sorensen and WE. Stewart, J Chem. Eng. Data, 19 (1974) 169; Data Compilationof Pure CompoundProperties,Design Institute for Physical Property Data, AIChE, New York, NY (2000).
76 ..
2.
Transport and rate processes
Table 2.8 Diffusivities in some liquid mixtures Liquid pair; solute A (concentration in g mol/1)-solute B
T (K)
DAB × 105 (cm2/s)
Ammonia (3.5)-water Ammonia (1.0)-water Ethanol (3.75)-water Ethanol (2.0)-water Ethanol (0.05)-water Chloroform (2.0)-ethanol Sodium chloride (0.05)-water Sodium chloride (0.2)-water Sodium chloride (1.0)-water
278 288 283 289 283 293 291 291 291
1.24 1.77 0.50 0.90 0.83 1.25 1.26 1.21 1.24
A-B
T (K)
XA
DAB × 105 (cm2/s)
Chlorobenzene-bromobenzene
283.3 283.3 283.3 283.3 313.1 313.1
0.0332 0.2642 0.5122 0.7617 0.0332 0.2642
1.007 1.069 1.146 1.226 1.584 1.691
Water-n-butanol
303.2 303.2 303.2
0.131 0.222 0.358
1.240 0.920 0.560
298.15 298.15 298.15 298.15 298.15
0.026 0.266 0.408 0.680 0.880
1.076 0.368 0.405 0.743 1.047
Ethanol-water
Source: P.A. Johnson and A.L. Babb, Chem. Rev., 56 (1956) 387; P.W.M. Rutten, Diffusion in Liquids, Delft University Press, Delft, The Netherlands (1992); A. Vignes, I.E.C. Fundam., 5 (1966) 189; M.T. Tyn and W.E Calus, J. Chem. Eng. Data, 20 (1975) 310.
The molecular theory yields the self-diffusivity of component i at low density
Dii -
2 4'n'miKT 1 3rr
~
-
-
,rrd2
(2.71)
p
Equation (2.71) can be compared with Eq. (2.46) for the thermal conductivity of gases, and with Eq. (2.19) for the viscosity. For binary gas mixtures at low pressure, D O• is inversely proportional to the pressure, increases with increasing temperature, and is almost independent of the composition for a given gas pair. For an ideal gas law P = c R T , and the Chapman-Enskog kinetic theory yields the binary diffusivity for systems at low density
D 0. = 0.0018583 ~/T3 (1/Mi + 1~My )
(2.72)
2 Po'ijaDij
where 1
O"U = "~(0" i + O ' j ) ,
~
~
K
=
Si
--
,
12ij = f
KT
K
Here D,y in cm2/s, o-0.in A, T in K, and P in atm. The dimensionless quantity DDij is the collision integral for diffusion, and is a function of the dimensionless temperature KT/eo..The parameters o-• and e~ are those appearing in the LennardJones potential between molecules i andj.
2.4
77
Transportphenomena
Example 2.12 Estimation of diffusivity in a gas mixture at low density Estimate the diffusivity of benzene in air at 25°C and 1 atm. Solution: Assume that the system pressure is low. Use the Chapman-Enskog equation for the binary diffusivity at low density (Eq. 2.72):
Oij
=0.0018583
~/T3 (1/A4i -+-l / J~,[j) pcr2 f~~.:
The data and critical properties are
Species Benzene (A) Air (B)
M
(2-(A)
e/K (K)
Tc(K)
Pc (atm)
78.11 28.97
5.443 3.617
387 97
562.6 132
48.6 36.4
CrAB = ~(CrA + crB) = 4.53A, eAB = K KT
298
eAB
193.7
= 193.7
-1.538
The approximate collision of integral f~0 =flKT/eij)= 1.187 DAB = 0.0018583 X/2983(1/28"97 + 1/78.11i = 0.0854 cm2/s (1)(4.53)21.187 From Table 2.7, experimental diffusivity is = 0.0962 cm2/s, and deviation is --- 11.2%.
2.4.15
Effect of Temperature and Pressure on Diffusivity
At low pressures, we use the following expression developed from a combination of kinetic theory and corresponding states approach to estimate the effect of temperature on diffusivity:
T D~/=a
, /
1 + 1 MC ~
(PciPcj)I/3 (TciTcj ) 5/12 P
(2.73)
Here D~j is in cm2/s, P in atm, and T in K, and the constants a = 2.745 × 10-4 and b - 1.823 for nonpolar gas pairs, excluding helium and hydrogen, and a = 3.64 × 10-4 and b = 2.334 for pairs consisting of water and a nonpolar gas. Equation (2.73) predicts the data at atmospheric pressure within an average deviation of less than 10%. A corresponding states plot of the self-diffusivity DAA,, which is the interdiffusion of labeled molecules of A at the low-pressure limit, is shown in Figure 2.7. The reduced self-diffusivity that is CDAA, at pressure P and T divided by CDAA,at the critical point is plotted as a function of the reduced pressure and reduced temperature. Figure 2.7 shows that CDAA,increases sharply with increasing temperature, especially for liquids. The values of CDAA, decrease toward a low-pressure limit at each temperature. We can use the following empirical relation for estimating the critical self-diffusivity between i and labeled species i* ( cDii. )c:
)1/2
(cD.,)c
- 2.96 X IO-6 ~
Mi
+~
Mi *
Pc2/3
Tic/6
(2.74)
78
2.
Transportand rate processes
/ /
-¢
2
1.5
II ~o
~ 1.0
L~ •~. 0.8
~ £~ow-pressure
~
limit
....
~"pr
/y;t7
Vapor / , ~ 1 ~ ~
= 10
p'~pr=5
"" .
.
.
.
0.6
"~
Two- I phase I 0.4 region#
/
t I
/Sat-u'rated liquid 0.2
i t
0.6
6.8 1.0 1.5 2 3 Reduced temperature, T r = T / T c
4
5
Figure 2.7. Change of reduced self-diffusivity with reduced temperature and reduced pressure [J.J. van koef and E.G.D. Cohen, Physica A, 156 (1989) 522; B.I. Lee and M.G. Kesler, AlChE J., 23 (1975) 510].
where c is in mol/cm 3, Dii in cm2/s, Tc in K, and Pc in atm. For binary diffusion of chemically dissimilar species at low pressure, we use (cDij)c -- 2.96 X 10 -6
~
+~
Mi
Mj.
)1/2(PciPcj)1/3 (TciTcj )1/12
(2.75)
Here the c-multiplied diffusion coefficients are used because their dependence on pressure and temperature is simpler, and they are frequently used in mass transfer calculations. We may calculate the diffusion coefficient at a specified temperature and a specified pressure from a known value by using the following relation:
a~ ( T2 ' P2 ) -- Dij ( TI ' P1) ~
~-Tll)
~-~D(T2)
For pressures below 25 atm and away from critical values, Eq. (2.76) yields reliable corrections. If gas species 1 is diffusing through a mixture of known composition, then the diffusion coefficient obtained by
(2.76)
Ol_mixture is
1
Ol-mixture
(Y2/D12)+(Y3/D13)+...+(Yn/Dln)
(2.77)
where Y2 is the mole fraction of species i in the mixture estimated on a component-i-free basis, and obtained from * _
Y
2
Y2 -- Zj=z yj
=
Y2
Y2 + Y3 + " "
+ Yn
(2.78)
2.4 Transportphenomena
79
Example 2.13 Estimation of diffusivity in a gas mixture at low pressure Estimate the diffusivity of carbon dioxide in benzene at 318 K and 1 atm. Solution: Assume that the pressure is low. Use Eq. (2.73) to find the binary diffusivity at low pressure:
~-'I'-~
Dij-a x/TcTT~/
M,
1
1/2
)1/3
(PciPcj (TciTcj ) 5/12
M,
P
The critical properties are Species Carbon dioxide (A) Benzene (B)
M
Tc (K)
Pc (atm)
44.01 78.11
304.2 562.6
72.8 48.6
1/2
(TcATcB)5/12=
151.55, ( 1 • ~,M A
1 MB
= 0.188,
( P c a e c B ) 1/3 - -
15.2
with a = 2.745 x 10 -4 and b = 1.823 for nonpolar gas mixture a (T/4TciTc/)
b=
2 . 6 1 6 × 1 0 -4 .
With P = 1 atm DAB = (2.616 × l0 --4 )(0.188)(15.23)(151.55) = 0.0739cm2/s Experimental diffusivity from Table 2.7 is DAB = 0.0715 cmZ/s, and deviation is 3.44%.
Example 2.14 Estimation of diffusivity in a gas mixture of isotopes Estimate the diffusivity of C 1 4 O in ordinary CO at 225 K and 172.5 atm. The measured value of DAA, -- 0.109 cm2/s at T = 194.7 K and P = 1 atm. Solution:
Assumptions- Assume that the critical value (DAA,)c obtained at P = 1 atm pressure can be used. For CO the critical properties are Tc = 133 K and Pc = 34.5 atm. As we have a measured value, use Figure 2.7 with the reduced temperature Tr = 194.7/133 = 1.46 and the reduced pressure Pr = 1/34.5 = 0.116 to read the approximate value of (CDAA,)r = 1.4. The value of concentration is c = critical value is
(CDAA*)C =
P/RT =
CDAA (CDAA*)r
6.25 x 10-5 mol/cm 3 (R = 82.05 atm cm3/(mol K)). Therefore, the
_ (6.25×10-5)(0.109)
= 4 . 7 6 3 × 1 0 .6 mol/(cms)
1.43
At the required reduced temperature and pressure" Tr = 225/133 - 1.7 and Pr = 172.5/34.5 = 5. If the approximate reduced value o f ( C D A # ) r - 1.5 mol/(cm s); the predicted value is CDAA* = (CDAa*)c (cDaA)r- (4.763 × 10-6)(1.5)= 7.145 × 10 -6 cm2/s The predictions may not be satisfactory as the critical value is estimated at a low pressure of 1 atm. The critical value may also obtained from Eq. (2.74)
(cDii*)c
= 2.96 X 10 -6
/
1
Jr" 1
Mi Mi*
/¸j2
pc~/3 _ 2.96 × 10 -6 ~ - J r Tl/6c, 28.01
~
3(}.01
1/2 34.52/3 - 3 . 6 4 8 × 1 0 -6 mol/(cms)
1331/6
80
2.
Transport and rate processes
Therefore, the predicted value is CDAA* -- (CDAA*)c
(cDAA*)r-- (3.648
× 10-6)(1.5) -- 5.472 × 10 -6 mol/(cms)
Example 2.15 Estimation of diffusivity in a gas mixture Estimate the diffusivity of an air-carbon dioxide mixture at 200 K and 103 atm. The measured value of DAB = 0.177 cm2/s at T = 317.3 K and P = 1 atm. Solution: Assume that the critical value (DAA.)C obtained at P = 1 atm pressure can be used. The critical properties are Species Carbon dioxide (A) Air (B)
M
Tc (K)
Pc (atm)
44.01 28.97
304.2 132
72.8 36.4
As we have a measured value, use Figure 2.7 with the reduced temperature T 317.3 Tr = ~/TcATc------~ = 4304,2(132 ) = 1.58 and the reduced pressure P
1 Pr = 4PcAPc------~ = 436.4(72.9 ) = 0.019 We can use Figure 2.7 to read the approximate reduced value of (CDAB)r = 1.65. The value of concentration: c = P/RT= 3.84 × 10-Smol/cm 3 (R = 82.05 atmcm3/(mol K)). Therefore, the critical value is CDAA*
(CDAA*)c =
_ (3.84 ×10-5)(0.177)
(CDAA*)r
= 4.119 × 10 -6 mol/(cm s)
1.65
At the required reduced temperature and pressure
T~
200 x/304.2(132 ) = 1.0, P~
103 ,]36.4(72.9) = 2.0
From Figure 2.7 we read the approximate reduced value of (CDAB)r = 0.87 mol/(cm s). The predicted value is CDAB -- (CDAB)c (CDAB)r -- (4.110 × 10-6)(0.87) -- 3.570 X 10 -6 cm2/s
The predictions may not be satisfactory as the critical value is estimated at a low pressure of 1 atm. The critical value may also obtained from Eq. (2.74) / 1/2
(°DAB)c = 2.96 X 10 -6 _ ~ 1 nt" _ ~ 1 MA
MB
)1/3
(PcA PcB
= 4.05 x 10 -6 mol/(cm s)
(TcATcB) 1/12
Therefore, the predicted value is
CDAB=
(CDAB)c (CDAB)r = (4.05 X 10-6)(0.87) = 3.52 X 10 -6 mol/(cm s)
2.4
81
Transportphenomena
Example 2.16 Estimation of diffusivity of a component through a gas mixture Estimate the diffusivity of carbon dioxide (CO2) through a gas mixture of benzene and methane with the known mole fractions given in the following table. The mixture is at 300 K and 2 atm. Welty et al. (1984)
Yi
T (K)
0.15
318
0.55 0.30
318 273
Species Carbon dioxide (1) Benzene (2) Hydrogen (3)
D1j (cm2/s)
P (atm)
0.0715 0.550
1 1
Solution:
Assume that Eq. (2.76) is used in temperature and pressure corrections of the experimental diffusivities by ignoring the collision integral correction since the temperatures are close to each other:
Dij ( T2 ' P2 ) = Dij ( TI ' P1) -~2
--~1)
Therefore, we have 3 0 0 ~ 3/2
D12 (300 K,2 atm) - 0.0715
D13(300K,2atm ) - 0 . 5 5
(,)(3oo] ~ ~,~]
= 0.0327 cm2/s
-0.316cmZ/s
We need to calculate the compositions of benzene and hydrogen on a carbon dioxide-free basis:
Ybenzene
=(
)
0.55 - 0.647 1 -- 0.15
and
Yhydrogen =
/ 0.3 ) 1 -- 0.15
= 0.353
Using these estimations in Eq. (2.77), we have Dl_mixtur e --
= 0.323 cm 2/s (0.647/0.0327) + (0.353/0.316)
No experimental values are available for comparison.
2.4.16
Diffusion in Liquids
Even the binary system diffusivities in liquid mixtures are composition dependent. Therefore, in multicomponent liquid mixtures with n components, predictions of the diffusion coefficients relating flows to concentration gradients are empirical. The diffusion coefficient of dilute species i in a multicomponent liquid mixture, Dim , may be estimated by Perkins and Geankoplis equation 1
Dim -
(2.79)
0.8 ~ , x i DijlX °8 (i 4=j ) /J'm .j=l
where XJ is the mole fraction of species j, Dim the effective diffusion coefficient for a dilute species i in the mixture in cmZ/s, D O.the infinite dilution binary diffusion coefficient of species i in species j, and ~m and/XJ the viscosities of mixture and pure species j in cP, respectively. A modified version of the Vignes equation may be used to represent the composition effect on the liquid diffusion coefficient I~
D!j = _ [ ( D ° #
0
j)x/+(D./i/~i
)x i
]
(2.80i
82
2.
Transport and rate processes
where D ° is the infinite diffusion of species i in solventj and F the thermodynamic factor
FI'naAI l'naB 0 In XA r,P
(2.81)
0 In x B I T,P
where a is the activity of species A. The value of F is close to unity for gases away from the critical conditions, while it is a necessary correction for liquids. For diffusion in liquids, we mainly rely on empirical expressions. For example, the Wilke-Chang equation predicts the diffusivity for dilute mixtures of species of i inj by
DO.=7.4×10 -8 TX/~~.
(2.82)
Here V~is the molar volume of the solute i in cm3/(mol) at its normal boiling point (for some values of V~,see Tables 2.9a and b),/z the viscosity of solvent in/xP, q9 an association parameter for solvent, and T the absolute temperature in K. Recommended values of the association parameters are 2.6 for water, 1.9 for methanol, 1.5 for ethanol, and 1.0 for benzene, ether, heptane, and other unassociated solvents. Equation (2.82) should be used only for dilute nondissociating solutions. Table 2.9a Atomic volumes at the normal boiling point Element
Atomic volume (cm3/(mol))
C H O (except as below) O in aldehydes, ketones O in methyl esters O in methyl ethers O in ethyl ethers O in ethyl esters O in higher esters O in higher ethers
14.8 3.7 7.4 7.4 9.1 9.9 9.9 9.9 11.0 11.0
O in acids (-OH) O joined to S, P, N N doubly bonded N in primary amines N in secondary amines
12.0 8.3 15.6 10.5 12.0
Element
Atomic volume (cm3/(mol))
Br C1 C1 in RCHC1R C1 in RC1 (terminal)
27.0 21.6 24.6 21.6
F I S P Ring, three-membered as in ethylene oxide Ring, four-membered Ring, five-membered Ring, six-membered Naphthalene ring Anthracene ring C1 in RCHC1R
8.7 37.0 25.6 27 -6 -8.5 -11.5 -15 -30 -47.5 24.6
Source: G.L. Bas, The Molecular Volumesof Liquid Chemical Compounds, David McKay Co., New York (1915).
Table 2.9b Molar volumes at the normal boiling point Compound
Hydrogen, H 2 Oxygen, 0 2
Nitrogen, N2 Bromine, Br2 Chlorine, C12 Iodine, I2 Carbon monoxide, CO Carbon dioxide, COz Carbonyl sulfide, COS
Molar volume (cm3/(mol)) 14.3 25.6 31.2 53.2 48.4 71.5 30.7 34.0 51.5
Compound
Sulfur dioxide, SO 2 Nitric oxide, NO Nitrous oxide, N20 Ammonia, NH3 Water, H20 Hydrogen sulfide, HzS Air
Molar volume (cm3/(mol)) 44.8 23.6 36.4 25.8 18.9 32.9 29.9
Source: G.L. Bas, The Molecular VolumesofLiquid ChemicalCompounds,David McKay Co., New York (1915).
2.4
83
Transportphenomena
The Stokes-Einstein equation can also be used to estimate the diffusivity of binary liquid mixtures
Dii = ~
KT
(2.83)
4~Tr/xj
where r/. is the radius of diffusing species i and ~j the viscosity of solventj. Another relation is by Tyn and Calus D//= 8.93 × 10 -8
V/
(Vo-) i
T
(2.84)
where Vs.is the molar volume of solvent at the normal boiling point in cm3/mol, and o/and crj the surface tensions in dyn/cm = 10.3 N/m 2. This relation should not be used for diffusion in viscous solvents for which the viscosities are above 20-30 cP. Binary and multicomponent diffusions are different in nature; in binary diffusion flow, species i is always proportional to the negative of the concentration gradient of species i. In multicomponent diffusion, however, other situations may arise: (i) in reverse diffusion, a species diffuses in a direction opposite to the direction imposed by its own concentration gradient; (ii) in osmotic diffusion, a species diffuses although its concentration gradient is zero, and (iii) in the case of a diffusion barrier, a species does not diffuse even though its concentration gradient is nonzero. The Wilke-Chang equation may be modified for a mixed solvent case
Dim =
7.4 ×10 -8 Tx/-qtM
(2.85)
~mV/0"6
where 0M = ~ x/0j Mj (i 4=j) /=1
Example 2.17 Estimation of diffusivity in a dilute liquid mixture Estimate the diffusivity of ethanol in a dilute solution of ethanol-water at 25°C. The density of ethanol is 0.79 g/cm 3 and the viscosity of water at 25°C is 0.95 cP. Solution: Assume that the mixture is a dilute solution. Use the Wilke-Chang equation:
Dii = 7.4×10_8 T ~ j M / •
~ V/0.6
The molar volume of ethanol at its boiling point can be estimated from the Rackett equation
v = VcZ~1-r~>')~
(2.863
The critical properties for ethanol are (Table B 1) Tc = 513.9K, Vc = 167cm3/(mol), Tb = 351.4K(boilingatlatm), and Tr = Tb/Tc =0.6838 From the Rackett equation we have V = 59.8 cm3/(mol). The association parameter for solvent water is 4; = 2.6. Using the Wilke-Chang equation, we find D e w - 7.4× 10-8 [ 298x/216(18) ] 0.95(59.8)o. 6 =1.364×10 -s cm2/s The experimental value (Table 2.8) at 298.15 K and Xe = 0.026 is 1.076 × 10.5 cm2/s, so that the estimation has ---26% relative error.
84
2.4.17
2.
Transportand rate processes
Diffusion in Electrolyte Solutions
On dissociation of a salt, ions start to diffuse in a solution. Without an electric potential effect, however, the diffusion of a single salt is treated as molecular diffusion. For dilute solutions of a salt, the Nernst-Haskell equation is used to estimate the diffusivity coefficient DO = RT(1/n+ + 1/n_ )
(2.87)
F 2 (1/A° + 1/A°)
where R is the gas constant (8.314 J/(mol K)), 1/h ° and 1/h ° the limiting ionic conductances (at zero concentration) ((A/cm 2) (V/cm) (g equiv/cm3)), n + and n_ the valences of cation and anion, respectively, and F the Faraday constant (96,500 C/(g equiv)). Table 2.10 lists the values of limiting ionic conductances at 298 K for some ionic species. The values of 1/h ° and 1/h ° at a temperature T other than 298 K need a correction term of T/(334~w), where/~w is the viscosity of water at T in cP. The mutual diffusivities of some inorganic salts in aqueous solutions are tabulated in Table 2.11.
2.4.18
Diffusion in Colloidal Suspensions
Consider a dilute suspension of spherical particles A in a stationary liquid B. If the spheres are sufficiently small, yet large with respect to the molecules of stationary liquid, the collisions between the spheres and the liquid molecules B lead to a random motion of the spheres. This motion is called the Brownian motion. Dilute diffusion of suspended spherical colloid particles is related to the temperature and the friction coefficient ~:by KT DA B --
KT
67r/ZBRA
(2.88)
where R A is the radius of the sphere. The reciprocal of the friction coefficient, sc, is called the mobility.
2.4.19
Diffusion in Polymers
For a dilute solution of polymer A in a low molecular weight solvent B, the polymer molecules are modeled as beadspring chains. Resistance in the motion of beads is characterized by a friction coefficient ~. As the number of beads is proportional to the polymer molecular weight M, we have
Table 2.10 Limiting ionic conductances in water at 298 K ((A/cm 2) (V/cm) (g equiv/cm3)) Anion
h°
Cation
h°
OHCIBrINO~ CLO~ HCO~ HCOj CH3COj CICHzCO~ CNCHzCO~ CH3CHzCO ~CH3(CH2)2CO2 C6H5CO J HC20~ (1/2)C2 O2(1/2)SO42(1/3)Fe(CN) 3(1/4)Fe(CN) 4-
197.6 76.3 78.3 76.8 71.4 68.0 44.5 54.6 40.9 39.8 41.8 35.8 32.6 32.3 40.2 74.2 80 101 111
H+ Li + Na + K+ NH~Ag + Ti + (1/2)Mg 2+ (1/2)Ca 2+ (1/2)Sr 2+ (1/2)Ba 2+ ( 1/2)Cu 2+ (1/2)Zn 2+ (1/3)La 3+ ( 1/3)Co(NH3)63+
349.8 38.7 50.1 73.5 73.4 61.9 74.7 53.1 59.5 50.5 63.6 54 53 69.5 102
Source: H.S. Hamed and B.B. Owen, The Physical Chemistry of Electtvlytic Solutions, ACS Monogr. 95 (1950).
2.4
Transportphenomena
85
1 DAB
(2.89)
-- .__
x/M
Tables 2.12-2.14 show some values of diffusion coefficients in solids and polymers. In a flow of dilute solution of polymers, the diffusivity tensor is anisotropic and depends on the velocity gradient. The Maxwell-Stefan equation may predict the diffusion in multicomponent mixtures of polymers. Phenomenological systems show that in relatively slow processes, the conjugate flow J is largely determined by frictional forces, and is linearly related to the conjugate force X J-
(2.90)
LX
where the coefficient L is a proportionality factor, which is not necessarily constant but independent of both J and X.
Table 2.11 Mutual diffusion coefficients of some inorganic salts in aqueous solutions Solute
T (°C)
Concentration (mol/1)
DAB × 105 (cm2/s)
HC1
12
0.1
2.29
H2SO 4 HNO3
20 20
NH4C1 H3PO 4 HgC12 CuSO4 AgNO3 CoC1, MgSO4 Ca(OH)2 Ca(NO3), LiC1 NaOH NaC1
20 20 18 14 15 20 10 20 14 18 l5 18
0.25 0.05 0.25 1.0 0.25 0.25 0.4 0.17 0.1 0.4 0.2 0.14 0.05 0.05 0.4
1.63 2.62 2.59 1.64 0.89 0.92 0.39 1.28 1.0 0.39 1.6 0.85 1.12 1.49 1.17
18
0.8 2.0 0.01
1.19 1.23 2.20
0.1 1.8 0.4 0.8 2.0 0.046 0.2
2.15 2.19 1.46 1.49 1.58 1.49 1.43
KOH
KC1
18
KBr KNO 3
15 18
Source: International Critical Tables, McGraw-Hill, New York (1926-1930).
Table 2.12 Diffusivities in some solids A-B
7"(K)
DAB (cm2/s)
He-SiO2
293.2
2.4-5.5 × l0 ~o
H2-SiO2
773.2
0.6-2.1 × 10 s
H2-Ni Bi-Pb Hg-Pb
358.2 438.2 293.2 293.2
Sb-Ag A1-Cu
293.2 293.2
3.5 x 10 2~ 1.3 × 10 3o
Cd-Cu
293.2
2.7 × 10 ~5
Source: R.M. Barrer, Diffusion in and Through Solids, Macmillan, New York (1941 ).
1.16 × 10.5 × 1.1 × 2.5 ×
10-s 10 8 10 ~' 10 ~5
86
2.
Transport and rate processes
Table 2.13 Diffusion coefficients in some solid polymers: DAB X 106 (cm2/s)
Polymer
He
Polyethylene terephthalate (glassy crystalline) Polycarbonate (Lexan) Polyethylene, density 0.964 Polyethylene, density 0.914 Polystyrene Butyl rubber Polychloroprene (neoprene) Natural rubber Silicone rubber, 10% filler (extrapolated) Polypropylene, isotactic Polypropylene, atactic Polyethyl methacrylate Butadiene-acrylonitrile (Perbunan) Polybutadiene Polyvinyl acetate (glassy)
H2
0 2
1.7
0.0036 0.64
3.07 6.8 10.4 5.93 21.6 53.4 19.5 41.6 44.1 11.7 9.52
CO 2
4.36 1.52 4.31 10.2 67.1 2.12 5.7
0.00054
0.021 0.170 0.46 0.11 0.81 0.43 1.58 17.0
0.0048 0.124 0.372 0.058 0.058 0.27 1.10
0.11 0.43 1.5 0.051
0.030 0.19 1.05
4.5 9.6 2.10
CH 4
0.00017
0.057 0.193
0.89
0.0019
Source: J. Crank and G.S. Park (Eds), Diffusion in Polymers, Academic Press, New York (1968); D.W. van Krevelen, Properties of Polymers, 3rd ed., Elsevier, Amsterdam (1990). Table 2.14 Diffusivities in some molten and thermally softened polymers
Gas
Diffusivities × 106 (cm2/s) Polyethylene
Helium Argon Krypton Monochlorodifluoromethane Methane Nitrogen Carbon dioxide
Polypropylene
17.09 9.19
10.51 7.40
4.16 5.50 6.04 5.69
4.02 3.51 4.25
Polyisobutylene
Polystyrene
12.96 5.18 7.30 2.00 2.04 3.37
0.42 0.348 0.39
Source: R.G. Griskey and P.L. Durill,AIChE J, 12 (1966) 1147; R.G. Griskey and EL. Durill,AIChE J., 15 (1967) 106; R.G. Griskey, Mod. Plast., 54 (1977) 158; J.L. Lundberg, M.B. Wilk and M.J. Huyett,J Appl. Phys., 1131 (1960); J.L. Lundberg, M.B. Wilk, and M.J. Huyett, J Polym. Sci., 57 (1962) 275; J.L. Lundberg, M.B. Wilk, and M.J. Huyett, Ind. Eng. Chem. Fundam., 2 (1963) 37; J.L. Lundberg, E.J. Mooney, and C.E. Rodgers, J Polym. Sci.,A-2(7) (1969) 947; D.M. Newitt and K.E.J. Weale, Chem. Soc. (Lond.), 1541 (1948).
2.5
THE MAXWELL-STEFAN EQUATIONS
For multicomponent diffusion in gases at low density, the Maxwell-Stefan equations provide satisfactory approximations when species i diffuses in a h o m o g e n e o u s mixture
xi:n i XiX ('i
(2.91)
where xi = ci/c, c is the mixture concentration, and D~j the binary diffusivities of species i in j, and there are (1/2)n(n - 1) o f binary diffusivities required for n-component system. For gases at low and high densities, liquids, and polymers, the M a x w e l l - S t e f a n equation can be used with diffusivities called the M a x w e l l - S t e f a n diffusivities, which can be related to the Fick diffusivities through the t h e r m o d y n a m i c correction factor for nonideal systems. In a simple limiting case, a dilute species i diffuses in a homogeneous mixture, Nj --~ 0, and Eq. (2.91) in y-direction becomes
dx--L = - J i £ xj dy j=l cD~.
(i 4: j )
(2.92)
2. 7
87
Electric charge flow
If we define a diffusivity of species i in a mixture by D i m - -Ji/(dxi/dy), we have
xj -/=1D~/
Dim -
(i 4: j)
(2.93)
These simple relations are applied to some ternary systems (see Chapter 6).
2.6
TRANSPORT COEFFICIENTS
The one-dimensional flows of momentum, energy, and mass at constant densities are
d ~xv = - u - - ( P v v ) dx d qx = - o ~ - - ( p C p T ) dx d
JAr =--DAB -7--(PA) ax
(2.94)
(Cp = constant)
(2.95)
(2.96)
Therefore, momentum transport occurs because of a gradient in momentum concentration, energy transport is due to a gradient in energy concentration, and mass transport is the result of a gradient in mass concentration. These three transport processes show analogies in their formulations. However, these analogies do not apply in two- and three-dimensional transport processes, since ~-is a tensor quantity with nine components, while Ja and q are vectors with three components. The mass diffusivity Dij, the thermal diffusivity c~ = k/pCp, and the momentum diffusivity or kinematic viscosity ~' = Ix~p, all have dimensions of (length)Z/time, and are called the transport coefficients. The ratios of these quantities yield the dimensionless groups of the Prandtl number, Pr, the Schmidt number, Sc, and the Lewis number, Le P r - u _ Cpix
Sc-
Le-
(2.97)
u D~i ce
D!/
pDi/ -
k ~ pCpO(j
(2.98) (2.99)
These dimensionless groups of fluid properties play important roles in dimensionless modeling equations of transport processes, and for systems where simultaneous transport processes occur. The close interrelations among mass, momentum, and energy transport can be explained in terms of a molecular theory of monatomic gases at low density. The continuity, motion, and energy equations can all be derived from the Boltzmann equation for the velocity distribution function, from which the molecular expressions for the flows and transport properties are produced. Similar derivations are also available for polyatomic gases, monatomic liquids, and polymeric liquids. For monatomic liquids, the expressions for the momentum and heat flows include contributions associated with forces between two molecules. For polymers, additional forces within the polymer chain should be taken into account.
2.7
ELECTRIC CHARGE FLOW
According to the Ohm's law, the flow of electricity that is the current I is directly proportional to electric potential difference or applied voltage AO 1_±4'
(2.100)
88
2.
Transport and rate processes
where R is the resistance of the medium to the current. The value of resistance is influenced by the medium configuration, and for many materials is independent of current. When an electric field is applied, the free electrons experience acceleration in a direction opposite to that of the field, and the flow of charge is called the electric current. Quantum mechanics predicts that there is no interaction between an accelerating electron and atoms in a perfect crystal lattice. Since the current reaches a constant value after the electric field is applied, there exist frictional forces, which counter the acceleration. The frictional forces are the result of the scattering of electrons by imperfections in the crystal due to impurity atoms, dislocations, and vacancies. Thermal vibrations of the atoms may also cause frictional forces. The frictional forces cause the resistance that may be described by the drift velocity Vd and the mobility of an electron, /Ze. The drift velocity represents the average electron velocity, while the electron mobility indicates the frequency of scattering phenomena, and has units of mZ/(v s). Thus, we have (2.101)
Vd = ].Zee
where e is the electric field intensity, and is defined as the voltage difference between two points divided by the distance I separating them A6 1
(2 102)
The conductivity of most materials may be expressed in terms of number of free electrons n per unit volume, absolute magnitude of the electric charge of an electron (]e I = 1.6 × 10-19 C) in Coulomb, and the mobility of electrons ~=nlel~
(2.103)
e
The electric conductivity specifies the electric character of the material. Solid materials, in three groups of conductors, semiconductors, and insulators, exhibit a wide range of electric conductivities. Metals have conductivities on the order of 107 (12 m) -1, insulators have conductivities ranging between 10-l° and 10-2° (1~ m) -1, and the conductivities of semiconductors range from 10-6 to 104 (12 m) -1. The resistivity p is independent of specimen geometry and related to resistance R by RA
p-
(2.104)
1 where A is the cross-sectional area normal to the direction of the current. For most metals and their alloys, resistivity increases with temperature due to the increase in thermal vibration and other irregularities, such as plastic deformations, which serve as electron-scattering centers. Resistivity also changes with composition for alloys. Electric conductivity is defined as the reciprocal of resistivity in ~ m -
1
(2.105)
P Using Eqs. (2.104) and (2.105), Ohm's law becomes Je = O'e
(2.106)
where Je is the current flow (density) that is the current per unit of specimen area I/A. Table 2.15 shows the units of electric parameters. Table 2.15 Units and symbols of electric parameters Quantity Electric charge (C) Electric potential (V) Electric field strength (V/m) Electric current (A) Resistance (~) Resistivity (~1 m) Conductivity (12 m) -1 Current flux (density) Electron mobility
Symbol
SI
Q #s e I R p oJe /ze
C kg m2/(s 2 C) kg 1TI/(S2 C) C/s kg mZ/(s C 2) kg m3/(s C 2) s C2/(kg m 3) C/(s m 2) m 2 (V s) = s C/kg
2.9
2.8
Chemical reactions
89
THE RELAXATION THEORY
For describing transport phenomena, we use one of the constitutive equations of Newton's law of viscosity, Fourier's law, or Fick's law; each one of these relates flow to conjugate thermodynamic driving forces. The conservation laws for momentum, heat, and mass transfer lead to parabolic equations of change, which suggest that the velocity of the propagation of an external disturbance such as a thermal disturbance at any point in the transfer medium is infinite. This can be seen in Figure 2.3, when the surface of the semi-infinite solid material suddenly is brought to T1 from initial uniform temperature To. The solution of temperature profile shows that at time t = 0, T = To, but for t > 0 the material temperature is T(y, t) everywhere, implying that the change in surface temperature is felt everywhere in the material. This phenomenon is explained by the hypothesis of heat flow relaxation, which states that Fourier's law is an approximation of a more exact equation called the Maxwell-Cattaneo equation
0q Ot
q = -kVT-rq
(2.107)
where rq is the relaxation time of heat flow. Analogous equations for the irreversible flows of momentum and mass can also be expressed. For example, for mass transfer, an identical equation to Eq. (2.107) is obtained from the nonstationary version of the Maxwell-Stefan equation, and for momentum diffusion, a similar equation is obtained from the Maxwell equation for viscoelastic fluids. The relaxation time for heat transfer is around rq = 10-12 s for metals, 10-9 S for gases at normal conditions, and 10-~ to 10-13 s for typical liquids. Relaxation times can be greater in rarefied gases, viscoelastic liquids, capillary porous bodies (% = 10-4 s), liquid helium ( % - 4.7 × 10-3 s), turbulent flows (rq = 10-3 to 103 s), and dispersed systems. Combining the thermal energy conservation with Eq. (2.107) yields --c~
V2T -
iJt
~
(2.108)
c2 0 t 2
where c~ is the thermal diffusivity and Co the propagation speed of the internal wave
o-i
Tq
(2.109)
Equation (2.108) is a hyperbolic type, and its solution for the semi-infinite solid medium suggests that two regions exist in a solid; the first region is called the disturbed region where the heat diffusion occurs, and the second region is undisturbed. Fourier's law of heat transfer predicts heat diffusion everywhere in the medium. However, as soon as the surface temperature changes, the wall heat flow q(0, 0) does not start instantaneously, but rather grows gradually. The heat flow rate depends on the current relaxation time and not the relaxation in state. For example, chemical reaction phenomena may illustrate the state relaxation, and heat and viscous stress relaxations and also current relaxation in electric circuits associated with a change in the magnetic energy may illustrate the current relaxation. The wall heat flow reaches a maximum and decreases in time, since temperature gradient at the wall decreases. Therefore, the Fourier's and Fick's laws are inappropriate for describing short-time effects, which may be theoretically important although the relaxation times are typically very small. No exact general criterion is available when it is necessary to include the relaxation terms in the equations of change; however, relaxation terms are necessary for viscoelastic fluids, dispersed systems, rarefied gases, capillary porous mediums, and helium, in which the frequency of the fast variable transients may be comparable to the reciprocal of the longest relaxation time.
2.9
CHEMICAL REACTIONS
Chemical reaction rate depends on the collisions of molecules, per second per unit volume. Since the number of collisions of a species is proportional to its concentration, the chemical reaction rate is proportional to the product of concentrations (mass action law). Thus, for a single homogeneous elementary chemical reaction ulA(g) + u2B(g ) = u3C(g ) + u4D(g)
(2.110)
90
2.
Transport and rate processes
the flow of reaction (velocity) Jr refers to the difference between the forward rate Jrf = kfCAC B and backward rate 4b = kbCcCD J r ~-~ J r f -- Jrb m_ kfCACB -- kbCcCD
(2.111)
where kf and kb are the forward and backward reactions rate constants, respectively. The ratios of mole changes of reacting and produced species are related to the extent of reaction e defined by -dN----~A = Pl
dN-------~B- d N c - dND - d e 92
P3
(2.112)
94
where 9i are the stoichiometric coefficients, which are positive for products and negative for reactants. For a single homogeneous reaction, a generalized reaction rate Jr is a scalar value, and can be expressed in terms of the extent of reaction Jr -
dE
(2.113)
dt
The affinity A is (2.114)
A=-~_~vi~i
where/.L i is the chemical potential of component i. For the chemical reaction system given in Eq. (2.110), the affinity becomes A = P l e A + P2/-~B --(P3/-~C + 94/-~D )
(2.115)
At constant temperature and pressure, the affinity of the chemical reaction is the negative of the change of the Gibbs free energy
116 T,P
If the value of A is greater than zero, the reaction moves from left to fight; if it is smaller than zero, the reaction proceeds from right to left; when A = 0, no reaction takes place.
2.10
COUPLED PROCESSES
When two or more processes occur simultaneously in a system, they may couple (interfere) and cause new induced effects. In a multicomponent system, there are five main driving forces that cause the transport of mass with respect to the mean fluid motion: (i) concentration gradient, (ii) pressure gradient, (iii) temperature gradient, (iv) electric field or an electric potential gradient, and (v) external forces affecting the various chemical substances unequally, such as magnetic effects. In a multicomponent fluid, we have flows of momentum, energy, mass, and electric current, each resulting from an associated thermodynamic driving force. There may be a contribution to each flow stemming from each driving force in the system. This is the result of coupling that can occur between flow-force pairs, which are tensors of equal order or differ by 2 in order. For example, momentum flow is a tensor of order 2, while mass flow or heat flow is a vectorial process. Table 2.16 tabulates some coupled processes, which are briefly discussed below.
2.10.1
Electrokinetic Effects
In 1808, Rous, a colloid chemist, observed that imposing an electric potential difference across a porous wet clay led not only to the expected flow of electricity, but also to a flow of water. He later applied hydrostatic pressure to the clay and observed a flow of electricity. This experiment displayed the electrokinetic effect and demonstrated the existence of coupled phenomena where a flow may be induced by forces other than its own driving force. Therefore, the electric current is evidently caused by the electromotive force, but it may also be induced by the hydrostatic pressure. When two
2.10
91
Coupled processes
Table 2.16 Direct and coupled transport phenomena Flow
Thermodynamic force
Heat
Thermal
Chemical
Electric
Hydraulic
Thermal conduction,
Chemical osmosis,
Peltier effect,
Thermal filtration
Jco = -ci(rKV Tllh, where
Jq = -k~V T
Fluid
Thermal osmosis,
Jq = Lqe( ~ / T ) ,
K is the hydraulic coefficient, Hh the osmotic pressure head (IIh = II/pg), (r the coefficient of osmotic efficiency Dufour effect
where E is the electric field
Electric osmosis
Advection, Jadv =-ciKVh
Diffusion, Ji = -D~Vc,
Electrophoresis
Hyper filtration,
JTq = CikTV T, where kr is the thermoosmotic permeability (m=/(K s)) Thermal diffusion (Soret effect),
Solute
Jco = - c icrKV h,
JTD -De sc iV T,
where h is the hydraulic head
=
Current
where s is the Soret coefficient Seebeck (Thompson effect)
Diffusion and membrane potential
Electric conduction
Rouss effect: thermokinetic effect
chambers containing electyrolytes are separated by a porous wall, an applied potential generates a pressure difference called the electrosmotic pressure. Also, mass flow may generate an electric current called the streaming current. Gradients of electric potential and pressure govern the behavior of ionic systems, selective membranes, and ultracentrifuges. In electrokinetic phenomena, induced dipoles can cause separations, such as dielectrophoresis and magnetophoresis, which may be especially important in specialized separations. Diffusion potential is the interference between diffusion and electric conduction in an anisotropic crystal where heat conduction occurs in one direction caused by a temperature gradient in another direction.
2.10.2
Thermoelectric Effects
In a nonisothermal system, an electric current (flow) may be coupled with a heat flow; this effect is known as the thermoelectric effect. There are two reciprocal phenomena of thermoelectricity arising from the interference of heat and electric conductions: the first is called the Peltier effect. This effect is known as the evolution or the absorption of heat at junctions of metals resulting from the flow of an electric current. The other is the thermoelectric force resulting from the maintenance of the junctions made of two different metals at different temperatures. This is called the Seebeck effect. Temperature measurements by thermocouples are based on the Seebeck effect.
2.10.3
Multicomponent Mass Flow
In a binary mixture, diffusion coefficients are equal to each other for dissimilar molecules, and Fick's law can determine the molecular mass flows in an isotropic medium at isothermal and isobaric conditions. In a multicomponent diffusion, however, various interactions among the molecules may arise. Some of these interactions are (i) diffusion flows may vanish despite the nonvanishing driving force, which is known as the mass transfer barrier, (ii) diffusion of a component in a direction opposite to that indicated by its driving force leads to a phenomenon called the reverse mass flow, and (iii) diffusion of a component in the absence of its driving force, which is called the osmotic mass flow.
2.10.4
Coupled Heat and Mass Flows
Another well-known example is the coupling between mass flow and heat flow. As a result, an induced effect known as thermal diffusion (Soret effect) may occur because of the temperature gradient. This indicates that a mass flow of component A may occur without the concentration gradient of component A. Dufour effect is an induced heat flow caused by the concentration gradient. These effects represent examples of couplings between two vectorial flows. The cross-phenomenological coefficients relate the Dufour and Soret effects. In order to describe the coupling effects, the thermal diffusion ratio is introduced besides the transport coefficients of thermal conductivity and diffusivity.
92
2.
Transport and rate processes
Table2.17 Excursion into history (Straub, 1997) 1687 1736 1749 1750 1755 1736-1819 1811 1821-1828 1822 1822-1838 1824 1842 1847 1848 1850 1852 1864 1865 1872 1873 1872-1957 1901 1905 1931 1941
2.10.5
Viscosity Mass point Hydrodynamic pressure Newton's basic law of motion Ideal equation of motion Steam engine Propagation of heat in solids Viscous equation of motion Equation of heat conduction Navier-Stokes equation Camot cycle, reversibility Conservation of energy Conservation of force Absolute temperature First law of thermodynamics Second law of thermodynamics Electromagnetic field Entropy Entropy (statistical) Equation of state for real fluids Thermodynamic affinity Statistical mechanics Mass-momentum-energy relations Reciprocal relations Dissipative structures
Newton Euler
James Watt Baron-Jaseph Fourier Cauchy Fourier Navier-Saint-Venant Sadi Carnot Robert Mayer Hermann von Helmholtz Kelvin Rudolf Clausius Kelvin Maxwell Rudolf Clausius Boltzmann Van der Waals De Donder Gibbs Einstein Lars Onsager Prigogine
Coupled Chemical Reaction and Transport Processes
The coupling between chemical reactions and transport in biological membranes, such as the sodium and potassium pumps, is known as active transport, in which the metabolic reactions cause the transport of substances against the direction imposed by their thermodynamic force of mainly electrochemical potential gradients.
2.10.6
Coupled Phenomena and Thermodynamics
Using the formulation of nonequilibrium thermodynamics, the cross-effects are mathematically described by the addition of new terms into the phenomenological laws. For thermal diffusion, for example, a term proportional to the temperature gradient is added to the right hand side of Fick's law. These relations need certain phenomenological coefficients, which can be related to the transport coefficients such as thermal conductivity, ordinary and thermal diffusion coefficients, and Dufour coefficient. Therefore, the measured transport coefficients will help determine the phenomenological coefficients. Coupled phenomena are experimentally verified, and they are part of a comprehensive theory of nonequilibrium thermodynamics and phenomenological approach. Studies on the coupled processes in biological, chemical, and physical systems are attracting scientists from various fields. Table 2.17 shows some of the historical developments in thermodynamics, transport phenomena, and rate process.
PROBLEMS 2.1
Estimate the viscosity and thermal conductivity of carbon monoxide using the Chapman-Enskog model at 1 atm and 250, 300, and 400 K and compare with the experimental (Welty et al. (1984)) values in the table below:
T (K) 250 300 400
/z × 105 (Pa s)
k × 102 (W/(m K))
1.5408 1.7854 2.2201
2.1432 2.5240 3.2253
93
Problems
2.2
Estimate the viscosity and thermal conductivity of hydrogen using the Chapman-Enskog model at 1 atm and 250, 300, 400, 600, and 800 K, and compare with the experimental values (Welty et al. (1984)) in the table below: T (K)
# × 10(' (Pa s)
k (W/(m K))
7.919 8.963 10.864 14.285 17.40
0.1561 0.182 0.228 0.315 0.384
250 300 400 600 800
2.3
Estimate the viscosity and thermal conductivity of carbon dioxide using the Chapman-Enskog model at 1 atm and 250, 300, and 400 K and compare with the experimental values (Welty et al. (1984)) in the table below: T (K)
# × 105 (Pa s)
k (W/(m K))
1.2590 1.4948 1.9318
1.2891 1.6572 2.4604
250 300 400
2.4
(a) Estimate the thermal conductivity of oxygen at T = 350 K and P = 1 atm using Eq. (2.48). At 350 K and 1 atm, the thermal conductivity is 0.0307 W/(m K) and Cp = 7.586 cal/(mol K) (Welty et al. (1984)). (b) Estimate the thermal conductivity at the same conditions using Eq. (2.47) and compare the result with the result from part (a).
2.5
Estimate the thermal conductivity of 02 at T = 300 K and P = 30 atm. The thermal conductivity of oxygen at 300 K and 1 atm is 0.02657 W/(mK) (Welty et al. (1984)).
2.6
Estimate the viscosity and thermal conductivity of helium using the Chapman-Enskog model at 1 atm and 422.2,477.8, and 533.3 K and compare with the experimental values (Welty et al. (1984)) in the table below: 7' (K)
# × 105 (Pa s)
k (W/(m K))
422.2 477.8 533.3
2.5299 2.7532 2.9466
0.1834 0.1973 0.2111
2.7
Estimate the viscosity of carbon monoxide at T - 4 0 0 K and P = 35 atm using the reduced viscosity chart in Figure 2.4.
2.8
Estimate the viscosity of a mixture described in the table below at T - 330 K and P = 25 atm using the reduced viscosity chart in Figure 2.4: Species
Mole fractions y;
N. H~ CO Air CO2
0.20 0.10 O.20 0.25 0.25
_
2.9
Estimate the viscosity of the following gas mixture at 293 K and 1 atm using the data (Welty et al. (1984)) given in the following table: Species Air Carbon dioxide
Yi
M
k (W/(m K))
/x (mPa s)
0.75 0.25
28.97 44.01
0.02624 0.01601
0.0181 0.0146
94
2.10
2.
Transport and rate processes
Estimate the radial direction heat flow through a hollow sphere described in the following figure: qout
r=~ T=
2.11
Estimate the thermal conductivity of helium at T = 500 K and P = 1 atm and compare with the experimental value of 5.07 cal/(cm s K) (W.K. Saxena and S.C. Saxena, J. Phys. D Appl. Phys., 1 (1968) 1341).
2.12
Estimate the thermal conductivity of the following table:
0 2at 200 and 300 K at low pressure. Use the data (Welty et al. (1984)) in T (K)
Cp(cal/(molK))
200 300
2.13
2.14
6.971 7.039
Estimate the thermal conductivity of the following gas mixture at 293 K and 1 atm using the data (Welty et al. (1984)) given in the following table: Species
Yi
M
k (W/(mK))
/z (mPa s)
Air Carbon dioxide
0.5 0.5
28.97 44.01
0.02624 0.01601
0.0181 0.0146
Two large vessels containing binary mixtures of gases A and B are connected by a truncated conical duct, which is 2 ft in length and has internal diameters at the ends of 8 and 4 in., respectively. One vessel contains 80mo1% of gas A, and the other 30mo1% of A. The pressure is 1 atm and temperature is 32°E The diffusivity under these conditions is 0.702 ft2/h. Disregarding the convection effects: (a) Calculate the rate of transfer of A. (b) Compare the results that would be obtained if the conical duct was replaced with a circular duct with a diameter of 6 in. Cone ~r z ~ - - - ~ o
Cylinder ~ l r ~ z
...~ut ...
2.15
Estimate the diffusivity of acetic acid in a dilute solution of acetic acid-water at 12.5°C. The density of acetic acid at its boiling point is 0.973 g/cm 3 and the viscosity of water at 12.5°C is 1.22 cP.
2.16
Estimate the diffusivity of ethanol in a dilute solution of ethanol-water at 15°C. The density of ethanol is 0.79 g/cm 3 and the viscosity of water at 15°C is 1.2 cP.
2.17
Estimate the diffusivity of ethylbenzene in water in a dilute solution at 293 K. The density of ethylbenzene at its normal boiling point (at 1 atm) is 0.761 g/cm 3 and the viscosity of water at 293 K is 1.0 cP.
2.18
Estimate the diffusivity of acetic acid, i, in a mixed solvent with 21 mol% ethanol, e, and 79 mol% water, w, at 298 K. The viscosities of ethanol and water are/~e = 1.1 cP and/~w - 0.894 cP at 298 K.
95
Problems
2.19
Estimate the diffusivity of carbon monoxide in nitrogen at 373 and 194.7 K and 1 atm. The experimental values of diffusivities are DAB - - 0 . 3 1 8 at 373 K and D A B - 0.105 at 194.7 K.
2.20
Estimate the diffusivity of carbon dioxide in air at 276.2 and 317.3 K and 1 atm. The experimental values of diffusivities are (Griskey, 2002)
2.21
T (K)
DAB (cm2/s)
276.2 317.3
0.142 0.177
Estimate diffusivity of carbon dioxide A in sulfur dioxide B at 263 and 473 K and 1 atm. The experimental values of diffusivities are (Griskey, 2002) T (K)
DAB (cm2/s)
263 473
2.22
0.064 0.195
Estimate the diffusivity ofhexane in air at 294 and 328 K and 1 atm. The experimental values of diffusivities are Griskey (2002) T (K)
DAB (cm2/s)
294 328
2.23
2.24
0.080 0.093
Estimate the diffusivity of C1402 in ordinary CO 2 at 300K and 145atm. The measured value of DAA~ = 0.125 cmZ/s at T - 312.8 K and P - 1 atm (Griskey (2002)). Estimate the diffusivity of an air-hexane mixture at 390K and 66atm. The measured value of DAB -- 0.093 cm2/s at T = 328 K and P - 1 atm (Griskey (2002)).
2.25
Estimate the diffusivity DAB for the system carbon dioxide and dinitrogen oxide N20 at 273.2 K and 1 atm. The reported experimental diffusivity at these conditions is DAB = 0.096 cmZ/s (Bird et al. (2002)).
2.26
Estimate the diffusivity DAB for the system nitrogen N2 and ethane C2H 6 at 298.2 K and 1 atm. The reported experimental diffusivity at these conditions is DAB = 0.148 cmZ/s (Bird et al. (2002)).
2.27
Estimate the diffusivity DAB for the system air and hexane C6H14 at 328 K and 1 atm. The reported experimental diffusivity (Griskey, 2002) at these conditions is DAB -- 0.093 cmZ/s.
2.28
Estimate the diffusivity DAB for the system methane CH 4 and hydrogen H 2 at 298.2 K and 1 atm. The reported experimental diffusivity at these conditions is DAB = 0.726 cmZ/s (Bird et al. (2002)).
2.29
Estimate and compare the diffusivities DAB for the systems propane C3H8 and normal butane n-C4H10, and propane C3H8 and iso-butane i-C4Hj0 at 378.2 K and 1 atm. The reported experimental diffusivities (Bird et al. (2002)) at these conditions are DAB -- 0.0768 cm2/s and 0.0823 cruZ/s, respectively.
2.30
Estimate the diffusivity of carbon monoxide through a gas mixture of ethylene, hydrogen, and nitrogen with the known mole fractions given in the following table. The mixture is at 295 K and 2 atm. Species Carbon monoxide ( 1) Ethylene (2) Hydrogen (3) Nitrogen (4)
y~
T (K)
D!/(cm2/s)
P (atm)
0.05 0.20 0.30 0.45
273 273 288
0.151 0.651 0.192
1 1 1
96
2.
Transport and rate processes
REFERENCES R.B. Bird. WE. Stewart and E.N. Lightfoot, TransportPhenomena, 2nd ed., Wiley, New York (2002). R.G. Griskey, Transport Phenomena and Unit Operations. A CombinedApproach, Wiley, New York (2002). R.C. Reid, J.M. Prausnitz and B.E. Poling, The Properties of Gases and Liquids, 4th ed., McGraw-Hill, New York (1987). J.R. Welty, C.E. Wicks and R.R. Wilson, Fundamentals of Momentum, Heat and Mass Transfer, 3rd ed., Wiley, New York (1984).
REFERENCES FOR FURTHER READING W.M. Deen, Analysis of Transport Phenomena, Oxford University Press, New York (1998). I. Prigogine, Introduction to Thermodynamics of Irreversible Processes, Wiley, New York (1967). S. Sieniutycz and P. Salamon (Eds), Flow, Diffusion and Rate Processes, Advances in Thermodynamics, Vol. 6, Taylor & Francis, New York (1992). A.H.P. Skelland, Diffusional Mass Transfer, Krieger Publishing Company, Malabar (1985). J.M. Soler, J. Contam. Hydrol., 53 (2001) 63. D. Straub, Alternative Mathematical Theory of Non-equilibrium Phenomena, Academic Press, New York (1997).
3 FUNDAMENTALS OF NONEQUILIBRIUM THERMODYNAMICS 3.1
INTRODUCTION
Physical systems identified by permanently stable and reversible behavior are rare. Unstable phenomena occurring macroscopically result from inherent fluctuations of the respective state variables. Near global equilibrium, the fluctuations do not disturb the equilibrium; the trend toward equilibrium is distinguished by asymptotically vanishing dissipative contributions. In contrast, nonequilibrium states can amplify the fluctuations, and any local disturbances can even move the whole system into an unstable or metastable state. Equilibrium states can be classified by correlations having a typical average range of action of about 10-10 m. Correlations existing far from equilibrium can extend to macroscopic distances on the order of 1 cm or more. This feature is an important indication of the qualitative difference between equilibrium and nonequilibrium states. Kinetic and statistical models often require more detailed information than is available to describe nonequilibrium systems. Therefore, it may be advantageous to have a phenomenological approach with thermodynamic principles to describe natural processes. Such an approach is the formalism used in nonequilibrium thermodynamics to investigate physical, chemical, and biological systems with irreversible processes. In the formalism, the Gibbs equation is a key relation since it combines the first and second laws of thermodynamics. The Gibbs relation, combined with the general balance equations based on the local thermodynamic equilibrium, determines the rate of entropy production. Quantifying entropy production helps in analyzing the level of energy dissipation during a process, and in describing coupled phenomena. Phenomenological laws may describe many common irreversible processes with broken time symmetry. An irreversible phenomenological macroworld and a microworld determined by linear and reversible quantum laws should be related to each other. Prigogine and his colleagues attempted to unify the basic micro and macroscopic descriptions of matter. The first attempts to develop nonequilibrium thermodynamics theory occurred after the first observations of some coupled phenomena of thermal diffusion and thermoelectric. Later, Onsager developed the basic equations of the theory, and Casimir, Meixner, and Prigogine refined and developed the theory further. This chapter outlines the principles of nonequilibrium thermodynamics for systems not far from global equilibrium. In this region the transport and rate equations are expressed in linear forms, and the Onsager reciprocal relations are valid. Therefore, sometimes this region is called the linear or Onsager region and the formulations are based on linear nonequilibrium thermodynamics theory. In this region, instead of thermodynamic potentials and entropy, a new property called entropy production appears. The formulation of linear nonequilibrium thermodynamics has proven to be valid and useful for a wide range of transport and rate processes of physical, chemical, and biological systems.
3.2
LOCAL THERMODYNAMIC EQUILIBRIUM
A local thermodynamic state is determined as elementary volumes at individual points for a nonequilibrium system. These volumes are small such that the substance in them can be treated as homogeneous and contain a sufficient number of molecules for the phenomenological laws to be applicable. This local state shows microscopic reversibility that is the symmetry of all mechanical equations of motion of individual particles with respect to time. In the case of microscopic reversibility for a chemical system, when there are two alternative paths for a simple reversible reaction, and one of these paths is preferred for the backward reaction, the same path must also be preferred for the forward reaction. Onsager's derivation of the reciprocal rules is based on the assumption of microscopic reversibility.
98
3.
Fundamentalsof nonequilibrium thermodynamics
The reversibility of molecular behavior gives rise to a kind of symmetry in which the transport processes are coupled to each other. Although a thermodynamic system as a whole may not be in equilibrium, the local states may be in local thermodynamic equilibrium; all intensive thermodynamic variables become functions of position and time. The definition of energy and entropy in nonequilibrium systems can be expressed in terms of energy and entropy densities u(T~k) and s(T, Nk), which are the functions of the temperature field T(x) and the mole number density N(x); these densities can be measured. The total energy and entropy of the system is obtained by the following integrations
S = ~{s[T(x)],[Nk(x)]}dV
(3.1)
U = ~ {u[T(x)], [Nk(x)]}dV
(3.2)
From the internal energy density u(x) and entropy density s(x), we obtain the local variables of (Ou/Os)v,xk = T(x), --(Ou/OV)s,Uk = P, and (Os/ONk)u = -t~(x)/T(x). The densities in Eqs. (3.1) and (3.2) are dependent on the locally well-defined temperature. In a nonequilibrium system, therefore, the total entropy S is generally not a function of the total entropy U and the total volume V. Also, the classical thermodynamic equations such as the Gibbs and the Gibbs-Duhem equations
i=1
i=1
are valid in a multicomponent medium. For a large class of nonequilibrium systems, thermodynamic properties such as temperature, pressure, concentration, internal energy, and entropy are locally well-defined concepts. Prigogine expanded the molecular distribution function in an infinite series around the equilibrium molecular distribution functionj~ f = f0 + fl + f2 + "'"
(3.5)
Equation (3.5) is valid not only for an equilibrium system, but also for a nonequilibrium system that is characterized by the equilibrium distribution function of (J~ +J]) representing a nonequilibrium system sufficiently close to global equilibrium. Prigogine's analysis applies only to mixtures of monatomic gases, and is dependent on the ChapmanEnskog model. The domain of validity of the local equilibrium is not known in general from a microscopic perspective. The range of validity of the local thermodynamic equilibrium is determined only through experiments. Experiments show that the postulate of local thermodynamic equilibrium is valid for a wide range of macroscopic systems of common gases and liquids, and for most transport processes if the gradients of intensive thermodynamic functions are small and their local values vary slowly in comparison with the local state of the system. For chemical reactions, the reactive collision rates are relatively smaller than overall collision rates. The change in an intensive parameter is comparable to the molecular mean free path, and energy dissipation rapidly damps large deviations from global equilibrium. The local equilibrium concept is not valid in highly rarefied gases where collisions are too infrequent. The extension of equilibrium thermodynamics to nonequilibrium systems based on the local equilibrium assumption is a well-accepted practice in nonequilibrium thermodynamics.
3.3
THE SECOND LAW OF THERMODYNAMICS
Let us consider the system shown in Figure 3.1. In region I, an irreversible process occurs. Region II is isolated and contains region I. Equilibrium is attained everywhere in region II. One possibility is that region I is a closed system that can only exchange heat with region II. The first law relates the internal energy change dU I to a quantity of heat gained from region II, 6q I, and quantity of work 6 WI is performed on region I, so that we have dU I = ~qi + dW I. According to the second law, region I obeys the general inequality dS = ~qrev > -~q -
T
T
(3.6)
3.3
The second law of thermodynamics
T,P,p II
Figure 3.1.
99
@
A closed system with a subsystem I.
where dS, the entropy gain, is determined by 6qrev, the heat that would be absorbed in a reversible change. Since we are considering real (irreversible) processes, TdS is always greater than 6q, which is the actual heat absorbed. It follows that in practice, region I fails to absorb the maximum amount of heat, which theoretically might be transformed into work. Then for real processes, we have dS - 6q + (3q' T T
(3.7)
Here 6q is the actual uptake of heat and 6q' is the additional heat that would have been absorbed from region II for a reversible change, and hence it is a positive quantity. In an actual change of state, the entropy increment dS contains the quantity of entropy 6q'/T. This may be, for example, due to a mixing process or a chemical reaction within region I. The relation of Carnot-Clausius gives the change of the entropy of a closed system dS - des + d i S -
6eq + diS T
(3.8)
Equation (3.8) identifies the two contributions: (i) entropy inflow from the environment deS, and (ii) entropy change inside the volume under consideration diS. The value of diS in every macroscopic region of the system is positive for irreversible changes, while it is zero for reversible changes dis > 0 (irreversible changes)
and
diS = 0 (reversible changes)
(3.9)
The des may be due to a flow of internal energy, convection entropy flow transported along with the macroscopic flow of the substance as a whole, or the entropy flow caused by diffusion of the individual components. The quantity des may be positive, negative, or zero in a special case. For a closed, thermally homogeneous system de S _ 6eq _ dU + PdV T T
(3.10)
where 6eq is the elementary heat due to thermal interaction between the system and the environment. The concept of a macroscopic region describes any region containing enough molecules for microscopic fluctuations to be disregarded. The second law formulation given in Eq. (3.8) may be interpreted as a local formulation, while the second law formulation of classical thermodynamics may be interpreted as a global formulation. With the local formulation, we can analyze various irreversible processes and interactions between them within the same nonequilibrium system. Explicit calculation of diS as a function of the appropriate variables is essentially the basis of nonequilibrium thermodynamics.
3.3.1
Entropy Change of an Ideal Gas
For an ideal gas, we can obtain the change of entropy in terms of volume (or pressure), temperature, and the number of moles. For a unit mass of fluid undergoing a mechanically reversible process in a closed system, the first law yields dU = 6qrev - PdV
(3.11)
100
3.
Fundamentalsof nonequilibrium thermodynamics
On the other hand, the differential form of enthalpy is d H = d U + P d V + VdP
(3.12)
When we eliminate d U between these relations, we get (3.13)
6qrev = d H - VdP For an ideal gas, we have the following relations
d H = Cp,idealdT, V -
RT p
(3.14)
R dP
(3 15)
Dividing Eq. (3.13) by temperature, we get 6qrev --C
T
dT
~p'idealT -- 7
By integrating Eq. (3.15) between temperatures T1 and T2, and pressures P1 and P2, the change of entropy becomes
=
ideal
In terms of volume we have
AS = r2 Cu,ideal T
Equations (3.16) and (3.17) consist of state properties that are independent of the process path, and are general relations for an ideal gas.
3.3.2
Entropy Change of Phase Transformation
Heat exchanged during a phase change under constant temperature is another way to estimate the value deS = 6q/T. For a solid-liquid transformation after a flow of molar heat of melting ~ / m at a melting temperature Tm and at constant pressure, we have
ASm
_
~J-/m rm
(3.18)
Similarly, we can estimate the change of entropy for vaporization using the molar heat of vaporization AHv at boiling point Tb ASv - ~ / v
Vb
(3.19)
Table 3.1 displays the entropy changes of melting and vaporization for some pure substances. The entropy of vaporization is proportional to the ratio of the degree of randomness in the vapor and liquid phases. For a pure component, ASv consists of translational, rotational, and conformational motion of molecules. The translational effect is the largest contribution to the entropy of vaporization.
3.3
101
The second law of thermodynamics
Table 3.1 Enthalpies and temperatures of phase change of some compounds at P = 1 atm Compound
Tm (K)
M-/m (kJ/mol)
ASm (kJ/(mol K))
Tb (K)
2ff-/,~(kJ/mol)
ASv (kJ/(mol K))
H2 02 N2 CO2 NH3 CS2 eel 4 H20 CH3OH C2HsOH
14.01 54.36 63.15 217.0 195.4 161.2 250.3 273.15 175.2 156.0
0.12 0.444 0.719 8.33 5.652 4.39 2.5 6.008 3.16 4.6
0.009 0.008 0.011 0.038 0.029 0.027 0.010 0.022 0.018 0.029
20.28 90.18 77.35 194.6 239.7 319.4 350.0 373.1 337.2 351.4
0.46 6.82 5.586 25.23 23.35 26.74 30 40.656 35.27 38.56
0.023 0.076 0.072 0.130 0.097 0.084 0.086 0.109 0.105 0.110
Source:
G.W.C. Kaye and T.H. Laby (Eds.) Tables of Physical and Chemical Constants, Longman, London (1986).
We can estimate the change in the entropy of vaporization at azeotropic temperature when the heat flow is known at azeotropic pressure
ASaz -
AH v
(3.20)
raz
We may estimate the heat of vaporization for azeotropic mixtures from the Lee-Kesler correlation, with some suitable mixing rules A H v - R T c [6.096448-1.2886T r +1.0167T] + w ( 1 5 . 6 8 7 5 - 1 3 . 4 7 2 1 T
r +2.615T])]
(3.21)
where
--
--
n
i=1
,
(.0 ---
X i (.0 i ,
Tr
i=1
2i=lxiVci
rc
Here, xi is the mole fraction of component i, n is the number of components, Tci and Vci are the critical temperature and volume, and wi is the acentric coefficient for species i. In Eq. (3.21), Tb is the normal boiling point in Kelvin at atmospheric pressure, R = 1.987 cal/(mol K), and M-Iv is in cal/mol. Table 3.2 shows the entropy of vaporization of some binary and ternary azeotropic mixtures obtained from the Lee-Kesler correlation.
3.3.3
Entropy Change of Expansion of a Real Gas
For an irreversible expansion of a real gas at constant temperature due to a heat reservoir, the change of entropy flow is deS - 6 q / T , where 6q is the heat flow between the gas and the reservoir to maintain the constant temperature. The increase of entropy during the expansion is
dis_
Ap
dV =
Pgas -- episton
T
(3.22)
dV
T
where Ppiston is the pressure on the piston. Therefore, the change of entropy becomes
d S - d e S + d~S - 6q + T
The t e r m ( P g a s -
Ppiston) dV
Pgas -- Ppiston
dV
T
is the uncompensated heat as named by Clausius.
(3.23)
3.
102
Fundamentalsof nonequilibrium thermodynamics
Table 3.2 Entropy of vaporization for some binary and ternary azeotropic data at 1 atm Binary species
X1
Tb (K)
AHv (J/mol)
ASv (J/(mol K))
Water(l) Chloroform Ethanol Ethyl acetate n-Butanol Nitromethane Acetonitrile Pyridine
0.160 0.096 0.312 0.753 0.511 0.307 0.768
329.2 351.3 343.5 365.8 356.7 349.6 367.1
32155.3 40683.6 34821.8 41464.4 39643.5 36358.3 41514.4
97.68 115.81 101.37 113.35 111.14 104.00 113.09
Methanol(I) Acetonitrile Acrylonitrile Toluene Ethyl acetate
0.231 0.724 0.883 0.684
336.6 334.5 348.1 335.4
36717.8 38605.1 39778.0 37992.7
109.08 115.41 114.27 113.28
Ethanol(I) Acrylonitrile Ethyl acetate Benzene Hexane
0.445 0.462 0.448 0.332
343.9 344.9 341.4 331.8
38065.2 37557.3 37647.5 34691.9
110.69 108.89 110.27 104.56
Ternary species
xl
X2
Tb (K)
AHv (J/mol)
ASv (J/mol)
Water( 1)-chloroform(2 ) Methanol Ethanol Acetone
0.066 0.129 0.163
0.698 0.795 0.353
325.4 328.4 333.5
34088.7 32954.9 32303.9
104.76 100.35 96.86
Water(I)--ethanol(2) Benzene Hexane
0.233 0.112
0.228 0.274
338.0 329.5
36847.4 34658.3
109.02 105.18
Source: Y. Demirel, Thermochim. Acta, 339 (1999) 79.
Example 3.1 Total entropy change of an air flow in a nozzle Air enters a nozzle at 400 K and 60 m/s and leaves the nozzle at a velocity 346 m/s. The air inlet pressure is 300 kPa, while the pressure at the outlet is 100 kPa. Heat lost in the nozzle is 2.5 kJ/kg. Determine the total entropy change if the surroundings are at 300 K. Solution:
Assume that air is an ideal gas. The nozzle operates at steady state. The properties of air from Table E4" T1 = 400 K
Hi = 400.98 kJ/kg
S1 = 1.9919 kJ/(kg K)
Energy balance for a nozzle at steady-state conditions yields Eout =/~in
(3.24)
The enthalpy at the outlet is v22- v~ H2 = H1 - 0out -
~
/1 kJ/kg m2/s k 2 g) = 340"42 2 = 4 0 0 . 9 8 - 2 . 5 - 3462 k- 602J ( 1000
Therefore, the conditions at the outlet are: T2 = 340 K, $2 = 1.8279 kJ/(kg K)
3.3
103
The second law of thermodynamics
From the entropy balance, we have: Total entropy change = entropy changes of the air + entropy change of the surroundings AStota 1 = ASai r q-ASsurr Z~Sai r - - S 2 - S 1 -
Rln P2 = 1.8279-1.9919-(0.287 kJ/(kg K)) lnk, ~
= 0.151kJ/(kg K) (3.25)
ASsurr _ qsurr,in _ kJ/_______~s 2.5 = 0.00833 kJ/(kg K)
ro z~Stota 1 -
ASair +
3ooK ASsurr =
0.151 + 0.00833 - 0.1593 kJ/(kg K)
Example 3.2 Total entropy change in a polytropic compressing of methane We compress methane from an initial state at 100 kPa, 300 K, and 20 m 3 to 250 kPa and 400 K. The compression process is polytropic ( P V ~ constant). The average heat capacity of methane is Cp,av- 40.57 J/(mol K). Estimate the supplied work and the total entropy change if the surroundings are at 290 K.
Solution: Assume that the methane is an ideal gas. The heat capacity is constant. The number of moles of methane is n--
P1V1 _
Rr~
1 × 105 Pa(20 m 3) 8.314(m 3 Pa/(molK)) 300K
= 801.86 moles
The mass of methane is" m = n M W = 801.86 (16)(1/1000)= 12.829kg The heat capacities are: Cp = 40.57 J/(mol K) and Cv - C p - R = 40.57-8.314 = 32.256 J/(mol K) (for an ideal gas) and the ratio is
y = Cp Cv
_
1.257
The entropy change of methane is
'methan :
32 0 12
The volume of methane after the compression is
V2=v T2 P'
- 20 m 3 400 ( 1.0 × 10 s j 300 2.5×105 =10"66m3
For this polytropic process, we have
PV(~ - c°nstant --' ~
~,V2)
and
{25x,05! (20/ 1.0×105
=
i0.66
-+a=1.456
(3.26)
104
3.
Fundamentals of nonequilibrium thermodynamics
The surface work of the polytropic compression becomes
Win__I2vlPdV__tlRTl((P21(a-1)/a I a _ l ( ~-~l )
-1
/ -
(3.27)
1)/1.456
W.ln 801.861.456__1(8.314)(300) ([(1102"5;OT
(3.33)
where de = - d N 1 = dN2. Here the flow of mass from one region to another is accounted for by the extent of reaction de, although no real chemical reaction takes place. The rate of entropy production is
p _ d i S - _(/x2 - / X l ) de dt T -~ > 0
3.3.7
(3.34)
Electrical Conduction
The rate of entropy production due to electrical conduction is d i S _ i)1
dt
T
(3.35)
where the product ~I represents the heat generated because of potential difference ~ and current I. This heat is also called Ohmic heat per unit time. Here the flow is the electric current and the corresponding force is O/T; the linear phenomenological equation is expressed by
~b I - L~ -T
(3.36)
where Le is the phenomenological coefficient. From Ohm's law ~ = IR, where R is the resistance, and hence,
Le - T/R. 3.3.8
Electrochemical Reactions
The rate of entropy production due to electrochemical reactions is
diS dt
A de T dt
(3.37)
where ~/is the electrochemical affinity defined by z~[i -- Ai -nt-2 i F ( O l - ~2 ) = ~61 - ~2
(3.38)
Here z/is the electrovalency of ionic species i, F is the Faraday number, which is the electrical charge associated with 1 mol ion of a species with an electrovalency of 1, and ~1 is the electrical potential at position 1. The term/2 i is the electrochemical potential of species i, and defined by
~ci - tz i + ziF6
(3.39)
106
3.
Fundamentals of nonequilibrium thermodynamics
The level of electrical current is related to the extent of the electrochemical reaction by
d~
I = ziF--;7 = ziFJ r
(3.40)
tit
For an isolated system, deS = 0 and diS > 0. However, for an open system, we have
deS =
dU + PdV T
+ (deS)matter and diS > 0
(3.41)
Systems that exchange entropy with their surroundings may undergo spontaneous transformation to dissipative structures and self-organization. The forces that exist in irreversible processes create these organized states, which range from convection patterns of Bbnard cells to biological cycles.
3.3.9
Rate of Energy Dissipation
The loss of energy is directly proportional to the rate of entropy production because of irreversible processes in a system. The loss of energy may be estimated based on the temperature of the surroundings of the system To, and we have
,oss-- zT0(dis) k dt
= (kg) (K) (kJ/(kg s K)) = kW
(3.42)
As Eq. (3.42) indicates, the surrounding conditions represent a state where the process reaches equilibrium at which the thermodynamic driving forces vanish. The value of energy Eloss is the rate of energy dissipated to the surroundings.
Example 3.3 Energy dissipation in a nozzle Steam enters a nozzle at 30 psia and 300°F, and exits as a saturated vapor at 300°E The steam enters at a velocity of 1467 ft/s, and leaves at 75 ft/s. The nozzle has an exit area of 0.5 ft 2. Determine the rate of energy dissipation when the environmental temperature is TO= 500 R. Solution: Assume that there are no work interactions, the potential energy effects are negligible, and the nozzle operates at steady state. The properties of steam: State 1: Superheated steam P~ = 30psia
T~ = 760 R
H 1 = 1189.0 Btu/lb m
S1 = 1.7334 Btu/(lb m R)
State 2: Saturated vapor T2 = 760R
H 2 = l179.0Btu/lb m
S2 = 1.6351Btu/(lb m R)
V2 =6.466ft3/lbm
The energy balance for a nozzle at steady-state conditions yields
Eout -- Ein
qout By estimating the steam flow rate, we can determine the heat loss from the nozzle:
1 rh = ~ A2v2
1 6.466(ft3/lbm)
(0.5 ft 2)(75 ft/s) =
5.8 lbm/s
3.3
107
The second law of thermodynamics
Therefore, the heat loss is
qout m/H2 H, (/out = -5.8[1179.7-1189.0 +
752 - 14672 ( 1Btu/lbm 2
~ 25,037 ft2/s 2
) = 302.55 Btu/s
The entropy balance contains the nozzle and its surroundings, and we have deS
diS
dt
dt
dS
-
dt
-0
Rate of net entropy + Rate of entropy production = Rate of change of entropy
qou, ) -+-Spr°d - -
th(S 1 - S 2 ) - - ~ a
Sprod -- 5.8(1.6351-1.7334)-+
0
302.55 500
(3.43) - 0.03501Btu/(s R)
The energy dissipated is /~loss -- ToSprod = 500(0.03501)= 17.506Btu/s
(3.44)
Example 3.4 Energy dissipation in a compressor Air enters a compressor at 15 psia and 80°E and exits at 45 psia and 300°E The inlet air velocity is low, but increases to 250 ft/s at the outlet of the compressor. The power input to the compressor is 250 hp. The compressor is cooled at a rate of 30 Btu/s. Determine the rate of energy dissipation when the surroundings are at 540 R. Solution: Assume that there are no work interactions, the potential energy effects are negligible, and steady flow occurs in the compressor. The properties of air can be obtained from Table E4 of Appendix: State 1: P~=15psia
T1 = 5 4 0 R
H l=129.06Btu/lb m
Sl=0.60078Btu/(lb mR)
=45psia
T2 = 7 6 0 R
H 2=182.08Btu/lb m
S2=0.68312Btu/(lb mR)
State 2:
The energy balance for a compressor at steady-state conditions yields /~7out = Ein Win +/41 H 1 + T
-~/°ut--/41
Using the energy balance, we can estimate the mass flow rate
H2+
108
3.
Fundamentals of nonequilibrium thermodynamics
W i n - Oout
- th( n2 - nl + v2-) 2v2
(250 hp)
0.7068 Btu/s]_ ( 2502 ]-@ ) 30 Btu/s - rh 182.08-129.06 + 2
1Btu/lb m ) ) 25037 ft2/s 2
rh = 2.7 lbm/s
The entropy balance contains the compressor and its surroundings, and we have deS + diS _ dS
dt
dt
dt
-0
Rate of net entropy + Rate of entropy production - Rate of change of entropy \
AS-
t)out / +
-- 0
ro )
(
,out/ =
r0
By taking into account the variable heat capacity, we have - A S = rh(S 2 - S 1 - R l n ~ ) - A S = 2.71bm/s(0.68312-0.60078-(0.06855Btu/lbm) l n ( 4 5 ) ) = 0.019Btu/(s R)
Sprod -- -- A,~ + !)out = 0 . 0 1 9 + 30 = 0 . 0 7 4 5 Btu/(s R)
TO
540
The energy dissipated is /~loss -- ToSprod -- 540(0.0745)- 40.26 Btu/s
Example 3.5 Energy dissipation in an adiabatic mixer In a mixer, we mix a hot water (stream 1) at 1 atm and 90°C adiabatically with cold water (stream 2) at 15°C. The hot water flow rate is 60 kg/h. If the warm water (stream 3) leaves the mixer at 30°C, determine the rate of energy dissipation if the surroundings are at 300 K.
Solution: Assume that the kinetic and potential energy effects are negligible, and this is a steady process. The properties of water from the steam tables in Appendix D: Stream 1: Hot water P~ = 100kPa
T1 =90°C
H 1 = 376.9kJ/kg
S1 = 1.1925kJ/(kgK)
P2 = 100 kPa
T2 = 15°C
H 2 = 62.94 kJ/kg
S2 = 0.2243 kJ/(kgK)
T3 = 30°C
H 3 = 125.7 kJ/kg
S3 = 0.4365 kJ/(kgK)
Stream 2: Cold water
Stream 3: Warm water P3 = 100 kPa
3.3
109
The second law of thermodynamics
The mass, energy, and entropy balances for the adiabatic mixer are Mass balance: thout -- thin ~ thl -t- th2 - th3
(3.45)
Energy balance:/~out
--/~in ~ thl Hi + th2H2 = th3 H3
(3.46)
Entropy balance: Sin - gout -+" Sprod = 0 --~ thiS 1 -Jr-th2S2 - th3S3 -t- Sprod = 0
(3.47)
Combining the mass and energy balances, we estimate the flow rate of the cold water thl H1 + th2 H2 = (thl + th2 ) H3
th2 = th~( H13 - H3 H 2 ) = 60 kg/h( 1376"9-125"7 2 5 . 7 - 6 2 . 9 4 ) = 2401 "53 The mass flow rate of the warm water is: th~ =thl +th: = 60.0 + 240.153 = 300.153 kg/h The rate of entropy production for this adiabatic mixing process is Sprod = th3S 3 -- thiS 1 - t h 2 S 2
Sprod = 300.153(0.4365) -- 60.0(1.1925) -- 240.153(0.2243) = 5.6 kJ/(h K) The energy dissipated because of mixing is: /~loss - -
~)'~prod
--
300(5.6) = 1680.0 kJ/h
Example 3.6 Energy dissipation in a mixer In a mixer, we mix a saturated steam (stream 1) at 110°C with a superheated steam (stream 2) at 1000 kPa and 300°C. The saturated steam enters the mixer at a flow rate 1.5 kg/s. The product mixture (stream 3) from the mixer is at 350 kPa and 240°C. The mixer loses heat at a rate 2kW. Determine the rate of energy dissipation if the surroundings are at 300 K. Solution: Assume that the kinetic and potential energy effects are negligible, this is a steady process, and there are no work interactions. The properties of steam from the steam tables: Stream 1: Saturated steam T1 = 110°C
H l = 2691.3kJ/kg
S 1 = 7.2388kJ/(kgK)
Stream 2: Superheated steam P2 = 1 0 0 0 k P a
T2 - 3 0 0 ° C
H 2 =3052.1kJ/kg
S2 - 7 . 1 2 5 1 k J / ( k g K )
2945.7 kJ/kg
S 3 = 7.4045 kJ/(kg K)
Stream 3: Superheated steam P~ = 350 kPa
T3 = 240°C
H 3 =
The mass, energy, and entropy balances for the mixer are:
Mass balance: thout -- thin--~ thl -t-th2 = th3 Energy balance:/~out
--/~in thlHl + th2H2 = 0out + th3H3
Entropy balance: Sin - Sour + '~prod = 0---~ thiS1 -Jr-t h 2 S 2 - th3S3 - qout _jr_ " = 0 T0 Spr°d
110
3.
Fundamentalsof nonequilibrium thermodynamics
Combining the mass and energy balances, we estimate the flow rate of the super heated steam
Oout --/hlH1 -F/h2H2 - (/hi mr-/h2)H3 /h2 - qout - - / h l (Hi - H 3 ) = 2 kW - 1.5 kg/s (2691.3 - 2945.7)kJ/kg = 3.605 kg/s H 2 -H 3 3052.1-2945.7 The mass flow rate of the warm water is:/h3 =/hi nt-/h2-- 1.5 + 3.605 = 5.105 kg/h. The rate of entropy production for this adiabatic mixing process is
Sprod --/h3S3 - / h i S1 -/h2 $2 q- qout
r0
2 kJ/s
o
Sprod = 5.105kg/s(7.4045)- 1.5 kg/s (7.2388)- 3.605 kg/s (7.1251) + ~
300
- 1.262 kJ/(s K)
The energy dissipated because of mixing is
L'loss -- T0Sprod = 300(1.262)= 378.76 kW
E x a m p l e 3.7 E n e r g y dissipation in a t u r b i n e A superheated steam (stream 1) expands in a turbine from
5000 kPa and 325°C to 150 kPa and 200°C. The steam flow rate is 10.5 kg/s. If the turbine generates 1.1 MW of power, determine the rate of energy dissipation if the surroundings are at 300 K. Solution: Assume that the kinetic and potential energy effects are negligible, this is a steady process, and there are no work interactions. The properties of steam from the steam tables: Stream 1" Superheated steam P1 = 5000kPa
T1 = 325°C
H 1 = 3001.8kJ/kg
S1 = 6.3408kJ/(kgK)
T2 = 200°C
H 2 =2872.9kJ/kg
S2 = 7.6439kJ/(kgK)
Stream 2: Superheated steam P2 = 150kPa
The mass, energy, and entropy balances for the mixer are
Mass balance:/hout = thin
Energy balance:/~out -"/~in ~/hill1 = 0out -]- Wout "]-/h2H2 Entropy balance: Sin - Sour + Sprod "- 0--}/hiS1 -/hiS2 - qout ..]_Sprod -- 0
r0
We estimate the heat loss from the energy balance ~¢out = -/~out +/hi(H1 - H2) = - 1100 kJ/s + 10.5 kg/s (3001.8- 2872.9) = 253.45 kJ/s
3.3
The second law of thermodynamics
111
And the entropy production from the entropy balance is 253.45 -- 14.526 kW/K kg/s (7.6439 - 6.3408) + ~ 300
Sprod -- ghl ($2 - S1) + qout __ 10.5
r0
The amount of energy dissipated becomes: 300(14.526)= 4 3 5 8 . 0 5 k W
/~loss -- ToSprod --
3.3.10 Stationary State Intensive properties that specify the state of a substance are time independent in equilibrium systems and in nonequilibrium stationary states. Extensive properties specifying the state of a system with boundaries are also independent of time, and the boundaries are stationary in a particular coordinate system. Therefore, the stationary state of a substance at any point is related to the stationary state of the system. In a stationary state the total entropy does not change with time, and we have dS
deS diS + -0 dt dt
dt
(3.48)
The term deS/dt is the reversible entropy change in time as a result of an entropy flow between the system and its surroundings. On the other hand, diS/dt represents the rate of entropy production inside the system. Equation (3.48) shows that the entropy exchange with the surrounding must be negative at stationary state
deS dt
--~
diS
< 0
dt
(3.49)
Therefore, the total entropy produced within the system must be discharged across the boundary at stationary state. For a system at stationary state, boundary conditions do not change with time. Consequently, a nonequilibrium stationary state is not possible for an isolated system for which deS/dt = 0. Also, a steady state cannot be maintained in an adiabatic system in which irreversible processes are occurring, since the entropy produced cannot be discharged, as an adiabatic system cannot exchange heat with its surroundings. In equilibrium, all the terms in Eq. (3.48) vanish because of the absence of both entropy flow across the system boundaries and entropy production due to irreversible processes, and we have deS/dt = diS/dt = dS/dt = O. For the total entropy to be constant the entropy flowing out of the system must be equal to the entropy entering the system plus the entropy generated within the system: d i S + (J~,, in - J s out ) -- 0 dt
'
'
(3.50)
From Eqs. (3.48) and (3.50), we have des d[
-
(Js,in - Js,out) < 0
(3.51)
So the stationary state is maintained through the decrease in entropy exchanged between the system and its surrounding. Entropy change inside an elementary volume by irreversible phenomena is the local value of the sum of entropy increments. By the second law of thermodynamics, the entropy production diS is always positive for irreversible changes and zero for reversible changes.
3.3.11 EntropyChange in aTwo-Compartment System Assume that region I in Figure 3.1 is an open system than can exchange both matter and energy with region II. The total entropy change in the system is dStotal -- d S I + d S II
(3.52)
112
3.
Fundamentals of nonequilibrium thermodynamics
From Eq. (3.8) we have d S I = de S~ + d i S I
(3.53)
d S II - - d e S I I = - - de SI
(3.54)
so that
since the entropy change in region II is solely due to exchange with region I. From Eqs. (3.52) to (3.54), it is clear that d S t o t a 1 ~ di SI
(3.55)
Thus di SI represents the total increase of entropy in the environment due to processes taking place in region I. It can be shown that the entropy production is associated with a loss of free energy or the capacity to do work. At constant temperature and pressure, the Gibbs free energy G measures the maximum work capacity, and the changes of Gibbs free energy in each region are dG I = dU I + PdV I - TdS I
(3.56)
dGII - d U H + P d V II - T d S II
(3.57)
Summing these two equations, and remembering that the volume and internal energy are constant in an isolated system, we have d G t o t a 1 = d G ~ + d G ~ = _ TdStota 1
(3.58)
Since for any real process d S t o t a 1 is necessarily positive, the free energy of the entire system decreases. The rate of decrease of the Gibbs free energy is of interest.
Example 3.8 Entropy production in a composite system Consider a composite system consisting of subsystem I enclosed inside subsystem II. The whole system containing subsystem I and II is isolated. However, in subsystems I and II some irreversible processes may take place. Discuss the total entropy production in the whole system. The second law of classical thermodynamics predicts that: clS = d S I + d S II >~ O. From Eq. (3.9) we may have the following two possible phenomena for each subsystem: (a) We may have: di $I > 0
and
diSII> 0
This phenomenon shows that in every macroscopic region of the system, the entropy production is positive and hence both processes are irreversible, which leads to: d S = d S I + d S II >~ 0 (b) We may also have: di $I >> 0
and
diS II < 0
This phenomenon shows that decrease or absorption of entropy in subsystem II may be compensated by a larger entropy production in subsystem I. This is possible only if subsystems I and II are coupled by some suitable coupling mechanisms leading to: d S = d S I + d S II>- O. With thermodynamic coupling a process in subsystem II may progress in a direction contrary to that determined by its own thermodynamic force. Some biological reactions represent coupled reactions for which the total entropy production is positive.
3.4
BALANCE EQUATIONS AND ENTROPY PRODUCTION
Balance equations of extensive quantities describe a change in a system (except in rare gases and shock waves). These balance equations also contain intensive parameters specifying the local state of a continuous medium. Intensive parameters described by the macroscopic properties of the medium are based on the behavior of a large number of particles.
3.4 Balanceequations and entropy production
113
It is necessary to consider the mechanics of a continuous medium to determine the thermodynamic state of a fluid. The properties of a fluid can be determined that are at rest relative to a reference frame or moving along with the fluid. Every nonequilibrium intensive parameter in a fluid changes in time and in space.
3.4.1
Total Differential
Consider the temperature as a function of time and space T = T (t, x, y, z); the total differential of T is expressed as
OT OT dT - .OT dt . + . dx +. ~ d y 3t 3x 3v
+
OT
dz
(3.59)
Oz
Dividing the total differential by the time differential, we obtain the total time derivative of T
dT _ OT + OT dx + _OT _ dv. + -OT - - - dz dt Ot Ox dt Oy dt Oz dt
(3.60)
The partial time derivative of T, (OT/Ot), shows the time rate of change of temperature of a fluid at a fixed position at constant x, y, and z
dT
OT
dt
Ot
(3.61)
If the derivative in Eq. (3.61) vanishes, then the temperature field becomes stationary. The terms dx/dt, dy/dt, and
dz/dt are the components of the velocity of the observer relative to the velocity of the fluid.
3.4.2
Substantial Derivative
If the velocity of the observer is the same as the mass average velocity of the fluid v with components v~, v,,, and v__, then the rate of temperature change is given by
DT
-
Dt
3T
OT
+ ~'.. ~
i~t
Ox
OT
+ v,.-
- Oy
OT
+ v_ - - -
(3.62)
- Oz
or
DT
OT
Dt
Ot
+v.VT
(3.63)
The special operator, DT/Dt is the substantial time derivative, and represents the time rate of change if the observer moves with the substance. A scalar or a vector function expressed in terms of O/at can be converted into the substantial form; for a scalar function T = T(x, y, z, t), we have
DT
O(pT) + Ot
Dt
=p
l
OT
+
aT
-7--+ v r - - +
Ot
-
3x
Ox
+
33' i~T
v~, - - + v _ - -
0~'
-
-
3z
~T)
+T
Oz
(Op
--+ at
O(pv,.) 3x
" +
O(pv,,)O(pv:)) Oy
+
3z
(3.64)
The second term in the second line of Eq. (3.64) is the equation of continuity and vanishes, so that in vector form we have
p
DT Dt
-
O(pT) Ot
~-(V.p vT)
(3.65)
This equation is valid for every local quantity, which may be a scalar, an element of a vector, or an element of a tensor.
114
3. Fundamentalsof nonequilibrium thermodynamics
3.4.3 Conservation Equation An extensive quantity E for a fluid in volume V can be expressed in terms of the specific quantity e, and we have
E = ~ pe dV
(3.66)
V
The partial time derivative of E pertaining to the entire body is equal to the total differential
OE Ot
dE dt
d ; pe dV = ~ O(pe) dV dt v v Ot
(3.67)
In Eq. (3.67) the quantity pe is determined per unit volume when the observer is at rest. The amount of substance entering through an elementary surface area dA per unit time is p v. dA, where dA is a vector with magnitude dA and pointing in a direction normal to the surface. Along with the substance flow there is a convection flow (p ve), and the amount transported per unit time is - f ( p u e ) . dA. The conduction flow Je is a vector with the same direction as the flow, and the amount transported per unit time by means of conduction without a flow of substance is - fJe" dA. The rate of energy production inside the elementary volume of substance at a given point is ~e -
dE dVdt
(3.68)
For the entire volume at rest relative to the coordinate system, the balance equation per unit time is expressed as
dEdt -!O(pe)ot d V = - f
A
+f
dV
(3.69)
Using the Gauss-Ostrogradsky theorem, Eq. (3.69) can be written over the entire volume
dEdt_!O(pe)otdV =-;[~'(pev)]dV-;(V'Je)dVv v +;~ev dV
(3.70)
From Eq. (3.70), the local balance equation for a fixed observer becomes
O(pe) Ot
--V'(pev)-V'Je +~e
(3.71)
The local balance equation for properties subject to a conservation law is called the conservation equation, which is given for e as follows
O(pe) - -V.(pev)- V'J e Ot
(3.72)
If the system is in a stationary state, the extensive property E does not change with time dE/dt- 0, and we have V.(J e + pev) = 0
(3.73)
Equation (3.73) shows that the net amount of E exchanged through the boundary must be zero, and the divergence of the sum of the conduction and convection flows governed by a conservation law is equal to zero in the stationary state. For the values e = 1, Je = 0, and @e -- 0, Eq. (3.71) becomes
Op _ _ V.(p v) = - p ( V . v) - v. Vp Ot
(3.74)
3.4
Balance equations and entropy production
115
The local balance equations for an observer moving along with the fluid are expressed in substantial time derivative form. From Eq. (3.71), we can express the substantial time derivative of e by
p De
+~c
--V.J
(1)
l ~ji.[Fi_T~(_~)]~ 0
(3.152)
1 -r" (Vv) 's > 0
(3.153)
+ T- i=1
where t • (Vv) = "r' • (Vv) 's + z(V.v). The tensor (Vv)' is the sum of a symmetric part (Vv) 's and antisymmetric part (Vv) 'a, and the double dot product of these is zero.
3.5.1
Rate of Entropy Production
The time derivative of entropy production is called the rate of entropy production, and can be calculated from the laws of the conservation of mass, energy, and momentum, and the second law of thermodynamics expressed as equality. If the local entropy production, ~, is integrated over the volume, it is called the volumetric rate of entropy production
p_diS dt - I ~ P d V =
v
IZJiXi dV v i
(3.154)
This integration enables one to determine the total entropy production. When phenomena at the interface between two phases are considered, the amount of entropy produced is taken per unit surface area. Nonequilibrium thermodynamics estimates the rate of entropy production for a process. This estimation is based on the positive and definite entropy due to irreversible processes and of Gibbs relation
TdS = dU + PdV - ~ [&idNi
(3.155)
Entropy depends explicitly only on energy, volume, and concentrations because the Gibbs relation is a fundamental relation and is valid even outside thermostatic equilibrium. For an isotropic medium, the dissipation function or entropy rate can be split into three nonnegative parts /~/0
*-*o
/~/1
+ ' 1 + ' 2 - ~_~JiXi + ~ J i ' X i i=1
i=1
/12
+~Ji
"X i
(3.156)
i=1
where no is the number of scalar, n 1 the number of vectorial, and n 2 the number of tensorial (rank two) thermodynamic forces. The choice of thermodynamic forces and flows must ensure that in the equilibrium state when the thermodynamic forces vanish (X/= 0), the entropy production must also be zero. In contrast to entropy, the rate of entropy production and the dissipation function are not state functions since they depend on the path taken between the given states. 3.5.2
Dissipation
Function
From the rate of entropy production and the absolute temperature, we derive the dissipation function ~ , which is also a positive quantity
= TOp - T Z X i J i >_ 0
(3.157)
3.5 Entropy production equation
125
The increment of the dissipation function can be split into two contributions d~=dx~+dj~
(3.158)
Where d x ~ = TY_.,iJidX i and d j ~ = TY__,iXidJ i , When the system is not far away from global equilibrium, and the linear phenomenological equations are valid, we have d x ~ = dj~ = d ~ / 2 , and a stationary state satisfies d~_-- 0
(3.164)
Since only the n-1 diffusion flows are independent, we have n--1 Ji "VT~; = Z Ji" VT (/J'~ -- ~ ; ) i=1 i=1
Introducing this equation into Eq. (3.164), we have ,, ~1 = a q ' V
1
-T/=1
j i . V T ( i ~ ; _ tZn) > 0
(3.165)
Therefore, the thermodynamic driving force of mass flow becomes ,
xi
=
(3.166)
127
3.6 Phenomenologicalequations
We relate the dissipation function to the rate of local entropy production using Eqs. (3.151)-(3.153) xIJ' = T(~ = r ( %
+ ~ 1 + ( I ) 2 ) = xt/'o + xI$1 + xI*2
(3.167)
If the dissipation function identifies the independent forces and flows, then we have l
(3.168)
*0 = 7 (V" v ) - E A / J r / > - 0 j=l n-I tt
*
*
~ l - - - J q . V l n T - ' ~ . . , ji'VT(l, Zi --~n)Z>O
(3.169)
i=1
q~2 - a"" (Vv) 's _> 0
(3.170)
We can modify Eq. (3.169) using the following transformation of Eq. (3.144)
TV
]&i
T
/&i V T 4- Vepi = V l~ ~ - tJ~--j-iV T T T
- Fi - V I i i - ~
(3.171)
and •
n-1
•
Ji" V/'ti = E Ji" t~7(/"Li --/{L; ) i=1
(3.172)
i=I
Therefore, Eq. (3.169) becomes n-I qff , - - J , "V T - E Ji . v(/x7 - / x ; ) > _ 0 i--1
(3.173)
As shown by Prigogine, for diffusion in mechanical equilibrium, any other average velocity may replace the centerof-mass velocity, and the dissipation function does not change. When diffusion flows are considered relative to various velocities, the thermodynamic forces remain the same and only the values of the phenomenological coefficients change. The formulation of linear nonequilibrium thermodynamics is based on the combination of the first and second laws of thermodynamics with the balance equations including the entropy balance. These equations allow additional effects and processes to be taken into account. The linear nonequilibrium thermodynamics approach is widely recognized as a useful phenomenological theory that describes the coupled transport without the need for the examination of the detailed coupling mechanisms of complex processes.
3.6
PHENOMENOLOGICAL EQUATIONS
In nonequilibrium systems, spontaneous decaying phenomena toward equilibrium take place. When systems are in the vicinity of global equilibrium, linear relations exist between flows J; and thermodynamic driving forces Xk Ji = LikXk
(3.174)
where the parameters Li~Tare called the p h e n o m e n o l o g i c a l coej.ficients. For example, Fourier's law relates heat flow to the temperature gradient, while Fick's law provides a relation between mass diffusion and concentration gradient. The temperature and concentration gradients are the thermodynamic forces. The Fourier and Fick laws consider a single force and a single flow, and are not capable of describing coupled heat and mass flows. Choice of a force Xi conjugate to a flow J; requires that the product JiXi has the dimension of entropy production. The validity of Eq. (3.174) should be determined experimentally for a certain type of process; for example, linear relations hold for an
128
3.
Fundamentals of nonequilibrium thermodynamics
electrical conductor that obeys Ohm's law. Fluctuations occurring in turbulent flow deviate relatively little from the local equilibrium state. If a nonequilibrium system consists of several flows caused by various forces, Eq. (3.174) may be generalized in the linear region of the thermodynamic branch (Figure 2.2), and we obtain
Ji = Z LikXk
(3.175)
k
These equations are called the phenomenological equations, which are capable of describing multiflow systems and the induced effects of the nonconjugate forces on a flow. Generally, any force X/can produce any flow Ji when the cross coefficients are nonzero. Equation (3.175) assumes that the induced flows are also a linear function of nonconjugated forces. For example, ionic diffusion in an aqueous solution may be related to concentration, temperature, and the imposed electromotive force. By introducing the linear phenomenological equations, into the entropy production relation, • = ~JX, we have
OP-- ~ LikXiX k ~ 0 i,k=l
(3.176)
This equation shows that the entropy production is a quadratic form in all the forces. In continuous systems, the base of reference for diffusion flow affects the values of transport coefficients and the entropy due to diffusion. Prigogine proved the invariance of entropy for an arbitrary base of reference if the system is in mechanical equilibrium and the divergence of viscous tensors vanishes. Equation (3.176) leads to a quadratic form
f~--~tiiX?-.[- ~ Lik[-Lki XiXk ~0 i=1 i,k=l 2
(i 4: k)
(3.177)
or the following matrix form [Lll L12 ...L1 n
*-
X1
~.LikXiXk-[X1X2...Xn]
IL~I...Lz.2..::.'.L.z" X2 >-0
i,k=l
LLml Lm2""Lnn
(3.178)
Xn
A necessary and sufficient condition for • >--0 is that all its principal minors be nonnegative
Lii Lik = LiiLkk -- LikLki >--0 L~ Lkk
(3.179)
If only a single force occurs, Eq. (3.178) becomes
dp= LiiX2 >_0
(3.180)
and then the phenomenological coefficients cannot be negative Lii > O. For a system at metastable equilibrium, we may have J / = 0, • = 0, and Xk 4: O. Otherwise, all forces and flows are independent, and the inequality sign holds in Eq. (3.180).
3.6.1
Flows
Mass flow, heat flow, and chemical reaction rate are some examples of the "flows" J/. The thermodynamic "forces" X/ of the chemical potential gradient, temperature gradients, and the chemical affinity cause the flows. The affinity A is
Aj =-~VijlX i
(j = 1,2,...,/)
(3.181)
i=1
where vii is the stoichiometric coefficient of the ith component in the flh reaction, n the number of components in the reaction, and 1 the number of reactions.
3.6
129
Phenomenological equations
The flows may have vectorial or scalar characters. Vectorial f l o w s are directed in space, such as mass, heat, and electric current. Scalar flows have no direction in space, such as those of chemical reactions. The other more complex flow is the viscous flow characterized by tensor properties. At equilibrium state, the thermodynamic forces become zero and hence the flows vanish
Ji,eq (Xi
(3.182)
= 0) = 0
As an example, the diffusion flow vector J; for component i is the number of moles per unit area A per unit time t in a specified direction, and defined by J; -
1 dN i A
(3.183)
dt
Considering a small area dA at any point x, y, z perpendicular to average velocity vi, in which v i is constant, the volume occupied by the particles passing dA in unit time will be v~dA. If the concentration per unit volume is ci then the total amount of the substance is
N i ~--cividA
(3.184)
The local flow, which the amount of substance passing in a unit area per unit time, is
Ji --CiVi
(3.185)
Generally, these three quantities 'Ji, Ci, and v i are the functions of the time and space coordinates. If the area dA is not perpendicular to the flow vector, we consider a unit vector i, perpendicular to dA, whose direction will specify the direction of the area (3.186)
d A = idA
The scalar product vi" dA gives the volume dV, which is multiplied by the local concentration ci to find differential flow dQi , which is the amount of the substance passing an area at any angle with the velocity vector vi dQi
=
Ji
" dA
(3.187)
= civ i • d A
For a volume enclosed by a surface area A, the total amount of species i leaving that volume is Q, = I J i .dA
(3.188)
A
The divergence of the flow Ji is
V.J;-
OJi x '
Ox
+
OJi
'Y +
Oy
OJi -
'~
(3.189)
Oz
Here J;.x, Ji,y, and Ji._- are the Cartesian coordinates of the vector Ji. As the volume V and the product Ji" dA are scalars, the divergence is also a scalar quantity. A positive divergence means a source of component i, while a negative divergence indicates a sink, and at points of div Ji = 0, there is no accumulation and no removal of material. Transformation of the surface integral of a flow into a volume integral of a divergence is
I V'JidV l"
- IJi "dA
(3.190)
A
The divergence of the mass flow vector pv is used in the continuity equation Op
0t
- - ( V . pv)
(3.191)
130
3.
Fundamentals of nonequilibrium thermodynamics
Similarly the local equivalent of the law of conservation of mass for an individual component i is OCi - -
Ot
-(V'Ji)
(3.192)
Equation (3.192) cannot describe a flow process for a reacting component. Another conserved property is the total energy, and in terms of local energy density Pe for each point in the system, we have 0Pe - -(V. je ) Ot
(3.193)
where Je is the energy flow. The total entropy of a system is related to the local entropy density Sv
S = f sv dV v
(3.194)
dS _ r os v dV Ot ~ Ot
(3.195)
The total entropy changes with time as follows
The entropy flow is, on the other hand, is the result of the exchange of entropy with the surroundings
deS - -j" Js" dA -- -j" (V.js) dV Ot A V
(3.196)
An irreversible process causes the entropy production • in any local element of a system, and the rate of total entropy production is
diS - f ~ dV dt v
(3.197)
From Eqs. (3.196) and (3.197), the total change in entropy becomes
dS_deS
diS
dt
dt
dt
(3.198)
Inserting Eqs. (3.195) to (3.197) into Eq. (3.198), we obtain
f OSv dV = -[. (V.js) dV + [. • dV Ot v v v
(3.199)
Therefore, for any local change, an irreversible process in a continuous system is described by
OSv - - V . js + ~ Ot
(3.200)
Equation (3.200) is the expression for a nonconservative change in local entropy density, and allows the determination of the entropy production from the total change in entropy and the evaluation of the dependence of • on flows and forces. Stationary state flow processes resemble equilibria in their invariance with time; partial time differentials of density, concentration, or temperature will vanish, although flows continue to occur in the system, and entropy is being produced. If the property is conserved, the divergence of the corresponding flow must vanish, and hence the steady flow of a conserved quantity is constant and source-flee. At equilibrium, all the steady-state flows become zero.
131
3.6 Phenomenologicalequations
At stationary state, the local entropy density must remain constant because of the condition Osv/Ot= 0. However, the divergence of entropy flow does not vanish, and we obtain q~ = V.j~
(3.201)
Equation (3.20 l) indicates that in a stationary state, the entropy produced at any point of a system must be removed by a flow of entropy at that point. In the state of equilibrium, all the flows including the flow of entropy production vanish, and we obtain the necessary and sufficient condition for equilibrium as • =0
(3.202)
Equation (3.200) may be useful in describing the state of a system. For example, the state of equilibrium can be achieved for an adiabatic system, since the entropy generated by irreversible processes cannot be exchanged with the surroundings.
3.6.2 Thermodynamic Forces For thermodynamic vectorial forces, such as a difference in chemical potential of component i, proper spatial characteristics must be assigned for the description of local processes. For this purpose, we consider all points of equal /.Li as the potential surface. For the two neighboring equipotential surfaces with chemical potentials/.L i and ].Li -+- d/xi, the change in IX~with number of moles N is Otxi/ON, which is the measure of the local density of equipotential surfaces. At any point on the potential surface, we construct a perpendicular unit vector with the direction corresponding to the direction of maximal change in/x~. With the unit vectors in the direction x, y, and z denoted by i, j, and k, respectively, the gradient of the field in Cartesian coordinates is grad/x; = i O]&i + i
Ox
Ol&i + k OIJbi Oy
Oz
(3.203)
A thermodynamic driving force occurs when a difference in potential exists, and its direction is the maximal decrease in ~;. Consequently, at the point x, y, z, the local force X causing the flow of component i is expressed by Xi =-grad/x i
(3.204)
For a single dimensional flow, Eq. (3.204) becomes
Ot~i Xi - ' - - - i ~ Ox
(3.205)
From the definition of the chemical potential, we have
Ol~i
O
Ox
Ox
ONi
=-~i
Ox
(3.206)
where -dG shows the free energy available to perform useful work, 6 W, and the differential of work with distance, 6 W/dx, is a force. Therefore, Xi is a force per mole of component i, causing a flow in the direction of the unit vector. The overall thermodynamic force that is the difference in chemical potential for the transport of the substance between regions 1 and 2 in discontinues systems is the integral of Eq. (3.191) 2 f Xidx l
20
-if
i'zi a OX
dx
-
i (/.Li,1 -/dq, 2 ) -= ~/.L i
(3.207)
I
Here, ~]J'i is a difference in potential, while Xi is a conventional force used in classical mechanics. Electric potential q~that causes a current at the point x, y, z lead to the definition of electric force Xe Xe =-grad where Xe is the force per unit charge, or the local intensity of the electric field.
(3.208)
132
3.
Fundamentals of nonequilibrium thermodynamics
When we consider the difference in electric potential between two points instead of local electric forces, the quantity of electromotive force AO is defined in a single direction by 2 dl//
a~b = - i I ~ clx = i (~1 -- ~2 ) ldx
(3.209)
Other types of forces of irreversible processes may be derived similarly. In general, the flows and forces are complicated nonlinear functions of one another. However, we can expand the nonlinear dependence of the flows J/and the forces Xi in a Taylor series about the equilibrium
-Ji = Ji,eq(Xj = 0)+ ~ ( OJi ] j=l~OXj )eq xj
1 ~ (02Ji] 2 ( 022 )eq xj Jr'""
"~- ~.T j = l
- X i "- Xi,eq(Jk --0)+ ~ ( Oxi ] 1 ~(02Xi,) 2 ~ 0~k)eq Jk +-~..k=l OJ2 )eq J~ +"" k=l
(3.210)
(3.211)
If we disregard the higher order terms, these expansions become linear relations, and we have the general type of linear phenomenological equations for irreversible phenomena
Ji = ~LikXk
(i, k= 1,2,...,n)
(3.212)
i=1 ?7
Xi = Z KikJ k
(3.213)
k=l
Equation (3.212) shows that any flow is caused by all the forces, whereas Eq. (3.213) shows that any force is the result of all the flows present in the system. The coefficients Lik and Kik are called the phenomenological coefficients. The coefficients Lik are the conductance coefficients and Kik the resistance coefficients. The straight coefficients with the same indices relate the conjugated forces and flows. The coefficients with i 4=k are the cross coefficients representing the coupling phenomena. According to the principle of Curie-Prigogine, vector and scalar flows are able to couple only in an anisotropic medium. This theory has important consequences in living cells.
3.7
ONSAGER'S RELATIONS
Onsager's reciprocal relations state that, provided a proper choice is made for the flows and forces, the matrix of phenomenological coefficients is symmetrical. These relations are proved to be an implication of the property of "microscopic reversibility", which is the symmetry of all mechanical equations of motion of individual particles with respect to time t. The Onsager reciprocal relations are the results of the global gauge symmetries of the Lagrangian, which is related to the entropy of the system considered. This means that the results in general are valid for an arbitrary process. The cross-phenomenological coefficients are defined as (3.214)
(3.215)
t,
)
Ji-o
3.8
Transformation of forces and flows
133
The phenomenological coefficients are not a function of the thermodynamic forces and flows; on the other hand, they can be functions of the parameters of the local state as well as the nature of a substance. The values of Lik must satisfy the conditions L;; > 0
1
Lii Lkk > ~ (Lik + Lki
(i = 1,2,...,n)
)2
(i =/:k; i, k = 1, 2,..., n)
(3.216)
(3.217)
or
Kii > 0 (i -
1 ): KiiKkk > -4 (Kik + K~q
1,2,...,n)
(i 4= k; i,k = 1,2 .... ,n)
The matrix of phenomenological coefficients Lki and Kk/are related by K = L-1 where L-~ is the inverse of the matrix L. In a general matrix form in terms of the conductance Lij and resistance Kij coefficients, Eq. (3.178) becomes - XTLX = jTKj
(3.218)
This relation suggests that the local rate of entropy production is a quadratic form in all forces and in all flows if the cross coefficients differ from zero.
3.8
TRANSFORMATION OF FORCES AND FLOWS
Consider a system in which the thermodynamic forces are independent, while the flows are linearly related
0 - £ YkJk
(3.219)
k=l
If the constants are nonvanishing, y~ :~ 0, then we have the flow for component n
j~ = _ ~ k=l
/
Yk Jk
(3.220)
Using Eq. (3.220) in the local entropy production equation, we have
=
Jk Xk = ~ Jk Xk -- Y_LX,, k=l
k=l
Yn
(3.221)
Equation (3.221) has n-1 independent forces ( X k - (vk/y~)Xn) and n - 1 independent flows of Jk, therefore, the phenomenological equations are
Ji - £ LikXk (i,k = 1,2,...,n)
(3.222)
k=l or
Ji - ~,~ L'ik Xk - Y__LX , k--1
Yn
(i,k = 1,2,...,n- 1)
(3.223)
134
3.
Fundamentals of nonequilibrium thermodynamics
Onsager's reciprocal rules lead to (L~ =
Li'k).Substituting Eq. (3.223) in Eq. (3.220), we have Jn - - - - £ L~. i,k=l
Xk-
YiYky2X,,)
(3.224)
The comparison of phenomenological coefficients yields
Yk
n-1
n-1
Lik--Ltik' Lin----ZEikk=l \-~n)' L"i=-ZL'ik~,TSn)k=l
Yk Lnn = ~" r,k k=l
Equations (3.224) and (3.225) show that the Onsager reciprocal relations remain valid From Eq. (3.219), we have
0 = ~ YkLik k=l
0=
~.~YiLik
YiYk
~--~n 2 )
Lik = Lki (i, k =
(3.225) 1,2, ..., n).
(3.226)
i=l
Here, only (2n-l) equations are independent. From Eq. (3.226), we have
0 = ~ yiYkLik
(3.227)
i,k--1 These results prove that the Onsager reciprocal relations remain valid when homogeneous relationships relate the flows to each other.
3.8.1 Two-Flow Systems For a two-flow system, we have the phenomenological equations in terms of the flows J1 = LllX1 + L12g2
(3.228)
J2 -- L21X1 + L22X2
(3.229)
L22 x1-qg
(3.230)
From these relations we can derive the forces L12
L21 Lll X2 =-]L----~J1 + - ~ J 2
(3.231)
We can also write the phenomenological equations in terms of the forces X 1 = K l l J 1 Jr-K12J 2
(3.232)
X 2 = K21J 1 + K22J 2
(3.233)
The following relations link the phenomenological coefficients of Lik to Kik __ L22. L12. L21. Lll Kll--~' K12 = ILl' K21 = ILl' K22 - I L l
where [LI is the determinant of the matrix: ILl =
LllLz2- LIzLzl
(3.234)
3.8
135
Transformation of forces and flows
From Eq. (3.234) and Onsager's relations, we have
L22 ZllZ22 -Z12L21
L22 Zl 1L22 -(/-.12) 2
zt2 ZllZ22 -L12L21
K12 --K21 --
=
Zl 1L22 -- (LI2) 2
/-,11 Zl 1L22 -- (L12) 2
Lll K22
L21
ZllZ22 -LI2L21
Example 3.10 Relationships between t h e conductance and resistance phenomenological coemcients F o r a three-flow system, to derive :the relationships between the conductance and resistance phenomenological coefficients, Consider the linear phenomenological equations relating forces to flows with resistance coefficients,
KllJ 1+ K i2J2 + K13J 3 :
::
:
.........
....
& = K21J1 :,::: K22J2 +K23J3
:
After applying the Onsager relations to the linear matrix solutions, we: have
.
.
.
(K12
J1-
.
.
::
D
.
: :
.
.
..
.
.
.
.
3:- K23K13)i ..... x~ D : .... ' .
.
.
.
.
.
. .
J2 = L21X1 + ~ 2 X 2 + ~3X3 .
.
.
.
.
.
.
.
. .
.
.
.
. .
.
.
.
. .
.
.
.
.
.
. . ..
.
.
.
.
.
.
.
. .
.
.
. . .
.
. . ..
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. ..
.
.
. . .
.
. .
.
:
:
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
. .
.
. .. .
.
.
.
.
.
.
.
.
.
.
.
.
:
. .
. . .
.
. .
.
.
..
.
.
. .
. .
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. . .
.
..
.
.
.
.
. .
. . .
.
.
.
:
3
.
.
. .
.
.
.
.
.
.
.
.
D ~:
< ÷ .
. .
. .
.
.
: .
. .
.
.
. .
.
.
.
~
.
. .
.
.
. .
.
:::
.... . .
. .
. .
.
.
.
. : ....
.
.
.
.
. .
.
. .
. .
.
.
.
.
.
.
.
. .
.
.
.
.
.
.....
2 ~ K12K23
. .
.
. .
.......
. . . . . . . . . .
J3- ,,
. . . . .
.
. .
.
.
.
.
.
.
.
.
.
~ l & + ~2X2 + L33X3
.
.
.
.
.
.
.
+ (X~lX23 - x~x~ " .
.
.
.
.
.
.
..
.
.
.... .
.
.
. .
.
.
. . .
.
. .
.
.
.
.
.
.
. .
.
. .
. .
. .
. .
. .
.
.
. .
. .
.
.
.
.
.
.
.
:
J3
. .
.
.
.
.
. . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. .
.
.
. .
.
.
.
-'gliK33 D .
.
.
.
.
.
.
..
.
.
. .
.
.
.
.
.
.
.
.
.
.
.
. .
.
.
.
:
.
. .
....
.
.
.
.
. .
.
:
.
.
. . .
.
.
.
.
.
. . . .
.
.
. .
.
.
.
.
.
. .
. .
.
.
.
.
.
. .
. . .
. .
. .
j 2 - (K12K33-K13K23) ~l_ 1_ • ::i D " .
. .
.
. .
.
.
: .
.
.
.
. .
.. . ..
.
.
.
.
.
.
.
. .
.. .
.
. ::
.
.
. .
.. .
.
: . .
:
(K13K22 ~ K23K12) : D :
.
.
.
.
:
......
.
.
.
.
.
....
K12
3 .
.....
)
+(
K:~
D
:
::. . . .... 3K 23-K[2K-K11K23 2 33 : . . . . . +K22K33 . . . where D - -g~ag22 +. .:2K12K1
::
.
.
.
.
.
:::;
. ... . . .
.
.
Example 3.11 ~ a n s f o r m a t i o n of phenomenological equations: dependent flows Transform the t h e ~ o d y namic forces and flows w h e n the forces are independent, while the flows are ~linearly dependent in a two-flow system: 0 = y J l + J2. : : i The local entropy production is : ...... ..... . . . . . . . ............. ?
.
.
.
.
.
:
~
=
JIXI + J2X2
:
:i
:i(3.235)
136
3.
Fundamentalsof nonequilibrium thermodynamics
Using the linear relation between the flows, 0 = Y J1 +
J2,
Eq. (3.235) becomes
= J~& - yJ1X2 = Jl(Xl
- yx2)
(3.236)
The linear phenomenological equations are J l = LI~& +&2X2
(3.237)
J2 = L21X1 + L22X2
(3.238)
However, from Eq. (3.236), the modified phenomenological equation for J1 is J1 = L ~ I X ~
(3.239)
J2 = - y J i = - y / 4 ~(Xl - yx2 )
(3.240)
where X[ = X1 -yX2 Then, the second flow is
Using the linear relation between the flows in Eqs. (3.237) and (3.238), we have - YJ1 -- - y L l l X 1 - yL12 X2
(3.241)
J2 = 521Xl -I- Z22X 2
(3.242)
Subtracting Eq. (3.242) from Eq. (3.241), we get yLll -+-L21 = 0----~L21 -- - y L l l
522
yL12 nt- L22 -- 0---+ L12 = - - -
(3.243) (3.244)
Substituting Eq. (3.244) into Eq. (3.237), we find J1 -- L i l Y 1 - L22 X 2
(3.245)
Y
Comparing Eq. (3.245)with Eq. (3.239), we have J1 -- Z l l X l - L22 X2 J( "-/-JII(X1 "+-y X 2 )
Y Therefore, L22
L l l - y2
(3.246)
From Eqs. (3.243) and (3.244), we have L22 Ll2 =
y2L11 -
Y
- yL11
Y
(3.247)
Equation (3.247) shows that Onsager's reciprocal relations are satisfied in the phenomenological equations (Wisniewski et al., 1976).
3.8
Transformation of forces and flows
137
Example 3.12 Transformation of phenomenological equations: dependent forces Transform the thermodynamic forces and flows when the flows are independent, while the thermodynamic forces are linearly dependent in a two-flow system: 0 = yX~ + X2. The local entropy production is rip = J1X1 -+-Y2 X 2
Using the linear relation between the thermodynamic forces, 0 - yX1 + X2, the local entropy production becomes (# -- Yl X l - Y J2 X l "- X1 ( J1 - Y J2 )
(3.248)
The linear phenomenological equations in terms of the resistance coefficients are X1 = KllJ1 + K12J2
(3.249)
X 2 = K21J 1 + K22J 2
(3.250)
However, From Eq. (3.248), the modified phenomenological equation for the force Xa is X 1 = K ' J ' = K ' (Yl - Y J2 )
(3.251)
Then, the second thermodynamic force is
x2
(3.252)
= - y K ' ( J~ - y J 2 )
Using the linear relation between the forces in Eqs. (3.249) and (3.250), we have y X 1 = y K 11Jl + YK12J 2
(3.253)
- Y X 1 - K21J1 + K 2 2 J 2
(3.254)
Subtracting Eq. (3.254) from Eq. (3.253), we get y K l l + K21
= 0-+-
K21
-
Kll
Y )'Kle + K22 = 0 - * K12 = -
~2
(3.255)
(3.256)
Substituting Eq. (3.256) into Eq. (3.249), we find K.. XI = K11J1 - 77/-z J2 Y
(3.257)
Comparing Eq. (3.257) with Eq. (3.251), we have K22 Xt - K"lt - YK'J2 = KllJl -~J2 Y
and K11 -
K22
v2
and
K12 = K21
(3.258)
These results show that Onsager's relations are satisfied in Eqs. (3.249) and (3.250) since the dependency of the forces to the flows are linear (Wisniewski et al., 1976).
138
3.
Fundamentals of nonequilibrium thermodynamics
Example 3.13 Transformation of phenomenological equations: dependent flows and forces Transform the phenomenological equations when the flows and forces are linearly dependent: 0 = z J1 + J2 and 0 = yX1 + X2. The local entropy production is ~ = J1X1 +J2X2 = J1X1 + ( - z J 1 ) ( - Y X 1 ) = J 1 Y l ( l + z y )
(3.259)
The linear phenomenological equations are J1 = L11X1 "+-L12X2 -- L ' X '
(3.260)
J2 "- L21X1 + L22X2
(3.261)
1 =(l+zy) -T
(3.262)
where X' is defined by X'=(l+zy)X
=
The phenomenological coefficient is L ' = L~I
(3.263)
Using the linear flows and forces in Eqs. (3.260), (3.261), and (3.262), we have J1 = L' (1 + zy) 321
(3.264)
J2 = - z L ' (I + zy) X1
(3.265)
J1 - - ( L l l - yL12)X1
(3.266)
J2 -- (L21 - yL22) Yl
(3.267)
From Eqs. (3.260) and (3.261), we obtain
Comparing Eqs. (3.264) and (3.265) with Eqs. (3.266) and (3.267), we find ( L l l - yL12) X 1 = L ' ( l + z y ) X 1 - - + ( l + z y ) L ' = I_ql-YL12
(3.268)
(L21- yL22) X1 = - L ' z ( l + z y ) X 1 --+-(l + z y ) L ' z = L21-YL22
(3.269)
In a two-flow system, there are two degrees of freedom in choosing the phenomenological coefficients. With the linear relations of flows and forces, there is one degree of freedom that is L12 -- L21, and L22 is proportional to L' L22 = wL'
(3.270)
With Eq. (3.270), the relations in Eqs. (3.268) and (3.269) become L21 - yLze = - L ' z ( 1 + zy) L21- y(wL') = - L ' z ( 1 + zy) L21 = - L ' (z + zZy) + ywL'
(3.271)
L12 -- L21 -- - L ' (z + z2 y - yw) and
&l - y & 2 = L' ( 1 + z y ) LI~ = L'(l+zy)- yL'(z+z2Y- yw) Lll = L ' ( 1 - z Z y 2 +y2w)
(3.272)
3.10
139
Heat conduction
Since the local entropy production is positive if L i i > O , L' =L{1 > 0 , L' = L{I > 0 and Lii > 0 restrict w to positive values, and we have
2
and L l l L 2 2 - L 1 2 > O ,
the conditions
1 - z 2 y 2 + y2 w > 0 - . y 2 w > z 2y2 _ 1 z2y 2 1 w > . y2. . . yZ
2
~w>z
1 -- y2
(3.273)
Inequality (3.273) leads to w > z 2 > 0 (Wisniewski et al., 1976).
3.9
CHEMICAL REACTIONS
For an elementary step reaction, we may relate the flow Jr and the affinity A to the forward Jrf and backward Jrb r e a c t i o n rates as follows
Jr - Yrf -- drb
(3.274)
R T In drf Jrb
(3.275)
A
=
If we solve these equations together, we obtain the reaction (velocity) flow Jr - Yrf ( 1 - e -A/RT )
(3.276)
Close to the thermodynamic equilibrium, where A / R T 0
(3.286)
Here,i/represents the heat flow rate. The rate of entropy production per unit volume is the entropy source strength
dt d V
ATZdt------~x= -
T q
~
-~x > 0
T+dT q+6q
x
x+dx
Figure 3.2. Heat conduction in an isotropic rod.
(3.287)
3.11
141
Diffusion
where A is the area, Jq the heat flow Jq = q/Adt, and Xq = dT/dx the thermodynamic force due to the finite temperature difference AT. Three-dimensional heat conduction in an isotropic solid is
1
dP = - J q - ~ V r = Jq "V
{1 ! =--T.VlnT
(3.288)
In Eq. (3.288), Jq is the heat flow, V(1/T) is the inverse temperature gradient representing the thermodynamic force for heat conduction, and Jq/T- Js is the entropy flow.
3.11
DIFFUSION
The local entropy production for diffusion of several substances per unit volume is
+--
J,.V --T-
Based on the entropy production, linear phenomenological equations for an isothermal flow of substance i become Ji = - Z - - Lik ~ V/xk
(3.290)
1
It is clear from the Gibbs-Duhem equation that not all the forces V(/xz/T) are independent. For example, for a twosubstance system at constant pressure and temperature, we have 0 = C1V~I +C2V/Z 2
(3.291)
The condition for no volume flow corresponding to no change in volume due to diffusion is
0 = J1V1 + J2V2
(3.292)
where V; is the partial molar volume for substance i. Therefore, for a two-substance system Eq. (3.289) becomes
~=_1
j1_ T
j2 c2
V~l~=--7/"
1+ Y
Jl.V~l V2c2 T
(3.293)
where V/x1 = (Otxl/Oc~)Vcl. Then, the linear phenomenological equation is
(3.294) Comparing this equation with Fick's lawJ~ = -D~Vcl, we have
J1---~
L/
1+
V2c2
~
Ocl
/
Vc I = - D lVc 1
(3.295)
Therefore, the diffusion coefficient is related to the phenomenological coefficient by
-- I"
1+V2c2 } t OC1 )
(3.296)
142
3.
Fundamentals of nonequilibrium thermodynamics
For diffusion flow of substance 1 in a dilute solution, we have Da -
LllR
(3.297)
x1
since ~1 = / z ° ( P , T) + RTln (C1/C) = ~O(p,T) + RTln (Xl) , where c is the concentration of the solution.
3.12
VALIDITY OF LINEAR PHENOMENOLOGICAL EQUATIONS
If a system is not far from global equilibrium, linear phenomenological equations represent the transport and rate processes involving small thermodynamic driving forces. Consider a simple transport process of heat conduction. The rate of entropy production is
+=-U E>0
(3.298)
The corresponding linear relation between the heat flow and the thermodynamic force is
Jq= _ L q q ( d T )
(3.299)
Equation (3.299) is identical to Fourier's law of heat conduction, k = Lqq/T 2. The validity of Eq. (3.299) is the same as the validity of Fourier's law, and the equation is valid when the relative variation of temperature is small within the mean free path distance A in the case of gases
AOT 0
T
(3.301)
Considering a homogeneous chemical reaction S - P, the corresponding affinity is A = ~s -/Xp
(3.302)
For a mixture of perfect gases, the chemical potential is/x =/x ° + RT In C. We can relate the chemical potentials to the chemical equilibrium constant and the affinity by
RT In K(T) : -~__,l,i tx° (T)
(3.303)
i
A
=
- ~ vi/x° i
-
R T ~ vi In Ci = RT In K(T) i
(3.304)
(Cp/Cs)
From the kinetic expression, we have
Jr=Jrf-Jrb=kfCs-kbCp=
kfCs ( 1 - 1) K-~sP
(3.305)
where the indices f and b refer to forward and backward reactions.
exp/
(3.306)
3.13
Curie-Prigogine principle
143
Equation (3.306) is a general and nonlinear relation between reaction flow and affinity. However, when the reaction is close to equilibrium, we have
~T]
0), the extremum in Eq. (3.343) is a minimum. Such a state is called a stationary state ofjth order.
Example 3.18 Minimum entropy production in an elementary chemical reaction system Consider a monomolecular reaction, for example, the following isomerization reaction. R< 1 ..X< 2 >B
(3.344)
In this open reaction system, the chemical potentials of reactant R and product B are maintained at a fixed value by an inflow of reactant R and an outflow of product B. The concentration of intermediate X is maintained at a nonequilibrium value, while the temperature is kept constant by the reaction exchanging heat with the environment. Determine the condition for minimum entropy production. The entropy production per unit volume is ~ = 4 Jrl + A2 Jr2 > 0 T ?-
(3.345)
Where A1 and A 2 are the affinities for reactions 1 and 2. The linear reaction flows with vanishing cross coefficients are A1 Jrl = L22 ~A2 Jrl = LI 1 --T--
(3.346)
As the chemical potentials ~R and/..LB are fixed by the flow conditions, we have a constant total affinity A A = (/xR - / x x) + (/xx - / x B) =/x R - / x B = 4 + A2
(3.347)
3.15
Minimumentropy production
149
At the stationary state, we have (3.348)
Jrl = Jr2 After inserting Eq. (3.347) into Eqs. (3.345) and (3.346), we get
,=
A1Jr, + ( A-AI~) Jr2~0 T T
A1 Jrl =/-11 7 -
Jr2 = L22
A-4
(3.349)
(3.350)
After combining Eqs. (3.349) and (3.350), we get
A2 ( A - A1)2 ~(A1) =/-11-~k- L22 T2
(3.351)
The entropy production is at a minimum with respect to the affinity of reaction 1
OdP(A1) =/-ql 2A1 2 ( A - A1) OA------~ - ~ - L22 T2 = 0
(3.352)
Al_ A2 /-ql ~ L22 T = J r l - J r 2 = 0
(3.353)
Therefore, we have
Equation (3.353) proves that with the linear reaction flows, the entropy production is minimized at nonequilibrium stationary state where the reaction velocities are equal to each other Jr1 - Jr2.
Example 3.19 Minimum energy dissipation in heat conduction Use the minimum entropy production principle to derive the relation for nonstationary heat conduction in an isotropic solid rod. For an isotropic rod, we have Os
p - - = - V. J, + ~ Ot p
Os Ot
-
p Ou T Ot
(entropy balance) (3.354)
10T T Ot
-
where ,Is = ,Iq/T. From Eq. (3.337), we have
6X i =/:0 6J i = 0 6 f (dO- O)j, dV = 0
(3.355)
V
where Lqq V
4,=-5With the Gauss-Ostrogradsky theorem, we have
6
p-~t-qt
dV+6 Ji
Js'dA=O A
(3.356)
150
3.
Fundamentalsof nonequilibrium thermodynamics
For the isotropic rod with constant temperatures at the boundaries, Eq. (3.356) yields
v
T
(3.357)
Ot
By using the absolute inverse temperature as the variable subject to change, Eq. (3.357) becomes
+ E qq lllll,llll v
T
o
Ot
(3.358)
This variational equation based on Eq. (3.355) is equivalent to a differential heat conduction equation in the following form p C v OT _ - V . Ot
I(1t] ZqqV
= V .
,33,9
V
(3.360)
The Lagrangian of the variational problem is Lq
p C v OT TOt
Lqq 2
A Euler-Lagrange equation for the -variational problem of 6Iv L q d V may be obtained by considering the differential heat conduction equation, and we have OLq
3
3.15.1
0
OLq
. ~ Oxi 0 [O(1/T)/Ox i ] = 0
O(1/T)
(3.361)
Entropy Production in an Electrical Circuit
In electrical circuits, electrical energy is converted into heat irreversibly in resistors and capacitors, and entropy is produced. When there is an electrical field, the change of energy is d U -- TdS - p d V + ~_, t£i dN i + Z Fzil[ti dNi i i
(3.362)
where F is the Faraday constant, and z i the ion number. The product FzidN i represents the amount of charge transferred. When we transfer the charge dI from a potential ~bl to a potential ~2, then the rate of entropy production is d i S _ _ ~2 - i~1 ~ F z i dNi _ _ i~2 - ~1 d I dt T i dt T dt
(3.363)
In Eq. (3.363), the difference (~2--~tl) is the voltage across the element, while dI/dt is the electric current. For a resistor, using the Ohm law V = (~2-~1) - IR, where R is the resistance, the rate of entropy production is diS - VI _ R I 2 dt
T
T
> 0
(3.364)
In Eq. (3.364),/U 2 is the Ohmic heat rate produced by a current through an element, such as a resistor. For a capacitor with capacitance C, the rate of entropy production is diS - V c I - Vc d I _
dt
r
r dt
C
-T
dVc
d-T
(3.365)
151
3.15 Minimumentropy production
where d V c - - d I / C is the voltage decrease when we transfer charge of dI. We can modify Eq. (3.365) as follows
dt
- -T dt
2
(3.366)
- T d t --~
where the terms (CV2/2) - (I2/(2C)) represent the electrostatic energy stored in a capacitor. Only for an ideal capacitor there is no entropy production and no energy dissipation. For an inductance, the rate of entropy production is
dt
-
~£
(--~) = -
r dt
-
_> 0
r
(3 367)
where the energy stored in an inductance (in the magnetic field) is LI2/2. The voltage is V - -LdI/dt. The phenomenological equations for resistance, capacitance, and inductance are as follows V V I - L R --z-, I - Lc -2, I 1
and
I =--L L
V
(3.368)
T
where LR, Lc, and LL are the phenomenological coefficients, which may be related to resistance. Using Ohm's law, we have
1 R
-
LR
, R-
T
T
and
T
R-
Lc
(3.369)
LL
Example 3.20 Minimum entropy production in electrical circuits Determine the conditions that minimize the entropy generation in electrical circuits with n elements connected in series. Assume that the voltage drop across the circuit is kept constant. The entropy production is diS _ V - --I dt T
(3.370)
where V is the voltage across the element (t)2 - ~1) and I the current passing through the element. The phenomenological law is
Vj Ij - L/J T
(3.371)
Since the voltage drop across the circuit is kept constant, we have
V = ~ V/
(3.372)
J The total entropy production for the n circuit elements is
p _ 4~s _ Vl I~ + -=v2 + . . . + mv,,i n d~ r i 12
(3.373)
T
]2
After combining Eqs. (3.373), (3.371), and (3.372), we get
( v - G) p-
dis _ Lll dt ~
+
L22
J +"" + Lnn
T2
(3.374)
152
3.
Fundamentalsof nonequilibrium thermodynamics
To minimize the rate of entropy production, we use OP/OVj with n - 1 independent values of Vj, which leads to /1 = / 2 . . . . .
In
(3.375)
Therefore, in a circuit element, the entropy production is minimized if the current through the n elements is the same. In an electrical circuit, the relaxation to the stationary state is very fast, and nonuniform values of/are not observed.
PROBLEMS 3.1
Derive the relationships between the conductance type ofphenomenological coefficients Lik and the resistance type of phenomenological coefficients K/y in a three-flow system.
3.2
Consider a monomolecular reaction in Example 3.18, and determine the condition for minimum entropy production when the rate of entropy production is expressed in terms of the concentration. In this open reaction system, the chemical potentials of reactant R and the product B are maintained at a fixed value by an inflow of reactant R and an outflow of product B. The concentration of intermediate X is maintained at a nonequilibrium value, while the temperature is kept constant by exchanging the heat of reaction with the environment. Determine the condition for minimum entropy production.
3.3
Consider the following sequence of reactions R< 1 ) X l (
2 .X2(
3 )...(
n-1 ) X n _ l (
n >p
Identify the states at which the entropy production will be minimal. 3.4
Consider the following synthesis reaction H 2 4- Br2 = 2HBr This results from the following intermediate reactions Br2 < 1 >2Br H2+Br< 2 ; H B r + H H + B r 2 ." 3 , H B r + B r The affinity of the net reaction is maintained at a constant value by the flows of H2 and Br2. One of the reactions is unconstrained. Show that the stationary state leads to minimal entropy production.
3.5
Consider one-dimensional heat conduction in a rod with a length of L. Obtain the function that minimizes the entropy production.
3.6
Consider an elementary reaction of A = B, and calculate the change of Gibbs free energy when/3 = Q/K changes from 0.1 to 10.
3.7
For a three-component diffusion system, derive the relations between the diffusion coefficients and the phenomenological coefficients under isothermal conditions.
3.8
Transform the thermodynamic forces and flows when the forces are independent, while the flows are linearly dependent in a two-flow system: 0 = J1 + YJ2.
3.9
Transform the thermodynamic forces and flows when the flows are independent, while the thermodynamic forces are linearly dependent in a two-flow system: 0 = X1 + yX2.
3.10
Transform the phenomenological equations when the flows and forces are linearly dependent and the forces are linearly dependent: 0 = J1 4- zJ2 and 0 =)(1 + yX2.
Problems
153
3.11
A steam enters a nozzle at 500 kPa and 220°C, and exits at 400 kPa and 175°C. The steam enters at a velocity of 200 m/s, and leaves at 50 m/s. The nozzle has an exit area of 0.2 m 2. Determine the rate of energy dissipation when the environmental temperature is T0 = 300 K.
3.12
A steam enters a nozzle at 4000 kPa and 425°C with a velocity of 50 m/s. It exits at 286.18 m/s. The nozzle is adiabatic and has an inlet area of 0.005 m 2. Determine the rate of energy dissipation if the surroundings are at To = 300 K.
3.13
A steam enters a nozzle at 3200 kPa and 300°C with a velocity of 20 m/s. It exits at 274.95 m/s. The nozzle is adiabatic and has an inlet area of 0.01 m 2. Determine the rate of energy dissipation if the surroundings are at T o - 300K.
3.14
A compressor receives air at 15 psia and 80°E The air exits at 40 psia and 300°E At the inlet the air velocity is low, but increases to 250 ft/s at the outlet of the compressor. The power input to the compressor is 350 hE The compressor is cooled at a rate of 200 Btu/s. Determine the rate of energy dissipation when the surroundings are at 540 R.
3.15
In a mixer, we mix a hot water at 1 atm and 80°C adiabatically with a cold-water stream at 25°C. The flow rate of the cold water is 20 kg/h. If the product leaves the mixer at 50°C, determine the rate of energy dissipation if the surroundings are at 295 K.
3.16
In a mixer, we mix a hot water at 1 atm and 86°C adiabatically with cold-water stream at 25°C. The hot water flow rate is 60 kg/h. If the warm water leaves the mixer at 35°C, determine the rate of energy dissipation if the surroundings are at 300 K.
3.17
In a mixer, we mix liquid water at 1 atm and 25°C with a superheated steam at 325 kPa and 200°C. The liquid water enters the mixer at a flow rate of 70 kg/h. The product mixture from the mixer is at 1 atm and 55°C. The mixer loses heat at a rate of 3000 kJ/h. Determine the rate of energy dissipation if the surroundings are at 300 K.
3.18
In a mixer, we mix liquid water at 1 atm and 20°C with a superheated steam at 1350 kPa and 300°C. The liquid water enters the mixer at a flow rate 70 kg/h. The product mixture from the mixer is at 1 atm and 55°C. The mixer loses heat at a rate of 3000 kJ/h. Determine the rate of energy dissipation if the surroundings are at 290 K.
3.19
Steam expands in a turbine from 6600 kPa and 300°C to a saturated vapor at 1 atm. The steam flow rate is 9.55 kg/s. If the turbine generates a power of 3 MW, determine the rate of energy dissipation if the surroundings are at 298.15 K.
3.20
Air enters a nozzle at 400 K and 60 m/s and leaves the nozzle at a velocity of 250 m/s. The air inlet pressure is 300 kPa and the air leaves the nozzle at 100 kPa. If the nozzle loses 2.2 kJ/kg, determine the total entropy change if the surroundings are at 290 K.
3.21
(a) At steady state, a 4-kW compressor is compressing air from 100 kPa and 300 K to 500 kPa and 450 K. The airflow rate is 0.02 kg/s. Estimate the rate of entropy change. (b) If the compression takes place isothermally by removing heat to the surroundings, estimate the rate of entropy change of air if the surroundings are at 290 K.
3.22
Derive the following isentropic relation tbr ideal gases with constant specific heats.
T2-(P2] (7..,/~/TI where 7 is the ratio of heat capacities at constant pressure to heat capacity at constant volume. 3.23
Refrigerant tetrafluoroethane (HFC-134a) enters the coils of the evaporator of a refrigerator as a saturated vapor liquid mixture at 240 kPa. The refrigerant absorbs 100 kJ of heat from the interior of the refrigerator maintained at 273.15 K, and leaves as saturated vapor at 240 kPa. Estimate the total entropy change.
3.24
Methane gas is compressed from an initial state at 100 kPa, 280 K, and 10 m 3 to 600 kPa and 400 K. The compression process is polytropic ( P V ~ = constant). The average heat capacity of methane is Cp,av= 40.57 J/(mol K). Estimate the total entropy change if the surroundings are at 300 K.
3.25
Hydrogen gas is compressed from an initial state at 100 kPa, 300 K, and 5 m 3 to 300 kPa and 370 K. The compression process is polytropic ( P V ~ = constant). The average heat capacity of hydrogen is Cp,av= 29.1 J/(mol K). Estimate the total entropy change if the surroundings are at 290 K.
154
3.
Fundamentals of nonequilibrium thermodynamics
REFERENCES Y.A. Cengel and M.A. Boles, Thermodynamics. An Engineering Approach, 4th ed., McGraw-Hill, New York (2002). S.R. De Groot, Thermodynamics of Irreversible Processes, North-Holland Publishing, Amsterdam (1966). G.A.J. Jaumann, Wien. Akad. Sitzungsberichte (Math-Nature Klasse), 129 (1911) 385. D. Kondepudi and I. Prigogine, Modern Thermodynamics From Heat Engines to Dissipative Structures, Wiley, New York (1999). I. Prigogine, Introduction to Thermodynamics of Irreversible Processes, Wiley, New York (1967). S. Wisniewski, B. Staniszewski and R. Szymanik, Thermodynamics of Nonequilibrium Processes, D. Reidel Publishing Company, Dordrecht (1976).
REFERENCES FOR FURTHER READING E Kock and H. Herwig, Int. J. Heat Fluid Flow, 26 (2005) 672. H.C. Ottinger, Beyond Equilibrium Thermodynamics, Wiley, New York (2005). A. Perez-Madrid, Physica A, 339 (2004) 339. A. P6rez-Madrid, J. Chem. Phys., 123 (2005) 204108-1. J.M. Rubi and A. Perez-Madrid, Physica A, 264 (1999) 492. J.M. Rubi and A. Perez-Madrid, Physica A, 298 (2001) 177. D.P. Ruelle, Proc. Natl. Assoc. Sci., 100 (2003) 3054. I. Santamaria-Holek, A. Perez-Madrid and J.M. Rubi, J. Chem. Phys., 120 (2004) 2818.
4 USING THE SECOND LAW: THERMODYNAMIC ANALYSIS 4.1
INTRODUCTION
We all widely utilize aspects of the first law of thermodynamics. The first law mainly deals with energy balance regardless of the quality of that part of the energy available to perform work. We define first law efficiency or thermal efficiency as the ratio of the work output to total rate of heat input, and this efficiency may not describe the best performance of a process. On the other hand, the second law brings out the quality of energy, and second law efficiency relates the actual performance to the best possible performance under the same conditions. For a process, reversible work is the maximum useful work output. If the operating conditions cause excessive entropy production, the system will not be capable of delivering the maximum useful output. In the last 30 or so years, thermodynamic analysis had become popular in evaluating the efficiency of systems. Thermodynamic analysis combines the first and second laws of thermodynamics, and makes use of second law analysis, exergy analysis, and pinch analysis. Second-law analysis can identify the sources and quantity of entropy production in various processes in a system. Exergy analysis describes the maximum available work when a form of energy is converted reversibly to a reference system in equilibrium with the environmental conditions; hence, it can relate the impact of energy utilization to environmental degradation. Pinch analysis aims for a better integration of a process with its utility in reducing energy cost. On the other hand, the equipartition principle states that a process would be optimum when the thermodynamic driving forces are uniformly distributed in space and time. Thermodynamic analysis aims at identifying, quantifying, and minimizing irreversibilities in a system. Such analysis is of considerable value when efficient energy conversion is important. This chapter discusses second law analysis, exergy analysis, and pinch technology, providing some examples.
4.2
SECOND-LAW ANALYSIS
The mathematical statement of the second law is associated with the definition of entropy S, dS = 6qrev/T. Entropy is a thermodynamic potential and a quantitative measure of irreversibility. For reversible processes, dS is an exact differential of the state function, and the result of the integration does not depend on the path of change or on how the change is carried out when both the initial and final states are at stable equilibrium. The entropy of a closed adiabatic system remains the same in a reversible process, and increases during an irreversible process. A system and its surrounding create an isolated composite system where the sum of the entropies of all reversible changes remains the same, and increases during irreversible processes. The product of thermodynamic forces and flows yields the rate of entropy production in an irreversible process. The Gouy-Stodola theorem states that the lost available energy (work) is directly proportional to the entropy production in a nonequilibrium phenomenon. Transport phenomena and chemical reactions are nonequilibrium phenomena and are irreversible processes. Thermodynamics, fluid mechanics, heat and mass transfer, kinetics, material properties, constraints, and geometry are required to establish the relationships between physical configuration and entropy production and to minimize entropy production. Generally, we may minimize entropy production through a set of modifications in design and operating conditions. The second law of thermodynamics is applicable to all physical, chemical, and biological processes, as well as to heat and work conversions. The second law can quantify the thermodynamic equivalence of heat to work through exergy
156
4.
Usingthe second law: Thermodynamic analysis
and availability analysis, and hence it can provide specific insights into the design, operating conditions, or the retrofitting of an existing process. Some concepts and properties of entropy are: • Processes follow certain directions and paths, that must yield positive entropy production. This principle might force chemical reactions to proceed without reaching completion. • Entropy production is a measure of dissipated useful energy and degradation of the performance of a process; the level of dissipation depends on the extent of irreversibilities. • Entropy is a nonconserved property; it is conserved during an ideal reversible process only. • A reversible adiabatic process is isentropic, meaning that a substance will have the same entropy values at the beginning and end of the process. Systems such as pumps, turbines, nozzles, and diffusers are nearly adiabatic operations and are more efficient when irreversibilities, such as friction, are reduced, and hence operated under isentropic conditions. • Isentropic efficiency of a turbine rh at steady state is defined as the ratio of the actual work output Wactof the turbine to the work output of isentropic operation Ws
mact 7~t- ms
(4.1)
• Isentropic efficiency of a compressor r/c is the ratio of isentropic work to actual work input
~c -
Ws Wact
(4.2)
• Entropy does not exist in various forms. Second-law analysis can determine the level of energy dissipation from the rate of entropy production in the system. The entropy production approach is especially important in terms of process optimality since it allows the entropy production of each process to be determined separately. The map of the volumetric entropy production rate identifies the regions within the system where excessive entropy production occurs due to irreversible processes. Minimizing of excessive irreversibilities allows a thermodynamic optimum to be achieved for a required task. Estimation of the trade-offs between the various contributions to the rate of entropy production may be helpful for attaining thermodynamically optimum design and operation.
4.2.1
Entropy Balance
In every nonequilibrium system, an entropy effect exists either within the system or through the boundary of the system. Entropy is an extensive property, and if a system consists of several parts, the total entropy is equal to the sum of the entropies of each part. Entropy balance is Change in total entropy = Total entropy i n - Total entropy out + Total entropy produced Entropy balance in the rate form is given by
ASsystem--(Sin- Sout) nt-Sprod
(4.3)
where A shows the net change within the system. The first term on the fight in Eq. (4.3) shows the rate of net entropy exchange between the system and its surroundings, which may be by heat and/or mass (Figure 4.1). The rate of entropy production cannot be negative; however, the changes in entropy of the system may be positive, negative, or zero. For a reversible process, the entropy production is zero, and the entropy change of a system is equal to the net entropy transfer. The entropy of an isolated system during an irreversible process always increases, which is called the increase of
entropy principle. Heat and mass flows can transfer entropy. Entropy transfer through the boundary represents the entropy gained or lost by a system during a process. No entropy is transferred by work. According to the first law of thermodynamics, there is no difference between heat and work. According to the second law, however, energy exchange accompanied by entropy transfer is the heat transfer, and energy exchange that is not accompanied by entropy transfer is the work.
4.2
157
Second-law analysis
Sin System Mass >_0 Mass
-j
Heat ]
~Sout
~
Figure 4.1. Mechanism of entropy transfer for a general system. The general entropy balance relations for a control volume are given in terms of the rate of entropy change due to the heat transfer, mass flow, and entropy production
/~'-~cv-- Z 0--Qi~r-£/0 _nLZ fhinSin- Z FhoutSout-{-Sprod
(4.4)
where q0 and To are the environment's heat and temperature. For a general steady-state flow process, the rate form becomes
'~prod- 2 FhoutSout-- Z fhinSin- - 00 - ~ 0i r0
(4.5)
r
Second-law analysis can account for the quality of energy. This may lead to possible improvements in energy converting processes, and the effective use of resources. Some second law guidelines are: • • • • • • • • • •
Avoid excessively large or small thermodynamic driving forces in processes. Minimize the mixing of streams with large differences in temperature, pressure, or chemical composition. Do not discharge heat at high temperature into the environment. Do not heat refrigerated streams with hot streams. When choosing streams for heat exchange, try to match streams where the final temperature of one stream is close to the initial temperature of the other. Extremely large or small amounts of flows may not be easy to manage efficiently. When exchanging heat between two streams, the exchange is more efficient if the heat capacities of the streams are similar; otherwise, consider splitting the stream with the larger flow heat capacity. Hot or cold sources with temperatures far from the ambient temperature are useful. Minimize the throttling of fluid flow, steam, or other gases. Use exergy balance or exergy loss calculations to evaluate the utilization of energy and as a guide for process modifications.
These guidelines may be useful in designing and optimizing the processes such as power plants, heat exchangers, and other thermal systems. Suitable trade-offs between the use of energy and capital may be required by identifying and eliminating design parameters and operating conditions that cause excessive entropy production.
4.2.2
Throttling
Sometimes, fluids flow through a restriction, such as an orifice, a valve, or a porous medium, and a pressure drop occurs adiabatically. If the changes in kinetic and potential energies are negligible, the flow process is a throttling process, which causes no change in enthalpy at the inlet and outlet, and we have M-/= 0. Some properties of throttling processes are: • • • •
In an ideal gas, enthalpy is a function of temperature only, and temperature remains constant. At moderate pressure and temperature, a throttling process causes a drop in temperature for most real gases. If wet steam is throttled to a considerably low pressure, the liquid evaporates and the steam becomes superheated. The throttling of a saturated liquid causes vaporization (or flashing) of some of the liquid, which produces saturated vapor and liquid.
158
4.
Usingthe second law: Thermodynamic analysis
Example 4.1 Lost work in throttling processes n-Butane gas with a flow rate of 25 mol/s is throttled from 15 bar and 450 K to 1 bar in a steady-state flow process. Determine the final temperature and the lost work. Assume that the surroundings are at 298.15 K.
Solution: Assume that kinetic and potential energy effects are negligible To = 298.15K,
R = 8.314 J/(mol K)
This example uses the Lee-Kesler generalized correlation for the reduced enthalpy estimations (see Tables F5-F8) in a throttling process. The reduced properties lead to enthalpies
82 -- N0,ideal + I
Cpdr + I4~
To
and H 1 = N0,idea 1 + I Cpdr + I-I~ To
(4.6)
By using the throttling property of AH = 0 and the above equation, we have z~-I = O = C p, av ( T2 - T1) + H R2 - H R
(4.7)
where Cp,avis the average heat capacity between 7'1 and T2. At the outlet conditions, the n-butane gas is ideal, and hence H2R = 0. Therefore, Eq. (4.7) becomes Te = T1 + HR
(4.8)
Cp, av The critical properties for n-butane are: ire = 425.1 K, Pc = 37.96 bar, and the acentric factor to = 0.2. The reduced properties and heat capacity are:
rr-
T1 - 1.058, P~ = P__I= 0.395,
Cp =
R(1.935 + 0.00369T)
Pc
Using the generalized correlation, we have
: RT Pr
_ Tr
dTr
-T r ~
=-2830.12J/mol
(4.9)
where B0=0.083_~=0.422 -0.3026 and B 1 =0.139 rrl.6
0.172 - 0.00326 T4-2
(4.10)
As T2 is not known, an initial value Cp is calculated at T1 = 450 yielding Cp = 151.38 J/(mol K). With the known value of Cp, Eq. (4.8) yields T2 = 431.6 K. As the temperature difference is not large, the average value of Cp at the average temperature value is: 450+431.6 Tav
= 440.8,
Cp,av =
151.38 J/(mol K)
Therefore, the value of temperature at the outlet is Tz=T1 +H1R
.....
Cp,av
450
2830.12 --=431.3K 151.38
4.2
159
Second-law analysis
The value of entropy change of throttling is
/
-Rln(P21-S R
iP, )
= 18.377 J/(mol K)
(4.11)
where dB : - R P r t, dTr +
dBldTr) =
-2.2858 J/(mol K)
(4.12)
For a flow rate of 25 mol/s and To = 298.15 K, the lost work or dissipated energy is Eloss =
nToAS =
136.97 kW
Example 4.2 Dissipated energy in an adiabatic compression In an adiabatic compression operation, air is compressed from 20°C and 101.32 kPa to 520 kPa with an efficiency of 0.7. The air flow rate is 22 mol/s. Assume that the air remains ideal gas during the compression. The surroundings are at 298.15 K. Determine the thermodynamic efficiency rhh and the rate of energy dissipated Eloss.
Solution: Assume that kinetic and potential energy are negligible, and the system is at steady state. Basis: 1 mol/s air To - 298.15 K R = 8.314 J/mol K Cp/R= 3.355 + 0.575 × 10 -3 Twhere Tin K andR = 8.314 J/molK (Table B3) n = 22 mol/s For an isentropic operation (AS = 0), the final temperature Tzs is y-1
(a)
where
-
Y 1
y
_
R
and
3'-
Cp
Cp Cv
As the value of Cp is temperature dependent, Eq. (a) needs iterations. A value greater than 293.15 K may be an initial temperature. After iterations, we have Tzs = 460.59 K An average heat capacity may be estimated from
C r,av =
v,
(b) = 29.695 J/(mol K)
The work required under isentropic operation Ws is
~H s - Ws - Cp,av(TZs -
T~) - 4972.17 J/tool
Actual work mact is Wact - Ws - 7 1 0 3 . 1 1 J/mol rtc
160
4.
Usingthe second law: Thermodynamic analysis
"With the value of actual work, we can estimate the actual temperature by M-/ T2 = T I + ~ Cp,av Eq. (c) requires iterations,
as
(c)
Cp,avis dependent on the value of T. After iteration, we have T2 = 531.0K
Using Eq. (b), we have the new value of Cp,av= 29.863 J/mol K. The change of entropy is
AS=Cp,av ln(-~12) - R l n ( @ l ) = 4.414 J/mol K The value of ideal work is Widea 1 --
A H - ToAS = 5867.74 J/mol
The thermodynamic efficiency becomes T/th -- Wideal -- 0 . 8 2 6 Wact
The dissipated energy is ~+loss= h ToAS = 28.95 kW
4.2.3
Heat and Fluid Flow
Researchers and engineers extensively utilized second law analysis in the field of heat and fluid flow. Bejan developed the basic approach, methodology, and applications. In two-dimensional Cartesian coordinates, the local rate of entropy production per unit volume in a convective heat transfer is Sprod d x
dy =
qx + (Oqx/OX)dx qy + (Oqy/Oy)dy qx qy r +(OT/Ox)dx dy+ T + (OT/Oy) dy d x - ~ Td y - - - d rx + s+--clx Ox
vx+
I I(
+
Ox
dx
p + - - d x dy Ox
(4.13)
I(
O(ps) dx dy Ot
o+ ay
The first four terms on the right of the above equation account for the entropy transfer due to heat transfer, the next four terms represent the entropy convected into and out of the system, and the last term represents the rate of entropy accumulation in the control volume. Dividing Eq. (4.13) by dxdy, the local rate of entropy production becomes
Spr°d--T -
Ox
Oy
---f-2 qx --~x + qy
+ P --O-/+Vx --~x+ Vy
+S
(4.14)
Here, the last term on the right vanishes based on the mass conservation principle
Dp Dt
~+pV.v=0
(4.15)
4.2
161
Second-law analysis
where D / D t is the substantial derivative• Therefore, in vectorial notation, we have the volumetric rate of entropy production •
1
1
(4.16)
ds
@rod -- ~ - V ' q - - ~ q ' V T
+ p dz-7
From the canonical relation du - Tds - P d ( l / p ) , we obtain p
ds dt
-
p du
P
T dt
p T dt
dp
(4.17)
The first law of thermodynamics expressed locally in the convection of a Newtonian fluid is dH --=-V.q-P(V.v)+r" P dt
Vv
(4.18)
Introducing the above equation into Eq. (4.17) and combining it with Eq. (4.16), we have the following equation for an incompressible flow" •
1
1
Sprod = - 7 2 ( q ' V T ) + r ( ' r
(4.19)
" Vv)
The term ('r • Vv) represents the conversion of mechanical energy into the viscous dissipation heating. This heat source can be considerably high in flows with large viscosity and large velocity gradients, such as high-speed flights, rapid extrusion, and lubrication.
•
--
+ 2 # *'["*'7
Oxj
)]2
-/.tO
Ox i
where/x is the viscosity and ® the viscous dissipation function (in S-2). When the index i takes on the values 1, 2, 3, the velocity components Vx, v,,, v_ and the rectangular coordinates xi become x, y, z. Using the Fourier law q = - k V T and Eq. (4.20), Eq. (4.19) becomes • Sprod =
k ( V T ) 2 -+--/x--~
(4.21)
For a two-dimensional Cartesian coordinate system, the above equation can be expressed as aT
Spr°d -- 7
0N
+ --
ov
+
tx
7
2
Ov x
k Ox )
+2
Ov v
o),
+
Ov x
oy
+
Ov),
/2}
(4.22)
Ox
The above equation shows that the local irreversibilities are due to heat and viscous effects. Entropy production is positive and finite as long as temperature and velocity gradients exist.
E x a m p l e 4.3 T h e r m o m e c h a n i c a l c o u p l i n g in a C o u e t t e flow b e t w e e n p a r a l l e l plates Couette flow provides the simplest model for the analysis of heat transfer for flow between two coaxial cylinders or parallel plates (Figure 4.2). The Couette flow is important in lubrication, polymer flows, and food processing. The tangential annular flow is a model for a journal and its beating in which one surface is stationary while the other is rotating, and the clearance between the surfaces is filled with a lubricant oil of high viscosity. For such a system, the viscous energy dissipation appears as a heat source term in the energy equation, which is necessary to predict the temperature distribution in the narrow gap of a Couette device. Heat transfer and friction in a Couette flow causes entropy production and loss of useful energy.
162
4.
Using the second law: Thermodynamic analysis
For planar Couette flow, the rate of entropy production of an incompressible Newtonian fluid is
•
, (,,)2 (4.23)
Figure 4.2 shows a Couette flow of a fluid of constant density p, viscosity/~, and thermal conductivity k between parallel plates. The bottom plate is at rest, while the top plate is moving at a constant velocity Vl. The upper and lower plates are at uniform temperatures T1 and T2, respectively. The equation of motion for fully developed flow in the x-direction is
-Tyy ~ Tyy = - ~
(4.24)
The boundary conditions are v = •1 at y = H, and v = 0 at y = 0. We can integrate Eq. (4.24) twice to obtain the dimensionless velocity profile U (Figure 4.3) U = v Y[1 + A(1- Y)] v1 where A =
(-dP/dx)HZ/2~Vl and
Y=
(4.25)
y/H.
V=vl,T=T1 H
0
v=O,T=T2
Figure 4.2. The plane Couette flow. Reprinted with permission from Elsevier, Y. Demirel, Int. J. Heat Mass Transfer, 43 (2000) 4205.
-2
l
0.5
0.4
7 Figure 4.3. Dimensionless velocity field Ufor the plane Couette flow for H = 0.005 m. Reprinted with permission from Elsevier, Y. Demirel, Int. J. Heat Mass Transfer, 43 (2000) 4205.
4.2
Second-law analysis
1 63
For the case of (-dP/dx) = 0, the velocity is linear across the fluid. For a negative pressure drop the velocity is positive, and for a pressure increase, the velocity can become negative leading to backflow. At the point of reversal dv/dy - 0 at y = 0. This occurs when -dP/dx - -2/2v1/H 2. The velocity gradient dv/dy from Eq. (4.25) is
dv dy
_
(-dP/dx)
(H
2y) + v l
-
2/2
H
(4.26)
The energy equation for laminar and hydrodynamically developed flow is
d2T - /2 ( dv ] 2 @2 k~ dy)
(4.27)
with the boundary conditions of T = T2 at y = 0, and T = T1 at y = H. Substitution of Eq. (4.26) into Eq. (4.27) yields the temperature distribution
E 1
0 - T - T2 _ (- dP/dy)Y [all4 (1 - y3) _ 2bH 3(1- y2) + 6cH 2 (1 - Y)] + Y 1+ ~ Br(1 - Y) T1 - T 2
12k(T1 - T 2)
1
(4.28)
where a=
(-dP/dy) /2
; b=
(-dP/dy)H /2
+
2v 1
; c=
(-dP/dy)H 2
m
+vl; B r =
4/2
/2v2 k(T1 -T2)
Here, Br is the Brinkman number, which is a ratio of the viscous heating to the heat conducted through the gap of Couette device. From Eq. (4.28), we obtain the temperature gradient
dT
dy
(-dP/dY)[aH3(l_4Y 3 ) - 2 b H 2 ( 1 - 3 y 2 ) + 6 c H ( 1 - 2 Y ) ] + T 1 - T 2 I I + I H ~ B r ( 1 - 2Y) ] 12k
(4.29)
Figure 4.4 shows the temperature profile for T2 = 300 K and - 2 . 0 < Br < 8.0. The rise of temperature in the middle part of the Couette device is considerably large for high values of Br. Inserting Eqs. (4.26) and (4.29) into Eq. (4.23) yields an expression for the volumetric entropy production rate for a Couette flow.
8 Br
6
-2
y 0 . 6 ~ 1 Figure 4.4. Dimensionless temperature profile 0 for the plane Couette flow for Vl = 5.0 m/s and H = 0.005 m. Reprinted with permission from Elsevier, Y. Demirel, Int. J. Heat Mass Transfer, 43 (2000) 4205.
164
4.
Using the second law: Thermodynamic analysis
Example 4.4 Thermomechanical coupling in a circular Couette flow For a circular Couette flow (Figure 4.5), the entropy production rate for an incompressible Newtonian fluid held between two coaxial cylinders is k (dT) 2 ~[ d (~_)] 2 Sprod= 7 ~ . ~ ) q-y r~r •
(4.30)
The circular Couette flow between concentric cylinders is in the 0-direction only, and satisfies Vr =Vz = O, Vo = vo(r), and T = T(r). The inner cylinder is stationary while the outer cylinder rotates with an angular velocity w. Assuming a steady and laminar flow without end effects, the velocity distribution is v o _ 1 (r 2 - r i 2)
(4.31)
Wro R(d- d) where R = r/ro. With Eq. (4.31), we determine the second term on the fight of Eq. (4.30) as follows:
4w2 :ri4 T F4(F2-/~2)
(4.32)
- ~
T r~r
With surface temperatures of To and Ti for the outer and inner cylinders, respectively, the temperature profile is given by
0
(in.
To-~i-(B+I)
n21n n
1-~nn)+B
1/
R2
(4.33)
with the dimensionless quantities of n _ _ _4~ _ /L/,W2r2 n = rl" B = Br )2 " and Br = ro ' (1-n 2 ' k(To - T i )
(4.34)
where Br is the Brinkman number for the annulus. Equation (4.33) satisfies the boundary conditions of 0 = 0 at R - n, and 0 - 1 at R - 1. The temperature gradient can be obtained from Eq. (4.3 3)
d T _ ( T O _ Ti ) dr
2Bro r3
+
B
B+I)
~ - ~ rn 2 Inn r Inn
(4.35)
Substituting Eqs. (4.32) and (4.35) into Eq. (4.30), we can determine the rate of entropy production for the tangential annular flow.
v=wro, T = T o
\
v=O,T=T~
Figure 4.5. The circular Couette flow. Reprinted with permission from Elsevier, Int. J. Heat Mass Transfer, 43 (2000) 4205.
4.2
1 65
Second-law analysis
The first terms of Eqs. (4.23) and (4.30) show the entropy production due to the heat transfer Sprod,kT, while the second terms show the entropy production due to the fluid friction Sprod,6p; hence, the rate of entropy production expression has the following basic form:
Sprod = Sprod,kT + Sprod,kao
(4.36)
The irreversibility distribution ratio is Be = Spr°d'AT
(4.37)
Sprod
and is called the Bejan number Be. Be = 1 is the limit at which all irreversibility is due to heat transfer only. Irreversibility due to heat transfer dominates when Be >> 1/2, while Be Pr > 0.6, and 3000 > Sc > 0.6, we have Cf = 0.0592Re -°'2
(4.74)
h = 0.296 ReO.SprO.33 _k
(4.75)
X
(4.76)
k m =0.296Re°8Sc °33 Di X
Substituting Eqs. (4.69)-(4.76) into Eq. (4.68) and performing the integration, the entropy production rate can be obtained for a laminar flow on a flat plate
•
w
Sprod -- 1.456~- ReT.°5
7,2 +
R8c-O.]33
,, R/pr°33
j2
MiT~
M2iCi,~Di
k
I )
(4.77)
+0.664 Re °5 ~ u ~ v ~
We can obtain a similar expression for the entropy production for a turbulent flow from Eqs. (4.68) and (4.74)-(4.76) . W ReTO. s• Sprod - 28.15 7
[ q2 ) pr-0.33 JiqR ~ + MiT~
k
j/2RSc-°33 ] M2Ci,~Di
(4.78)
where W and L are the width and length of the plate, D the mass diffusivity coefficient, and Sc and Pr the Schmidt number and Prandtl number, respectively. By setting 0Sprod/ORe - 0, we determine the optimum Reynolds number, and hence the minimum entropy production at known values q and J;. For a laminar flow over a plate, we have "~ ' 033 2 2.19 [(1/L2)((q'/T~)+(JiqR/MiT~))(Pr-°33/k)+(JiRSc- • / L M i Ci,~Di) ] ReL,op t =
p=u=v2/T=
(4.79)
174
4.
Usingthe second law: Thermodynamic analysis
For a turbulent flow over a plate, the optimum Reynolds number becomes
(1/t2)[((q2/T2) + (JiqR/MiT~))(Pr-0"33/k)+ (j2 RSc-O.33/M2Ci ~Di)1}0.625 ReL,opt =62.69
2 p~v~v~/T~
'
(4.80)
For a cylindrical geometry, the rate of entropy production becomes Spr°d = 0"462 1LRel)°466 [( -~+MiT~ q2 JiqR]
Pr-°33 k
~ +
j2RSc-°'33 ] ~ D
i
(4.81)
+2.743Re~754(p~v~v2) .......
.....
For cylindrical geometry, the following empirical relations, which are valid for 40 < Re < 1000, are used: Cf = 5.484 ReD0246
(4.82)
h = 0.689 Re~466pr 0"33 k D
(4.83)
km= 0.689Re~466Sc0.33De
(4.84)
D
The expression for the optimum Reynolds number is
ReD,opt = 1.5 7
{ (1/L2)[((q2/T2) + (JiqR/MiT~))(Pr-°33/k) + (JiRLSc-°33/Mz Ci ~Di)]t
p~v~v2/T~
0.82 (4.85)
The optimal Reynolds number defines the operating conditions at which the cylindrical system performs a required heat and mass transport, and generates the minimum entropy. These expressions offer a thermodynamically optimum design. Some expressions for the entropy production in a multicomponent fluid take into account the coupling effects between heat and mass transfers. The resulting diffusion fluxes obey generalized StefanMaxwell relations including the effects of ordinary, forced, pressure, and thermal diffusion.
Example 4.8 Chemical reactions and reacting flows The extension of the theory of linear nonequilibfium thermodynamics to nonlinear systems can describe systems far from equilibrium, such as open chemical reactions. Some chemical reactions may include multiple stationary states, periodic and nonperiodic oscillations, chemical waves, and spatial patterns. The determination of entropy of stationary states in a continuously stirred tank reactor may provide insight into the thermodynamics of open nonlinear systems and the optimum operating conditions of multiphase combustion. These conditions may be achieved by minimizing entropy production and the lost available work, which may lead to the maximum net energy output per unit mass of the flow at the reactor exit. One of the ways to reduce energy costs in the chemical process industry is to increase process reversibility by increasing equipment size. Engineers have to make a trade-off between the energy and area costs. Other tradeoffs are possible between the system output and transfer area and the system output and energy consumption. The equipartition of forces principle suggests that the trade-offs would be optimum for those processes with uniformly distributed thermodynamic driving forces over the transfer area. The local entropy production of a reacting mixture in a system with gradients in temperature T, and chemical potentials/z i, is given by
• Spr°d---JqV
-'-f1 i JiVtxi,r- j Jrj T
(4.86)
4.2
175
Second-law analysis
Here, Jq is the total heat flow, Ji the mass flow of component i, and Jr/the reaction rate (flow) of reaction j. For chemical reactions, linear phenomenological equations are l
jr/ - - Z L i
J
•
AG/ •
"
(4.87)
T
With a homogeneous reaction and mechanical equilibrium (VP = 0), consider a reactor consisting of a large number of n subsystems with equal volumes and the same reaction taking place in all subsystems. We assume that the subsystems have a uniform composition and temperature. The reaction flow in subsystem k is Jrk and the driving force is A G~/T. The total system is a nonhomogeneous reactor with variations in temperature and composition
Spro&min £OPkVk £Lk(~kGk) 2 k k
VI(
(4.88)
~_~JkVk = - ~ Lk - 7 - - V k = constant
(4.89)
_._
~
~
For a specified total reaction rate, AGa.
The Cauchy-Lagrange method of constant multipliers yields
oZ.,%
O(AGk/T) + A O(AGk/T) = 2L k
+ ALk - 0
(4.90)
Thus, we obtain AG k - - ' ~ T 2
(4.91)
The above equation implies that for a given total reaction rate and a given total volume, entropy production is minimal when the driving force AG/Tis equal in all n subsystems. According to the linear duality theory, the results of the optimization will be the same if we maximize the total reaction speed for a given entropy production. Therefore, a thermodynamically efficient reactor has a uniform AG/T in all parts of the reactor volume. This result is independent of the local variations in the reaction rate. Another consequence of Eq. (4.91) is that if we arrange the n subsystems in time instead of in space, then the collection of subsystems constitutes the reaction path of a batch reactor where Vk is the volume of subsystem k. For a specified conversion and time, we should minimize the sum of.Jk(AGk/T)Vk. This minimization leads to results similar to Eq. (4.91), and supports the principle of equipartition of forces. Hence, for a given total conversion and reaction time, minimum entropy production results when the driving force AG/T is equal in all n time intervals. Similarly, the conversion is maximum for a given entropy production and reaction time when the driving forces are uniform. The Gibbs-Helmholtz relation is
~(1/T)
p
(4.92)
If N/-/is constant, integrating the above equation from an equilibrium temperature Teq to the optimal temperature Topt yields
~G°pt - AH ( 1 1) Topt Topt Teq
(4.93)
There is no constant of integration due to the boundary condition that both AG/T and A(1/T) are zero at equilibrium. However, N/-/will be temperature dependent most of the time. For example, in producing ammonia from hydrogen
176
4.
Usingthe second law: Thermodynamic analysis
and nitrogen, the goal is to maximize the output of ammonia at the exit. An approximately constant ATbetween the optimal path and the equilibrium temperature provides the optimal temperature profile, which reduces the exergy loss by --~60% in the reactor. The equipartition of forces principle for multiple, independent rate-controlled reactions and multiphase and coupled phenomena, such as reactive distillations, may lead to the improved use of energy and reduced costs (Sauar et al., 1997).
4.3
EQUIPARTITION PRINCIPLE
The equipartition of forces principle combines the nonequilibrium thermodynamics approach with the CauchyLagrange optimization procedure. The principle shows that the best trade-offs between entropy production and transfer area in transport processes are possible when the thermodynamic driving forces are uniform over the transfer area. For example, in a rate-controlled chemical reaction, the distribution of ~ G / T should be uniform through space and time in the reactor system (the term ~G is the change in the Gibbs energy for a reaction). For example, mathematical models show that a cascade drying process with uniform driving force across every stage yields a substantial decrease in energy consumption. Some options for achieving a thermodynamic optimum are to improve an existing design so the operation will be less irreversible and to distribute the irreversibilities uniformly over space and time. This approach relates the distribution of irreversibilities to the minimization of entropy production based on linear nonequilibrium thermodynamics. For a transport of single substance, the local rate of entropy production is (4.94)
= JX
where J is the local flow of the substance and X the conjugate driving force. Assuming that the linear phenomenological relations hold between the flow and force, we have (4.95)
J = LX
where L is a phenomenological coefficient assumed to be constant and positive. The total entropy production is the integral of • over the time and space variables P = [[*dVdt = L[[ XZdVdt
JJ
JJ
(4.96)
The total flow is the integral over time and space of the local flow J = L f l X d V d t = L VtXav
JJ
(4.97)
where Xav is the average driving force. The total entropy production from Eq. (4.96) is
Pav-- JXav
(4.98)
The difference between the general case and the average value is (Hohmann, 1971)
P-Pav -= L [ ~ y 2 d V d t - ( Y a v ) 2 V t ] = tVt[(X2)av - (Xav) 2 ]
(4.99)
The square bracket on the fight of the above equation is the difference between the mean square and the square mean of the force distribution, and is the variance of X. We therefore have P-Pay > 0 Vt
(4.100)
Pav (equipartioned) < P (arbitrary)
(4.101)
or
The entropy production Pav of a process with a uniform driving force is smaller than that of a nonuniform situation with the same size, and duration of the same average driving force, and the same overall load J. Equations (4.94) and (4.95) show that the local flow and the local rate of entropy production • will be constant when X is uniform.
4.3
1"/7
Equipartition principle
Therefore, the equipartition of forces is analogous to equipartition of flows or of entropy production. The relations of flow and the entropy production in matrix forms are a - [L]. X - Xs'J-
(4.102)
X T'[LI'X
(4.103)
where [L] is the symmetric matrix of phenomenological coefficients due to the Onsager reciprocal relations. The total entropy production P and the total flow J (specified duty) are
P- ff xf.lLl.X.dVdt
(4.104)
a - f l [L]. XdVdt - Vt[L]. Xav
(4.105)
where Nayis the average driving force vector, and elements of it are the averages of the individual driving forces. Using the average force, the total entropy production becomes P~v = J" XaSv The excess entropy production P
-
(4.106)
is obtained as
Pay
P - P a y - If( XT "[L]'X-XaTv "[L]'Xav)dVd'
(4.107)
The above equation can also be expressed as
If ( e l ' e - e f a v ' e a v ) d V d t - [ ( e s ' e ) a v - e a v 'we a v ] Vt
P-R~v -
(4.108)
where e = [R] • X, and [L] is the positive matrix and may be decomposed into a product of matrices [ L ] - [R] T .[R]
(4.109)
The quantity of excess entropy production is positive by the Cauchy-Schwartz inequality (similar to the inequality in Eq. (4.101)), indicating that P > PaySince the minimization of entropy production is not always an economic criterion, it is necessary to relate the overall production and distribution of entropy to the economic analysis by considering various processes with different structures and operating configurations. One optimum requires a uniformly distributed entropy production rate in a heat exchanger, mixer, or separator. Consider the example of countercurrent and cocurrent heat exchangers shown in Figure 4.11. Temperature profiles
Wl
1
_-
Z
=_
2
K
Z
2
J I .....
,,] __
~]
F-
1 [
I
1F 1
1
countercurre11[
]
COCurrent
Figure 4.11. Heat exchangers with countercurrent and cocurrent operations.
178
4.
Usingthe second law: Thermodynamic analysis
show that the driving force AT, or 1/AXT,is more uniformly distributed in the countercurrent than in the cocurrent flow operation. This is the basic thermodynamic reason why a countercurrent is better than a cocurrent operation. The duty of the exchangers depends on the flow rate and inlet and outlet temperatures T1 and T2 of cold streams. The duty is the amount of heat transferred from the hot fluid to cold fluid. The heat exchangers are identical except for the flow arrangements. The cocurrent exchanger will require a higher flow rate and/or higher temperature of hot fluid, and hence the operating cost will be higher than that of the countercurrent exchanger. Alternatively, the cocurrent exchanger will require a larger heat transfer area for a specified flow rate and inlet temperature of the hot fluid, which will require a greater initial investment. Therefore, the countercurrent exchanger may minimize either operating or investment costs compared with the cocurrent exchanger.
Example 4.9 Entropy production in separation process: Distillation Distillation columns generally operate far from their thermodynamically optimum conditions. In absorption, desorption, membrane separation, and rectification, the major irreversibility is due to mass transfer. The analysis of a sieve tray distillation column reveals that the irreversibility on the tray is mostly due to bubble-liquid interaction on the tray, and mass transfer is the largest contributor to the irreversibility. A distillation column requires a large amount of heat in the reboiler and discharges a similar amount of heat at the condenser, and hence resembles a heat engine that requires optimum operating conditions. Heat causes a required separation of the components of a feed stream into products. Second-law analysis may be an effective tool for identifying the possible improvements in distillation column design by determining the entropy production due to thermodynamic inefficiencies in a column. The lost work profiles may quantify the inefficiencies in terms of the pressure drop, mixing, heat and mass transfer, and coupling between heat and mass transfer. The thermodynamic optimization of a distillation column leads to uniform irreversibility distributions. Such an optimization may be achieved through a set of internal and external column modifications, such as altering the feed condition or feed stage location, and using intermediate exchangers to reduce irreversibility in sections with large driving forces and to increase irreversibility in sections with small driving forces. Nonequilibrium molecular dynamics simulations show that the assumption of local equilibrium in a column with heat and mass transfer is acceptable. The dissipation function in a binary distillation is (Ratkje et al., 1995; Sauar et al., 1997) 2
= TSproO=
-JqVlnT- EJiVIXi,v
(4.110)
1
On the other hand, the rate of entropy production is Sprod =
-Jq
VT
1 _2
T2
T
Z1 JiVl&i,T
(4.111)
At constant pressure V/.t;,r = V/xc is the concentration-dependent part of the chemical potential gradient. Through the Gibbs-Duhem equation, we can relate the chemical potentials of heavy "h" and light "1" components in the gas phase as follows: V/x~ =
Yl V/xf Yh
(4.112)
where Yl and Yh are the mole fractions in the gas phase of the light and heavy components. From Eqs. (4.110) and (4.112), we obtain Sprod -
where
-Jq --~--
Vtx--LT
(4.113)
Jd (in m3/(m2 h)) is the relative mass flux across the interface Jd
J1
Jh
Yl
Yh
(4.114)
4.3
Equipartition principle
179
The phenomenological equations that follow from Eq. (4.113) are
Jq = - Lqq -VT ~ - LqlY1 V/~ T
(4.115)
Jd = -- Llq ~VT - LllYl V/rx___L
(4.116)
where Lji are the local phenomenological coefficients, which can be determined from experiments. For isothermal conditions, the phenomenological coefficients for mass transfer are
T ) (4.117)
LI1 = _ Yl V/~l
~XT=0
Using the chemical force for mass transfer VIZl _
Yl T
Jd
Llq V T
Lli
L11 T2
(4.118)
we obtain an expression for the heat flow
Jq=- Lqq-Lql Lu )---~+--~1Jd
(4.119)
On the other hand, Fourier's law of heat conduction without mass transfer is
(Jq )J,/=o = -kVT
(4.120)
where the thermal conductivity k becomes /-qq) 1
k = Lqq - Lql LI---~ T2
(4.121)
The total rate of entropy production for a stage is '~prod -- f fI)vdV v
(4.122)
The entropy production rate is determined with quasi-steady-state calculations. The following constant gradients in the gas phase at each stage are used: VT-
AT Ax
(4.123)
7/xl -
A/x1 2ix
(4.124)
By assuming that Yl and T are approximately constant, and using Eq. (4.122), the entropy production for a stage becomes •
1 AT r
Spr°d -- -- r 2 z2LlcJ
Yl Akh f
J qdV - T A x j J ddV
(4.125)
180
4.
Using the second law: Thermodynamic analysis
where d V = dAdx, A is the contact area, and x the distance. The integrals are the heat and mass transfer per unit time in the mixture volume, respectively, and can be calculated using Eqs. (4.115) and (4.116). Equation (4.122) determines the rate of entropy production. The flow on a stage Jd can be calculated from the diffusion coefficients in the gas phase and from the energy balance. In an adiabatic distillation, the heat flow across the interface contains the latent heat and heat conducted away from the interface kAT
(Jh + J1)Hn = (Jh + J1)Hn-1 + JhAHh + JlZ~J]l - ~Ax
(4.126)
If the enthalpy of the mixtures at stages n and n - 1, Hn and Hn_ 1, are similar, and the temperatures of these stages are close to each other, we have Jl~/1
~ --Jh~/h
(4.127)
This means that the heat of vaporization is approximately equal to the heat of condensation for the mixture. From Eqs. (4.118) and (4.127), the flow on a stage is expressed by
Jd=Jl(I + l zxH1 Yh ~J-/h
(4.128)
Fick's law predicts the diffusion of the light component in the gas phase (4.129)
J1 = - D Acl kx
where D is the diffusion coefficient of the light component and AC1 the concentration difference of the light component across Ax. The concentration difference is ACl -- Cl,g - Cl,g
(4.130)
where the concentration in the gas phase at the total pressure PT is PT Cl,g = Yl R T
(4.131)
* PT
(4.132)
At the liquid-vapor interface, we have
Cl,g = Yl R T
where the mole fraction Yl is the inlet composition of the liquid. Inserting Eqs. (4.128) and (4.129) into Eq. (4.117), and assuming constant driving forces, we express the phenomenological coefficient of the mass transfer as follows: 1.11_
T
[ (11 1 1 D Aq
YlA/xl
-~
Yh AHh
(4.133)
m
The average coefficient/-ql is dx fill -- a 0( ~ ) 2
(4.134)
The phenomenological coefficients obtained from Eq. (4.133) may vary considerably from enriching section to stripping section.
4.3
181
Equipartifion principle
The chemical driving tbrce on a stage has inlet and outlet concentrations as boundary conditions. In the enriching section below the feed plate, a flow Jd from liquid to vapor occurs while in the stripping section, the direction of flow is from vapor to liquid. For the column with specified inlet and outlet compositions, the entropy production rates are
~1 = I L1','X(dAdx' ~2 = I I~',2x2dAdx V
(4.135)
V
where the net separation flow Jd,1, for example, for stage 1 is . j j ,l dA dx = [. I~ l,l X l dA clx V
(4.136)
V
where Xis the chemical force in Eqs. (4.135) and (4.136). A specified level of separation determines the boundary for the forces, and an increase in the force in one stage must lead to a reduction in another stage. The sum of the entropy production rates is
Sprod,1 -+-Spro&2 = I (Lll,1x2 + Lll,222) dAdX V
(4.137)
The total flow Jj is given by (4.138) V
V
V
An increase in the flow for a given entropy production rate and a reduction in the entropy production rate for a specified separation are desired; the yield y is defined as the benefit-cost ratio in an economic sense, and given by Jd,~_LllX; _ 1 Y--
~i
L,1X~
Xi
(4.139)
When the derivative ofy with respect to Xi is higher in one stage than in another, the ratio of flow to entropy production increases by increasing the driving force, and hence the entropy production rate is adjusted by increasing or reducing the force. We can maximize the separation output by redistributing the forces between the stages. Such a distribution results from the following differentiation: d(1/Xl_____~)= d(1/X2 ) dX1 dX2
(4.140)
X 1- X2
(4.141)
The above equation yields
The equality of forces is independent of the individual values of the phenomenological coefficients. This means that the variation of the entropy production rate along the column follows the variation of the phenomenological coefficient Lll. The reversible operation is possible when X1 and )(2 approach zero and y increases toward infinity. Therefore, the practical way to improve second law efficiency is to apply the relationship between dX1 and dX2. For a constant Jd, we obtain m
m
dX1 = - Lll'---LdX2
Lll,2
The above equation relates the driving forces of the two stages (Tondeur and Kvaalen, 1987).
(4.142)
4.
182
4.3.1
Usingthe second law: Thermodynamic analysis
Separation Work
In a distillation column, we supply heat at a higher temperature source in the reboiler, and then discharge at a lower temperature in the condenser (Figure 4.12). Assuming the column to be a reversible heat engine, the net work available from the thermal energy is (Ognisty, 1995)
4,43 where To is the ambient temperature. The temperature corrections describe the maximum fraction of theoretical work extracted from thermal energy at a particular ambient temperature. The minimum separation work Wminrequired for separation is the net change in availability A (A = H - ToS) - W m i n - - ~ d s = A p r o d - Afeed
(4.144)
The change of availability of separation is the difference between the work supplied by the heat and the work required for separation, which contains the work lost due to irreversibilities ~ d s -- Wheat -- Wts
(4.145)
where Wts is the total work necessary for the separation. Minimizing the work lost due to irreversibility will minimize the total heat needed for separation. Efficiency based on the second law of thermodynamics Thh is mmin
(4.146)
.
't~th --
dPdt Wmin + V :o
Heat and mass transfer in a distillation column are coupled, and if the temperature field or chemical force is specified in the column, the other force would be defined. Maximum second law efficiency results from minimizing the entropy production rate with respect to one of the forces. For example, if the contribution of mass transfer is dominant, we should try to minimize the change of the entropy production with respect to the chemical force. The main effects through which work is lost are pressure drops due to fluid flows, heat transfer between fluids with different temperatures, and mass transfer between streams that are not in equilibrium: • The work lost due to a high-pressure drop (as high as 10psi) is considerable at the condenser and reboiler. The pressure drop is relatively smaller through the trays (0.1 psi or less). A large pressure difference affects the distance
I
"
Tc
....,',',',] Condenser
Distillate
Feed
=,... v
l qR
Bottoms
TR I Reboiler Figure 4.12. Distillation column as a heat engine between reboiler and condenser.
4.3
Equipartition principle
183
from equilibrium and causes the large temperature difference and hence utility costs between the condenser and reboiler to increase. • The work lost due to heat transfer results from differences in temperature between the inlet streams of liquid and vapor on each tray, and is usually a large contributor to the total lost work. Enthalpy profiles display the heat transfer on each tray. Since the heat and mass transfer are coupled, any changes in heat transfer through heating and cooling modifications will change the internal mass balances. A modification in mass transfer will have similar effects on heat transfer properties in the column. If a cheap power source is available, intermediate exchangers may be feasible, although the number of trays will need to increase due to the operating line being closer to the equilibrium curve. Lowering the duties of the reboiler and condenser reduces the overall flow in the column, and results in a smaller diameter column design. Improving thermodynamic efficiency leads to a basic trend of taller and more slender columns. • Large amounts of lost work due to mixing and mass transfer mainly occur around the feed trays. The mixing may take place between streams with widely different compositions. Amounts of heavier components decrease above the feed tray and lighter components diminish below the feed tray. From the thermodynamic perspective, we may adjust the location of feed tray to counterbalance the lost work. Often, the feed location is determined at the minimum utility loads and tray count or simply by taking into account light- and heavy-key component compositions. The relative cost of the heating and cooling media may influence the location of the feed stage. If a very cold feed enters a stage, then a large amount of heat exchange is necessary below the feed stage to strip the light components. When the heat transfer rises considerably around the feed tray location, feed preconditioning may be useful to unload the top or bottom sections of columns. Preconditioning the feed is less expensive than interheating or intercooling. Heat profiles and heat transfer lost work plots can be used together to determine if feed preconditioning is necessary. • In many multicomponent mixture separations by distillation, components can display large concentration changes within the column, which cause considerable lost work. One of the ways of overcoming this obstacle is to remove the key components from the feed. As seen in Figure 4.13, light nonkey components can be removed by using an absorber, and the bottom products of the absorber provide the feed to the main distillation column. The heavy nonkey components are removed by using a prestripper, and the excess products of the stripper become the feed of the main distillation column. As seen in Figure 4.13, heat pumping, vapor compression,
Absorber
Main column ~- Liglhts Lights Light key
Feed
_l
-
Feed
Light bottoms ....
Bottoms
I_.
FHeavy bottoms (b)
Stripper
Main column (a)
t
-D fl ¸
C
-~D C
~,B ~B
(c)
(d)
(e)
Figure 4.13. Prefractionation arrangements: (a) removing light keys with absorber; (b) removing heavy keys with stripper; (c) heat pumping; (d) vapor recompression; (e) reboiler flashing. B: bottom product, D: distillate, V: valve.
184
4.
Usingthe second law: Thermodynamicanalysis
and reboiler flashing may also be useful. These modifications will reduce the load of the column to prevent bottlenecking and reduce the required number of stages. • For a sieve tray distillation column, we may calculate the entropy production for heat, mass, and momentum transfer accounting for the movement of a bubble through a moving liquid pool. Some variables are orifice diameter and weir height for the required separation characteristics of components in vapor-liquid phases. Mass transfer and work done against liquid during bubble growth and the drag on bubbles are the major causes of entropy production on a tray. The bubble-liquid interactions are the major contribution of irreversibility, while the effects of the interaction of the flowing liquid with the tray internals are negligible. Bubbles forming at sieve trays may cause entropy production, as most of the heat and mass transfer occurs before bubble detachment. Bubble growth after detachment is small, while viscous drag on bubbles also contributes to irreversibility. The effect of the weir height is more significant than the effect of the sieve-hole diameter. Increasing the weir height shows a monotonic increase in the entropy production, while an increase in sieve-hole diameter is associated with maximum entropy production. This diameter range mainly depends on the properties of the mixture on the tray.
4.4
EXERGY ANALYSIS
Mass and energy are never lost in any physical transformation process. Energy remains constant but changes its form during a process. To determine what is lost in resource transformation processes, we need to utilize the second law of thermodynamics, which states that a part of accessible work potential is always lost in any real process. A certain amount of the total energy is not available to do useful work. For example, the same amounts of total energy may have different capacities to cause a change because of the varying available energy. The available energy is a measure of a process's maximum capacity to cause a change. The capacity exists because the process is in a nonequilibrium state.
4.4.1
Exergy
Exergy is the maximum amount of work theoretically available by bringing a resource into equilibrium with its sur-
rounding through a reversible process. Therefore, exergy is a function of both the physical properties of a resource and its environment. In all real processes, exergy loss always accompanies exergy transfer. The maximum work output of any process occurs if the process proceeds reversibly toward equilibrium with the environment (dead state or reference state). The actual work output is much smaller due to process irreversibility. The work loss in a continuous process is the difference in the exergy before and after the process. At the dead state, both the system and its surroundings possess energy but no exergy, and hence there is no spontaneous change within the system or the surroundings. Available energy, A = H - ToS, or exergy is a measure of the departure from the ambient or dead state. As shown in Figure 4.14, in a heat exchanger, the temperature of the hot stream decreases, and its availability goes down, while the temperature of a cold stream increases, and availability increases. So, a heat exchanger transfers available energy from the hot stream to the cold stream, and some of the available energy is lost to allow the heat transfer processes to occur within a finite time and cost. Exergy is an extensive property and a thermodynamic potential. In contrast to energy, exergy is not conserved and decreases in irreversible processes. Exergy is a broadly useful concept both in engineering and in proper resource management for reducing environmental destruction. Exergy expresses simultaneously the quantity and quality of energy; quality is the ability to produce work under the conditions determined by the natural environment. If we discharge the waste product of the process into the environment, external exergy loss occurs due to the deviation of
A
~,
Hot
A
70
Hot
7o ,,..._ r
y X
Figure 4.14. Heat transfer above and at the ambient temperature To.
X
4.4
185
Exergyanalysis
thermal parameters and the chemical composition between the product and the components of the environment. The thermal state and chemical composition of the natural environment represent a reference level (dead state) for the calculation of exergy. Exergy is a unifying concept of many forms of energy, such as heat, mechanical work, and chemical energy. We can derive the exergy Ex relation from the energy and entropy balances for the composite system shown in Figure 4.15 Ex - (Et - Uo)+ P o ( V - Vo)- T o ( S - So)
(4.147)
where E t is the total energy (E t - U + KE + PE) and U, V, and S denote the internal energy, volume, and entropy of the system, respectively. The terms with subscript '0' are the values of the same properties when the system was at the dead state. The terms KE and PE are the kinetic and potential energies, respectively. Some properties of exergy are" • Exergy is measured with respect to the environment; therefore, it is attributed to the composite system. If the environment is a reference state with zero exergy, then exergy becomes a property of the system. • If the system is not at the dead state, then it will undergo a spontaneous change. • The value of exergy loss cannot be negative. • Exergy decreases due to irreversibilities in the system. If a system undergoes a spontaneous change to the dead state without a device to perform work, then exergy is completely lost. • Exergy is the minimum theoretical work input necessary to change the system from the dead state to the specified sate. Specific exergy ex based on a unit mass is given by V2 e x -- ( U -- lto ) + PO ( I .... Vo ) - To ( s - so ) + --£- + g z 2
(4.148)
The kinetic energy (v2/2) and potential energy (gz) are relative to the surroundings and contribute fully to the magnitude of exergy. Using Eq. (4.147), the change in exergy between two states of a closed system is EX2 -- E X l :
(E 2 -
E l ) + Po(V2 - V1 ) - To(S 2 - S1)
(4.149)
where Po and TOshow the pressure and temperature of the surroundings.
4.4.2
Environment
Since exergy is a measure of the departure of the state of the system from that of the environment, it relates the system to the environment. When a system is in thermal, mechanical, and chemical equilibriums with the environment, there are no processes taking place and the system is at the dead state. At dead state, the system has no motion and elevation relative to coordinates in the environment. Only after specifying the environment can we estimate a value for exergy. Specifying environment usually refers to some portion of a system's surroundings. For example, in estimating exergy values, the temperature and pressure of the environment are usually the standard state values, such as 298.15 K and
/
Surroundings TI,.PI, .~'.~" Heat j/~'" . . . . . -'oS-." Work
/ /
\
", system
/'
Figure 4.15. Combined system.
~"/
/
4.
186
Usingthe second law: Thermodynamic analysis
101.31 kPa. Sometimes, the standard state values are the average values of the ambient temperature and pressure of a location where the process under consideration takes place. The environment is composed of large numbers of common species within the Earth's atmosphere, ocean, and crust. The species exist naturally. They are in their stable forms and do not take part in any chemical or physical work interactions between different parts of the environment. We mainly assume that the intensive properties of the environment are unchanging, while the extensive properties can change because of interactions with other systems. Coordinates in the environment are at rest with respect to each other, and relative to these coordinates, we estimate kinetic and potential energies. In the natural environment, however, there are components of states differing in their composition or thermal parameters from thermodynamic equilibrium state. These components can undergo thermal and chemical processes. Therefore, they are natural resources with positive exergy. Only for commonly appearing components can a zero value of exergy be accepted. A correct definition of the reference level is essential for the calculation of external exergy losses. The most probable chemical interaction between the waste products and the environment occurs with the common components of the environment.
4.4.3
Exergy Balance
The decrease of exergy of a system during a process can be expressed as Change in the total exergy = Total exergy i n - Total exergy o u t - Total exergy loss The exergy balance consists of internal exergy losses. Irreversible processes may cause the distribution ofexergy losses within the volume, and the partition of exergy losses may help in understanding the thermodynamic performance of the system. The exergy balance of a closed system (Figure 4.16) between states 1 and 2 is Ex 2 - Ex 1 =
+ Sprod
¢5q) -
=
(4.150)
6q - W
--F
(4.151)
b
where W and q denote work and heat transferred between the system and its surroundings, respectively, Tb is the temperature on the system boundary, and Sprod shows the entropy production by internal irreversibilities. For deriving the exergy balance for heat and work streams, first we multiply the entropy balance by the temperature To and then subtract it from the energy balance, and we obtain
gx2
_ gx1
.__ ~2
1 - -~b 6q - [ W - Po (V2 - V1)] - ToSprod
(4.152)
The above equation is analogous with the entropy balance of the second law. The first term in this expression shows the exergy transfer accompanying heat when the temperature at the heat transfer medium is not constant.
EXin
t
w°k:i Mass
Heat
-'-
Heat "-I ---~
Mass Work
Exou t
Figure 4.16. Mechanism of exergy transfer for a general system.
187
4.4 Exergyanalysis
For a steady-state flow, the energy balance is
Z
(hH+Jt+~)-
(nH+4+Ws)= 0
2
out of system
(4.153)
into system
The exergy balance for a steady-state system shows the exergy loss
~_~
hEx + gt 1- T °
into system k
~s
-+"~4/~s - out ofZ system
[
hEx+gt 1 - - ~
+I/Vs =L'Xloss
(4.154)
where Ws is the shaft work. The rate of loss exergy Exloss represents the overall thermodynamic imperfections, and is directly proportional to the rate of entropy production due to irreversibilities in a process. As the exergy loss increases, the net heat duty has to increase for the process to occur. Consequently, smaller exergy loss means less waste heat or thermodynamic imperfections. At absolute temperature T, the exergy transfer accompanying heat transfer becomes
(l )q
15')
The exergy transfer accompanies work EXwork = W -- W~ur~(for boundary work)
(4.156)
EXwork = W (for other forms of work)
(4.157)
In accordance with the second law, the exergy loss is positive in an irreversible process and vanishes in a reversible process. The change in exergy of a system can be positive, negative, or zero. When the temperature of a process where heat transfer occurs is less than the temperature of the environment, the transfer of heat and exergy flows in opposite directions. Work and the accompanying exergy transfer can be in the same or opposite directions. For an isolated system, there is no transfer of exergy between the system and its surroundings, and hence the change of exergy is equal to exergy loss (4.158)
EXl°ss = T°Spr°d
This equation shows the decrease of exergy principle, which states that the exergy of an isolated system always decreases for irreversible processes, and remains constant for a reversible process. This is similar to the increase of entropy principle, and is a statement of the second law. Exergy balance can also be expressed in exergy rate form
dEx dt
To
-
/
dV
i ! ¸ ,--;-- o;
-
-
loss
(4.159)
If we consider the exergy of a change from a given reference state (where exergy is zero), the work attainable in a real process would be
W = E x - T0AStota1
(4.160)
If the total entropy change vanishes, as in a reversible process, exergy defines an upper limit to the work that is extractable from any process. If heat is transferred between two reservoirs with temperatures T and :To,the exergy becomes
-T°ASt°t~' = To T o -
= q 1-
(4.161)
The above equation is a generalization of the Carnot relation. The ratio between the exergy and the heat Ex/q is called the exergyfactor. When T < To, there is a lack of energy in the system; the value of Ex/q greatly increases for
188
4.
Usingthe second law: Thermodynamic analysis
low temperatures. When T approaches absolute zero, 0 K, then Ex/q approaches infinity; at higher temperatures, Ex/q moves closer to unity. Therefore, exergy reflects the quality of energy; heat or cold is more expensive and valuable when it is needed the most. Since the exergy depends on the state of the environment, waste heat carries a higher exergy in winter than in summer. The exergy of sensible heat with temperature T is expressed as (4.162) If a power generation unit discharges 2000 MW of waste heat at T = 310 K from cooling water at a local ambient temperature To = 300 K, the above equation shows that the waste exergy discharged with the slightly warmed water is ----33MW. This waste energy causes a temperature increase in the local environment, which could gradually change the local ecology. The exergy of light relates to the exergy power per unit area of black body radiation 6x
[ 1()4
~x-O 1+~
--~
(4.163)
where e=o-T 4 is the energy power emission per unit area and tr ~ 5.67 × 108 W/(K 4 m 2) the Stefan-Boltzmann constant. Since the Earth receives sunlight with/'Sun = 6000 K, for the environmental temperature /'Earth= 300 K the exergy factor becomes
[l+ll:oool4300] 093336000
(4.164)
The exergy of material substances can be calculated if the pressure P and the temperature T are constant and equal to ambient conditions P0 and To, and the exergy is Ex = ~ ni(ix i - tXio )
(4.165)
i
where/J-,i and ].1,io are the chemical potentials of substance i in its present state and in its environmental state, respectively, and ni the number of moles. The chemical potential/zi is defined in terms of concentration c [Zi -- ],1,0 nt- R T In c i
(4.166)
where/x ° is the standard state chemical potential. Substitution of Eq. (4.166) into Eq. (4.165) yields EX "-" E t'li (l'LOi -- ]'zOo) + RTO E gli In c--!-/ i i Cio
(4.167)
where/z ° and Cio are the chemical potential and concentrations of component i, respectively, at environmental conditions (dead state). For a single-component system, Eq. (4.167) becomes
Combustion reactions often cause extensive exergy loss. Exergy calculations show that the entropy production can cause the loss of considerable potential work due to a reaction. An electrochemical membrane reactor or a fuel cell could reduce exergy loss considerably. For pure components, the chemical exergy consists of the exergy that can be obtained by diffusing the components to their reference concentration ci0 with a partial pressure of Pio. For an ideal gas, we obtain Ex = n R T o In p.P~
(4.169)
4.4
189
Exergyanalysis
where Pi and Pi0 refer to the partial pressure of the gas in the emission and in the environment, respectively. Equation (4.169) shows that exergy use may have ecological and environmental effects.
4.4.4 Flow Exergy As flow processes are common in industry, exergy of the mass flow crossing the system boundary is important. The main components of exergy are kinetic exergy, potential exergy, physical exergy, and chemical exergy. We define the kinetic and potential exergies by the kinetic and potential energies calculated in relation to the environment. Physical exergy results from the deviation of temperature and pressure from the environmental values. Chemical exergy results from the deviation of the composition from the composition of the environment. In an open system,.flow exergy is exergy transfer due to mass flow and flow work, and the specific flow exergy exf is
exf -- rh
(h-ho)-To(S-So)+---~+gz
(4.170)
where h and s represent the specific enthalpy and entropy, respectively, at the inlet or outlet; h0 and So represent the respective values at the dead state. The flow work rate is rh(Pv), where 1/7 is the mass flow rate, P the presure, and v the specific volume at the inlet or exit. The exergy rate balance for a control volume is
dt
-
.
- ~
q/ -
- P°
dt
/
+ Zi lhexfi - Ze lheXfe - EXl°ss
(4.171)
where the first three terms on the right represent the rate of exergy transfer, and the last term is the rate of exergy loss. The term qj shows the heat transfer rate through the boundary where the instantaneous temperature is ~., while the term Wcv shows the energy transfer rate by work other than flow work. The terms mexfi and mexfe denote the exergy transfer rates accompanying mass flow and flow work at the inlet i and exit e, respectively. For a control volume at steady state, exergy rate balance becomes
0=
1--~/ •
q/
(4.172)
.
i
e
This equation shows that the rate of exergy transferred into the control volume must exceed the rate of exergy transferred out, and the difference is the exergy destroyed due to irreversibilities. Exergy concepts for some steady-state processes are: • Energy remains the same in the throttling valve, while exergy is destroyed because of the expansion of the fluid. • Exergy is destroyed by irreversibilities associated with pressure drops due to fluid friction and stream-to-stream heat transfer due to temperature differences. • In a steam power plant, exergy transfers are due to work, heat, and exergy loss within the control volume. • In a waste heat recovery system, we might reduce the heat transfer irreversibility by designing a heat recovery steam generator with a smaller stream-to-stream temperature difference, and/or reduce friction by designing a turbine with a higher efficiency. • A cost-effective design may result from a consideration of the trade-offs between possible reduction of exergy loss and potential increase in operating cost.
4.4.5 Exergetic(Second Law) Efficiency Energy supplied by the heat transfer, qin, is either utilized, %, or lost to the surroundings, q., and the thermal efficiency, r/, is qu
r/-
. qin
(4.173)
190 The exergetic efficiency,
4.
T]th,
Usingthe second law: Thermodynamic analysis
iS (exergy recovered)/(exergy supplied)
'Oth = 1"]
1-
To/ru )
1-
To~Tin
(4.174)
Generally, the value of exergetic efficiency is less than unity even when r/= 1. Exergy use would increase as the temperature of utilization of energy approaches the temperature of inlet energy. The rate of exergy loss accompanying the heat loss 01 is (1- To/T1)gll , and depends on the operating temperature. Exergetic efficiency expressions can take different forms, including the following for engineering various steady-state processes: • A turbine with adiabatic operation
Wt
T/t =
(4.175)
/h(exf, in -- eXf, out)
where Wt shows the work produced by the turbine. • A compressor or pump with work input 1//and adiabatic operating conditions
th(eXf, out -- exf, in) n~
-- --
(4.176)
(--rP)
• A heat exchanger not mixing at adiabatic conditions with both streams at temperatures above To rhc (eXf, o u t - exf, in ) c
(4.177)
"0th = th h (exf, in _ eXf, out )h
where rhc and rhh represent the mass flow rates of cold and hot streams, respectively. • An adiabatic mixer with streams 1 and 2 entering and stream 3 leaving the system rhh =
th2 (exf,3 - e x f , 2 )
(4.178)
rhl (exf, 1 - e x f , 3 )
• For an adiabatic chemical reaction at constant pressure, the enthalpy remains constant. The loss in exergy is given by the exergy of reactants exl and the exergy of the reaction products ex2 W 1 = e x 1 - e x 2 = T O (s1 - s 2 )
(4.179)
and "Oth - -
W, r0 (sl 1- ' = 1ex 1
$2 )
(4.180)
ex 1
For a combustion reaction taking place in a well-insulated chamber with no work produced, exergetic efficiency becomes Tith
=
1 Exl°ss Ex F
(4.181)
where E x F is the rate of exergy entering with the fuel and Exloss the exergy loss. The primary exergy load Exp,i is a fraction of the total primary exergy. The transformed exergy load Ext,i is the ratio of the transformed exergy to the total primary exergy. The relationship between the individual efficiencies '0i and the overall efficiency r/is r / = E [Exp,i'r/i - Ext,i ( 1 - 'qi)] i
(4.182)
191
4.4 Exergyanalysis The primary exergy loads have the constraint
Z Exp, i =
(4.183)
1
i
Equation (4.182) shows that by increasing a local efficiency r/i or decreasing a transformed exergy load Ext, i, the overall efficiency rt of a process may increase as long as this does not cause any opposite and larger effect through a change in other parameters. An effective way of improving the overall exergy is to increase the primary load of the units with the largest efficiencies at the expense of those with the lowest efficiencies. The exergy efficiency by the second law is Exout Tit h - -
(4.184)
Exin
Intrinsic efficiency T/in takes into account the transiting exergy Extr EXou t --
Extr _ Exp
rim . . . . . . . . . . . . . .
Exin - Extr
Exc
(4.185)
The transiting exergy is the part of the exergy entering a unit operation; it traverses without undergoing any transformation and exergy loss. The terms Exc and Exp are the exergies actually consumed and produced, respectively. In Eq. (4.182), intrinsic efficiency is used.
Example 4.10 Thermodynamic efficiency in a power plant A steam power plant produces 65 MW electricity with an efficiency of 70%. It uses steam (stream 1) at 8200 kPa and 550°C. The discharged stream (stream 2) is at 75 kPa. If the expansion in the turbine is adiabatic, and the surroundings are at 298.15 K, determine: (a) The maximum work output; (b) The thermodynamic efficiency. Solution: Assume that kinetic and potential energies are negligible, and the system is at steady state. (a) Basis" 1 kg/s steam with the properties from the steam tables (Appendix D) H 1 = 3517.8kJ/kg, S 1 - 6. 8648 kJ/(kg K) at ~ = 550°C, P1 = 8200kPa 82, V =
7.3554 kJ/(kg K), $2, L = 1.213 lkJ/(kgK) at P2 = 75 kPa(saturated steam)
He, v = 2663.0 kJ/kg, H2, c = 384.45 kJ/(kg K) at P2 = 75 kPa (saturated steam') T~t=- 0.7, To = 298.15 K, R = 8.314 J/(mol K) If the turbine operates at isentropic conditions, then we have S 2 = S 1 < 7.3554 kJ/(kg K) Therefore, the discharged steam is wet steam, and the vapor fraction Xs (the quality at isentropic operation) is 6.8646-1.2131 x~= 7.4570-1.2131
= 0.905
The discharged steam enthalpy at isentropic conditions Hzs is H2s = H2, L (1 - Xs) + H2, v x s = 384.45(1 - 0.905) + 2663(0.905) = 2446.8 kJ/kg The enthalpy change at isentropic conditions is ~ / s = H2s - H1 = - 1070.98 kJ/kg
192
4.
Usingthe second law: Thermodynamic analysis
The actual enthalpy change is AH = "qtAHs = - 7 4 9 . 6 7 The actual enthalpy of the discharged steam is H2 = H1 + AH = 2768.11 kJ/kg > H2,v Therefore, the final state is superheated steam at 75 kPa and between 125 and 15°C. Using the linear interpolation of the values from the steam table, we have T2 -- 144.88°C, S2 = 7.7254 kJ/(kg K) The steam rate is -65,000 rh = ~ = 86.70 kg/s H2- H1 The maximum work with AS = $2 - $1 is Wmax = t h ( ~ -
ToAS ) = - 8 7 , 2 5 3 . 3 2
kW
(b) The thermodynamic efficiency is
W Tit - - - Wmax
4.4.6
-65,000.0
= 0.744
-87,253.32
Exergy Analysis Procedure
The thermodynamic analysis of an existing operation consists of three parts. The first part mainly assesses the thermodynamic performance of the current operation. If found to be necessary, in the second part, targets for modifications are identified to reduce the cost of operation. The third part involves the assessment of the thermodynamics and economic effectiveness of the modifications. Exergy analysis can help in all three parts above. The main steps of exergy analysis are: • • • • • • • • • •
4.5
Define the system boundary of processes to be analyzed. Define all the assumptions and the reference conditions of temperature and pressure. Choose the thermodynamic methods for property and phase equilibrium estimations. Consider possible heat recovery and heat integration strategies for all the processes analyzed. Obtain a converged solution using a simulator for the mass and energy balances. Estimate the rate of exergy flows for material and heat streams crossing the system boundary. Determine the total exergy losses. Determine the thermodynamic efficiency. Use exergy loss profiles to identify the regions performing poorly. Identify improvements and modifications to reduce the cost of energy and operation.
APPLICATIONS OF EXERGY ANALYSIS
Example 4.11 Energy dissipation in countercurrent and cocurrent heat exchangers The two most commonly used heat exchangers are countercurrent and cocurrent at steady-state flow conditions as shown in Figure 4.17. Estimate the energy dissipated from these heat exchangers if the surroundings are at 290 K. Consider the data below: Cocurrent Countercurrent:
Thl = 365 K, Th2 = 315 K, Tcl= 280 K, To2 = 305 K Thl = 365K, Th2 -" 315K, Tcm = 280K, Tc2 = 355K
4.5 Applications of exergy analysis
193
Thl
Thl Th2
Tc2
Tc2
Y
Tcl J
mh2
Tcl
J
q
q
Cocurrent
Countercurrent
Figure 4.17. Temperature profiles for heat exchangers operated in cocurrent and countercurrent modes.
The temperature difference between the hot and cold streams is 10 K at the end of the heat exchange. Assuming that kinetic and potential energies are negligible, we apply the following general energy balance by disregarding work interactions"
fhhCph(Th2 -- Thl ) -Jr-thcCpc(%2
(4.186)
-%1 ) - 0
where rhh and rhc are the flow rates of hot and cold streams, which may be related to the temperature changes of hot and cold streams when Cph-- (/pc: ivhc - Thl -- Th2 /h h
(4.187)
Tc2 - Tel
The total rate of entropy change for the hot and cold streams is (4.188)
A S - ;hhASh Jr-/~/cASc
If the pressure changes of the streams are negligible, and the heat capacities of both streams are constant and equal to Cp, then we have total entropy change A , ~ -- th h
Cp
/
In Th2 -~ t~/c In rhl
th h
c2/
(4.189)
Tcl
Applying Eqs. (4.187) and (4.189) for cocurrent and countercurrent operations, we find: Cocurrent heat exchanger I: th c _ fhl -- Th2 _ 365 - 315
rhh
A~o I = fh h
Cp
Tce -Tcl
(
305-280
=2.0
(3,5
305
In Th2]/.,hl-4-"thh *hc ln.Tc I - rhhCp In 365 + 2.0 I n 280
/~Xloss,i --
~()AS
1 -
- 0.0237rhhC p
2 7 3 . 1 5 K ( A S t ) - 6.473rhhC p
Countercurrent heat exchanger I|" rhc _ Th~ - The
3 6 5 - 315
Jr//h
355-
To2 - Tcl
= 0.666
280
ASu=rhhCl,(lnTh2+rh--!ClnfC2j-rhhCp(ln315+O.6661n 355 Thl
mh
Tel
365
280
~+Xloss,iI - ToASI1 - 273.15 K ( A S . ) - 2.932rhhC p
= 0.0107thhC p
4.
194
Usingthe second law: Thermodynamic analysis
The ratio of exergy losses yields EXloss,iI ~ =
/~Xloss,I
2.932 rhhCp
=0.453
6.473rhhC p
This ratio shows that the rate of energy dissipated in the cocurrent heat exchanger is almost twice the dissipation in the countercurrent heat exchanger. Although the heat exchanged between the hot and cold streams is the same, the countercurrent operation is thermodynamically more efficient.
4.5.1
Power Generation
Steam power plants produce electricity with rather low thermal efficiency. An increase in efficiency leads to savings in fuel costs and minimizes environmental effects. The two basic approaches in increasing the thermal efficiency of a cycle are: (i) design a process that transfers heat to the working fluid at high temperature in the boiler and (ii) design a process that transfers heat to the working fluid at low temperature in the condenser. These may decrease the temperature differences, and hence the level of irreversibility.
Example 4.12 Exergy analysis of a power plant A steam power plant operates on a simple ideal Rankine cycle (see Figure 4.18). The turbine receives steam at 698.15 K and 4200 kPa, while the discharged steam is at 40 kPa. The mass flow rate of steam is 3.0 kg/s. In the boiler, heat is transferred into the steam from a source at 1500 K. In the condenser, heat is discharged to the surroundings at 298 K. Determine the energy dissipated at each state. Solution: Assume that the surroundings are at 298 K and the kinetic and potential energy changes are negligible rhs = 3.0kg/s, V1 = 0.001022m3/kg P3 = 4100kPa, P4 = 40kPa H 3 = 3272.3 kJ/kg, S3 = 6.845 kJ/(kg K) H4, v = 2636.9 kJ/kg, H 1 = H4, L = 317.65 kJ/kg Sa,v = 7.6709 kJ/(kg K), S1 =
S4,L,
84, L --
1.0261 kJ/(kg K)
T0 = 2 9 8 K , TH = 1 5 0 0 K , Tc = 2 9 8 K
~mm,=
Figure 4.18. The effects of superheating the steam to higher temperatures and reducing the condenser pressure on the ideal Rankine cycle.
4.5
195
Applications of exergy analysis
Wp,in = Vl(P2, p1)= 0.001022(4100,50) ( llkPam3 kJ)
=4.14kJ
H 2 = H 1 + mp,in = 321.79 kJ/kg
(4.190)
(4.191)
Because this is an isentropic process $3 = $4 and S1 = $2. We estimate the quality of the discharged wet steam (S3,v < S4,v) after passing through the turbine: 6.845-1.0262 X4s =
= 0.875
(4.192)
7.6709-1.0261
H 4 = 317.65(1-0.875)+ 2636.9× 0.875-2356.6kJ/kg
S4 = S 3 =1.0261 × ( 1 - 0 . 8 7 5 ) + 7.6709 × 0.875 = 6.8402 kJ/(kg s) qin = H3 - H2 = 2 9 5 0 . 5 1 kJ/kg qout = H4 - H1 - 2 0 3 8 . 9 5 kJ/kg
Wnet=qin -- qout = 911.56 kJ/kg
rl = 1- qout
=
0.309
(4.193)
(4.194)
qin
The exergy balance for 3.0 kg/s working fluid yields EXloss,12 = 0 (isentropic process)
EXloss,23=thToIS 3 - S 2 + (~3q4i4n3) .)5 8=k W T u
EXloss,3 4 = 0 (isentropic process)
•
/
qout) =914.76kW
EXloss,41=rhTo S 1 - S 4 nt-~C
The exergy losses or the work losses are 79% and 21% in the boiler and condenser, respectively. In a Rankine cycle, exergy losses are due to irreversibilities occurring during heat transfer with finite temperature differences in the boiler and condenser. In order to decrease exergy losses, the temperature differences should be made smaller. Regeneration may help to decrease the temperature differences.
4.5.2
Improving the Efficiency of Power Plants
Some modifications to improve the efficiency of steam power production are
1. Modification of operating conditions of the condenser and boiler: Lowering the operating pressure of the condenser reduces the temperature of the saturated steam within the condenser. Therefore, the heat flow from the condenser to the environment will be at a lower temperature. The lower limit of the condenser pressure is the saturation pressure at the temperature of the cooling medium. For example, if the cooling water is at 20°C, and the temperature difference for the heat transfer is 10°C, the steam temperature must be above 30°C, and the condenser pressure must be above 4.24 kPa, which is the saturation pressure of water at 30°C. At the same time, we have to control the quality of the discharged steam from the turbine: steam with a high level of liquid water and of low quality lowers the
196
4.
Usingthe second law: Thermodynamic analysis
efficiency of the turbine and may corrode the turbine blades. The level of superheating in the boiler controls the quality of the discharged steam. Superheating the steam to high temperature increases the temperature at which the heat flows into the boiler (decreasing the temperature difference between the heat source and boiler), and increases the turbine efficiency and quality of the discharged steam. Figure 4.18 shows the effects of superheating the steam to higher temperatures and reducing the condenser pressure of the ideal Rankine cycle on a T-S diagram. The gray area underneath 3-3' is the increase in the work due to superheating the steam to a higher temperature, while the area underneath 1-4 is the increase in the work due to reducing the condenser operating pressure. However, the area underneath 2-2' shows the heat input increase, which is considerably smaller. Increasing the operating pressure of the boiler increases the boiling point and the average temperature for heat flow into the boiler. The operating pressures may be as high as 30 MPa (4500psia) in many power plants. The temperature of a furnace (heat source) is --~1350°C, which makes the energy in the furnace very valuable. This energy is used to produce steam at --~200-250°C. This process is irreversible, and hence causes a large amount of work loss. 2. Reheating the steam" Reheating enables the expansion of the steam in various stages instead of a single expansion process. Mainly, reheating increases the steam quality to protect the material. In an ideal reheat Rankine cycle with two-stage expansion, for example, the steam is expanded to an intermediate pressure isentropically in the high-pressure turbine section, and sent to the boiler to be reheated. In the low-pressure turbine section, the reheated steam is expanded to the condenser operating pressure. The reheating increases the average temperature at which the steam is heated, and hence a single reheating may increase the cycle efficiency by --~4%. A double reheating is common. Increasing the number of reheatings to more than 2 may not be feasible, as the additional increase in efficiency may be marginal. The optimum intermediate pressure in the reheating is about one-fourth of the maximum cycle pressure, while the reheating temperature is close or equal to the turbine inlet temperature.
Example 4.13 Simple reheat Rankine cycle in a steam power plant A simple ideal reheat Rankine cycle is used in a steam power plant (see Figure 4.19). Steam enters the turbine at 9000 kPa and 823.15 K and leaves at 4350 kPa. The steam is reheated at constant pressure to 823.15 K. The discharged steam from the low-pressure turbine is at 10 kPa. The net power output of the turbine is 65 MW. Determine the thermal efficiency and the work loss at each unit. Solution: Assume that the surroundings are at 298 K, the kinetic and potential energy changes are negligible, and this is a steady process. V1= 0.00101 m3/kg
T
3
5
t t
/
s Figure 4.19. Simple reheat Rankine cycle.
4.5
197
Applications of exergy analysis
P3 =9000 kPa, H 3 =3509.8kJ/kg, S 3 =6.8143 kJ/(kg K) P6 = 10 kPa, H6, v = 2584.8 kJ/kg, H 1 = H6, L = 191.81 kJ/kg $6,v = 8.1511 kJ/(kg K), $6,L = 0.6493 kJ/(kg K) T4=698.15K , P4=4350kPa, H 4 =3268.5kJ/kg, S 3 = S 4 = 6 . 8 1 4 3 k J / ( k g K ) Ts=823.15K, Ps=4350kPa, H s - 3 5 5 5 . 2 k J / k g ,
Ss=S6=7.1915kJ/(kgK)
T 0 - 2 9 8 K , TH=1600K, T c = 2 9 8 K 1 kJ ) Wp,in - V1(Pz - P~) = 0.00101(9000 - 10) 1kPa m 3 = 9.08 kJ/kg
H 2 = H 1 + Wp,in = 2 0 0 . 8 9
kJ/kg
Because this is an isentropic process $3 = $4 and $1 = $2. We estimate the quality of the discharged wet steam (S3,v < S4,v) after passing through the turbine: 7.1915-0.6493 X6s =
8.1511-0.6493
=0.872
H6s =191.81(1-0.872)+ 2584.8×0.872 = 2278.69 kJ/kg The turbine work output is: Wout =
H6) = 1517.80 kJ/kg
(H 3 - H4) + (H 5 -
kJ/kg
q23,in = H 3 - H 2 = 3 3 0 8 . 9 1
- H4 = 286.70
kJ/kg
qout = H 6 - H1 = 2 0 8 6 . 8 9
kJ/kg
q45,in --
qin = ( H 3
H5
- H2 ) +
(Hs
- H4 )= 3595.61
Wnet = Wou t - Wp,in -
_Wnet
T~t m w qin
kJ/kg
1508.72 kJ/kg
-n 0 . 4 2
The thermodynamic analysis is: S1 = S 2 =0.6493 kJ/(kg K), S3 = S 4 =6.8143, S 5 = S 6 -7.1915 kJ/(kg K) The exergy balance with a basis of rh = 1kg/s of working fluid yields: EXlo~s,12 = 0 (isentropic process)
(-- q23,in) ] = 1 2 2 0 . 8 9 k W
r.
)
198
4.
Usingthe second law: Thermodynamicanalysis
Table 4.1 Distribution of exergy losses in Example 4.13
Process
&loss (kW)
Percent
1-2 2-3 3-4 4-5 5-6 6-1 Cycle
0 (isentropic) 1220.89 0 (isentropic) 59.00 0 (isentropic) 137.31 1417.20
0 86.1 0 4.2 0 9.7 100
°
EXloss,34 -- 0 (isentropic process)
JEXloss,45--l'hTo(S5-$4-+.
q45'in-------~)) (-- W T -- H59.00 k
EXloss,56 = 0 (isentropic process)
%ut)
/~Xl°ss'61--/'hT° S 1 - $ 6 +-~--c --137.31kW
Table 4.1 shows the distribution of exergy losses at each process. In this ideal reheat Rankine cycle, the steam from the first part of the high-pressure section is expanded and reheated in the boiler until it reaches the boiler exit temperature. The reheated steam is expanded through the turbine to the condenser conditions. The reheating decreases the moisture within the discharged steam. The exergy losses or work losses are 86.1% and 9.7% in the boiler and condenser, respectively. The exergy loss in the reheating step is low (4.2%).
Example 4.14 Actual reheat Rankine cycle in steam power generation A reheat Rankine cycle is used in a steam power plant (see Figure 4.20). Steam enters the high-pressure turbine at 9000 kPa and 823.15 K and leaves at 4350 kPa. The steam is reheated at constant pressure to 823.15 K. The steam enters the low-pressure turbine at 4350 kPa and 823.15 K. The discharged steam from the low-pressure turbine is at 10 kPa. The net power output of the turbine is 65 MW. The isentropic turbine efficiency is 80%. The pump efficiency is 95%. Determine: (a) The thermal efficiency; (b) The work loss at each unit. Solution: Assume that the surroundings are at 298 K, the kinetic and potential energy changes are negligible, and this is a steady process. The steam data are: V1 = 0.00101 m3/kg T3=823-15K, P3 =9000kPa, H3=3509.8kJ/kg, S3=6.8143kJ/(kgK) P6=10kPa, H6,v=2584.8kJ/kg,
Hl=H6,L=191.81kJ/kg
$6,v = 8.1511 kJ/(kg K), $6,L = 0.6493 kJ/(kg K) T4 = 698.15 K, P4=4350kPa, H4s=3268.5kJ/kg,
S3=S4s=6.8143kJ/(kgK)
4.5
Applications of exergy analysis
~/
3
199
5
2
J
Figure 4.20. Actual reheat Rankine cycle.
T s = 823.15K, P5 = 4350kPa, H s = 3555.2kJ/kg, $5= S6s= 7.1915 kJ/(kg K) To = 298K, TH = 1600K, Tc = 2 9 8 K (a) In this actual reheat Rankine cycle, the steam expands in two stages. In the first stage, the steam expands to 4350 kPa and is sent to the reboiler where it is reheated at constant pressure to 823.15 K. In the second stage, the steam expands and the discharged steam at 10 kPa is sent to the condenser. Therefore, the total heat input and total turbine work output become qin = q23 + q45 = (H3 - H2) + (H5 - H4) mou t = W34 + W56 =
(H 3 - H4) + (H 5 - H6)
lkJ Wp,in -
VI(P2 - P 1 ) = 0 . 0 0 1 0 1 ( 9 0 0 0 - 1 0 )
~
lkPam3
=9.55kJ/kg
H 2 - H 1 + Wp,in = 200.88 kJ/kg Because this is an isentropic process $3 = S4s and S~ = S2s, Ss,v < S6,v we estimate the quality of the discharged wet steam after passing through the turbine"
~t =
H3- H4
-+ H 4 - H 3 - Tit (H 3 - H4s ) = 3316.76kJ/kg
H 3 -H4s 7.1915-0.6493 X6s =
=0.872
8.1511 - 0.6493
H6s = 191.81(1- 0.872) + 2584.8 × 0 . 8 7 2 - 2278.7 kJ/kg H5- H6 37 t z
H 5 - H6s
---, H 6 - H 5 - l~t ( H 5 - H6s ) - 2533.99 kJ/kg < H6, v (= 2584.8 kJ/kg)
(4.195)
4.
200
Usingthe second law: Thermodynamic analysis
2354.0-191.81 =0.978 x6 = 2 5 8 4 . 8 - 1 9 1 . 8 1 S 6 = $6, L (1 -
x 6 ) -k- 86, V x 6 =
7.9918 kJ/(kg K)
(4.196)
The turbine work output is: Wout = ( H 3 - H 4 ) + ( H 5 - H 6 ) = 1214.24 kJ/kg q23,in = ( H 3 - H 2 ) = 3308.43 kJ/kg q45,in = ( H 5 - H 4 ) = 238.44 kJ/kg qin = (H3 - H 2 ) + ( H 5 - H 4 ) = 3546.87 k J / k g Wnet = Wout - Wp,in = 1 2 0 4 . 6 8 kJ/kg
_
Tit
_
Wnet
--0.339
qin
The rate of steam is rhs -
Ws Wnet
65,000
- ~
= 53.95 kg/s
1204.68
Table 4.2a shows the state properties of the actual reheat Rankine cycle.
Table 4.2a State properties of the actual reheat Rankine cycle in Example 4.14 State
T (K)
P (kPa)
H (kJ/kg)
S (kJ/(kg K))
1 2 2s 3 4 4s 5 6(x 6 = 0.978) 6s(X6s = 0.872)
319 321.27 321.15 823.15 718.94 698.15 823.15 319 319
10 9000 9000 9000 4350 4350 4350 10 10
191.81 201.36 200.98 3509.8 3316.76 3268.5 3555.2 2534.0 2278.7
0.6492 0.6776 0.6492 6.8143 6.8800 6.8143 7.1915 7.9918 7.1915
Table 4.2b Distribution of exergy losses in actual and ideal reheat Rankine cycles Process
Example 4.14 (actual reheat Rankine) Exloss (kJ/kg)
1-2 2-3
0.477 1212.54
Example 4.13 (ideal reheat Rankine)
Percent 0.03 72.4
Exloss (kJ/kg)
0
0
1220.89
86.1
3-4
19.57
1.2
4-5
48.41
2.9
5-6
238.50
14.3
0
6-1
154.10
9.2
137.31
Cycle
1673.55
100
Percent
0 59.00
1417.20
0 4.2 0 9.7 100
4.5
201
Applications of exergy analysis
(b) Exergy balance with a basis of 1 kg/s working fluid yields Exloss,12 = rh(Wp,a - Wp,s) - 9.5577 - 9.0799 = 0.477 kW
EXloss,23 = t//TO S 3 -- S 2 q-
( - qin,23) ] _ 1 2 1 2 . 5 4 k W
r.
)
Exloss,34 - & T 0 ( & -- ,5'3) -- 19.57 k W
•
{
EXloss,45 = th TO 5'5 -- S 4 Jr- ( - qin,45) ] _ 4 8 . 4 1 k W
rH
)
EXloss,56 - thT0 (S 6 - $5) - 2 3 8 . 5 0 k W
•
(
qout] -
EXloss,61 = th TO S 1 - S 6 + U
154.10 kW
Table 4.2a shows the state properties in actual reheat Rankine operations, while Table 4.2b compares the exergy losses of ideal reheat and actual reheat Rankine cycle operations• The total exergy loss increases from 1417.2 to 1673.55 kJ/kg in the actual operation• This shows an increase of 18.0% in the actual operation. In this actual reheat Rankine cycle, the steam expands in two stages. In the first stage, the steam expands to 4350 kPa and is sent to the reboiler where it is reheated at constant pressure to 823.15 K. In the second stage, the steam expands and the discharged steam at 10 kPa is sent to the condenser. The exergy losses or the work losses are 72.8% and 9.2% in the boiler and condenser, respectively. The exergy loss in the reheating step is low (2.9%). Reheating reduces the moisture in the turbine.
3. Regeneration: Increasing the boiler feed temperature by using expanding steam is possible in a regenerative cycle• Steam extracted at intermediate pressures from various parts of the turbine is used in countercurrent heat exchangers (closed heaters) to heat the feed. The steam leaves the condenser as a saturated liquid at the condenser operating pressure. The condensed water pressure is increased to the feed heater pressure through pumping. By adjusting the fraction of steam extracted from the turbine, one can produce a saturated liquid output at the heater operating pressure. After passing through the heater a pump increases the pressure of the water to the boiler pressure. In open feed heaters, the expanded steam from the turbine is mixed with water. Regeneration increases the efficiency and helps to deaerate the water and control the discharged steam flow rate. The efficiency will increase further if more heaters are used. As many as eight heaters may be used. However, the savings in boiler fuel costs corresponding to the increase in the efficiency of the cycle should exceed the cost of the heaters.
Example 4.15 Ideal regenerative Rankine cycle A steam power plant is using an ideal regenerative Rankine cycle (see Figure 4.21). Steam enters the high-pressure turbine at 8200 kPa and 773.15 K, and the condenser operates at 20 kPa. The steam is extracted from the turbine at 350 kPa to heat the feed water in an open heater. The water is a saturated liquid after passing through the feed water heater. The work output of the turbine is 70 MW. Determine the thermal efficiency and the work loss at each unit. Solution: Assume that the surroundings are at 285 K, the kinetic and potential energy changes are negligible, and this is a steady-state process. The steam data are: l~ - 0.001017 m3/kg =773.15K,
P~-8200kPa,
H 5-3396.4kJ/kg,
Ss-6.7124kJ/(kg K)
202
4.
Using the second law: Thermodynamic analysis
& T
5
2
qout S
Figure 4.21. Ideal regenerative Rankine cycle.
P~ = P7 = 20 kPa, H7, v = 2609.9 kJ/kg, /-/1 = H6, L = 252.45 kJ/kg $7,v = 7.9094 kJ/(kg K), $7,L = 0.8321 kJ/(kg K) P3=350kPa, H3,L=584.27kJ/kg, H3,v=2731.50kJ/kg V~=0.001079m3/kg,
S3,L=l.7273kJ/(kgK),
S3,v=6.9392kJ/(kgK )
To=285K, TH=1600K , Tc=285K In this ideal regenerative Rankine cycle, the steam extracted from the turbine heats the water from the condenser, and the water is pumped to the boiler. Sometimes, this occurs in several stages. The condensate from the feed water heaters is throttled to the next heater at lower pressure. The condensate of the final heater is flashed into the condenser
Wpl = V~( p 3 - p~)= 0.001017(350- 20) (
lkJ ) 1kPa m 3 = 0.335 kJ/kg
H 2 = H 1 + Wpl = 252.78 kJ/kg
Wp2 = V1(P3- P~)= 0.001079(8200- 350)( 1kPa lkJm 3 / = 8.47 kJ/kg H 4 = H 3 + mp2 = 592.74 kJ/kg Because this is an isentropic process, 5'5 = $6 = $7. We estimate the quality of the discharged wet steam at states 6 and 7: x6=
6.7124-1.7273 =0.956 6.9392-1.7273
4.5
Applications of exergy analysis
203
H 6 = 584.27(1- 0.956) + 2731.50 × 0.956 = 2638.06 kJ/kg 6.7124-0.8321 X7 =
=0.830
7.9094-0.8321
H 7 = 252.45(1 - 0.83) + 2609.9(0.83) = 2211.18 kJ/kg The fraction of steam extracted is estimated from the energy balance: t h 6 H 6 + th 2 H 2 = th 3 H 3
In terms of the mass fraction z = th6/th3, the energy balance becomes
zH 6 + (1 - z)H 2 = H 3 The mass fraction is z = H3 - H2 = 0.139 H 6- H 2
The turbine work output is: qin = H5 - H 4 - 2 8 0 3 . 6 6 kJ/kg
qout =(1 -- z)(H 7 - H~) = 1686.52 kJ/kg W n e t = qin -- qout = 1117.13 kJ/kg
Tit
-
-
1 - qout = 0.398
(4.197)
qin
The rate of steam is"
~hs m
Ws _ 70,000 Wnet
= 62.66 kg/s
1117.13
The exergy balance with a base of 1 kg/s yields EXloss,12 = 0
L'Xloss,23 = # / T O(S 3 - 82) =- 255.13 k W
EXloss,3 4 = 0
F__,Xloss,45-lhTo{S5-$4-+-(- k q i nW) )
- T9 2 1n. 3 5
gXloss,56 -- 0
•
EXloss,71 -/~/T0
i Sl -
87 nt- -q°ut/ ~ 0 = 10.63 k W
4.
204
Usingthe second law: Thermodynamic analysis
Table 4.3a State properties of the ideal regenerative Rankine cycle in Example 4.15
State
T (K)
1
319
2 2s 3 4 4s 5 6(x 6 = 0.978) 6s(X6s = 0.872)
321.27 321.15 823.15 718.94 698.15 823.15 319 319
P (kPa)
10 9000 9000 9000 4350 4350 4350 10 10
H (kJ/kg)
S (kJ/(kg K))
191.81 201.36 200.98 3509.8 3316.76 3268.5 3555.2 2534.0 2278.7
0.6492 0.6776 0.6492 6.8143 6.8800 6.8143 7.1915 7.9918 7.1915
Table 4.3b Distribution of exergy losses at each process in Example 4.15
Process
Ideal regenerative Rankine cycle ~bXloss(kW)
Percent
0 255.13 0 921.35 0 10.63 1187.12
0 21.5 0 77.6 0 0.9 100
1-2 2-3 3-4 4-5 5-6 7-1
Cycle
"x
{
/~Xloss,cycle = th TO / qout
rc
qin | = 1187.12 kW
rH)
Table 4.3a shows the state properties of the ideal regenerative Rankine cycle. Table 4.3b shows the distribution of exergy losses at each process. As seen from this table, the highest exergy loss occurs due to heat transfer in the boiler.
Example 4.16 Actual regenerative Rankine cycle A steam power plant is using an actual regenerative Rankine cycle (see Figure 4.22). Steam enters the high-pressure turbine at 11,000 kPa and 773.15 K, and the condenser operates at 10 kPa. The steam is extracted from the turbine at 475 kPa to heat the water in an open heater. The water is a saturated liquid after passing through the water heater. The work output of the turbine is 90 MW. The pump efficiency is 95% and the turbine efficiency is 75%. Determine the work loss at each unit. Solution: Assume that the surroundings are at 285 K, the kinetic and potential energy changes are negligible, and this is a steady-state process. The steam data are:
Ps=ll,000kPa, Hs=3362.2kJ/kg , S5=6.5432kJ/(kgK), Ts=773.15K Pl=P8=10kPa, H8,v=2584.8kJ/kg, Hs,L=191.83kJ/kg, Vl=0.00101m3/kg $7,v = 8.1511 kJ/(kg K), S7,t, = 0.6493 kJ/(kg K) P3=P6=475kPa, H3,L=631.29kJ/kg, H3,v=2745.30kJ/kg
4.5
205
Applications of exergy analysis
Boiler
Open Heater
@
3
I
I_.., 6 Condenser
2 P1
(a)
T
4
5
y
S
(b) Figure 4.22. (a) Schematic of actual regenerative Rankine cycle and (b) T-S diagram.
$3, L =1.8408kJ/(kgK),
$3,v =6.8365kJ/(kgK)
T/p--0.95, r/t=0.75 , To=285K , TIq=1700K , T c = 2 8 5 K In this actual regenerative Rankine cycle, the steam extracted from the turbine heats the water from the condenser, and the water is pumped to the boiler. Sometimes, this occurs in several stages. The condensate from the feed heaters is throttled to the next heater at lower pressure. The condensate of the final heater is flashed into the condenser
%1-~(P2- ~ ) -
0.00101(475-10) (lkJ j 0.95
1kPa m 3 = 0.494 kJ/kg
H 2 = H 1 +Wv~ =191.83+0.494=192.32kJ/kg
206
4.
Usingthe second law: Thermodynamic analysis
Table 4.4a State properties of the actual regenerative Rankine cycle in Example 4.16
State
T (K)
P (kPa)
H (kJ/kg)
1 2 2s 3 4 4s 5 6 6s (X6s = 0.941) 7 (x7 = 0.920) 7s (Xys - - 0.785)
319 319.5 319 423.07 428 428 773.15 457.83 423.07 319 319
10 475 475 475 11000 11000 11000 475 475 10 10
191.81 192.32 631.81 643.89 3362.20 2806.46 2621.21 2394.47 2071.90
S (kJ/(kg K)) 0.6493 0.6510 0.6493 1.8408 1.8451 1.8408 6.5432 7.0125 6.5432 7.5544 6.5432
0.001097(11,000-550)( lkJ / 0.95 1kPa m 3 = 12.08 kJ/kg
Wp2 = V1(P4 - P3) =
9 4 = 9 3 -Jr-mp2 = 6 3 1 . 2 9 +
12.08 = 643.37 kJ/kg
Because this is an isentropic process $5 = S6s = S7s, and we estimate the quality of the discharged wet steam at states 6s and 7s: X6s =
6.5432-1.8408 6.8365-1.8408
=0.941
H6s = 631.29(1- 0.941) + 2745.3(0.941) = 2621.18 kJ/kg From the turbine efficiency, we estimate the enthalpy of superheated vapor at state 6 H 5-H 6 Tit =
--~ 9 6 = 9 5 -- Tit ( 9 5 -- H6s ) = 2 8 0 6 . 4 6 k J / k g
H 5 - H6s
XTs =
6.5432-0.6493 8.1511-0.6493
=0.786
HTs =191.81(1-0.786) + 2584.8(0.786) = 2071.90 kJ/kg From the turbine efficiency, we estimate the enthalpy at state 7 H5- H7 Tit --
--+ H 7 = H 5 - Tit ( g 5 - H7s ) = 2394.48 kJ/kg
H 5 - H7s
Table 4.4a shows the state properties of the actual reheat regenerative Rankine cycle. The fraction of steam extracted is estimated from the energy balance m6H6 + m2H2 = rn3H3. In terms of the mass fraction z = rh6/rh3 , we have z H 6 + ( 1 - z ) I t 2 = H 3. The mass fraction is 2=
~H
3 -= H
2
H 6 -H 2
732.03-192.68
=
2776-192.68
The turbine work output based on 1 kg/s working fluid is" {)in = H 5 - H e = 2 7 1 8 . 3 0 k W
0.168
4.5 Applications of exergy analysis
207
Table 4.4b Distribution of exergy losses at each process in Example 4.16
Actual regenerative Rankine cycle
Process
1-2 34 4-5 5-7 7-1 Regeneration Cycle
Exl .... (kW)
Percent
0.0247 0.6043 883.86 262.24 195.22 34.29 1376.62
64.2 19.0 14.2 2.4 100
To = 285 K, TIt = 1700 K, and T~ = 285 K.
l)out = ( 1 -
z)(H 7 - H1) = 1832.34 kW
~'~net -- 6)in -- qout -- 8 8 5 . 9 6 k W
The thermal efficiency is ~t - 1 - c)°ut -
0.326
~/in T h e rate o f s t e a m is
rhs _ Ws _ 90,000 _ 101.57 kg/s Wnet 886.04 The exergy balance based on specified hot and cold sources yields EXloss,12 = ~/~/pla - ~'~/pls - 0.0247 kW .
.
°
EXloss,34 - Wp2 a - mp2 s =0.6043 kW
~2Xloss,45 - To [ th(Ss - S4) + (- qin) ) - 883.86 k W
TH
EXloss,57 = rnT 0 (zS 6 + (1 -
z)S 7 -
S 5) - 262.24 kW
qout ) - 195.22 kW EXloss,71 - TO /~h(S1 - $ 7 ) ( 1 - Z) + - ~ c
/~Xloss,regen -- fh ~)(S 3 - zS 6 - (1 - z)S2) - 34.29 kW
/~Xloss,cycl e __ ~ ) ( ~/out~c ~Hj--~)in 1376.62kW
Table 4.4b shows the distribution of exergy losses at each process based on T o - 285K, T H - 1700 K, and Tc - 285 K. The highest exergy loss occurs due to heat transfer in the boiler. The regeneration stage work loss is minimum.
208
4.
Usingthe second law: Thermodynamic analysis
Example 4.17 Ideal reheat regenerative cycle A steam power plant is using an ideal reheat regenerative Rankine cycle (see Figure 4.23). Steam enters the high-pressure turbine at 9000 kPa and 773.15 K and leaves at 850 kPa. The condenser operates at 10 kPa. Part of the steam is extracted from the turbine at 850 kPa to heat the water in an open heater, where the steam and liquid water from the condenser mix and direct contact heat transfer takes place. The rest of the steam is reheated to 723.15 K, and expanded in the low-pressure turbine section to the condenser pressure. The water is a saturated liquid after passing through the water heater and is at the heater pressure. The work output of the turbine is 75 MW. Determine the work loss at each unit. Solution: Assume negligible kinetic and potential energy changes, and that this is a steady-state process. The surroundings are at 285 K. In this ideal regenerative Rankine cycle, the steam extracted from the turbine heats the water from the condenser, and the water is pumped to the boiler. Sometimes, this occurs in several stages. The condensate from the heaters is throttled to the next heater at lower pressure. The condensate of the final heater is flashed into the condenser.
Boiler
)" Open Heater ~.~
3
Condenser 2
©
1
-
P2 (a)
T 5
7
(b) Figure 4.23. (a) Schematic of ideal reheat regenerative Rankine cycle and (b) T-S diagram.
4.5
209
Applications of exergy analysis
The steam data are: P 5 - 9000kPa, H 5 -3386.8kJ/kg, S 5 = 6.6600 kJ/(kg K), T5 - 7 7 3 . 1 5 K Pl=Pa-10kPa,
Ha,v-2584.8kJ/kg,
Hs,L=191.83kJ/kg, Vl=0.00101m3/kg
$8,v - 8.1511 kJ/(kg K), $8,L = 0.6493 kJ/(kg K) P3 - 8 5 0 kPa,
732.03 kJ/kg,
H3, L -
H3, V =
2769.90 kJ/kg
PT=850kPa, H7=3372.7kJ/kg, S7-7.696kJ/(kgK),
Tv-723.15kPa
P6=850kPa, ~,=450.0kPa, H6-2779.58kJ/kg T o - 2 8 5 K , TH=1600K, T c - 2 8 5 K Work and enthalpy estimations yield:
W p l _ V l ( ~ _ p 1 ) _ 0 . 0 0 1 0 1 ( 8 5 0 _ 1 0 ) ( 1kPa lkJm 3 ] - 0.848 kJ/kg
H 2 - H 1+ Wpl - 192.68 kJ/kg
W p 2 = V l ( P 3 - P 1 ) - O ' O 0 1 0 7 9 ( 9 0 0 0 - 8 5 0 ) [ 1kPalkJ1113) = 9.046 kJ/kg
H4
- H 3 + Wp2 -
741.07 kJ/kg
Because this is an isentropic process, $ 5 - $6 and $ 7 - $8, and we estimate the quality of the discharged wet steam at state 8: 7.696-0.6493 X8 =
8.1511-0.6493
= 0.940
H a = 191.83(1-0.94) + 2584.8(0.94) = 2439.63 kJ/kg The fraction of steam extracted is estimated from the energy balance m6H 6 + m2H 2 =m3H 3. In terms of the mass fraction z =/n6/rh 3 , the energy balance becomes zH6 + (1 - z)t12 - H3. The mass fraction is H~-H~ z
=
-
-
H~ - H 2
=
732.03-192.68 2779.58-192.68
=0.208
Table 4.5a shows the state properties of the ideal reheat regenerative Rankine cycle based on: T0 - 2 8 5 K ,
TH - 1 6 0 0 K ,
Tc - 2 8 5 K
The turbine work output with rh - 1kg/s working fluid is" qin --
rh[(H5 - H4 ) + (1 - z)(H v - H 6 ) ] - 3115.18 kW
4.
210
Usingthe second law: Thermodynamic analysis
Table 4.5a State properties of the ideal reheat regenerative Rankine cycle in Example 4.17 State
T (K)
P (kPa)
H (kJ/kg)
S (kJ/(kg K))
1 2 3 4 5 6 7 8(x8 = 0.94)
319 321.27 446.1 447.15 773.15 449.97 723.15 319
10 850 850 9000 9000 850 850 10
191.81 192.68 732.03 741.07 3386.80 2779.58 3372.7 2439.63
0.6493 0.6493 2.0705 2.0705 6.6600 6.6600 7.6960 7.6960
To = 285K, TH = 1600K, and Tc = 285K.
qreheat = t h ( g 7 - 9 6 ) = 593.12 k W
qout = rh(1 - z) (H 8 - H1) = 1779.14 k W ~rnet -- qin -- l)out -- 1 3 3 6 . 0 3 k W
The thermal efficiency is T~t "-- 1 - qout __ 0 . 4 2 8 qin
The rate of steam is
hS
.
Ws _ 75,000 . . . . Wnet
56.36 kg/s
1336.03
The exergy balance for each unit with rh = 1 kg/s working fluid is: /~Xloss,12 -= 0
Eyloss,26 = thT0 (83 -
zS6 - 0 - z)$2)=47.88 k W
EXloss,34 = 0
/~Xloss,45 = To ( t h ( S 5
_ 84 ) + (--TH)qin) ] =
785.68 k W
EXloss,56 = 0
EXl°ss,67= T° (th(ST - S6) + (- q67)) =195"82 EXloss,78 -- 0
•
/
qout/
EXloss,S 1 = T o rh(S 1 - $ 8 ) + - ~ - c
= 228.13kW
4.5
211
Applications of exergy analysis
Table 4.5b Distribution of exergy losses at each process in Example 4.17
Ideal reheat regenerative Rankine cycle
Process
Percent
/% .... (kW) 1-2 2-6 3-4 4-5 5-6 6-7 8-1
0 3.9 0 62.5 0 15.5 18.1 100
0 47.88 0 785.68 0 195.82 228.13
Total Cycle
1256.89 1256.89
The work loss in the reheat stage is: (-- £/reheat) _
EXloss,reheat -- TO /~/(S 7 - S 6 ) nt- _
r.
= 195.72 kW
The work loss in the regeneration stage is: EXloss,rehea t -- # / T 0 (S 3 - [ z S 6 + ( 1 -
z)S2] ) -
47.88 kW
The work loss for the whole cycle is
qin/
= 1256.7 k W
Table 4.5b shows the distribution of exergy losses at each process. The highest exergy loss occurs due to heat transfer in the boiler.
Example 4.18 Actual reheat regenerative Rankine cycle A steam power plant is using an actual reheat regenerative Rankine cycle (Figure 4.24). Steam enters the high-pressure turbine at 11,000 kPa and 773.15 K, and the condenser operates at 10 kPa. The steam is extracted from the turbine at 2000 kPa to heat the water in an open heater. The steam is extracted at 475 kPa for process heat. The water is a saturated liquid after passing through the water heater. The work output of the turbine is 90 MW. The turbine efficiency is 80%. The pumps operate isentropically. Determine the work loss at each unit if the surroundings are at 290 K. Solution: Assume negligible kinetic and potential energy changes, and that this is a steady-state process. The steam data are: V1 = 0.00101m3/kg
P1 - P9 = 10 kPa, H1, v - 2584.8 kJ/kg, H9, L - 191.83.45 kJ/kg Sl, v =8.1511 kJ/(kgK), $1,L - 0 . 6 4 9 3 kJ/(kgK)
P3-P6=475kPa,
H3, L = 6 3 1 . 2 9 k J / k g ,
H3,v=2745.30kJ/kg
$3, L = 1.8408 kJ/(kgK), $3, v = 6 . 8 3 6 5 k J / ( k g K )
212
4.
Usingthe second law: Thermodynamicanalysis
Boiler
) 8 ]_.d
Open Heater 3
Condenser
I
2 1
P2
~ P1
(a)
T
9s!
r
(b) Figure 4.24. (a) Schematic of actual reheat regenerative Rankine cycle and (b) T-S diagram.
Ts = 773.15 K, Ps=ll,000kPa,
Hs=3362.2kJ/kg ,
Ss=6.5432kJ/(kgK)
T~t=0.80 , T0=290K, TH=1700K, Tc=290K In this actual regenerative Rankine cycle, the steam extracted from the turbine heats the water from the condenser, and the water is pumped to the boiler. Sometimes, this occurs in several stages. The condensate from the water heaters is throttled to the next heater at lower pressure. The condensate of the final heater is flashed into the condenser Wpl = V1(p2- p1)= 0.00101 (475_ 10)( 1kPa lkJm 3 ) = 0.46 kJ/kg H 2 = H 1+ Wpl = 191.83 + 0.46 = 192.28 kJ/kg
Wp2=Vg(p4_P3)=O.O0109101,O00_475)( lkJm 3 ) = 11.48 kJ/kg 1kPa
4.5 Applications of exergy analysis
213
H 4 - H 3 +//Vp2 = 6 3 1 . 8 1 + 11.48 = 642.77 kJ/kg P6=2000kPa,
H7=3467.3kJ/kg,
S 7 = 7 . 4 3 2 3 k J / ( k g K ), T 7 = 7 7 3 . 1 5 K
From the turbine efficiency, we estimate the enthalpy at superheated vapor at state 6 _
H5
-
T ] t ~ - -
H6
H 6 = H 5 - r/t (H 5 - H6s ) = 2691.67 kJ/kg
H5 - H 6 s P7=2000kPa,
_
T]t~
HT-H8
Sss=S7=7.4323kJ/(kgK), T7 = 7 7 3 . 1 5 K ~ H s s = 3 0 4 5 . 5 4 k J / k g H 8 = H 7 --Tit ( H 7 - Hss) = 3129.89 kJ/kg ~ S s = 7.5951 kJ/(kg K)
H 7 -- Hss
At the condenser operating conditions of 10 kPa, we have $7 = Sss = S9s, and we estimate the quality of the discharged wet steam at state 9s"
X9s =
7.4323-0.6493 8.1511--0.6493
= 0.904
H9s = ( 1 - X9s)l 91.83 + (X9s)2584.8 = 2355.51 kJ/kg From the turbine efficiency, we estimate the enthalpy at state 9:
T~t =
H 7 -H
9
H 9 = H 7 --T~t ( H 7 - H 9 s ) = 2577.87 kJ/kg
H 7 - H9s
We also estimate the quality of the discharged wet steam at state 9:
x9 =
2577.87-191.83 2584.80-191.83
= 0.997
S 9 =(1 - x9)0.6493 + (x 9 ) 8 . 1 5 1 1 = 8.1287 kJ/(kg K) The fraction of steam extracted is estimated from the energy balance rh6H6 + rh2H2 = m 3 H 3. In terms o f the mass fraction z = rh6/rh3, we have zH6 + (1 - z)//2 = H3. The mass fraction is
z=
H3- H2 631 81 - 192.29 ~ = " =0.149 H8- H2 3129.89 - 192.29
Table 4.6a shows the state properties of the actual reheat regenerative Rankine cycle. The turbine work output based on rh = 1 kg/s working fluids" Olin = t h [ ( H 5 - H 4) + ( H 7 - H 6 ) ] = 3 0 7 7 . 3 3 k W
qout = lh(1 - z)(H 9 - HI) = 2029.05 kW
//Vnet - qin -- £/out -- 1048.28 k W
The thermal efficiency is
T~th - - 1 - - qout __ 0 . 3 4 0 £/m
4.
214
Usingthe second law: Thermodynamic analysis
The rate of steam is
rh s -
I'Vs
90,000 - ~ = 85.85 kg/s Wnet 1048.28
The exergy balance based on specified hot and cold sources yields
]E,Xloss,45 -- To (lh(S5 - S 4 -+-S 7 - &) +
EXloss,59
( - /k/ i4n )W ) = T1 0 5 1I. 0 2
=lhTo[zS 8 +(1- z)39 + 86 - 87 -
85] = 224.37 kW
JEXloss,91=rhTo((S1-Sg)(1-z)+q°ut) =184"53kwTC
EXloss,regen = thT0 (S 3 - z S
( /~Xloss,cycle = T0 [
\
8 -O-z)S2)=44.16kW
0out
rc
qin)
= 1504.09 kW
Table 4.6b shows the distribution of exergy losses at each process based on ~t ~-- 0 . 8 0 , T O ~- 290 K, TH = 1700 K, and Tc - 290 K. The table shows that the highest exergy loss occurs due to heat transfer in the boiler. The work loss in the regeneration stage is minimal.
Table 4.6a State properties of the actual reheat regenerative Rankine cycle in Example 4.18 State
T (K)
P (kPa)
H (kJ/kg)
S (kJ/(kg K))
1 2 3 4 5 6s 6 7 8s 8 9s(x7 = 0.904) 9(x 9 = 0.997)
319 319 423.07 423.07 773.15
10 475 475 11000 11000 2000 2000 2000 475 475 10 10
191.83 192.29 631.81 643.29 3362.20 2914.83 3004.30 3467.30 3045.54 3129.89 2355.51 2577.87
0.6493 0.6493 1.8408 1.8408 6.5432 6.5432 6.7003 7.6765 7.6765 7.5951 7.6765 8.1287
571.58 773.15
319 319
Table 4.6b Distribution of exergy losses at each process in Example 4.18 Process
4-5 5-9 9-1 Regeneration Total Cycle
Actual reheat regenerative Rankine cycle Exloss (kW)
Percent
1051.02 224.37 184.53 44.16 1504.09 1504.09
69.9 14.9 12.3 2.9 100 100
r/t = 0.80, To = 290 K, TH = 1700 K, and Tc = 290 K.
4.5
215
Applications of exergy analysis
4.5.3 Cogeneration The cogeneration process produces electric power and process heat from the same heat source. This may lead to the utilization of more available energy and the reduction of waste heat. The process heat in industrial plants usually needs steam at 500-700 kPa, and 150-200°C. The steam expanded in the turbine to the process pressure is used as process heat. By adjusting the steam rate, the steam leaves the process as a saturated liquid. The saturated liquid is pumped to the boiler. Cycles making use of cogeneration may be an integral part of large processes where the adjusted energy of the expanded steam from the turbine at intermediate pressure may be fully utilized in producing electricity and process heat simultaneously. The utilization factor for a cogeneration plant is the ratio of the energy used in producing power and process heat to the total energy input. The utilization factor is unity if the plant does not produce any power.
Example 4.19 Energy dissipation in a cogeneration plant A cogeneration plant is using steam at 8200 kPa and 773.15 K (see Figure 4.25). One-fourth of the steam is extracted at 700 kPa from the turbine for cogeneration.
I
Turbine
Boiler 1-z 7 1_. Process Heater
3;
@
4
I
Mixer I I
Condenser
2
1 P1
P2
(a)
rI 5
6]
4
(b)
Figure 4.25. (a) Schematic of ideal cogeneration plant and (b) T-S diagram.
216
4.
Using the second law: Thermodynamic analysis
The extracted steam is condensed and mixed with the water output of the condenser. The rest of the steam expands from 700 kPa to the condenser pressure of 10 kPa. The steam flow rate produced in the boiler is 60 kg/s. Determine the work loss at each unit. Solution: Assume that the kinetic and potential energy changes are negligible, and this is a steady-state process. The surroundings are at 290 K. The steam data are: P1 = P8 = 10 kPa, HI, v = 2584.8 kJ/kg, H1, L = 191.83 kJ/kg, V1 = 0.00101 m3/kg S1,v = 8.1511 kJ/(kg K), S1, L = 0.6493 kJ/(kg K) P3=P6=700kPa,
H 3 = 6 9 7 . 0 6 k J / k g , S3=l.9918kJ/(kgK),
z=0.25
V4 = 0.001027 m3/kg P 6 = 8 2 0 0 k P a , H 6 = 3 3 9 6 . 4 k J / k g , S6=6.7124kJ/(kgK), T 6 = 7 7 3 . 1 5 K T 0 = 2 9 0 K , TH=1700K, T c = 2 9 0 K In this cogeneration cycle, the steam extracted from the turbine is used as process heat. The liquid condensate from the process heat is combined with the output of the condenser
1kJ ] = 0.697 kJ/kg
Wpl = V1(P2 - P1 ) = 0.00101 (700 - 10) 1 kPa m 3
H 2 - H 1 +Wpl = 1 9 1 . 8 3 + 0 . 6 9 7 = 192.53 kJ/kg From the energy balance around the mixer, we have th3/th 6 =0.25 t h 4 H 4 = t h 2 H 2 + t h 3 H 3 , H 4 = th2H2 + t h 3 H 3 th 4
H4 =
45(192.53) + 15(697.06) 60
= 318.66 kJ/kg
T4 = 349.15K, V4 = 0.001027kg/m 3
Wpz=V4(Ps_p4)=O.O01027(8200_700) (
l k Jm 3 ) = 7.70 kJ/kg 1kPa
H 5 = H 4 + mp2 -- 326.36 kJ/kg--. S 4 = 1.0275 kJ/(kg K) Because this is an isentropic process, $6 = $7 = $8 = 6.7124 and P7 = 700 kPa ~ H7 = 2765.68 kJ/kg. We estimate the quality of the discharged wet steam at state 8: 6.7124-0.6493 X8 =
=0.808
8.151 l - 0.6493
H 8 = 191.83(1 - 0.808) + 2584.80.3(0.808) = 2125.35 kJ/kg
4.5
217
Applications of exergy analysis
Using the reference values for H 0 = 71.31 kJ/kg and So = 0.2533 kJ/(kg K) at 290 K, we estimate the exergy flows To(Si - So) and miAi, which are shown along with the other properties in Table 4.7a. The energy balance yields the fraction of steam extracted
A i = H i - 14o -
mtotal --- th6 (H6 - H7) + rh8 (H7 - Hs) = 66,634.44 kW mpi ~- ml~pl 4.
th4Wp2
--
493.51 kW
The net work output is: mnet- ~'total-
~/in
--
Z Wpi - 66,140.93 kW
ms(H6 - H5) = 184,202.24 kW
£/process = rh7 (m7 - H 3 ) = 3 1 , 0 2 9 . 3 0 k W
qout = rh8 (H8-H~) = 87,032.01 kW ~/net -- £/in -- £/out -- £/process --
66,140.93 k W
The exergy balance based on specified hot and cold sources is To = 290 K, Tu = 1700 K, and Tc = 290 K /~Xloss,mixer = rh 2 A 2 4. th 3 A 3 - th 4 A 4 - 7 4 0 . 8 0 k W
i
EXloss,boiler =rJ/5A5 - t h 6 A 6 4. 1---~HH ~/in = •
67,494.52kW
or /~Xloss,boiler --th5 ~0 (S 6
- Ss) 4. [ T-~) (- c)in) = 67,494.52 kW
Exloss,process=rhT(AT-A3) -
1 - - ~ cC-
qprocess=10, 4 9 4 . 6 9 k W
or
•
/ ol
Exloss,process = F/73T0 (33 - $7) 4- ~
@rocess -
10,494.69 kW
Table 4.7a State properties of the ideal cogeneration plant in Example 4.19 State
T (K)
P (kPa)
H (kJ/kg)
S (kJ/(kg K))
A (kJ/kg)
rnA (kW)
1 2 3 4 5 6 7 8 (x~ = 0.808)
319 319.15 423.07 349.50 351.05 773.15 457.83 423.07
10 700 700 700 8200 8200 700 10
191.83 192.52 697.06 318.66 326.36 3396.40 2765.68 2125.87
0.6493 0.6493 1.9918 1.0275 1.0275 6.7124 6.7124 6.7124
5.68 6.38 121.58 22.83 30.53 1451.95 821.23 181.42
255.60 286.96 1823.77 1369.93 1832.08 87117.06 12318.47 8164.15
218
4.
Usingthe second law: Thermodynamic analysis Table 4.7b Distribution of exergy losses at each process based on Example 4.19
Process
Mixer Boiler Process Expansion Condenser Total Cycle
Cogeneration plant Exloss (kW)
Percent
740.80 67494.52 10494.69 0 7908.55 86638.58 86638.58
0.8 77.9 12.1 0 9.1 100
To = 290 K, TH = 1700 K, and Tc = 290 K.
/~Xloss,condenser -- th 8 (A 8 - A1) -
1-
qout = 7908.55 kW
or
&loss,condenser -- th8 T0 (S1 - - 8 8 ) +
(~-c°)0out= 7908.55 kW
EXloss,mrbkne -- th6 & - / h 7 A7 - / h s A8 - Wtotal - 0 kW
EXloss,cycle__thTc To(qoutq
qprocess TC
qTH in) = 86,638.58k w
Table 4.7b shows the distribution of exergy losses at each process based on To = 290 K, TH = 1700 K, and Tc = 290 K. The table shows that the highest exergy loss occurs due to heat transfer in the boiler. The work loss in the regeneration stage is minimal.
Example 4.20 Energy dissipation in an actual cogeneration plant A cogeneration plant uses steam at 900 psia and 1000°F to produce power and process heat (see Figure 4.26). The steam flow rate from the boiler is 16 lb/s. The process requires steam at 70 psia at a rate of 3.2 lb/s supplied by the expanding steam in the turbine. The extracted steam is condensed and mixed with the water output of the condenser. The remaining steam expands from 70 psia to the condenser pressure of 3.2 psia. If the turbine operates with an efficiency of 80% and pumps with an efficiency of 85%, determine the work loss at each unit. Solution: Assume that the kinetic and potential energy changes are negligible, and this is a steady-state process. The surroundings are at 540 R. The steam data are: th 3 th 6 =161b/s, z = ~ = 0.2 th 6
P1 =P8 = 3.2 psia, H1,v =1123.6 Btu/lb, H1,L =111.95 BtuAb, 1/1= 0.01631 ft3Ab 5'1,v = 0.2051Btu/(lb R), $1,L = 1.8810 Btu/(lb R)
4.5
219
Applications of exergy analysis
Boiler
Process I~ Heater
I Condenser
/
3 ;
2
1
T
Mixer
4
~.~
~
-'= P1
P2 (a)
T
5
6 7
24
Ib)
Figure 4.26. (a) Schematic of actual cogeneration plant and (b) T-S diagram.
P3 = P6 = 70psia, H 3 = 272.74 Btu/lb, S 3 =0.4411Btu/(lb R) V4 - 0.0175 ft3/lb P6-900psia, H6-1508.5Btu/lb, S6=l.6662Btu/(lbR), T6=1000°F To = 540 R, TH = 3000 R, Tc = 540 R The pump efficiency is % = 0.85. The turbine efficiency is r h = 0.80. In this cogeneration cycle, the steam extracted from the turbine is used in process heat. The liquid condensate from the process heat is combined with the output of the condenser Wpl = V1( P 2 - P1)= 0.01631(70- 3.2) ( 1Btu ) % 0.85 5.4039psiaft 3 = 0.237Btu/lb
220
4.
Using the second law: Thermodynamic analysis
H 2 -- H 1 + mpl -- 111.95 + 0.237 = 112.18 BtuAb From the energy balance around the mixer, we have th3/th 6 =0.2: ivh4 H 4 = th 2 H 2 nt- th 3 H 3 , /-/4 = th2H 2 + th 3 H 3 rh4
He
0.8(112.18) + 0.2(272.74)
= 144.29 Btu/lb ~ S 4 =0.2573 Btu/(lb R)
T4 = 635.87R, V4 = 0.0010911b/ft 3
Wp2 = V4/°5 - P4 = 0 . 0 1 7 5 0 0 0 - 70) ( 1Btu ~p 0.85 5.4039 psia ft 3
= 3.16 BtttOb
H 5 = H 4 + Wp2 = 147.46 Btu/lb Because these are isentropic processes, $6 = S7s = S8s = 1.6662 Btu/(lb R) and P7 = 70 psia. We have H7s = 1211.75 Btu/lb. We estimate the quality of the discharged wet steam at state 8s is: 1.6662-0.2051 X8s
=0.871
1.8810-0.2051
H8s = 111.95(1-0.871)+ 1123.6(0.871) = 993.93 Btu/lb Using the turbine efficiency, we estimate the enthalpies at states 7 and 8 H 6 -H 7
'0t -- - - - + H H 6 -H7s
7 = H 6 -'qt (H6-H7s
) = 1271.10 Btu/lb
H8 = H6 - ~t ( H 6 - Hss) = 1096.85 BmAb The steam quality at state 8 is 1096.85-111.95 =0.973 xs = 1 1 2 3 . 6 0 - 1 1 1 . 9 5 Using the reference values for H0 = 38.0 Btu/lb and So = 0.0745 Btu/(lb R) at 70°E we estimate the exergy flows .4, = H i - H o - T o ( S i - S o ) a n d thi~, which are shown along with the other properties below. Table 4.8a shows the state properties of the cogeneration plant. The turbine work output is" !)process = th3 (H7 - H3) -- 3194.75 Btu/h mtotal = T/t [th6 ( 9 6 - H7s) + th8 (H7s - Hss)] = 5975.17 B t u ~ Z ~/rPi = ghlWp1 + th4Wp 2 = 53.63 Btu/h
The net work output is" Wnet -- mtotal - ~ ~rpi = 6975.17 Btu/h 0in = rh5 (H6 - Hs) = 16(1508.5-147.46) = 21,776.64 Btu/h C)out= rhs (H 8 - H1) = 16(0.8)(1096.85-111.95) = 12,606.7 Btu/h
4.5
Applications of exergy analysis
221
Table 4.8a State properties of the actual cogeneration plant in Example 4.20 State
T (R)
P (psia)
H (Btu/lb)
S (Btu/(lb R))
A (Btu/lb)
1 2 3 4 5 6 7s 7 8s (xs~ = 0.87I) 8 (xs = 0.973 )
603.67 603.90 763.6 635.87 638.17 1459.67 820.72 937.38 603.67 603.67
3.2 70 70 70 900 900 7(1 70 475 10
111.95 112.18 272.74 144.29 147.46 1508.5 1211.75 1271.10 993.93 1096.85
0.2051 0.2051 0.4411 0.2573 0.2573 1.6662 1.6662 1.7384 1.6662 1.8366
3.37 3.61 36.72 7.53 10.69 610.93 314.18 373.53 96.36 107.22
#~A(Btu/h) 43.21 46.25 117.52 120.57 171.16 9774.91 1005.38 1195.30 1233.52 1372.41
The exergy balance based on specified hot and cold sources is: To = 540 R, TH = 3000 R, and Tc = 540 R /~Xloss,mixer =/4/2A 2 + th3A 3 -/4/4A 4 = 4 3 . 2 0 Btu/h
/~Xloss,boiler = ~/5 1 5 - th6 A6
+ (1- TT---°H ) 0in = 8253.10Btu~
or •
EYl°ss'b°iler =/h5T°(S6 - $ 5 ) - +
~H ( - q i n ) =
8253.10Btu~
/~TXloss,process = th7 ( A 7 - A3) - (1 - T~-) C)process = 1077.78 B t u ~
or
/~TXloss,process-/h3T0 (S 3
--$7)+-(T@)qprocess= 1077.78Btu/h
/~TXloss,condenser-- /4/8( & -- A 1 ) -
1--'~CC qout
=1329.19Btu~
or
/~Xloss,condenser = th 8T0 (S1 - 38) + (T~-) qout = 1 3 2 9 . 1 9 B t u ~
EXloss,turbine = rh6A6 - th7 "47 - th8A8 - ex and rl' < 1, but for secondary raw materials rI' > > 1, and for imported raw materials and fuels, we usually have rl > 1. For secondary raw materials, the exhaustion of nonrenewable natural resources is related to the consumption ofexergy for processing and transportation, and usually it is much smaller than the exergy of the materials under consideration. The inequality rI' > > 1 suggests that the utilization of secondary raw materials may be beneficial, since they substitute the semifinished products requiring a large amount of exergy for production. We may have ex > e for imported raw materials, fuels, and semifinished products if the exported goods are more advanced than the imported ones.
5.2.10
Exergy Destruction Number
The use of an augmentation device results in an improved heat transfer coefficient, thus reducing exergy destruction due to convective heat transfer; however, exergy destruction due to frictional effects may increase. The exergy destruction number N~ is the ratio of the nondimensional exergy destruction number of the augmented system to that of the unaugmented one
EXa
N,- - _-7-7
(5.43)
Ex s
where subscripts "a" and "s" denote the augmented and unnaugmented cases, respectively, and Ex* the nondimensional exergy destruction number, which is defined by E x * --
CXfd
mToCp
(5.44)
Here, exfd is the flow-exergy destruction, or irreversibility, and To the reference temperature. The system will be thermodynamically advantageous only if the N~ is less than unity. The exergy destruction number is widely used in secondlaw-based thermoeconomic analysis of thermal processes such as heat exchangers.
5.3
ECOLOGICAL COST
The production, conversion, and utilization of energy may lead to environmental problems, such as air and water pollution, impact on the use of land and rivers, thermal pollution due to mismanagement of waste heat, and global climate change. As an energy conservation equation, the first law of thermodynamics is directly related to the energy management impact on the environment. One of the links between the principles of thermodynamics and the environment is exergy, because it is mainly a measure of the departure of the state of a system from that of the equilibrium state of the environment. Performing an exergy analysis on the Earth's natural processes may reveal disturbances due to large-scale changes, and could form a sound base for ecological planning for sustainable development. Some of the major disturbances are: 1. Chaos due to the destruction of order is a form of environmental damage. Entropy is fundamentally a measure of chaos, while exergy is a measure of order; the exergy of an ordered system is greater than that of a chaotic system. 2. Resource degradation leads to exergy loss. A natural resource with exergy is in nonequilibrium state compared with the environment. 3. Uncontrollable waste exergy emission can cause a change in the environment. Exergy analysis may be an important tool to interrelate energy management, the environment, and sustainable development in order to improve economic and environmental assessments. Ecological cost analysis may minimize
286
5.
Thermoeconomics
the depletion of nonrenewable natural resources. Determining the exhaustion of nonrenewable natural resources connected with the extraction of raw materials and fuels from natural resources is not sufficient in fully understanding the ecological impact of production processes. The influence of waste product discharge into the environment should also be considered. Waste products may be harmful to agriculture, plant life, human health, and industrial activity.
5.3.1
Index of Ecological Cost
The exhaustion of nonrenewable natural resources is called the index of ecological cost. To determine the domestic ecological cost q~, the impact of imported materials and fuels is taken into account
where s is the index of harmful waste product, dnmis defined in Eq. (5.38), ~ the immediate gross consumption of the nonrenewable domestic natural resource k per unit of complex useful products containing a unit of the major product m, ~ m the exergy of harmful waste product s, x~ the destruction coefficient of the product n per unit of the exergy of waste product s, Yks the destruction coefficient of the nonrenewable natural resources k per unit of the exergy of waste product s, and z~ the multiplier of exergy consumption to eliminate the results of human health deterioration per unit of exergy of the waste product s. The destruction coefficients x and y are
x,s
~s'
Y~
~
(5.46)
where 6 ~ , is the exergy of the destroyed useful product n and 6~ k is the exergy decrease of the damaged natural resources. The coefficient xns should also take into account the reduction of agricultural and forest production. The global ecological cost can be calculated. The degree of the negative impact of the process on natural resources can be characterized by means of the ecological efficiency tie, and from Eqs. (5.18) and (5.45) we have
1% - -
q~
(5.47)
Usually, r/e < l, but sometimes values of r/e > 1 can appear if the restorable natural resources are used for the process. The transition from one form of exergy to another, for example, from chemical to structural, may create economic value. Self-organization is a production process, and exergy is necessary to build a structure with a value, that may not be measured and described by exergy.
5.3.2
Global Warming Potential
The global warming potential is a measure of how much a given mass of a chemical substance contributes to global warming over a given period of time. The global warming potential is the ratio of the warming caused by a substance to the warming caused by a similar mass of carbon dioxide. Therefore, the global warming potential of carbon dioxide is defined as 1.0, while water has a global warming potential of 0. Chlorofluorocarbon-12 has a global warming potential of 8500, while chlorofluorocarbon-ll has a global warming potential of 5000. Various hydrochlorofluorocarbons and hydrofluorocarbons have global warming potentials ranging from 93 to 12,100. These values are calculated over a 100-year period of time.
5.4
AVAILABILITY
One of the important definitions in finite-time thermodynamics is the definition of finite-time availability A given by
(5.48)
5.5 Thermodynamicoptimum
287
Here, ti and tf are the initial and final times of the irreversible process and To the environmental temperature. The maximization is carried out with the constraints imposed on the process. Equation (5.48) represents the second law of thermodynamics in equality form by subtracting the work equivalent of the entropy produced, which is the decrease in availability in the process. Availability depends on the variables of the system as well as the variables of the environment A = U + Po V - T o S - ~ txoiN i
(5.49)
Here, the temperature, pressure, and chemical potential are estimated at ambient conditions. For an optimal control problem, one must specify: (i) control variables, volume, rate, voltage, and limits on the variables, (ii) equations that show the time evolution of the system which are usually differential equations describing heat transfer and chemical reactions, (iii) constraints imposed on the system such as conservation equations, and (iv) objective function, which is usually in integral form for the required quantity to be optimized. The value of process time may be fixed or may be part of the optimization.
5.4.1
Essergy
The potential work of any system is given by E~ - E - roS + PoV- Y_..;oNi
(5.50)
i
where/x and N are the chemical potential and number of moles of substance i, E the total energy including all kinetic and potential energies in addition to internal energy, and the subscript "0" denotes the reference state representing the environment of the system. The term Ees is the essential energy in the form essential for work (power) production, so that Ees shows the essergy (essential energy). The corresponding flow of essergy ~0e~, excluding kinetic and potential energies for any uniform mixture of substances, is 4 ~ - H - ToS - ~ I~ioN i
(5.51)
i
5.5
THERMODYNAMIC OPTIMUM
Thermoeconomics formulates an economic balance through exergy cost and optimization. The minimization of entropy generation plays only a secondary role in thermoeconomics, mainly because economic performance is always expressed in economic values of money and price. Therefore, the thermodynamic optimization problem may not be expressed in terms of the problem of the minimization of irreversibility. For example, the problem of minimum overall exergy consumption may not be equivalent to the problem of minimum dissipation because of the disregarded exergy of the outgoing flows and changing prices of exergy unit. This problem mainly belongs to the areas of energy management and the cost of energy. Industrial systems consist of various resource consumption processes and supporting processes to supply and remove resources. The supporting processes may involve exergy loss and exergy transfer between resources, new resource upgrading, postconsumption recovery, and the dispersion and degradation of resources released to the environment. The contemporary theory of optimization can be used for analyzing these systems. The first approach is to optimize the system by adjusting the design and operating parameters through governing equations that describe internal changes, and by imposing control through system boundaries. The second approach aims to predict system behavior under a set of specified external conditions with governing equations derived from certain variational or extremum principles.
Example 5.4 Minimization of entropy production For a fixed design, the minimization of the rate of entropy production may yield optimal solutions in some economic sense. Such a minimization comes with certain set of constraints. For a single force-flow system, the local rate of volumetric entropy production is dp - 5v L X 2 d V
(5.52)
288
Thermoeconomics
5.
where L is the phenomenological coefficient and is not a function of the driving force X. The minimization problem is the optimization of the system with a finite size V, and the solution is the homogeneous distribution of the force over the system. Assuming a steady-state heat transfer operation with no momentum and mass transfer, the expression of total entropy production is (Tondeur, 1990)
: -I ~2 (qVT)dV : -k I v
dV
(5.53)
v
where the heat flux is obtained from the Fourier law (5.54)
q = -kVT
and k is the thermal conductivity assumed as a constant. The entropy production is a function of the temperature field. Then, the minimization problem is to obtain the temperature distribution T(x) corresponding to a minimum entropy production • using the following Euler-Lagrange equation:
Minimizing the entropy production fianction with the constraim expressed in Eq. (5.54), the above equation becomes
ZLr oxj - 7
=o
(5.56)
For a heat exchanger, a characteristic direction related to the temperature field is the direction Z(x) normal to the heat transfer area, and the above equation yields
~-~XIT 0-~-xT ]Z(x) =0
(5.57)
VT)
(5.58)
and we obtain
T
= constant Z(x)
The above equation shows that by keeping the driving force VT/T uniformly distributed along the space variables, entropy production will be minimum. For an optimum design, we may consider VT
AT
T
T
- constant
(5.59)
where AT= Th -- Tc or (p - 1)Tc, and the temperature gradient is a function of the temperatures Th and Tc of hot and cold streams, respectively Th = p T c
(5.60)
where p is a constant with value > 1. The following expression also produces a constant C: ~)Op
- Tc) 2 - U(ThVhVc = C = constant
(5.61)
5.6
Equipartitionand optimization in separation systems
289
where 6 is a small change, U the total heat transfer coefficient, and A the heat exchanger area
6~_6q(1
1)
Tc
6A-
and
Th '
6q g(Th
-Tc)
From Eqs. (5.60) and (5.61), we have (2 +
C/U) +[(2 +
P=
C/U)
2 -
4] 1/2
(5.62)
2
The energy balance is
(5.63)
whdTh = wcdTc
where 1/i2h and Wc are the products of heat capacity and hot and cold streams flow rates, respectively. From Eqs. (5.60) and (5.63), we have d~h dL
- - p --
Wc Wh
--
Th Tc
(5.64)
which are the matching conditions to minimize the entropy production in any heat exchanger. For example, for a specified heat exchanger area and hot stream input and output temperatures Ti and To, respectively, the minimum entropy production is obtained when We _ Ti _ constant % To
(5.65)
We can extend this approach for a network of heat exchangers.
5.6
EQUIPARTITION AND OPTIMIZATION IN SEPARATION SYSTEMS
Thermodynamic cost analysis relates the thermodynamic limits of separation systems to finite rate processes, and considers the environmental impact through the depletion of natural resources within the exergy loss concept. Still, economic analysis and thermodynamic analysis approaches may not be parallel. For example, it is estimated that a diabatic column has a lower exergy loss (39%) than that of adiabatic distillation; however, this may not lead to a gain in the economic sense, yet it is certainly a gain in the thermodynamic sense. The minimization of entropy production is not always an economic criterion; sometimes, existing separation equipment may be modified for an even distribution of forces or an even distribution of entropy production. Thermodynamic analysis requires careful interpretation and application. The results of thermodynamic analysis may be in line with those of economic analysis when the thermodynamic cost optimum, not the maximum thermodynamic efficiency, is considered with process specifications. Figure 5.3 shows pinch technology in terms of optimum hot and cold utilities by accounting for the investment costs and exergy cost. With an optimum approach temperature ATmin, the total cost may be optimized.
Example 5.5 Equipartition principle in separation processes: Extraction Since the minimization of entropy production is not always an economic criterion, it is necessary to relate the overall entropy production and its distribution to the economy of the process. To do this, we may consider various processes with different operating configurations. For example, by modifying an existing design, we may attain an even distribution of forces and hence an even distribution of entropy production. Consider a simple mixer for extraction. In minimal entropy production, size V, time t, and duty d are specified, and the average driving force is also fixed. We can also define the flow rate Q and the input concentration of the solute, and at steady state, output concentration is determined. The only unknown variables are the solvent flow rate and composition, and one of them is a decision variable; specifying the flow rate will determine the solvent composition. Cocurrent and countercurrent flow configurations of the extractor can now be compared with the
290
5.
Thermoeconomics
Hotutility O------
/
r~ 0
~
4
I-Iotutility
ATtain ~ - - - ' J / ~ Coldutility / J ~./f/~ Hotutility
\
.
r~
~~
O ~
.
.
.
C ~ l d ~ ~
~ ~"0---
Coldutility
Operatingcost Exergyloss Figure 5.3. Principleof pinch technology. same initial specifications (V, t, J, Q, c). Cocurrent operation will yield a larger entropy production P2 than the countercurrent operation P1, and investigating the implications of this on the decision variable is important. For a steady-state and adiabatic operation, we have the following relations from Tondeur and Kvaalen (1987). For processes 1 and 2 with the solvent flow rates of QI and Q2, we have
Q1As1 = - AS--bP1
(5.66)
Q 2 ~ s 2 -- - A s -[- P 2
(5.67)
where AS is the total entropy change, and AS 1 and AS 2 the changes in specific entropies of the solvent. Subtracting Eq. (5.67) from Eq. (5.66), we have
Qlz~,s1-Q2As2 = P1-P2
Pav2 = L(Xav2) 2(Vt)
(5.71) (5.72)
5.6
Equipartition and optimization in separation systems
291
Since P1 -- P2, combining Eqs. (5.71) and (5.72) yields Xa2vl ~> Xa2v2, and hence J1] > IJ2 • That is, the flow rate for equipartitioned process 1 is larger than that of process 2 at a given size, duration, and entropy production. In another operating configuration, we can compare the respective size and durations for specified duty and entropy production. Equations (5.71) and (5.72) are still valid, and we have P1 > Pav2 and J1 = J2, which yield P1 > Pav2
J1
(5.73)
J2
and thus Xav 1 > Xav 2
(5.74)
(Vt)l < (Vt)2
(5.75)
This result indicates that for a given flow and entropy production, the equipartitioned configuration is smaller in size for a specified operational time. Alternatively, it requires less contact time for a given size, and thus a higher throughput. To determine an economic optimum, we assume the operating costs are a linear function of the solvent entropy change and entropy production, and the investment costs are a linear function of the space and time of the process. The total cost is
C T : aP + b + czVt - I ~ (aLX2 + c'r)dVdt + b
(5.76)
where ~"is the amortization rate and a, b, and c the constants related to the costs. The integral in Eq. (5.76) is subject to the constraint of a specified flow
J = I ~ LXdVdt
(5.77)
The variational technique minimizes the total cost, and the Euler equation for variable X is given by 0
OX
(aLX 2 + cz + ALX) = 0
(5.78)
where A is a Lagrange multiplier. The above equation yields
2aLX + AL = 0 X =-
A 2a
- constant
(5.79) (5.80)
The obtained value of X that minimizes the total cost subject to J is a uniform distribution. This illustrates the economic impact of the uniform distribution of driving forces in a transport process.
Example 5.6 Thermoeconomics of extraction Consider a steady-state operation in which the forces are uniformly distributed; the investment cost Ci of a transfer unit is assumed to be linearly related to size V, and operating costs Co are linearly related to exergy consumption C v -- C i - Cif - a V
(5.81)
C o = Cof Jr- bz~kEx
(5.82)
where Cv is the variable part of the investment cost, Cif a fixed investment cost, Cof a fixed operating cost, and a and b the cost parameters. Exergy loss AExc is expressed as
292
5.
Thermoeconomics
AEx~ = AEx m + TocI3av
(5.83)
Here, To is a reference temperature (dead state) and Z ~ X m a thermodynamic minimum value. The total flow J = LVXav can be written by using Eq. (5.81)
J = LXavCv
(5.84)
a
Eliminating the constant (average) force Xav between Eq. (5.83) and the total entropy production (I)av = JAXav, we obtain
aJ 2 ~av -
(5.85)
LCv
Substituting the above equation into Eq. (5.81) and the latter into Eq. (5.82), a relationship between the operating and investment costs is obtained Co
ab ToJ2
: ~
LCv
at- Cof at-
bAExm
(5.86)
The optimal size is obtained by minimizing the total operating and investments costs, which are linearly amortized with the amortization rate z. CT(Ci) = zCi = Co. The minimum of CT is obtained as dCT/dCi = 0, and we have b,I, av
(Cv)op t
_
b(I'av
_
8Vop t
To
(5.87)
According to Eq. (5.87), the quantities bT0~av, which are related to irreversible dissipation and q'Vopt, should be equal in any transfer unit. Generally, operating costs are linearly related to dissipation, while investment costs are linearly related to the size of equipment. The optimum size distribution of the transfer units is obtained when amortization cost is equal to the cost of lost energy due to irreversibility. The cost parameters a and b may be different from one transfer unit to another; when a = b, then (I)av/Vop t is a constant, and the optimal size distribution leads to equipartition of the local rate of entropy production. The optimal size of a transfer unit can be obtained from Eq. (5.78)
(Cv)°pt:Ci'°pt-Cif :j(abT°] 1/2Ln" ( bT° 11/2
V°pt = J k ,
(5.88)
(5.89)
Distributing the entropy production as evenly as possible along space and time would allow for the design and operation of an economic separation process. Dissipation equations show that both the driving forces and flows play the same role in quantifying the rate of entropy production. Therefore, the equipartition of entropy production principle may point out that the uniform distribution of driving forces is identical to the uniform distribution of flows.
Example 5.7 Equipartition principle: Heat exchanger For a heat exchanger operating at steady state, the total entropy generation P is obtained by integrating over the surface area P =
L~y2dA A
(5.90)
5.6
Equipartition and optimization in separation systems
293
We consider that the duty of the exchanger is specified as qs qs - L I XdA
(5.91)
A
An average driving force over the surface area is obtained as 1 IXdA Xav = -A A
(5.92)
Thus, Eqs. (5.91) and (5.92) yield the specified duty qs qs = LAXav
(5.93)
The above equation shows that for a given surface area A and constant L, specification of the duty leads to the average driving force. Minimizing the integral in Eq. (5.90) subject to a constraint given in Eq. (5.91) is a variational problem, and the solution by the Euler equation in terms of the force is given by (Tondeur, 1990) 0 (X 2 + A X ) = 0 OX
(5.94)
where A is a Lagrange multiplier (a constant). The above equation is satisfied by X = -A/2, i.e., by a constant value of X. The second derivative yields 02
0)(22
(X 2 + AX)>0
(5.95)
The above equation implies that the extremum is a minimum. Thus, with a constant transfer coefficient, the distribution of the driving force that minimizes the entropy generation under the constraint of a specified duty is a uniform distribution. The minimal dissipation for a specified duty implies the equipartition of the driving force and entropy generation along the time and space variables of the process. When the linear phenomenological equations do not hold (Tondeur, 1990), we have P = I JXdV;
J-
Jo--
VJav with X = X ( j )
(5.96)
where j is the specific flow per unit volume. The constraints on these relations are JX>0;
X'-
dX dJ
>0;
J(X=O)=O
(5 97)
The Lagrangian expression is given by F ( J ) = P + A(J - J0 ) = I ( J X + AJ - AJav )dV
(5.98)
The Euler equation corresponding to an extremum of P is given by OF oJ
t'
(5.99)
- J ( J X ' + x + A)dV = 0
which yields JX' + X + A - ~0 Oj[(X+A)J ]
0
(5.100)
294
5.
Thermoeconomics
y
v
X
X
(a)
Figure 5.4.
(b)
Concave and convex relationships between the flow and the force.
The above equation shows that (X = h ) J = constant, and hence the solution yields J = constant and X = constant. Therefore, P is stationary when the flow, force, and entropy generation are uniformly distributed. The sign of the second derivative reveals whether this stationary value is a minimum or not 02
OJ 2 [(X + h)J] = X " J + X '
(5.101)
Since J and X' are positive, the quantity in the above equation (5.101) is always positive when X " > 0 which means when Xis a convex function of J. When the flow J is a linear or a concave function of the driving force X (02J/OX2 < 0) (Figure 5.4a), then equipartition of entropy generation corresponds to minimal dissipation. On the other hand, when X" is negative in the above equation, the sign of 02[(X+ J)J]/OJ 2 may be positive or negative, and may change along the process. When 02F/OJ 2 < O, the value of entropy generation is maximum. When the flow versus force curve is sufficiently convex (Figure 5.4b), nonuniformity may lead to an economic configuration. Such a situation may arise in an electrochemical cell that does not obey the Ohm's law. A strongly convex flow-force curve corresponds to ordered structures, which are dissipative and constantly require a supply of matter and energy from the outside. For example, B6nards cells occur during a natural convection in a fluid system heated from the bottom; after the difference between the surface temperature and the fluid bulk temperature exceeds a certain limit the system moves into the nonlinear region in the thermodynamic branch, and the fluid shows a structured state as long as the temperature difference is maintained. The generality of the equipartition principle should be investigated within flow-force relationships. In near equilibrium phenomena, linear flow-force relations are valid, and optimization criteria (e.g., for coupled heat and mass transfer)generally lead to a constant level of entropy generation along an optimal path provided that there is no constraint imposed on the parameters controlling the system. For systems far from equilibrium, the most stable configurations may correspond to unsteady, dissipative structures. Therefore, equipartition or stability should be considered in the economics of industrial systems or the evolution of natural systems. Equipartition may also help to improve existing designs and avoid flaws in the new design of processes.
Example 5.8 Characterization of the deviation from equipartition Consider a heat exchanger with surface area A and specified heat duty qs. The driving force is constant and equal to its average Xav _ qs
(5.102)
LA
The overall entropy production is
Po -- LfA X2dA
= L(Xav) 2 =
(5.103)
qsXav
Consider a real heat exchanger with the same area A, coefficient L, and qs. The average driving force is Xav. The difference in the entropy production between the real heat exchanger and the one with equipartition configuration is
-
]
(5.104)
5.6
Equipartition and optimization in separation systems
295
where the average driving force is defined in Eq. (5.92). The term (Y2)av is the mean of X 2 over the surface A. Therefore, the contents of square brackets represent the mean quadratic deviation from the mean and hence the variance s 2 of distribution X. The variance is a positive quantity, and we have
P - P o = LAs2(X) = LA[(X-Xav)2]av >0
(5.1o5)
The above equation suggests that the entropy production would be higher if the system deviates from the uniform distribution of the driving force. We can find an equivalent form to the above equation in terms of heat flow q P - Do _ L s 2 ( X )
A
--
1 $2
-L
(q)
(5.106)
where s2(q) is the variance of the distribution of the heat flow. Configurations that minimize s2(X) and s2(q) also minimize entropy production and lead to thermodynamically optimum configurations. Such thermodynamic analysis will contribute to the study of feasibility and economic analysis after relating the level of entropy production to engineering economics.
Example 5.9 Distribution of driving forces Consider two identical heat exchangers 1 and 2 operating at steady state with the same total entropy production P. Assume that the distributions of the driving forces are different and are characterized by s ( < s2
(5.107)
From Eqs. (5.105) and (5.107), we find
-Pol LA
Po2, since Pol and Po2 are the exchangers operating with an equipartition configuration (s 2 = 0). The heat exchangers may use different temperatures and flow rates for hot and cold streams. Equation (5.108) yields
(Xl,av) 2 > (X2,av) 2
(5.109)
IX,,avl > IXz,av[ and ]qll > q21
(5.110)
The above equation suggests that in heat exchanger 1, for example, the cold fluid would be heated more or the use of a larger cold flow rate is possible. Therefore, the heat exchanger with the smallest s 2 would achieve the largest duty and be more economic in practice. This simple analysis suggests that the distribution of entropy production may play a more important role than total entropy production.
Example 5.10 Variance and heat exchangers Consider two heat exchangers with the same heat duty and total entropy production. They have different heat transfer areas and different variances (Sl) 2 < ($2)2, and hence
Pol > ~Po2
A1
A2
(5.111)
Show that the heat exchanger with the smallest variance is more economic. Equation (5.111) leads to
(Xl,av) 2
(X2,av) 2
4
A2
(5.112)
296
5.
Thermoeconomics
Since the heat duties are the same, we have
q = LAl ( Xl,av ) = LA2 ( X2,av )
(5.113)
By multiplying Eqs. (5.112) and (5.113) side by side, we get (Xl,av) 3 > (X2,av) 3 :=~ Xl,av > X2,av
(5.114)
From Eq. (5.113), we obtain A 1< A2. Therefore, the heat exchanger with the smallest variance requires the smallest heat exchanger area and would be more economic.
Example 5.11 Hot fluid flow rate effect Consider two heat exchangers 1 and 2 operating at steady state and constant pressure with the same heat duty. The total entropy change of the cold fluid is the same for both heat exchangers and determined by the specified heat duty qs. There is no heat loss to the environment. The overall entropy balances for the heat exchangers are
rhlAs1 - P1 = AS
(5.115)
rh22~s2 - P2 = AS
(5.116)
where rh is the mass flow rate, As the specific entropy change of the fluid between output and input, P the total entropy production, and AS the total entropy change of the cold fluid. The heat duty is based on the enthalpy changes of the hot fluid Ah q = &lZ~hl = th2~h 2
(5.117)
If we assume that P1 < P2, and subtracting Eq. (5.117) from Eq. (5.116), we have /'hi ~s1 -/'h2 As2 =-- P1 - P2 < 0
(5.118)
Since the hot fluid becomes colder, As < 0, and we have
thl IZXSl[ >
rh2 IZ~S21
(5.119)
Combining Eqs. (5.117) to (5.119), we find
Ahl < Xh2 2~Sl
As2
(5.120)
On an enthalpy versus entropy diagram (Mollier diagram), the above equation shows the slopes of chords to the constant pressure curve between input and output conditions. The constant pressure curves are convex (02h/0s2). If the input conditions are the same for both exchangers, inequality (5.120) and Figure 5.5 show that ]Ahl] > ]2~h2]
(5.121)
and because of Eq. (5.118), we have rh1 < rh2 . Therefore, exchanger 1, having the smallest entropy production, requires a smaller flow rate of hot fluid. The condition (02h/Os 2) is always satisfied for pure fluids. For mixtures, however, this condition may not always be satisfied and should be verified.
5.6
Equipartition and optimization in separation systems
297
Constant pressure line
.......i/ inlet
Ah t
Ah 2 ............... ... /!
~ /:
y"
1......./
.... ~2........
()utlct 1 Figure 5.5. Mollier diagram for fluids 1 and 2 at constant pressure curve.
Example 5.12 Equipartition principle in an electrochemical cell with a specified duty We desire the electrode to transfer a specified amount of electricity Q over a finite time to
f'"Zdt= Iavt0
(5.122)
~O=-d0
where I is the instantaneous electric current and I~v its average over the time interval to. If Ohm's law holds within the cell, we have AE - R1
(5.123)
where AE is the electric potential difference between the electrodes and R the resistance. The power dissipated within the cell is P =
f tl) AEldt o
= R IO0 I2dt
(5.124)
As before, minimizing the power dissipated for a specified Q implies that I - constant. The deviation from this optimal configuration is similar to Eq. (5.106) P-Po
q
~
Rs2(I)--R
1 $2
(~E)
(5.125)
l0 The above equation shows that the steady state is less dissipative for a specified duty in a finite time. This conclusion is in line with the minimum entropy production principle of Prigogine. If we have multiple working electrodes or the electrode is divided into N different zones, which may work at different potentials, the overall specified duty becomes [o
Q--~-~ fo Idt - iavNt o i=l
(5.126)
Here, the discrete summation is over the N zones. The average Iav is defined over time and number of zones. Repeating the previous procedure and using Po as the entropy production corresponding to Iav at all times and in all zones (equipartition configuration), we have P - Po _ Rs2(i) Nt o
(5.127)
298
5.
Thermoeconomics
The above equation shows the minimal dissipation of power for a specified duty corresponding to an equipartition configuration of flow, driving force, and entropy production along the time and space variables of the process.
5.6.1
Thermoeconomics and Distillation Columns
Tables 4.17 and 4.18 show the performances of distillation columns with diabatic, adiabatic, and isoforce configurations. Diabatic and isoforce column operations reduce exergy losses as well as the amount of hot utility required considerably. These modifications may lead to a mass transfer unit more reversible but may require more transfer units and hence more column height and heat transfer area. For example, in a diathermal distillation column with heat exchangers at every stage, it is possible to adjust the flow ratio of the phases and thus the slopes of operating lines, and the driving forces along the column. This also affects the driving force distribution and entropy production. The task of a process engineer is to decide the target cost or costs to be optimized in a new design or an existing operation. Energy saving in distillation systems has attracted considerable innovative approaches. Such approaches incorporate the principles of thermodynamics and have reached an advanced stage through pinch analysis, exergy analysis, and the equipartition principle. Thermodynamic analysis considers the critical interrelations among energy cost, thermodynamic cost, and ecological cost. Thermodynamic analysis is becoming popular for other separation systems, such as supercritical extraction, desalination processes, hybrid vapor permeation-distillation, and cryogenic air separation. For example, the energy requirement analysis of common cycles used in supercritical extraction has utilized exergy losses and an optimum extraction pressure, which produces a minimum in exergy loss for specified temperature and separation pressure. Distillation columns should be optimized considering both capital cost and operating (energy) cost. The heuristics of using a reflux ratio of 1.03-1.3 times the minimum reflux ratio is in line with both the capital cost and the operating cost for binary distillation systems.
Example 5.13 Optimal distillation column: Diabatic configuration Consider a distillation column made of N distinct elements. The heat and mass transfer area can be defined separately. Defining an investment cost Ci for element i as a linear function of the size Ai, we have (5.128)
Ci -- ceiAi + fli
where a is proportionality cost factor and/3 a fixed cost (including the heat transfer area). We assume that the average operating cost Ci is a linear function of the exergy loss in the element i per year
Ci,av = "Yi [Exi + ToPi ] + 6i
(5.129)
where Yi is a proportionality cost factor and 6 i a fixed cost, while To is the reference temperature (environmental temperature). Here, the exergy loss is split into a thermodynamic minimum Ex i (e.g., minimum separation work), and an irreversible contribution ToPi, where Pi is the entropy produced in element i. In the diabatic distillation column, each element is small enough that equipartition of entropy production may be approximately achieved by adjusting the heat flows and thus the liquid and vapor flow rates. We assume that each element performs a specified duty of J/0. The total cost function Ct for all N elements is N
C t =Zfi,av-+-7"Ci
(5.130)
i
where z is the yearly amortization rate. Using the variational approach, we minimize a Lagrangian as N
N
~'~ = Z Ci,av + TCi + Z ~i (di - dio ) = O i i
(5.131)
This should satisfy the following conditions: -0 OA i
0212 and ~ > 0
(for alli)
(5.132)
5.6
Equipartition and optimization in separation systems
299
Using Xav = Jo/LA, Pi becomes
j2 (5.133)
L,A,
So, f~ becomes
f~ = ~Z~ rczi~ + r~i + yiExi + 6i + •
L,4
+ Ai (Ji - Jio )
(5.134)
The derivative with respect to A/yields a system of independent equations 0~ -
OAj
J J-r°2 =0 TOLj - -
2
LiAj
(5.135)
or
r =
/ToP:
~jAj
(5.136)
and we have
02D.
2r
A,
>0
(5.137)
Since r is a constant depending only on economic conditions, the right of Eq. (5.136) must be independent of the element j, and thus an invariant throughout the process. The product rjaAj is the annual cost related to irreversible energy waste in terms of exergy loss. These two quantities should be equal in any element. We may conclude that under these assumptions, the optimal size distribution of the elements requires the equipartition of the ratio TToP/aA on all the elements, and the cost of exergy loss is equal to the amortized proportional investment cost in that element. The equipartition principle is mainly used to investigate binary distillation columns, and should be extended to multicomponent and nonideal mixtures. One should also account for the coupling between driving forces since heat and mass transfer coupling may be considerable and should not be neglected especially in diabatic columns.
Example 5.14 Optimal feed state for a binary distillation Consider a binary distillation column with specified distillate and bottom compositions. The feed composition is 30 mol% of the more volatile component. Investigate the problem of conditioning the feed. Should the feed be in saturated liquid or saturated vapor state? Figure 5.6 shows the feed lines and operating lines for a saturated liquid feed and a saturated vapor feed. The feed line for liquid comes with an infinite slope as a vertical line starting from the feed location on the diagonal line on a McCabe-Thiele graph. On the other hand, the slope of the feed line for a saturated vapor is zero and represented by a horizontal line starting from the feed location. As shown, saturated liquid feed yields a better distribution of area between the equilibrium curve and the top and bottom operating lines compared with the area of a saturated vapor feed. The departure from equilibrium is more evenly distributed for the liquid feed and hence yields a better performance. For example, the trays are evenly distributed above and below the feed stage and the reflux ratio is smaller in the liquid feed. This configuration may be different for a different feed composition. 5.6.2
Retrofit of a Distillation Column
Retrofits are modifications of existing distillation columns to reduce the cost of operations by increasing the efficiency in energy utilization. Thermodynamic analysis is one method for determining the appropriate retrofits. Thermodynamic analysis mainly seeks modification targets for reducing thermodynamic losses due to heat and mass transfer, pressure drop, and mixing. For example, in a binary distillation, operating curves come closer to the equilibrium curve, and the reflux ratio approaches its minimum value. However, multicomponent distillation may be a more difficult
300
5.
Thermoeconomics
Y
Feed
x
Figure 5.6. State of the feed for a distillation column. The bold line is the operating line for a saturated liquid feed.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
t °"EF, t.............................................. t sect,o.5
-~ ,
,[ OFFGAS .... 1
HIERARCHY
I
SECTION1 HIERARCHY
---4 Asl
SECTION2
-t
I-->
I
SECTION3
407 ] - - ~
HIERARCHY
SECTION4 HIERARCHY
HIERARCHY
A ..................................................
A
I QTOP ~..................................................
..............................................................[QREB t..............................................................
Figure 5.7. Connection of the sections of the methanol plant with material and heat streams. problem; the sharpness of multicomponent separation is limited, and near reversible operating conditions may be difficult to achieve. To analyze the performance of an existing column quantitatively for exploring the energy saving potential, it is customary to construct the temperature-enthalpy and stage-enthalpy curves (called the column grand composite curves), and the stage exergy loss profiles. The column grand composite curves display the net enthalpies for the actual and ideal operations at each stage, and the cold and hot heat utility requirements, while the exergy loss profiles indicate the level of irreversibility at each stage, including the condenser and reboiler. Therefore, the area between the actual and the ideal operations in a column grand composite curve should be small, and exergy losses should be lower for a thermodynamically efficient operation. The column grand composite curves are constructed by solving the mass and energy balances for a reversible column operation. The stage exergy loss profiles are generated by the stage exergy balance calculations with a reference temperature.
Example 5.15 Retrofits of distillation columns by thermodynamic analysis The synthesis of methanol takes place in a tube reactor in section 3 in the methanol plant shown in Figure 5.7. The reactor outlet is flashed at 45°C and 75 bar, and the liquid product (stream 407) containing 73.45 mol% of methanol is fed into the separation section (see Figure 5.8), where the methanol is purified. Stream 407 and the makeup water are the feed streams to the section. Table 5.2 shows the properties and compositions of the streams in section 3. The converged simulations are obtained from the Redlich-Kwong-Soave method to estimate the vapor properties, while the activity coefficient
301
5.6 Equipartition and optimization m separation systems
,LA
H
--
i40N,EED, 1
,'
lii
il
Figure 5.8. Separation section of the methanol plant with subsystems used in thermodynamic efficiency estimations: ($1) column 1; ($2) column 2; ($3) columns 1 and 2. Table 5.2 Material and heat streams for the separation section of the methanol plant
Stream FEED4 MKWATER BTMS SIDE1 LIQ2 VAP1 OFFGAS METHANOL LIQ1 FEED2 FEED1 ql q2
n
ii7
T
H
I:t
S
Ex*
(kmol/h)
(kg/h)
(K)
(kJ/mol)
(MW)
(J/(mol K))
(kJ/mol)
-186.5 -355.0 -81.29 -1.254 -0.01011 -3.260 -5.243 - 124.7 -0.02267 -207.7 -216.2 15.299 18.900
-215.49 - 166.95 -142.06 -207.77 -224.70 -21.15 -32.75 -224.60 -237.28 - 192.77 -208.38 -
2655.32 444.21 1050.96 18.43 0.1560 33.80 70.25 1925.59 0.3414 2995.14 3029.28 -
76938.66 8002.62 18955.20 550.00 5.00 1388.90 2330.66 61700.40 11.10 81210.60 82610.60 -
318.15 313.15 393.17 359.77 348.00 305.91 318.86 348.29 305.91 359.00 323.15 377.00 409.00
-252.82 -287.73 -278.46 -244.99 -233.10 -347.23 --268.66 -233.07 -239.07 -249.61 --256.89 -
-188.44 -237.77 -235.93 -182.88 -165.97 -340.66 -258.69 - 165.97 -168.19 -- 192.00 - 194.68 3.199 5.122
*To= 298.15 K.
model NRTL and the Henry components method are used for predicting the equilibrium and liquid properties. Assessments of the performances of the existing columns, suggested retrofits, and the effectiveness of the retrofits with minimum or no change in column pressure and stage numbers are discussed below: C o l u m n 1: As the base case design in Table 5.3 shows, column 1 has 51 stages, and operates with a partial condenser with a duty of 1.3 71 M W at the top, and a side condenser with a duty of 8.144 M W at stage 2. It has no reboiler; however, it receives a side heat stream with a duty of 15.299 M W to the last stage from section 2 of the plant. The temperature profiles of both columns are shown in Figure 5.9. The temperature-enthalpy and the temperature and composition profiles may help in assessing the operation and determining the extent and position of side heating or condensing for the column. Figure 5.10a shows the temperature-enthalpy curve. There exists a significant area difference between the ideal and actual enthalpy profiles, which identifies the scope for side condensing. As the temperature change after stage 3 is very small, and a side condenser at stage 2 already exists, a second side condenser at stage 4 with a duty of 2.1 M W has been installed. As Figure 5.10b shows, the side condenser has reduced the area between the ideal and actual enthalpy profiles to some degree without increasing the number of stages. Figure 5.11 also shows that the actual vapor flow closely follows the thermodynamic ideal minimum vapor flow at stages 2 - 4 after the retrofitting. The duty of 2.1 M W is in the range of enthalpy difference between the hot duty of 15.299 M W and the total cold duty of 9.51 M W
302
5.
400 -
Thermoeconomics
J
I
I
390 - ] -o-- Column 1 _- Column2 380 -
J
J
370 360-
~ 340 -c 330 320 310 300 0
10 20 30 40 50 60 70 80 90 100 Stage
Figure 5.9. Temperature profiles for the columns in Example 5.15. Table 5.3 Operating parameters of designs 1 and 2 for column 1 Parameter No. of stages Feed stage Feed temperature (°C) Reflux ratio Condenser duty (MW) Distillate rate (kmol/h) Condenser temperature(°C) Side condenser 1 stage Side condenser 1 duty (MW) Stage 2 temperature(°C) Side condenser 2 stage Side condenser 2 duty (MW) Stage 4 temperature(°C) Heat stream (ql) duty (MW) Heat stream (ql) stage Heat stream (ql) temperature(°C) Boilup rate (kmol/h) Bottom rate (kmol/h) Bottom temperature(°C)
Design 1 (base case) 51 14 43.74 3.7 - 1.3717 34.141 32.75 2 -8.144 69.4 74.37 15.299 51 104 1551.284 2995.144 85.85
Design 2 (retrofitted) 51 14 65 4.56 - 1.691 34.140 32.75 2 -7.7 70.26 4 -2.1 74.36 15.299 51 104 1551.560 2995.144 85.85
(side condenser + partial condenser shown in Table 5.3). The existing side condenser duty is reduced to 7.7 from 8.144 MW, so that the new total duty of 11.49 MW is close to the previous total of 9.51 MW. After the retrofitting, therefore, the total cost would not change much, and the need for extra stages would be negligible as the heat changes sharply below the first side condenser. Since side heat exchangers are more effective at convenient temperature levels or stages for exchanging heat using cheaper utilities, care should be exercised in positioning them. Another approach may be based on the uniform distribution of the driving forces that cause the separation, leading to less entropy production and hence less exergy loss in the column where the coupling of heat and mass transfer may not be negligible. Figure 5.10 also displays a sharp change of the enthalpy on the reboiler side. The extent of the change determines the approximate feed preheating duty required, as the feed at 43.74°C is highly subcooled (Table 5.3). Therefore, a new heat exchanger (HEX, in Figure 5.8) with a duty of 1.987 M W is used as the second retrofit for the column, and the feed temperature has increased to 65 from 43.74°C. Figure 5.12 compares the enthalpies for the base case and retrofitted designs. The difference between the hot and cold duties is lower, and the actual and ideal profiles are closer to each other after the retrofits.
5.6
Equipartitionand optimization in separation systems
303
360
350-
340 -
330
320
310]
,
300 -2.5
i
0
2.5
5
7.5 10 12.5 Deficit, MW (a)
Enthalpy
,
15
,,
17.5
360
350
34()
~D
330
~D
32O
310 -.o-- Ideal Profile ] 300 -2.5
,
i
i
()
2.5
5
Enthalpy
i
i
7.5 10 i2.5 Deficit, MW (b)
,
15
17.5
Figure 5.10. Temperature-enthalpy curves for column l'(a) design 1 and (b) design 2.
The suggested retrofits also aim at reducing the irreversibility due to mixing of streams at different temperatures at the feed stage, which is at 80.18°C, and throughout the column. The exergy loss profiles of Figure 5.13 show that the reduction in exergy loss at the feed stage is - 6 0 % with the values of 0.3865 MW in design 1 and 0.1516 MW in design 2. However, the exergy loss at the partial condenser increases by 28%, and becomes 0.150 MW in design 2 instead of 0.117 MW in design 1. As Table 5.5 shows, the reduction in the total exergy loss or the recovered available energy is 21.5% with total column exergy losses of 0.837 and 0.656 MW in designs 1 and 2, respectively. Table 5.3 compares the base case design and the retrofitted design. Column 2: As the base case design in Table 5.4 shows, column 2 has 95 stages, and a total condenser with a duty of 281.832 MW. It operates with a high reflux ratio, and receives a side heat stream of 18.9 MW to the last stage from section 2 of the plant. One of the side products is the methanol stream described in Table 5.2, and drawn at stage 4 at 348.3 K. The second side product is drawn at stage 86 at 361.2 K. The temperature profile in Figure 5.9 shows that the temperature increases sharply after stage 84, and the separation system resembles a stripping column with the feed close to the bottom at stage 60. Figure 5.14a shows a significant area difference between the ideal and the actual enthalpy profiles above the feed stage representing the pinch, and hence suggests side reboiling at appropriate temperature levels to decrease the difference. The existing reboiler duty is 282.28 MW (Table 5.4). Besides that, there is a side product at stage 86 and a side heat input of 18.9 MW at stage 95. Therefore, the decision has been made to install two side reboilers
304
5.
Thermoeconomics
0 .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
,
,,,
,,,,
~
, .....
,,,
ag ID
gq
og > 1
6
L
11
16
21
26
31
36
41
46
51
31
36
41
46
51
Stage (a)
~g o
oo° > 6
11
16
21
26 Stage
(b) Figure 5.11. Vapor flow profiles of column 1" (a) design 1 and (b) design 2 in Example 5.15 (Table 5.3 describes designs 1 and 2).
Table 5.4 Operating parameters of designs 1 and 2 for column 2 Parameter No. of stages Feed stage Feed temperature (°C) Reflux ratio Condenser duty (MW) Distillate rate (kmol/h) Condenser temperature (°C) Reboiler duty (MW) Boilup rate (kmol/h) Bottom rate (kmol/h) Reboiler temperature (°C) Side reboiler 1 stage Side reboiler 1 duty (MW) Stage 87 temperature (°C) Side reboiler 2 stage Side reboiler 2 duty (MW) Stage 92 temperature (°C) Heat stream (q2) duty (MW) Heat stream (q2) stage Heat stream (q2) temperature (°C)
Design 1 (base case) 95 60 85.84 188765 -281.832 0.156 74.85 282.283 24890.68 1050.959 119.71 90.97
110.91 18.90 95 136
Design 2 (retrofitted) 95 60 85.84 188765 -281.833 0.156 74.85 52.292 4633.93 1049.66 120.02 87 180 93.35 92 50 110.94 18.90 95 136
5.6
Equipartition and optimization in separation systems
305
360 O
j
i
12
14
35O
:~ 34O
~D
sso
~" 3 2 0 --o2- D e s i g n 1 310
~Design
2
'
300 0
2
4
6
8
10
16
Enthalpy deficit, MW (a)
50
40 ---o-- D e s i g n 1 •
3O
~
+
Design 2 i
20
~
,
•
10
12
14
() 0
~
4
(5
8
Enthalpy deficit, MW (b)
Figure 5.12. Column grand composite curves for column 1 in Example 5.15: (a) temperature-enthalpy deficit curves and (b) stage-enthalpy deficit curves (Table 5.3 describes designs 1 and 2).
at stages 87 and 92 with the duties of 180 and 50 MW, respectively. Obviously, these two side reboilers are more economic as they operate at lower temperatures and require less expensive steams compared with the steam used in the existing reboiler. With the two side reboilers, the duty of the reboiler decreases to 52.3 from 282.3 MW. Extra stages due to the side reboilers would be minimal since the enthalpy rises sharply at each stage after stage 84. Figure 5.14b shows a considerable reduction in the area between the ideal and actual enthalpy profiles after the retrofits. Moreover, Figure 5.15 shows that the side reboilers have reduced the gap between the ideal and actual vapor flows between stages 84 and 95, where the stage temperatures change sharply. The enthalpy curves in Figure 5.16 also show that the retrofitted design is closer to ideal operation than design 1. Figure 5.17 compares the exergy loss profiles in designs 1 and 2. The base case design operates with rather large exergy losses at the feed stage and around the reboiler. The rest of the column has negligible exergy losses mainly due to the flat methanol concentration profile. The retrofits reduce the total exergy losses by --~41.3%, and hence save a considerable amount of the available energy. Table 5.4 compares the two designs of the colunm. Using data from Table 5.2, the minimum values of exergy for the required separation and thermodynamic efficiencies for designs 1 and 2 are estimated, respectively, and compared in Table 5.5. The estimations are based on the value T0 = 298.15. Figure 5.8 identifies the subsystems considered in Table 5.5. The reductions in the exergy losses range from 21.5% to 41.35%. The thermodynamic efficiencies have increased considerably in the retrofitted designs, although the efficiencies remain low, which is common for industrial column operations. For column 1, the efficiency increases to 55.4% from 50.6%, while the efficiency increases to 6.7% from 4.0% in column 2.
306
5.
Thermoeconomics
360 ¢
~_________._____._____._______.~
350 ~ [ ~N~,~[
1---O--Design 1[ [ ~ D e s i g n 2[
340 330 e-
320
310 300 0
, 0.1
, , 0.2 0.3 Exergy loss, MW
0.4
(a) 50 ,~ I
40
---O-- Design 1 Design 2
~30
I
20
0 0
0.1
0.2 0.3 Exergy loss, MW
0.4
(b)
Figure 5.13. Exergy loss profiles for column 1 in Example 5.15: (a) temperature-exergy loss profiles and (b) stage-exergy loss profiles (Table 5.3 describes designs 1 and 2).
Approximate economic analyses shown in Tables 5.5 and 5.6 compare the fixed capital costs of the retrofits with the savings in electricity due to the reduced exergy losses. The fixed capital cost consists of equipment, materials, construction, and labor. Table 5.6 shows the approximate values of fixed capital costs for the heat exchangers needed in the retrofits. The costs are estimated by using the current chemical engineering plant cost index of 420, and the approximate areas obtained from the individual duties. The energy saving estimations are based on the unit cost of electricity of US$ 0.060/(kWh) and a total 8322 h/year of plant operation. The costs of related retrofits and the yearly saved exergy equivalent of electricity for each subsystem are compared in Table 5.5, which shows that the retrofits are effective and save a considerable amount of energy in electricity per year. Column grand composite curves and exergy loss profiles enable process engineers to assess an existing operation, suggest retrofits if necessary, and determine the effectiveness of the retrofits. The suggested retrofits consist of an additional side condenser at stage 4 and feed preheating for column 1, and two side reboilers at stages 87 and 92, respectively, for column 2. The effectiveness of the retrofits has been assessed by the improved column grand composite curves and exergy loss profiles as well as by an approximate economic analysis. After the retrofits, actual and minimum vapor flow profiles have become closer. Also, the difference between the ideal and actual profiles of the enthalpies in the column grand composite curves has become smaller. The range of reductions in the total exergy losses is 21.5-41.3%, which leads to a considerable reduction in the available energy losses. The
5. 7
307
Thermoeconomics of latent heat storage
t11 ~ 0
tb i
t
.,
I /
1
I
=
7'-
Actual Profil© f
t
i,i
IL
............
-50
0
50
100
150
FalfllalpyDeficit (a)
200 MW
250
300
o
o
1 i--:-c~° __~ ÷
I 1 I Ideal Profile Actual Profile
.J"1
1
o
o
.J ee5 ,,
~;
~.,
:->-.~
T 1
0
0
50
100
.......
150
200
EnthalpyDeficit M W (b)
250
300
Figure 5.14. Temperature-enthalpy deficit curves for column 2 in Example 5.15: (a) design 1 and (b) design 2 (Table 5.4 describes designs 1 and 2).
thermodynamic efficiencies also increased considerably as thermodynamic imperfections decrease. The savings in electricity can recoup the initial cost of the retrofits in a short time. Government incentives for environmentfriendly designs may reduce the cost of the retrofits further. 5.7
THERMOECONOMICS
OF LATENT HEAT STORAGE
Latent heat storage is a popular research area with industrial and domestic applications, such as energy recovery of air conditioning, and under-floor electric heating by using a phase changing material. Figure 5.18 shows the charging and discharging operations with appropriate valves, and temperature profiles for countercurrent latent heat storage with subcooling and sensible heating. An optimum latent heat storage system performs exergy storage and recovery operations by destroying as little as possible of the supplied exergy. A charging fluid heats the phase changing material, which may initially be at a subcooled temperature Tso and may eventually reach a temperature Tsh after sensible heating. Therefore, the latent heat storage system undergoes a temperature difference of Tsh - Tsc, as shown in Figure 5.19. Heat available for storage would be qc - UA(ATu~)~ - thcCpc (Tci-Tco)
(5.138)
where U is the overall heat transfer coefficient, A is the heat transfer area, rnc is the charging fluid flow rate, Tci and Tco are the inlet and outlet temperatures of the charging fluid flow and z~Tlm the logarithmic mean temperature difference expressed by (Tci- Tsc)- (Tco - Tsh)
Tci - Tco
( Tci - T~c ) In T~o _ Tsh
NTU c
(5.139)
308
5.
Thermoeconomics
CD O g
.
.
~r
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
---c--- ThermodynamicIdeal MinimumFlow * Actual Flow
0
> 2 _/
.............................................. .
1
21
.
.
41
.
.
.
.
.
.
61
81
101
Stage (a) O . . . . . . . . . . . . . . . . .
C O
~..,
•
............................ 1
•
ThermodynamicIdeal MitKrnumFlow ActualFlow
t
,,
l
,
g--
{
O
tD
,aoo o~ o
! i
.2° Ph
I .......l............
. . . . . . . . . . . . . . . . . . . . . . . . . .
1
21
zl
, ..........
,,
,,
61
81
101
Stage (b)
Figure 5.15. Vapor flow profiles for column 2: (a) design 1 and (b) design 2 in Example 5.15 (Table 5.4 describes designs 1 and 2). where N T U = UA/(rhcCpc ) = (Tci- Tco)/Arlm is the number of transfer units. Equation (5.139) relates the value of NTU with temperature. Heat lost by the charging fluid qc will be gained by the phase changing material qs qc = qs = ms [Cps ( Tl - l'si ) + A H m q'- Cpl ( l"sh -- Th ) ]
(5.140)
where AHrn is the heat of melting, T1 and Th the lowest and highest melting points of the phase changing material, and @s and Cpl the specific heats of solid and liquid states of the phase changing material, respectively. The net rate of exergy E x of the charging fluid is
A/~x c - (/~Xco-/~Xci ) - th c
Cpc
( T c i - Tco ) - T O In
(5.141)
Exergy stored by the phase changing material is
Os where Ts is an average temperature of storage, which may be approximated by (Tsc + Tsh)/2. The first and second law efficiencies are r/=
actual heat stored maximum energy gain
_
Tci-
- ~
Tco
Tci-T s
(5.143)
5. 7
309
Thermoeconomics of latent heat storage
100 90 80 70 6O 50 40
i
----- Design 1
30
~
20
............
Design 2
,!
-~
10 0 0
50
100 150 200 Enthalpy deficit, MW (a)
100
I
!
250
300
i
90 80 70 60 ~D
50 40
Design 1 .........
3O
Design 2 1
i' 20
.
+
_
!
i . . . . . . . . . . . . .
I
{
........
!
10
0
50
100
150
200
250
300
Enthalpy deficit, MW (b)
Figure 5.16. Column grand composite curves for column 2 in Example 5.15: (a) temperature-enthalpy deficit curves and (b) stage-enthalpy deficit curves (Table 5.4 describes designs 1 and 2).
exergy of PCM
(T~i - Tc° ) 1 - T~h [ (Tci)j rbh - exergy of charge fluid = (Tci _ Tco ) - To In Tc°
If it is assumed that the phase changing material is totally melted and heated to a temperature the discharge flow qd is estimated by
qd -- U A A ( Tim )d --
thdCpd(Tdi -- Tdo )
(5.144)
Tsh ,
recovered heat by
(5.145)
310
5.
Thermoeconomics
400 390 380 370 [.., 360 Design ~1__ Design
350 340 0.00
t
1.00
2.00 3.00 Exergy loss, MW
4.00
(a) 10090- ~'L-~ - 807060
---o- Design 1 ~ Design 2
1
5040302010
oi
0.00
1.00
2.00 3.00 Exergy loss, MW (b)
4.00
Figure 5.17. Exergy loss profiles for column 2 in Example 5 15: (a) temperature-exergy loss profiles and (b) stage-exergy loss profiles (Table 5.4 describes designs 1 and 2).
Table 5.5 Assessments of the effectiveness of the retrofits System
Design 1 (base case) Exmin (MW)
(S1) Column 1 ($2) Column 2 (S1 + $2) Columns 1 + 2
Exloss (MW)
Design 2 (retrofitted)
llth (%)
Exmin (MW)
0.815
0.856
0.837
50.6
1.136
26.979
4.0
1.992
27.814
6.7
EXloss (MW)
'Tith (%)
Saved EXloss (MW)
Change Exloss (%)
FCC a of retrofits (US$)
Electricity savingb (US$/year)
0.656
55.4
0.179
21.5
183500
89578
1.135
15.847
6.7
11.133
41.3
409000
5558829
1.950
16.502
10.6
11.312
40.7
592500
5648407
(Exloss) total column exergy loss from the converged simulation by Aspen Plus with Soave-Redlich-Kwong, NRTL, and Henry component methods. aFCC" fixed capital cost. bElectricity equivalent of energy savings is based on a unit cost of electricity of US$ 0.060/(kWh).
5. 7
311
Thermoeconomics of latent heat storage
Table 5.6 Approximate fixed capital cost calculations for the retrofits Heat exchanger
Type
Duty (MW)
Preheater (HEX), column 1 Side condenser, column 1 Total for column 1
S/T b, fixed tube sheet S/T b, fixed tube sheet
Side reboiler 1, column 2 Side reboiler 2, column 2 Total for column 2
Floating head Floating head
Material
P (bar)
Area (m 2)
FCC ~
(us$)
1.9 2.1
5.0 1.5
Carbon steel Carbon steel
130 130
90500 93000 183500
180.0 50.0
2.0 2.0
Carbon steel Carbon steel
600 170
294000 115000 409000
~'Approximate fixed capital cost with the: chemical ~::ngineering plant cost index = 420. bS/T: shell and tube.
Solar Energy
Solar Energy Exp2c, cp2c
Exse, cse
1_ 4 ~a~ ca ]
Solar Air Heaters
<mfl "-I I ~
1
,
Latent Heat Storage 2
Ex3p, c3p
T
_.. Ex3a, c3a
Greenhouse
Exp2d, cp2d
Figure 5.18. Units of the latent heat storage system.
T
d '
/~xp 1, cpl
~
Tco
/~xp2c, cp2c Ts p¢,~,
,,
Tsc
Exp2d, cp2d
Ex3a, c3a
i
Figure 5.19. Approximate temperature profiles for a latent heat storage unit.
The net exergy change of the charging fluid would be
~L'X d -- (L'Xdi --/~Vdo ) -- mdCpd [(Tdi -- Tdo )-- TO In
rdo
(5.146)
The first and second law efficiencies are Tdo -- Tdi
(5.147)
Tdi -- Tsl
exergy given to discharge fluid 7Ith =
exergy of PCM
rd o -- Td ~ -- To I n
i r0 ~--
(Tdo-Tdi)(1--~sl)
(5.148)
312
5.
Thermoeconomics
All the temperatures are time dependent, and the charging and discharging cycles need to be monitored over the time of operation. Structural theory facilitates the evaluation of exergy cost and the incorporation of thermoeconomic functional analysis. Structural theory is a common formulation for the various thermoeconomic methods. It provides costing equations from a set of modeling equations for the components or units of a system (Figures 5.18 and 5.19). Structural theory needs a productive structure displaying how the resource consumption is distributed among the components of a system. The flows entering a component in the productive structure are considered fuels F and flows leaving a component are products P. The components are subsystems with control volumes as well as mixers and splitters. Therefore, the productive structure is a graphical representation of the fuel and product distribution. For any component j, or a subsystem, the unit exergy consumption eXc is expressed on a fuel/product basis by
eXcj -
Fj_Fj Ps Exs
(5.149)
For linear modeling, the average costs of fuels and products are defined by C*
-
OF° •
sin-Ore----f,
*
(5.150)
OF°
C jp - op j
where F o is the fuel to the overall system expressed as a function of the flow other related parameters. The total annual production cost Cr in $/(kW h) is N
N
j=l
j=l
mj or product pj, respectively, and the
(5.151)
where ci is the specific cost of product i in $/(kW h), CjF is the cost of fuel, and of componentj in kW and is expressed in terms of NTU using Eq. (5.139)
Exj : rhjCp[NTUjA~mj-T o ln(TJl]]
r0)j
Exj is the rate of exergy as a product
(5.152)
The optimum total production cost rate with respect to NTU is obtained from dC¢ - 0 dNTU
(5.153)
Including an ecovector to account for the exergoeconomic costs or environmental impact can extend the thermoeconomic approach. An ecovector is a set of environmental burdens of an operation, and can be associated with input flows; it includes information about natural resources, the exergy of these resources, and monetary costs. The external environmental costs associated with the environmental burdens may also be added into the ecovector. Extended exergy accounting includes the exergetics flowcharts for nonenergetic costs of labor and environmental remediation expenditures.
Example 5.16 Cash flow diagram for seasonal latent heat storage Figure 5.18 shows the three basic components of a latent heat storage system: solar air heaters, latent heat storage unit, and the greenhouse. Whenever the temperature in the greenhouse drops below a set point, a fan circulates the air from the greenhouse through the latent heat storage unit until the temperature reaches the required level. Costs are the amount of resources consumed to produce a flow or a product. When exergy is added into a flow, the cost of the flow leaving a component is equal to the cost of the flow entering plus the fuel value of added exergy. When exergy is removed, the fuel value of exergy is subtracted. The products and their average costs in the productive structure shown in Figure 5.18 are summarized below.
5. 7
313
Thermoeconomics of latent heat storage
Unit 1: Solar air heater system- Added exergy provided by the solar air heater system can be expressed in terms of NTU and z~Tlm using Eq. (5.152) as (5.154)
The airflow leaving the solar air heaters adds exergy, and therefore the cost is
Clp = Cli -+-C1v
(5.155)
where Cli is the cost of the flow entering unit 1 and CIF the fuel value of added exergy. The specific costs of warm air Ca and exergy Cex are
Clp
ClpF
Ca - /~Xlp- EXlp'
-- C1F %
(5.156)
A/~xl
where Clpv is the fuel value of the product leaving the solar air heating system.
Unit 2: Latent heat storage system- Figure 5.19 shows an approximate temperature profile within the storage unit. The removed exergy by the latent heat storage system during charging is
[NTUclmc In
(5.157)
The cost of the product after charging is
Ccp = Cci - Ccv
(5.158)
The specific costs of the product leaving the latent heat storage unit Cc and the removed exergy Cexc are
CC
m
Ccp _ CcpF /~Xcp ]~Xcp
CcF
,
2cExc
Cexc
°
(5 159)
Discharging flow adds exergy form the latent heat storage, and is given by
A/~x d = (/~Xdp --/~Xdi ) -- th3 Cp [ NTU d ATlmd-Toln(
TdO]
~,Tdi )
J
(5.160)
The cost of discharging flow is
Cdp = Cdi -t- CdF
(5.161)
The specific costs of discharging flow leaving the latent heat storage c d and the added exergy Cexd are
cd -
Cdp _ CdpF . --, EXdp EXdp
¢exd
-
CdF A]~Xd
(5.162)
Unit 3: Greenhouse- The exergy change within the greenhouse is A]~x3 - L'x3p -/~x3i - lCvl3CpINTU3z~Tlm3
t, T3i )
(5.163)
314
5.
Thermoeconomics
The exergy from the discharge flow is removed in the greenhouse, and the cost of the flow leaving the greenhouse becomes C3p = C3i - C3F
(5.164)
The specific costs of flow leaving the greenhouse Cg and the exergy removed Cexgare C3p __ C3pF Cg -- /~X3p /~X3p '
C3F ¢exg -- ~ X
(5.165)
3
The total cost of products of the three components would be CpT = Clp + Ccp + Cdp + C3p -- Ca/~Xlp Jr Cc/~Xcp Jr Cd/~Xdp Jr Cg/~X3p
(5.166)
The cost of a product for componentj is based on a fuel/product basis Cjp = Cjpv, SOthat the total cost of products is
= C,p + % + Cap + C3p = clp +
+
+ C3p
(5.167)
Cost optimization basically depends on the trade-offs between the cost of energy (fuel) and capital investment as seen in Figure 5.20. Fines for pollution and incentives for environment-friendly technologies may reduce the cost associated with exergy loss. The thermoeconomics of the latent heat storage system involves fixed capital investment, operational and maintenance cost, and exergy costs. The total fixed capital investment consists of (i) direct expenses, which are equipment cost, materials, and labor, (ii) indirect project expenses, which are freight, insurance, taxes, construction, and overhead, (iii) contingency and contractor fees, and (iv) auxiliary facilities, such as site development and auxiliary buildings. Table 5.7a shows the data used in the thermoeconomic analysis. Discounted cash flow diagram can determine the profitability criteria in terms of the payback period, net present value, and rate of return from. In the discounted cash flow diagram each of the annual cash flow is discounted to time zero for the latent heat storage system. The payback period is the time required, after construction, to recover the fixed capital investment. The net present value shows the cumulative discounted cash value at the end of useful life. Positive values of net present value and a shorter payback period are preferred. The rate of return is the interest rate at which all the cash flows must be discounted to obtain zero net present value. If rate of return is greater than the internal discount rate, then the latent heat storage system is considered feasible. Figure 5.20 shows the discounted cash flow diagram obtained from Table 5.7b using the data in Table 5.7a. A net present value ofUS$102,462.21 is obtained at the end of 15 years of useful life, which shows a profitable investment. The approximate discounted payback period is about eight years. The discounted rate of return is --~10.485%, which is greater than the internal interest rate of 8%. By changing the values of exergy costs, or the tax rate the cash flow diagram can be modified easily (Demirel and Ozturk, 2006).
Table 5.7a Economic data used for the thermoeconomic analysis of the seasonal heat storage system
Fixed capital investments for the components FCI1 + FCI 2 + FCI3 = US$ 200000 + US$ 200000 + US$100000 = US$ 500000 Cost of land: L = US$ 50000 Working capital: WC = 0.2(US$ 500000) = US$100000 Yearly revenues or savings: R = US$160000 Total cost of production (Cop) from Eqs. (5.166) and (5.167) C o p - - CpT = Clp -q- Ccp -+- Cdp -Jr-C3p = Clp F q- Ccp F + Cdp F Jr- C3p F -- US$ 55000 Clp F -- U S $
20000, Ccp F -- U S $ 1 5 0 0 0 ,
Cdp F = U S $ 1 0 0 0 0 , C3p F -- U S $ 1 0 0 0 0
Taxation rate: t = 35% Salvage value of the whole seasonal storage: S = US$ 50000 Useful life of the system: n = 15 years; depreciation over 10 years Discount rate: i = 8%
315
Problems
Table 5.7b Discounted cash flow estimations for the seasonal latent heat system n
FCI
D=
A t`'
Cop
R
Bc
DCF
CCF -50000
0
-50000
0
500000
0
-500000
-50000
1
-500000
0
500000
0
-600000
-555556
-605555.56
2
0
50000
450000
160000
55000
85750
73516.8
-532038.75
3
0
50000
400000
160000
55000
85750
68071.11
-463967.64
4
0
50000
350000
160000
55000
85750
63028.81
-400938.83
5
0
50000
300000
160000
55000
85750
58360.01
-342578.82
6
0
50000
250000
160000
55000
85750
54037.05
-288541.77
7
0
50000
200000
160000
55000
85750
50034.3
-238507.47
8
0
50000
150000
160000
55000
85750
46328.06
-192179.41
9
0
50000
100000
160000
55000
85750
42896.35
-149283.07
10
0
50000
50000
160000
55000
85750
39718.84
-109564.22
11
0
50000
50000
160000
55000
85750
36776.71
-72787.51
12
0
0
50000
160000
55000
68250
27103.01
-45684.50
13
0
0
50000
160000
55000
68250
25095.38
-20589.12
14
0
0
50000
160000
55000
68250
23236.47
2647.34
15
0
0
50000
160000
55000
68250
21515.25
24162.591
16
200000
0
50000
160000
55000
268250
78299.62
102462.20
~Depreciation- straight line method: D = (FCI - S)/n. bBook value: A = FCI - ED/,. CAfter tax cash flow - net profit + depreciation: B = (R - Cop - Dl V F R A C = 0 . M O L E - F L O W = 100. MOLE-FLOW C3 5. / IC4 10. / NC4 30. / IC5 20. / NC5 & 15. / NC6 20. B L O C K RADFRAC tL&DFRAC PARAM NSTAGE = 28 COL-CONFIG C O N D E N S E R = TOTAL FEEDS FEED 14 PRODUCTS DIST 1 L ! BOTTOM 28 L P-SPEC 1 4.4 < a t m > / 2 4 4.4 < a t m > COL-SPECS D : F - 0 . 4 4 M O L E - R R - 1 . 8 T-EST 1 308. < K > / 2 8 367. < K > EO-CONV-OPTI
5.6
Hydraulic analysis of the Aspen Plus simulator produces "thermodynamic ideal minimum flow" and actual flow curves for rigorous distillation column simulations. These types of calculations are performed for RADFRAC columns. Using the input summary given m Problem 5.5 construct the stage-flow curves. Assess the thermodynamic pertbrmance of the column.
5.7
Using the following input summary for tLA.DFRAC columns, construct the column grand composite curves and stage exergy profiles with the property methods of Peng-Robinson. Discuss the results. TITLE 'RADFRAC SIMULATION' IN-UNITS ENG COMPONENTS ETItAN-01 C 2 H 6 0 - 2 / 1-PRO-01 C 3 H 8 0 - 1 / ISOBU-01 C4H 100-3 / N-BUT-01 C4H 100-1 FLOWSHEET B L O C K D 1 IN=FEED OUT = DIS 1 BOT 1 B L O C K D2 IN=DIS 1 OUT = DIS2 BOT2 IN-UNITS ENG STREAM FEED S U B S T R E A M MIXED P R E S - 2 0 . VFRAC =0. M O L E - F L O W = 100. MOLE-FRAC ETHAN-01 0.25 / 1-PRO-01 0.5 / ISOBU-01 0.1 ! & N-BUT-01 0.15 B L O C K D 1 RADFtLA~C PARAM NSTAGE =41 HYDRAULIC = YES COL-CONFIG C O N D E N S E R = TOTAL FEEDS FEED 19 PRODUCTS DISI 1 L / BOT1 41 k P-SPEC 1 20. COL-SPECS D P - C O L = 0 . M O L E - D - 7 4 . 7 M O L E - R R = 3.65 SC-REFLUX DEGSUB =0. REPORT STDVPROF TARGET HYDANAL B L O C K D2 RADFRAC PARAM NSTAGE = 23 H Y D R A U L I C = YES COL-CONFIG C O N D E N S E R = TOTAL FEEDS DIS 1 12
318
5.
Thermoeconomics
PRODUCTS DIS2 1 L / BOT2 23 L P-SPEC 1 20. COL-SPECS DP-COL=0. MOLE-D=25. MOLE-RR=3.64 SC-REFLUX DEGSUB=0.
5.8
Use the following economic data and prepare a discounted cash flow diagram. Asses the feasibility of the investment on the latent heat storage system: Economic data used for the seasonal heat storage system Fixed capital investments for the components: FCI 1 + FCI2 + FCI 3 - US$200000 + US$200000 + US$200000 = US$600000 Cost of land: L = US$50000 Working capital: WC = 0.1 (US$600000)= US$120000 Yearly revenues or savings: R = US$150000 COp = CpT = Clp + Ccp nt- Cdp nt- C3p = Clp f nt- Ccp f nt- Cdp f + C3p f = U 8 5 5 0 0 0 0
Taxation rate: t = 30% Salvage value of the whole seasonal storage system: S = US$50000 Useful life of the system: n = 15 years; Depreciation over 10 years Discount rate i = 5.5%
5.9
Use the following economic data and prepare a discounted cash flow diagram. Asses the feasibility of the investment on the latent heat storage system: Economic data used for the seasonal heat storage system Fixed capital investments for the components: FCI~ + FCI2 + FCI3 = US$250000 + US$250000 + US$250000 = US$750000 Cost of land: L = US$50000 Working capital: WC = 0.2 (US$750000)= US$150000 Yearly revenues or savings: R = US$140000 COp = CpT = Clp nt- Ccp -Jr-Cdp nt- C3p = Clp f + Ccp f + Cdp f + C3pf = U 8 5 5 0 0 0 0
Taxation rate: t = 25% Salvage value of the whole seasonal storage system: S = US$50000 Useful life of the system: n = 12 years; Depreciation over 8 years Discount rate i = 6.5%
REFERENCES R.U. Ayres, Ecol. Econ., 26 (1998) 189. Y. Demirel, Sep. Sci. Technol., 39 (2004) 3897. Y. Demirel, Sep. Sci. Technol., 41 (2006) 791. Y. Demirel, Int. J. Exergy, 3 (2006) 345-361. Y. Demirel and H.H. Ozturk, Int. J. Energy Res., 30 (2006) 1001. R. Domanski and G. Fellah, Appl. Therm. Eng., 18 (1998) 693. B. Erlach, L. Serra and A. Valero, Energy Conversion Manage., 40 (1999) 1627. M.J. Moran and H.N. Shapiro, Fundamentals of Engineering Thermodynamics, 4th ed., Wiley, New York (2000). J. Szargut, Finite-Time Thermodynamics and Thermoeconomics, Eds. S. Sieniutcyz and E Salamon, Taylor & Francis, New York, 1990. D. Tondeur, Finite-Time Thermodynamics and Thermoeconomics, Eds. S. Sieniutcyz and P. Salamon, Taylor & Francis, New York, 1990. D. Tondeur and E. Kvaalen, Ind. Eng. Chem. Res., 26 (1987) 50. G. Tsataronis, Prog. Energy Combust. Syst., 19 (1993) 227. R. Turton, R.C. Bailie, W.B. Whiting and J.A. Shaeiwitz, Analysis, Synthesis, and Design of Chemical Processes, 2nd ed., Prentice Hall, Upper Saddle River (2003). A. Valero, C. Torres and M.A. Lozano, AES Vol. 9/HTD, Vol. 124, ASME, New York (1989).
REFERENCES FOR FURTHER READING J. Beusa and G. Tsatsaronis, Comp. Chem. Eng., 25 (2001) 359. M.A. Rosen and I. Dincer, Int. J. Energy Res., 27 (2003) 415. A. Valero, L. Serra and J. Uche, J. Energy Sources Technol., 128 (2006) 1, 9.
6 DIFFUSION 6.1
INTRODUCTION
We may describe multicomponent diffusion by (1) the Maxwell-Stefan equation where flows and forces are mixed, (2) the Chapman-Cowling and Hirschfelder-Curtiss-Bird approaches where the diffusion of all the components are treated in a similar way, and (3) a reference to a particular component, for example, the solvent or mass average (barycentric) definition. Frames of reference in multicomponent system must be clearly defined. Binary diffusion coefficients are often composition dependent in liquids, while they are assumed independent of composition for gases. Under mechanical equilibrium on a molecular scale, the exchange of momentum proceeds faster than the exchange of mass and heat for liquids. On the other hand, the molecular exchange of momentum, matter, and heat is on the same order as gases. The rate of exchange of transport processes is measured by the Schmidt number Sc and the Prandtl number Pr. Usually, the assumption of mechanical equilibrium in gases for heat and mass transfer is not reliable.
6.2
MAXWELL-STEFAN EQUATION
Maxwell-Stefan equations describe steady diffusion flows, assuming that sheafing forces for each species are negligible. As there are no velocity gradients assumed, the Maxwell-Stefan equations can be written in the forms of fluxes. For a ternary mixture of components 1, 2, and 3, the flow of component 1 in the z direction is dq dc 2 JI -- Dl l - 7 - + D 1 2 clz dz
(6.1)
Similar equations can be written for components J2 and J3. The coefficients Dll and D22 are the main coefficients; they are not self-diffusion coefficients. D~2 and D2~ are the cross-coefficients and assumed to be equal to each other for binary gas mixtures. Multicomponent diffusion by the Maxwell-Stefan equation is dx e a~
j=l c 2 D(j,
/ '/ CJ
ci
(6.2)
where c; is the concentration of species i, c is the total concentration, J; is the flow of species i, and D£- is the Maxwell-Stefan diffusivity. Maxwell and Stefan proposed a method to describe the diffusion in multicomponent gas and liquid mixtures of isotropic systems J -
LklPvt
(6.3)
l=1
where vl is the velocity of component l, P is the pressure, and Lkt is the Onsager reciprocal relations. Equation (6.3) shows the phenomenological equations used to describe diffusion where the Onsager reciprocal relations can be expressed in terms of the diffusion coefficients. Maxwell suggested an equation for dilute gases and Stefan suggested one for liquids, and hence Maxwell-Stefan equations cover both vapor and liquid phases. It is assumed that the diffusion results from equal and opposite forces that are proportional to the velocity differences of the components. The Maxwell-Stefan equations do not depend on choice of the reference velocity, and therefore they are a proper starting point for other descriptions of multicomponent diffusion. For ideal gas mixtures, diffusivities i, and D~l are
320
6.
Diffusion
independent of the composition, and equal to diffusivity D[t of the binary pair kl. In an n-component system, only n(n - 1)/2 different Maxwell-Stefan diffusivities are required as a result of the simple symmetry relations. Some advantages of the Maxwell-Stefan description of diffusion are as follows: 1. 2. 3. 4.
Diffusion is independent of choice of the reference velocity. The diffusion of all the components is treated equally. Diffusion is in agreement with the results of the kinetic theory of dilute monatomic gases. The Maxwell-Stefan diffusion coefficients represent binary diffusivities; for ideal and many nonideal mixtures, they are independent of the concentration of the species in the multicomponent mixtures.
Mass and molar diffusion are important in practice, and can be derived from the Maxwell-Stefan description of diffusion. The Maxwell-Stefan multicomponent diffusivities are obtained from the binary diffusivities, which are easy to measure.
6.2.1
Isothermal Systems
For an isothermal mixture, the dissipation function is l-1 £ air = _ p 2 Z
)2 Zkl (V l -- v k
>-- 0
(6.4)
k=l /=2
where v t - vk is the velocity difference between the species I and k. For a three-component system, Eq. (6.4) yields 2 XI'r=-p2(L12v21 nt-L13V~l-[-L23v~2) ~ 0
(6.5)
where vkt2 _ (Vk--Vt) 2. Eliminating v32 with v31 - v21, and dividing the result by v321leads to the following quadratic constraint
r/2
-(L12 -Jr-L23) v21 v31
r
+ 2L23 v21 -(/-q3 + L23) ~ 0 ~ v31
(6.6)
Necessary conditions for satisfying this inequality are - L 2 1 - L 2 3 - L22 -> O,
- L 3 1 - L 3 2 = L33 -> O,
-L12 -L13 - Lll -> 0
(6.7)
If Eq. (6.7) is satisfied, the constraint for the three-component system becomes L21L13 -nt- L23 (L12 nt- t13 ) ~ 0
(6.8)
Equation (6.8) is a sufficient constraint irrespective of the choice of independent velocity difference with --Lkt>--O for k 4: l, and shows that it is possible to have negative phenomenological coefficients without violating the condition of total positive dissipation. The value of-L23 does not need to be positive and is subject to the following constraint L12L13 -L23 z> ~ < 0 L12 -+-L13
(6.9)
by assuming that the other coefficients satisfy -L12 > 0 and -L13 > 0. The constraints for n-component system is generalized as --Lkk >-- 0 and LkkLtt - LZt >- 0
for n -> 3
(6.10)
Maxwell described diffusion by velocity differences, which yield forces from the friction between the molecules of different species. He considered a chemical potential gradient caused by friction, which is proportional to the concentration. The diffusion coefficient of Maxwell-Stefan can be defined as t
Dkt =--
XkXl PLkl
where xk - ck/c are the mole fractions. The following relations hold for the diffusion coefficients
(6.11)
6.2
321
MaxwelI-Stefan equation
D'kl >-- O, D~.+ - Dl'k,
and D£k --> 0
(6.12)
Thus, the Maxwell-Stefan diffusion coefficients satisfy simple symmetry relations. Onsager's reciprocal relations reduce the number of coefficients to be determined in a phenomenological approach. Satisfying all the inequalities in Eq. (6.12) leads to the dissipation function to be positive definite. For binary mixtures, the Maxwell-Stefan diffusivity has to be positive, but for multicomponent system, negative diffusivities are possible (for example, in electrolyte solutions). From Eq. (6.12), the Maxwell-Stefan diffusivities in an n-component system satisfy the following inequality xl ->0
(6.13)
l-I D~.l kvl
Negative Maxwell-Stefan diffusivities are allowed if they satisfy !
!
D[~ ~-- DkkDl¢
(6.14)
It is useful to express Eq. (6.14) in terms of D'a-I as follows Xk ~ --
Xl
XkX 1
k:, D~., k:, D'kl 6.2.2
(6.15)
>--0
D'k~
Nonisothermal Systems
For nonisothermal systems, diffusion flow becomes H
Jk
~" XkXl -- z__, (V l -- Vl,. ) -
Lkq T
- 1
grad T
(6.16)
+--t D[.t
If grad T 4=0, thermal diffusion flow appears, and is given by
Oh.vk - - p D [ grad T
(6.17)
T T
Here, the coefficient Da. is defined as the barycentric coefficients of thermal diffusion. Due to ~pkVk - 0
(6.18)
thermal diffusion coefficients satisfy the following constraint ~DT-0
(6.19)
k
From Eq. (6.16), the phenomenological coefficients are
_Lkq __ Lqk _ ~ xkx___L[ D[ D'kt
wl
(6.20) Wk
Using Eq. (6.20), the Maxwell-Stefan diffusion becomes
aa. = ~_, xkxl (v~ - v~ )
(6.21)
where vko is defined as
D[ grad T
~. - vk + - wk
V°
T
(6.22)
Equation (6.21) is considered a generalization of the Maxwell diffusion equation. Curtiss and Hirschfelder derived similar equations for dilute gases by using the kinetic theory of gases.
322
6.
Diffusion
Example 6.1 Maxwell-Stefan equation for binary mixtures For an application of the Maxwell-Stefan description of diffusion, we consider a binary isotropic mixture with components 1 and 2. To solve the mass balance equations, the diffusion flow has to be known. The binary diffusion flow without the electromagnetic field and external forces is given by J1 = - J 2
=
Pl
-~- [(grad ~1 )T,P q- (Vl -- V) grad P]
(6.23)
where Vl-V = (Wl + W2)V1--W1V1--W2V2 -" W2(V1--V2). Thermodynamic correction factor F is defined using the Gibbs-Duhem relation (6.24)
P ~. OXl )T,P
-- ~
~. OX2 ) T,P
The thermodynamic factor is a measure of deviation from ideal behavior, and equal to unity for ideal systems. Equation (6.24) can be rearranged as
_.•
t91
(grad ~I)T,P = --p
(.0&]
OXl ) T,P
grad x 1 = F grad X 1
(6.25)
Diffusion flow in Eq. (6.23) can be expressed in terms of F al = -a2
-- F
(grad x 1) +
pW 1W2 (V 1 - -
V2 )
grad P P
(6.26)
For a binary system, grad w 1 is related to gradxl by using the summation relationxl + x2 = 1 and M = M i x 1 + m2x2, and we obtain dw 1 =
d
X1 = ---~ dx 1 - x 1
;
and dw 1 - M1MM2 2 dXl
where M1 and M2 are the molar masses of components 1 and 2, respectively. From Eq. (6.27), we have
J1
M1M2 F grad w1 + PWlW 2 (V 1 -- V2)
( / gradp
P
(6.28)
From Eq. (6.21), we have V 2o -- V o1 _
D[2 J1
(6.29)
XlX2
Finally, the diffusion flow J1 becomes J1 = PlVl -- -PD12
/~~rd grad T grad W1 -- pO~' ,:,_a______~_pD( P T
(6.30)
where the binary diffusivity D12 is called the Fick diffusivity, which can be used in the barycentric description (mass average) as well as in the molar description. The binary pressure diffusivity D1P is defined by D1P -. PO{2 MM1 M e 2 w 1We (V1 - V2)
(6.31)
where D~2 = D12/1-'. Equation (6.30) shows that diffusion occurs due to concentration difference (ordinary diffusion), pressure difference (pressure diffusion), and temperature difference (thermal diffusion) without other external forces and electromagnetic field effects.
6.2
323
M a x w e l I - S t e f a n equation
The behavior of the Fick diffusion coefficient in nonideal systems may be complicated, while the MaxwellStefan diffusion coefficients behave quite well, and are always positive for binary systems. In nonideal binary systems, the Fick diffusivity varies with concentration. As seen in Figure 6.1, water-acetone and water-ethanol systems exhibit a minimum diffusivity at intermediate concentrations. Table 6.1 displays the dependency of binary diffusivity coefficients on concentration for selected alkenes in chloroform at 30°C and 1 atm. As the nonideality increases, mixture may split into two liquid phases at certain composition and temperature. A solution is a condensed phase of several components, which may be subject to strong intermolecular forces. Despite the fundamental differences between solutions and gases, some laws for solutions are analogous to those for gases. If the solution is sufficiently dilute, the osmotic pressure is described by an equation similar to that for an ideal gas, and ideal solutions are treated as a special case of ideal gas.
6.2.3
Ideal Solutions
The chemical potential of an ideal solution may be expressed by Izk
-
O
I~-
(6.32)
(T,P)+ RkTlnxk
where R k = R/Mk; R is the universal gas constant and Mk the molar mass. The behavior of dilute solutions may be close to that of ideal solutions depending on the nature of the solvent and dissolved substances. On the other hand, in electrolytes, deviations from ideal behavior may occur even in very dilute solutions due to the spatial range of the electromagnetic forces. Solutions of molecules with normal size and similar chemical structures can behave ideally over a large range of concentrations. The principal difference between ideal gases and ideal solutions is caused by molecular interactions
-.,.,.
2
O
0.5
,
i
,
0.2
I
I
I
0.4
J
I
0.6
i
0.8
1
XA (a)
"-"
4
E
% 3 g3
2
1
I
0.2
0.4
,
I
0.6
,
I
0.8
b
1
XA
(b) Figure 6.1. Concentration dependence of the Fick diffusivity for the binary mixtures: (a) water (A)--acetone (B) at 45°C, (b) water (A)methanol (B) at 40°C in terms of water mole fraction (Tyn and Caius, 1975).
324
6.
Diffusion
Table 6.1 Mutual diffusion coefficients as a function composition for binary mixtures of some alkanes in chloroform at 30°C and 1 atm Solute (1)
xl
D × 109 m2/s
n-Hexane
0.00413 0.2034 0.4028 0.6000 0.8009 0.9894
2.43 2.42 2.59 3.18 3.75 4.44
n-Heptane
0.00543 0.2007 0.4003 0.6006 0.7995 0.9895
2.23 2.21 2.38 2.85 3.27 3.52
n-Octane
0.00811 0.2016 0.4000 0.6024 0.7847 0.9870
2.03 1.94 2.13 2.44 2.83 3.08
3-Methylpentane
0.00814 0.2029 0.3998 0.6000 0.8009 0.9906
2.28 2.35 2.53 3.00 3.60 4.53
2,3-Dimethylpentane
0.00733 0.1507 0.3024 0.5015 0.7003 0.9883
2.08 2.09 2.20 2.54 2.80 3.55
2,2,4-Trimethylpentane
0.00641 0.2009 0.4005 0.6000 0.8012 0.9866
1.95 1.92 2.03 2.29 2.67 3.06
Source: R.L. Rowley, S.-C. Yi, D.V Gubler and J.M. Stoker, J Chem. Eng. Data, 33 (1988) 33.
in short time for gases and longer time for liquids. Therefore, the pressure effect on/x] (T, P) for ideal gases is different from the corresponding quantities for ideal solutions. This can be shown through the isothermal compressibility coefficient Kk
Kk
1 ( Ovk )
(6.33)
The compressibility of a liquid is practically independent of the pressure, and we have
vk (T,P)= vk (T, 0)(1-KkP ) The specific volume
vk does not depend on the concentration in an ideal system.
(6.34)
6.2
325
Maxwell-Stefan equation
For all ideal systems of gas and liquid mixtures, we have -~
/
'
7~
OP
= vk
(6.35)
T
From Eqs. (6.34) and (6.35), we get tx k ( T ) -
tx k ( T , P ) -
P
1---~KkP
vk
,
Isothermal compressibility for an ideal gas mixture K is given by 1/P, whereas for liquids the compressibility is negligible. The activities can be introduced to describe the deviations from the ideal behavior of solutions; the activities are expressed in terms of the activity coefficients.
6.2.4
Nonideal Solutions
For nonideal systems, intermolecular interactions may be simplified by introducing the activities into the diffusion potentials. Deviations from ideal behavior can be estimated by the Fick and Maxwell-Stefan diffusivities and the thermodynamic factor. The chemical potential/xx, can be expressed in terms of temperature, pressure, and activity a~ /x~: - ~[I (T, P) + RkT In a k
(6.37)
and the total differential is given by
d # k .... . O a k T.t'
OP ),l,,r
~, OT )p,,:
The partial differentiation of chemical potential with respect to the activity is obtained from Eq. (6.37), and given by ( 0#/;. ]
_ RkT
Oak )T,P
(6.39)
ak
For nonideal systems, Eq. (6.39) may be rearranged as __ P~ (grad/xx. )r.P ._. . .Pk . R~.T (grad a k )r,P = -xk - (grad a k )r,P P Pa~. ak
(6.40)
where Px_ = PkRkK and xk = Pk/P. Diffusion flow in terms of activity for nonideal systems is Jk - xk (grad a k )r,P + ( p x . v x . - w k ) grad P
a~.
(6.41)
P
The description of diffusion may be complex in mixtures with more than two components. Diffusion coefficients in multicomponent mixtures are usually unknown, although sufficient experimental and theoretical information on binary systems is available. The Maxwell-Stefan diffusivities can be estimated for dilute monatomic gases from Dik-~Dx-t when the Fick diffusivities are available. The Maxwell diffusivity is independent of the concentration for ideal gases, and almost independent of the concentration for ideal liquid mixtures. The Maxwell-Stefan diffusivities can be calculated from D~.l - Dkt F
(6.42)
The thermodynamic correction factor F can be expressed in terms of the activity using Eq. (6.39) F -x k
~ In a k Ox k
-
0 In a k Olnx k
(6.43)
326
6.
Diffusion
Using the definition of activity ak = xkYk, Eq. (6.43) becomes F=I+~
Oln yk Olnx k
(6.44)
The activity coefficients of nonideal mixtures can be calculated using the molecular models of NRTL, UNIQUAC, or the group contribution method of UNIFAC with temperature-dependent parameters, since nonideality may be a strong function of temperature and composition. The Maxwell-Stefan diffusivity for a binary mixture of water-ethanol can be considered independent of the concentration of the mixture at around 40°C. However, for temperatures above 60°C, deviation from the ideal behavior increases, and the Maxwell-Stefan diffusivity can no longer be approximated as concentration independent. For highly nonideal mixtures, one should consider the concentration dependence of the diffusivities.
6.2.5 Binary Systems For a binary mixture, if experimental diffusivities do not exist over the whole range of concentration, an interpolation of the diffusivities at infinite dilution DkzfOC is used. In calculating the diffusivities at infinite dilution by the StokesEinstein relation, we consider small isolated hard spheres, submerged in a liquid, that are subjected to Brownian motion. The friction of the spheres in the liquid is given by the Stokes' law; Einstein used the Stokes' law to calculate the mean-square displacement of a particle. The displacement increases linearly with time, and the proportionality constant is the Stokes-Einstein diffusivity kBT DO -
6rrr/j~
(6.45)
where r i is the radius of the particle i, kB is the Boltzmann constant, and fly is the viscosity of the solvent j. The generalized Stokes-Einstein diffusivity is modified to account for the particle sizes of solute and solvent, and is given by
where nc is around 3.5. For nonassociating and associating binary organic mixtures, the values of nc are 3.54, and 3.53, respectively; nc is 3.65 for self-diffusivities of organic species and 3.47 for polar organic species in water, where hydrogen bonding occurs. The radius is calculated from the van der Waals volumes. The estimates of the diffusivities with the modified Stokes-Einstein relation are comparable to or more accurate than those estimated with the Wilke-Chang relation. A modified version of the Vignes interpolation for concentrated binary solutions is adequate for nearly ideal systems, and is given by 1 ,t,DPk~ D 'kz = -~ kZ 1rt ~ ) x ~ ( D ,kll-< rl z ) x ,
(6.47a)
where rl is the viscosity of the mixture. The accuracy in estimating the diffusivity of moderately ideal mixtures is quite high. The simple linear interpolation with the viscosity correction correlates the diffusivities to within the experimental error 1 (DPk_~l r~d~ 1~qtxt) D'kl=-~ kt nkxk+~kt
(6.47b)
The interpolations of Eqs. (6.47a) and (6.47b) are satisfactory for nonideal nonassociating systems, however, for associating mixtures, they yield relatively large errors.
6.2.6 Ternary Systems Estimations for ternary diffusion are more complicated than for binary diffusion, and there is a lack of experimental data. To obtain estimates for ternary mixtures, the interpolation relations, given in Eqs. (6.47a) and (6.47b), are extended as Dkt,
--
)xi -1~ (D,k~l kZ ~k)xk (D,kZz-~lrlz)x, ~(D,i~l kl Tli
(6.48)
6.2
327
Maxwell-Stefan equation
,~ 1 /-~,I~1 r-~,i~l Dkl - ~ ( D~-~-~lrl~xk + "-'k/ rl/x / -Jr-l_)kl TliX i )
(6.49)
Estimations for nonideal mixtures need six diffusivities at infinite dilution and three diffusivities of the type Fltk--*l .L~kl • Negative diffusion coefficients can exist in ternary systems and are consistent with the nonequilibrium thermodynamics approach. Some of the molecular theories of multicomponent diffusion in mixtures led to expressions for mass flow of the Maxwell-Stefan fonn, and predicted mass flow dependent on the velocity gradients in the system. Such dependencies are not allowed in linear nonequilibrium thermodynamics. Mass flow contains concentration rather than activity as driving forces. In order to overcome this inconsistency, we must start with Jaumann's entropy balance equation DS p~ = -(V.s) + • Dt
(6.50)
where p is the density of the fluid mixture, S the entropy per unit mass, entropy production per unit volume. The operator D Dt
s
the entropy flow vector, and • the rate of
0 ---+(v.V)
(6.51)
Ot
is the substantial derivative. From the balance equations of mass, momentum, energy, and the Gibbs relation, one obtains explicit expressions for s and O. For multicomponent diffusion, and the mass flow expressions, we mainly use Fick's law and the Maxwell-Stefan forms • Using the symmetric difthsivity, in 1ength"/tlme, ' " we have cRTLij
D!) -
(6.52)
PiPi where p; is the density of species i, L!j are the phenomenological coefficients, and c is the total molar concentration P" c=Z; =ZM,
(6.53)
where M,. shows the molecular weight of component i. The diffusivity coefficients have the following properties D;) - D);
( i - 1,2 .... , n)
~ w;D!} = 0
(j = 1,2,..., n)
(6.54)
(6.55)
i=-I
There are n(n - 1)/2 independent diffusivities DIj, which are also the coefficients in a positive definite quadratic form, since according to the second law of thermodynamics, the internal entropy of a single process never decreases. In terms of these symmetric diffusivities, the mass flow becomes
J; - -D,rV In f - - p L D$@
(i - 1,2,..., n)
(6.56)
j--I
where D r is the generalized thermal diffusion coefficient in mass/(length)(time). The generalized driving force X, is given by
cRTX, - VP i - w i V P - piF~ + w i ~ p/Fi j=-I
(6.57)
328
6.
Diffusion
where F i is the force per unit mass acting on the ith species. Using Eqs. (6.56) and (6.57), we can express the mass flow in terms of general driving force, which is the Fick form
Ji = - D i T V l n T -
cRr
Dij V P j - w j V P - p j F
i +wj
j=l
(6.58)
PkFk k=l
It may be convenient to express Xi as a linear function of Ji, which is in the Maxwell-Stefan form
c R T Z C;,k Jk k4:i Pk
Ji : v e i _ w i V P _ ~ F i Pi .
_+_wi
pjFi-cRT~.~
j=l
k~i
Cik D~
D~i
Pk
Pi
(VlnT)
(6.59)
! t where Ci~ is the inverse diffusivity, and sometimes is expressed as C 0.! = xixk/Di~, and C/kare the Maxwell-Stefan diffusivities. Equations (6.58) and (6.59) contain the same information and are related through multicomponent diffusivities D~. and multicomponent inverse diffusivities C~.. For low-density gases, we have VPi = V(ciR7). For polymeric liquids, a similar form to Eq. (6.59) can be found from a molecular theory by replacing the pressure P and the partial pressure Pi with the total stress tensor and the partial stress tensor. The mass flow is related to the velocity gradient via the stress tensor, temperature, and concentration gradients.
6.2.7
Generalized Matrix Method
Fast and satisfactory mass transfer calculations are necessary since we may have to repeat such calculations many times for a rate-based distillation column model or two-phase flow with mass transfer between the phases in the design and simulation process. The generalized matrix method may be used for multicomponent mass transfer calculations. The generalized matrix method utilizes the Maxwell-Stefan model with the linearized film model for diffusion flux, assuming a constant diffusion coefficient matrix and total concentration in the diffusion region. In an isotropic medium, Fick's law may describe the multicomponent molecular mass transfer at a specified temperature and pressure, assuming independent diffusion of the species in a fluid mixture. Such independent diffusion, however, is only an approximation in the following cases: (i) diffusion of a dilute component in a solvent, (ii) diffusion of various components with identical diffusion properties, and (iii) diffusion in a binary mixture. For binary diffusion, there is only one independent flow, force or concentration gradient, and diffusion coefficient. On the other hand, multicomponent diffusion differs from binary diffusion because of the possibility of interactions among the species in mixtures of three or more species. Some of the possible interactions are (1) a flow may be zero although its zero driving force vanishes, which is known as the diffusion barrier; (2) the flow of a species may be in a direction opposite to that indicated by its driving force, which is called reverse flow; and (3) the flow of a species may occur in the absence of a driving force, which may be called osmotic flow. The theory ofnonequilibrium thermodynamics indicates that the chemical potential arises as the proper driving force for diffusion. This is also consistent with the condition of fluid phase equilibrium, which is satisfied when the chemical potentials of a species are equal in each phase. In the generalized Maxwell-Stefan equations, chemical potential gradients, which are the thermodynamic forces, are linear functions of the diffusion flows Xi = Z XiJm - xmJi - Z XiNm - xmNi R T vTt'/ci , , m cDim m cDim
(i = 1,2, ..., n - 1 ) , i 4: m
(6.60)
where x is the mole fraction, R is the universal gas constant, T is the temperature, and c is the total molar density. The D/m denotes the generalized Maxwell-Stefan diffusion coefficients, and according to the Onsager reciprocal rules, there are n(n - 1) of such coefficients, which exhibit the symmetry property Din , = Dmi. The first equality of Eq. (6.60) represents the diffusion irrespective of the reference velocity frame chosen, while the second requires one to define the reference velocity. In matrix notation, Eq. (6.60) becomes -c[FJ[B] -l(Vx) = (J)
(6.61)
where the elements of the matrix of thermodynamic factor [F] is defined by
I'ij - ~ij -~- xi 0 In 7i xj Olnx i
(i,j = 1,2, ..., n - l )
(6.62)
6.2
329
Maxwell-Stefan equation
where 6 Uis the Kronecker delta and equal to 1 for i = j, and 0 for i 4=j, and 3'; is the activity coefficient of species i. [B] is the matrix of diffusion coefficients, and defined by Bii = xi ..... .-{-n,~L ''Xm' O~'n m=l Dim
/' 't
ii ---taXi
t
Dij
(i = 1,2,..., n - 1), i # m
(i,j = 1,2,...,
n-l),
i 4= j
Di',,
(6.63a)
(6.63b)
Comparing the matrix notation of Fick's law (J) = -c[D](Vx) with Eq. (6.61) yields [D] = [B]-I[F]
(6.64)
Equation (6.64) provides a general method of predicting the elements of Fick's law of diffusion coefficient matrix.
6.2.8
Diffusion in Mixtures of Ideal Gases
Consider the problem of steady-state one-dimensional diffusion in a mixture of ideal gases. At constant T and P, the total molar density, c = P/RT is constant. Also, the Maxwell-Stefan diffusion coefficients Dim reduce to binary molecular diffusion Dim , which can be estimated from the kinetic theory of gases. Since Dim is composition independent for ideal gas systems, Eq. (6.61) becomes - c [ B ] - I (~Ty) -- ( J )
(6.65)
There are n - 1 independent relations in the Maxwell-Stefan formulation. For a one-dimensional diffusion in direction z, Eq. (6.65) becomes
dyi - ~
drl
m=l
YiJm -- YmJi kim
(i = 1,2, ..., n - 1), i 4: m
(6.66)
where r / = z/6 and kim = cDim/~. Here, r/is the dimensionless distance along the diffusion path, and kim is the transfer coefficient of the binary pairs of species i and m. Krishna and Standard (1976) suggested a general solution to the n-species system and defined the following parameters: (1) Matrix of dimensionless mass transfer rate factors [~] with the elements defined as
f~ii = ~Ni +
~ Nm
(i=1,2,
- ,
. n-l),
i4=m
(6.67)
m=l
~J=-Ni[
lkO.
k~l) (i,j
= 1,2, ..., n - l ) ,
i :/= j
(6.68)
(2) Column vector to with elements defined as
N~
(6.69)
(3) Square matrix [B0] with the following elements
Yoi ~ YOm
B°ii - kin -+- ,,=1 ~
(i = 1,2, ..., n - 1), i 4= m
(6.70)
330
6.
Diffusion
(1 1/ B°ij = - Y ° i
ko.
(i, j = 1,2, ..., n - 1), i 4: j
ki n
(6.71)
Using the parameters above, Eq. (6.66) may be converted into n - 1 dimensional matrix notation
d(y) dr/
- [~] (y) + (to)
(6.72)
l(d(Y) I
(6.73)
The diffusion flows are estimated from
(Jo) = -[Bo 1- [ , - ~ / , : o
The composition profiles can be obtained from Eq. (6.72), which also facilitates the estimation of the gradient (d(y)/drt)n=o; using the gradient in Eq. (6.73), we obtain
(J0) = -[k~] (y0 - y 6 )
(6.74)
Here, [1~] is called the matrix of high flux mass transfer coefficients, and is defined by [k~] = [B0]-I[~I t]
(6.75)
[ ~ ] : [ ~ ] { e x p [ ~ ] - [I]} -1
(6.76)
where [ ~ ] is the matrix of correction factors
The [I] is the identity matrix. When N i --~ 0, the matrix of correction factors [~] reduces to the identity matrix, and the matrix [1~] becomes [k0], which is defined as the matrix of zero flux mass transfer coefficients [k ° ] = [B ° ]-1
(6.77)
Using the diffusion fluxes J0i, the total molar fluxes are defined by N = J + N t (Y0)
(6.78)
where Nt is the total flux --1
Nt = _
znm : l ( P k - - P n ) J m m m=/=k Z" l)mYOm
(6.79)
m=l
Using the determinacy condition below, the total fluxes are estimated by a trial and error procedure.
~voN i = 0
(6.80)
i=1
where vi = 1 for equimolar countercurrent diffusion, and vi = 0 for diffusion of n - 1 components through a stagnant nth component.
Example 6.2 Diffusion in a ternary ideal gas mixture Methane is being absorbed from a mixture of argon and helium by a nonvolatile liquid in a wetted wall column operated at 25°C and 1 atm. The following boundary conditions and data may be used: System: methane (1)-argon (2)-helium (3)
6.2
MaxwelI-Stefan equation
331
Boundary conditions: z = 0(bulk gas), Y01 = Y02 = 0.4, Y03 = 0.2
(6.81)
z = 6 = 0.1 (interface), Y61 = 0.1, Y62 and Y63 unknown
(6.82)
Ordinary diffusion coefficients: Dye2 = 0.202, Dy,3 - 0.675, Dy23 = 0.729cm2/s Assume that a simple film model exists for the mass transfer, equilibrium is established at the gas-liquid interface, and the diffusion occurs at isobaric and isothermal conditions. Also assume that neither helium nor argon is absorbed, so that N2 = N3 = 0. Then, the Maxwell-Stefan equations for the diffusion of argon and helium are dy2 _ y2N1
dr/
(6.83)
k21
(6.84)
dy3 _ y3N1 dr/
k13
where the terms r/and k/j are defined in Eq. (6.66). By integrating Eqs. (6.83) and (6.84), and using the boundary conditions described in Eqs. (6.81) and (6.82) with the summation equation ~Y6i 1, we obtain =
(6.85) The solution to Eq. (6.85) yields the interfacial composition of argon and helium: Y~2 = 0.675 and Y,53 = 0.225
The rate of absorption of methane is N 1 = 4.29 × 10 .5 gmol/(cm 2 s) For the system, the flux of methane may also be estimated by defining an effective diffusivity Dl,eff J1
=
( gYl)
-CDl,eff ~. dz
(6.86)
where 1--y 1 Dl'eff - (Y2/D12 ) -+- (Y3/D13 )
We take the arithmetic average of Dl,eff as
Dl'eff --
Dl,eff (z = 0) + Ol,eff (z = ~) 2
(6.87)
So, Eqs. (6.83) and (6.84) can be integrated, and we have N 1 = 4.13 × 10-5 gmol/(cm 2 s). The newly obtained value of N1 is close to the previous value of N1. For a species absorbed from a mixture of inert gases by a nonvolatile liquid, the Dl,eff depends on the composition and diffusion coefficients, and may be generalized by 1-y 1
Dl,efe
~ ~=2(Y~/Dli)
(6.88)
332
6.
Diffusion
Example 6.3 Diffusion of species from a gas mixture to a falling liquid film In a wetted wall column, the falling liquid film consists of a mixture of acetone (1) and benzene (2). This falling film is in contact with a downward flowing gas mixture of acetone, benzene, and helium (3). The pressure, temperature, and thickness of the film are P - 1 atm, T - 28.1 °C, and 6 = 0.044 cm. The gas entering at the top of the column contains 10 mol% acetone and no benzene, and the composition of the vapor flow at the interface is Yll = 0.02, Y12 = 0.28, Y13 = 0.70 Y01 = 0.10, Y02 = 0.0, Y03 = 0.90
The diffusion coefficients are D12 -- 0.04, D13 -- 0.41, D23 -- 0.39(cm 2 s)
We assume that helium is insoluble in the liquid phase. The interfacial rates of diffusion of acetone and benzene can be calculated at the top of the column in an iterative manner using the following steps: Step 1: Calculate the matrix of zero flux mass transfer coefficients from the inversion of the matrix [B0], which is obtained from Eqs. (6.70), (6.71), and (6.77) °~Y/Col -" k13[k12 nt- f°l(k23 - k12)] ' S
(6.89)
,~yko, = k23Y01(k13 - k]2) 2 S
(6.90)
'~Y/g021 -'- k13Y°2 (k23 - k12 )
(6.9 1)
S k23 [k12 + )/'°2 (k23 - k12)]
ky°22 =
(6.92)
S
w h e r e S = Yolk23 Jr- Yo2k13 -k- Yo3k12
Step 2: For the first iteration, assume that the matrix of correction factors is the same as the identity matrix, and therefore the finite flux mass transfer matrix, given by Eq. (6.75), becomes [k;] = [Bo]-l[ko]
(6.93)
Step 3: Calculate diffusion fluxes Joi from Eq. (6.74) using the condition ~Joi = O. Step 4: Estimate the total molar fluxes Ni from Eq. (6.78) (6.94)
N 3 = 0 = J03 + Yo3Nt /
"x
Ni = J o i - [ Yoi ] J o 3
(i--1,2)
(6.95)
k, Y03 }
Step 5: Calculate the elements of the matrix of mass transfer rate factors [~] from Eqs. (6.67) and (6.68). Step 6: Calculate the elements of the matrix of correction factors [~] using Eq. (6.76), which may be obtained from the Sylvester expansion [~I~]-
A1
eAl-1
[(I~] -- A2[I ] +
A1-A 2
A2
e'~2-1
[CI~]-- AI[I] A2-A 1
(6.96)
where A1 and A2 are the eigen values of the matrix [~]. After completing the first iteration, the calculations are repeated starting from step 2 by updating the matrix of correction factors in a new finite flux mass transfer coefficient evaluation. When the two successively estimated values of Ni are close enough to each other, the iterations are
6.2
MaxwelI-Stefan equation
333
Table 6.2 Interfacial diffusion rates of acetone (1) and benzene (2) at the top of the column in Example 6.3
Nl•/C
N2t%/c
:¥,6/c
k,~, 6/c
k,._ 6/c
(mm:~-/s)
(mmdependent mutual diffusion coefficients and the matrix of thermodynamic factors. However, experimental data on r;mtual diffusion coefficients are rare, and prediction methods are satisfactory only for certain types of liquid mixtures. The thermodynamic factor may be calculated from activity coefficient models such as NRTL or UNIQUAC, which have adjustable parameters estimated from experimental phase equilibrium data. The group contribution method of UNIFAC may also be helpful, as it has a readily available parameter table consisting of many species. It, ihowever, reliable data are not available, then the averaged values of the generalized MaxwellStefan diffusion coefficients and the matrix of thermodynamic factors are calculated at some mean composition between x0; and x_~. Hence, the matrix of zero flux mass transfer coefficients [k0] is estimated by [ko] = [Bo]
....l[r]
(6.101)
The modified matrix of dimensionless mass transfer rate factors is defined as [@'] = [F]- ~[{I}]
(6.102)
The total fluxes N; are calculated as for ideai gases using the: fimtc timexmass trmxsfer coefficients defined by {6.103)
[kl} ]-- Ik o ] [{I}'] {exp [{I}']- [I] }--1
For nonideaI liquid mixtures, the generalized matrix method leads to only approximate solutions. The method is sensitive to the accuracy of the thermodynami~ factor.
6.3
DIFFUSION IN NONELECTROLYTE SYSTEMS
The linear phenomenological law' of diffusion tbr a binary system is given by
Jl - o~{ v l - v2 ) - 7
F1 -
/
(6.104)
ox I )
For a perfect gas or an ideal solution, we have {}
p,/ - Ac/{T,P)+RTlnq
(6.105)
Inserting Eq. (6.105) into Eq. (6.104) yields
L{ Opq) LRT{Oq cl(v 1 - v. ) -:/: V, J4, -~ - : ' i}x~ T c I Ox1
~Mlcl } RT
(6.106)
336
6.
Diffusion
Here, we can distinguish the following two separate systems: (i) For a uniform system, where Ocl/Oxl = 0, we have L
V1 --V 2 = ~ F 1 M
Tq
1
(6.107)
The coefficient of proportionality between the relative velocity (vl-v2) and the force F1M1 is called the mobility of component 1 B* and is defined by B*-
(6.108)
L Tc 1
(ii) For a system without external forces F 1 -- 0, we have q (v 1 _ v2)
L R T Oq T q Ox1
=
(6.109)
The coefficient of proportionality between the flow of diffusion cl(vl -v2) and the concentration gradient is the diffusion coefficient D-
LRT
(6.110)
T q
Comparing Eqs. (6.108) and (6.110) yields the Einstein relation between the mobility and diffusion coefficient
(6.111)
D = RTB*
For a system without an external force, Eq. (6.106) can be written as CI(V 1 _V2)__
L 0~1 O N 1 T ON1 0 x 1
(6.t12)
The phenomenological law defines the diffusion coefficient, D as Cl(V1 - v 2 ) = _DcON1
Ox1
(6.113 )
so that D = ~1 L O~I Tc ON1
(6.114)
This definition is equivalent to Eq. (6.110) for a perfect gas or for an ideal solution. Equation (6.114) shows that the diffusion coefficient is the product of the phenomenological coefficient L and the thermodynamic quantity (Otzi/ON1)/cT. The coefficient L is positive, and so is the quantity Otxl/ON1 for all ideal systems. This means that the diffusion coefficient is positive, and according to Eq. (6.113), the diffusion flow has the direction imposed by the existing concentration gradient. In some highly nonideal systems of partially or completely immiscible mixtures, such as water-butane and water-benzene, the quantity Oi~l/ON1 may be negative, corresponding to thermodynamic instability. Such systems may split into two liquid phases, and may have negative diffusion coefficients in the immiscible region. In contrast, the thermal conductivity is always positive. The diffusion coefficient is a product of two quantities, only one of which, the L, has a definite sign. 6.4
DIFFUSION IN ELECTROLYTE SYSTEMS
For electrostatic potentials and electric current of charged ionic species, we start with the fundamental Gibbs equation d U = TdS - 6 W
(6.115)
6.4
Diffusion in electrolyte systems
337
and reconsider the work term 6 W. Usually, the 6 W means the work of compression P d V and the work involved in changing the number of moles of the components (chemical work: - ~ t z i d N i ) . However, when we have a region with an electrostatic potential 48, a change in the charge de results in the electrostatic work, and hence Eq. (6.115) may be extended as follows d U = T d S - P d V + E t'tidNi + Ode
(6.116)
When the de is due to changes in the concentration of ionic species, we get (6.117)
de - y_, z i F d N i
and then Eq. (6.116) becomes d U - TdS - P d V + ~_~ t.tidN i + ~ , z i F O d N i i
(6.118)
i
Combining the last two summations, we have d U - T d S - P d V + E (tx~ + z~F6)dNi
(6.119)
i
Equation (6.119) indicates that the chemical work in electrolytes contains a chemical term txdN~ and an electrical term ziFg, dNi and the sum is called the electrochemical potential pci of the ionic species i [-*i - Izi + ziFO
(6.120)
If we have a phase in which the composition is identical at points 1 and 2 but ~1 =/=~2, then we have 1
#i-/x; -1
2
-2
#, - #i - ziF(Ol - ~2 )
(6.121)
(6.122)
Substituting Eq. (6.120) in Eq. (6.119), we get dU - T d S - P d V + E ficidNi
(6.123)
We may also use the change in the Gibbs free energy in terms of the chemical potential dG - - S d T + VdP + E fzidNi
(6.124)
or the Gibbs-Duhem relation, SdT-
VdP + E N i d ~ i - 0
(6.125)
Under isothermal and isobaric conditions, Eq. (6.125) reduces to ~__ Nidfic ~ - 0
(6.126)
i
It is often useful to express the electrochemical potential as a sum of explicit terms of activity and electrostatic potential as follows P~i - #~' ( T) + VP + R T In a i + z ; F ~
(6.127)
Thus, in the case of an ion distributed between two phases c~ and/3, the condition of thermodynamic equilibrium is /2~~-/2,./3
(6.128)
338
6.
Diffusion
Introducing Eq. (6.127) into Eq. (6.128) yields Atx° +(Vi~P '~ -Vi¢3P¢3)+ R T l n A a i + ziFAqt = 0
(6.129)
If, for example, c~ and/3 are aqueous phases separated by a membrane, then (/x°) ~ = ( o)/3 and V/~ = V/t~. So, Eq. (6.129) becomes V~AP + R T l n A a i + ziFA¢/ = 0
(6.130)
In most cases of interest, V;AP is negligible in comparison with the other terms, so that the condition of phase equilibrium across a membrane becomes (6.131)
a~__[_.= _ z i F A ~ R T In a/t3
For an ideal solution, the condition of phase equilibrium is (6.132)
c~__~_= _ z i F A O RT In c/~
Equation (6.132) may also written in base 10 logarithms
- M p - 2.3RT log
ziF
ci/3
: 58 log Zi
(mV at 20°C)
(6.133)
C~
Example 6.5 Diffusion in aqueous solutions Consider an aqueous solution with N1 moles of sodium chloride and N2 moles of calcium chloride. An increase in the concentrations of both salts by amounts dN1 and dN2 causes the following changes in the ionic concentrations dNNa = dN1, dNca = dN2, dNc1 = d N 1 + 2 d N 2
(6.134)
Introducing these relations into Eq. (6.123), we obtain d U = T d S - P d V +/.~NadNNa nt- ~ c a d N c a
+ ~CcldNcl
(6.135)
or
d U = TdS - P d V + (/'~Na at- ~C1 ) dN1 + (~Ca at- 2/2Cl) dN2
(6.136)
The corresponding chemical potentials of the electroneutral combinations are /'~Na at- ]~C1 --/'LNa + FO +/ZC1 -- F~p = ~Na at- ~C1
(6.137)
/'~Ca at- 2/2Cl = ~Ca at- 2F~b + 2/xCl - 2F~, = ~Ca + 2/xCl
(6.138)
and
The physical significance of these combinations stems from the dissociation equilibria NaC1 ¢:~ Na + + C1- and CaC12 ¢,. Ca 2+ + 2C1which are characterized thermodynamically by the following definitions NaCI -- ]~Na +/~C1 =/ZNaCl o + PVNaCl + R T In aNaac1
(6.139)
6.4
339
Diffusion in electrolyte systems
/'~CaCI~_ = ~Ca + 2/~C1 -- ~CaCI2 o + P Vcac12 Jr- R T In aca acl2
(6.140)
Therefore, for electrically neutral species, Eq. (6.136) becomes d U = TdS - P d V + (/&NaC1) dN1 + (~CaC12 ) dN2
(6.141)
Example 6.6 D i f f u s i o n a c r o s s a membrane The conditions of phase equilibria across a membrane separating two salt solutions a and 13 are /3 /U'NaC1 = [/'NaC1
(6.142)
/.L~aC12 -- /.Z/3CaC12
(6.143)
Let us assume that one side of the membrane contains a chloride salt of a macromolecule to which the membrane is impermeable. The other side contains a solution of CaCI2 alone. The concentration of the macromolecule is Cm, and the number of charged groups per molecule is v. The concentration of CaC12 in the solution containing the macromolecule is Cs, and the concentration in the other phase is Cs;3. For the equilibrium between phases a and/3, we have ~aCl2 if- VCac12P °~ Jr- R T l n c ~ a (C~I) 2 -- ~aC12 if- VCaC12P]3 q- RT lnccPa (Cc~I) 2
(6.144)
Since the pressure terms are negligible, this expression reduces to C~ a (C~I)2 _ CC~a(Cc~1)2
(6.145)
~ = l?Cm -+-2c s CCa - Cs~ and CC1
(6.146)
CC~a -- Cs/3 and Cc~a = 2Cs~
(6.147)
We know that
and hence Cs~(1}Cm if_ 2C 2 )2 =_ Cs]3(2Cs~ )2 = 4(Cs~ )3
(6.148)
Equation (6.148) is the well-known Donnan equilibrium of salt across a membrane in the presence of a polyelectrolyte, to which the membrane is permeable. It demonstrates the characteristic properties of the chemical potentials of neutral salts.
6.4.1 PhenomenologicalApproach in Electrolyte Systems The local dissipation function for a system with charged species is xp - - j , . V T - ~ JiV/2 + J,.A i
(6.149)
where A is the electrochemical affinity, and is given by A - -Z
vi#ci - - Z
vi(#i + ziFd/) - - Z
vi~i - F # ; Z viz i
(6.150)
Since the charge is conserved in the reaction, ~v;z; = 0, so that d -- - - Z
l?i[&i -- d i
(6.151)
340
6.
Diffusion
For an isothermal system excluding chemical reactions, Eq. (6.149) reduces to (6.152)
= -~_.JiV[zi
i Electrochemical potentials also obey the Gibbs-Duhem equation (6.153)
Z Cirri ~-"0 i
In an n-component system, there are n - 1 independent forces (-7/2/). Equation (6.153) is used to eliminate the force for the solvent, and we have ) n-1 n-1 Ci xI'r -- E Ji - - - Jw V(-]~i ) --- £ J d v (--]'~i) Cw i -7•
(6.154)
where j/a is the flow of solute relative to that of the solvent. For a solution of a single electrolyte dissociating into two ions, the dissipation function is (6.155)
xI* = - - J d V ~ l - JdV/22 The phenomenological equations relating the flows and forces defined by Eq. (6.155) are j f -- - L l l V ~ I - L12V~ 2
(6.156)
jd = _L21V/~I _ L22V/~2
(6.157)
If V ~ 2 -- 0 , then J1a = - E l 1V]~I, which indicates that Lll is the generalized mobility of the cation, since it is the proportionality coefficient relating the flow to its conjugate force. Then, J2a is not zero, but is given by J2a = -LalV/21, indicating that the diffusion of the cation causes a drag effect on the anions, and such interactions are determined by the coefficient L12 or L21. Equations (6.156) and (6.157) can be used in the special case of an electrical conductance measurement. This analysis is usually carried out under isothermal, isobaric, and uniform concentration (Vtx;= 0) for all species in the cell. The electric current I is driven by a potential difference between two nonpolarizable electrodes, and the local field intensity e is defined by = -Vqt
(6.158)
Then, the forces acting on a Zl-valence cation and a Zz-valence anion become
-zlFe
(6.159)
= ~/X2 .at-z 2 F V ~ =-z2Fe"
(6.160)
V~I = V/X 1 -t- ZlFV ~ = ~2
So, Eqs. (6.156) and (6.157) become jd = (Z1Lll + 22L12)Fg
(6.161)
jd = (ZlL12 + z2L22)Fe
(6.162)
For a monomonovalent salt such as NaC1 or KC1, for which zl
= -z2 =
1, we have
J f =(i[,11-Zl2)Fg
(6.163)
jd __ (L12 _ L22)F~
(6.164)
6.4
Diffusion in electrolyte systems
341
The electric current due to the transport of all ionic species is given by the sum over all the charges carried by the ionic flows I
- ~ ziFjdi
(6.165)
i=1
For a single electrolyte, we obtain I - z, FJ'l + + z?FJd2 - ( z ( L 1 ,
+ 2z, z2/_q2 + z 2 L 2 2 ) F 2 g ,
(6.166)
Due to the condition of electroneutrality, diffusion flows can be used in Eq. (6.166). The flows relative to the water velocity are jja _ c, (v, - v w) and JJ We also have cl = vlcs and c2 =
l?2Cs,
:
C2 (V 2 - - V w )
(6.167)
SO Eq. (6.166) becomes
I = z l F v l c s v 1 + z 2 F v 2 c s v 2 - v w F c s ( V i Z 1 + v2z2)
(6.168)
However, electroneutrality implies that vjz~ + v2z2 = 0, so I = ZlFJ 1 + z2FJ2, where J1 and J2 are the absolute flows, J1 =ClVl and J 2 =C2V2, respectively. Ohm's law holds for homogeneous, isothermal salt solutions, therefore, the relation between the current and the electric field intensity may be reduced to l=Kt
(6.169)
where K is the electrical c o n d u c t a n c e of the solution. Comparing Eqs. (6.165) and (6.169) indicates that (6.170)
K = L'F 2
where L' is given by L'-
Z2Lll q- 2zlz2L12 q- z2L22
(6.171)
Equation (6.171) shows the direct relation between the electrical conductance of the solution and the phenomenological coefficient. Similar relations are obtained by measuring the fraction of the total current that is carried by each ion, also under the conditions V/xi = 0. This fraction is called the H i t t o r f t r a n s f e r e n c e n u m b e r (ti) and is expressed by
, z(z
(6.172) I
V/~;=O
For the case of a single electrolyte, tl and t2 may be evaluated by introducing Eqs. (6.161) and (6.162) into Eq. (6.172) t 1 - zlFJ/~- -- zI(Z1Lll if-22/-"12) I L'
(6.173)
t2 - z2FJ2'J -- z 2 ( z l L I 2 + z2L22)
(6.174)
I
L'
It is apparent that the two transference numbers are not independent since tl + t2 = 1. Therefore, an additional expression is required to evaluate the three coefficients, Lll, L12, and L22. Such a relation may be obtained from the diffusion of the electrolyte. In this case, there is no electric current in the system, and the total transport of charge must vanish z, ji , + z2j~ _ I _ 0 F
(6.175)
342
6.
Diffusion
By introducing Eqs. (6.156) and (6.157) for ,I d and a2d, we obtain a relation between the forces acting on the two ions --Z1 (L1 lV/~ 1 -Jr-L12V~2 ) - z2 (L12V].~ 1 -+- L22V~2 ) -- 0
or using the general form of Eq. (6.139) have
191V/~1 nt- 192V/.~2 -- V ~ s ,
V~I=
and from the neutrality condition
(6.176) V1Z 1 -+- V2Z2 =0,
we
Z--~2( Z1L12L' 'k-Z2L22 VI'Zs
(6.177)
Z1 ( Z1Lll -k-z2L12)
(6.178)
192
Introducing Eqs. (6.177) and (6.178) into Eqs. (6.156) and (6.157), we obtain
,1zlz2(lL22 I 2=zlze( , lL22 I 192
L'
gl.Zs
191
L'
V/Zs
(6.179)
(6.180)
The flows of Jsd of the neutral salt is
jd _ Jd _ Jd z1z2 ( ZllZ22-I~2 ) 19--71 192 VlV2 L' Vies
(6.181)
Fick's law describes the diffusion of a neutral salt in a binary solution as (6.182)
Jd1 = - D V c s
Comparing the resulting Eqs. (6.181) and (6.182), we may express the diffusion coefficient in terms of the L/j as
ZlZ2(1L22 /
1211)2
L'
(6.183)
The electrical conductance, transference number, and diffusion coefficient provide the three relations from which the phenomenological coefficients can be determined, and for a monomonovalent salt we have
D
Kt(
/Xss
F 2
LI1 = ~ +
D
L22 -
Kt~
+ __F---7
(6.184)
(6.185)
/Zss D
LI2 = L21 -
Ktlt 2 F2
(6.186)
/Zss
Straight coefficients L ll and L22 are nearly linear functions of concentrations, while the cross-coefficient L 12 is highly dependent on concentration and becomes quite small at high dilution, where the interactions between the ions are minimal. For determining the properties of the phenomenological coefficients, it may be advantageous to consider the mobilities, which express the behavior of ions. This description of the behavior is similar to the one gained by considering the frictional coefficients in the case of membrane permeability. The mobility may be defined by using the explicit expressions for the flows under uniform chemical potentials
6.4
343
Diffusion in electrolyte systems
jd
__ c1 ( v 1 _ V w ) = VlCsOOlZ1F E
(6.187)
where VlCs=Cl is the concentration of ion 1, and ~ol is the ionic mobility. Equation (6.143) shows that tO 1 is the relative velocity of the ion per unit electrical force. It is the velocity acquired by a force of 1 dyne. In practice, the mobilities ui are defined as the velocity of the ions acquired in a field of e = 1 V/cm, and mobilities Ua and u2 of the cation and anion, respectively, are U1 = Z 10A 1F
and
-- z2 (.0 2 F
(6.188)
=--V2Csb/2 E
(6.189)
-//2
Therefore, from Eq. (6.187) we may write
Jd1 -- VlCsUle and J2d and the total electric current becomes
I = zaFjdl + z2Fa J = VlZlCsF(u 1 +U2)E
(6.190)
Therefore, in terms of the mobilities, the electrical conductance is given by (6.191)
K -- VlalCsF(Ul -+- u 2 )
It is often convenient to consider the equivalent conductance
-
~eq --
heq
F(u 1
instead of K
+ lg2 )
(6.192)
VlZ1Cs
Similarly, the conductance of a single ion can be defined as (6.193)
hi = Fum and A2 = Fu 2 So, heq = h 1 + h2, which is the well-known expression of Kohlrausch. For a cation, we can express the diffusion in terms of the mobility af
-- PlCsUl E -- ( Z 1 L l l q- z 2 L 2 2 ) F E
(6.194)
Therefore, we have __ (Z1Lll J r - z 2 L I 2 ) F U1 -C1
_ zZL11F _ ~ n t C1Z1
z1z2L12 F ~ C1Z 1
(6.195)
-- u12 -- ~ZlZ2L12F PlZlCs
(6 196)
-
We now define the reducedphenomenological mobility (uij)
Igl 1 --
z2
1F
and
Pl Z1Cs
where Ull is the reduced phenomenological mobility of ion 1, and/A12 is a measure of the interactions between ions 1 and 2, so that Ul -- Ul 1 --U12 a n d u 2 -- u22 - u 1 2
(6.197)
Introducing Eq. (6.197) into Eq. (6.191), we have K -- C l 2 1 F ( U l l - 2 u 1 2
-+- u22 )
(6.198)
The equivalent conductances become Aeq -- F ( U l l
- 2u12 -+- u 2 2 )
(6.199)
344
6.
Diffusion
AS - F(H11 -+-U12 )
(6.200)
A2 = F(u22 + u12)
(6.201)
The transference numbers are defined by Hll--H12
t1 =
(6.202)
Ull -- 2u12 + u22
t2 =
U22--H12
(6.203)
u]] -- 2u12 + b/22 Finally, the diffusion coefficient of the salt can be expressed in terms of the mobilities
ss/",,"22 cs
(6.204)
O =
From Eqs. (6.199) to (6.204), the reduced phenomenological mobilities are obtained as ull = F v~zlD + A2
u22
Cs/Zss
AeqF
F VlZlD ~
- A2 ~
Cs~ss
AeqF
H12 = F PlZlD
(6.205)
(6.206)
(6.207)
AIA2
CsP'ss
heqF
These expressions can be used to calculate u O. from known values of the other parameters. Calculations for NaC1 show that uli and u22 remain approximately constant over a relatively wide range of concentrations, while u~2 changes considerably.
6.5
DIFFUSION WITHOUT SHEAR FORCES
Following Kerkhof and Geboers (2005), an approximation of the Boltzmann equation for a multicomponent monatomic gas system is OVi _ /9/-~ --fli {Vi "VVi }-- r e / +
f n ~ [ DiE - P 2 XiXj ~ DT'i j=l D/~ Pi
" (vj
Dr'J V l n T + V • [2TIiSi]+P~_~ ~xixj
Pj
.-vi)
(6.208)
j:l
or
Acceleration force =-convected momentum change- partial pressure gradient + external forces - thermal diffusion force + shearing force + intermolecular friction force In Eq. (6.208), D~. are the Maxwell-Stefan diffusivities and D~ = D~, ~/is the Newtonian viscosity, and S is the rate of the deformation tensor defined by
1{
S i = ~ VV i "[-(VVi
)v- - -2~ ( V ' v i ) I }
(6.209)
6.5
345
Diffusion without shear forces
where I is the diagonal unit tensor. By neglecting shear forces and bulk viscosity effects at isothermal conditions, Eq. (6.208) reduces to OV i __ Pi--~----Pi{Vi'~Vi}--~Pi
n +PiFi +P~_. Xixj (Vj--Vi)
./=~
D:/
(6.210)
For a fixed coordinate system, we consider steady transport and have n
0 . - P i { V. i Vvi}.
VP,. +piFi + P ~ x i x j (Vj --Vi) j=l D~j
(6.211)
Here, the first term represents the change of the convection flow, which is small compared with the other forces, so that Eq. (6.211) becomes n
xix--j-j (Vj -- Vi)
0 -- - - V ~ . -F PiFi "Jr-P E
j=l D!~
(6.212)
When external force is absent, and using the flows defined by N/= vici, we find _ xiNi -xjNi
_
1 Vp i = Vc i
D!/!
,i=~
(6.213)
RT
Here, the system is isobaric and the total concentration is constant. Also, the total flux is constant in direction z, and we have ON_ " -
0
0z
(6.214)
The definition of molar average velocity 1
N
c .i=1
c
(6.215)
V u - - )[_civ / - -
leads all the other spatial derivatives to be
OVM,=
0VM.= --
0,
Oz
-
0,
Ox
OVu,:
-
0
(6.216)
Oy
So, we have ( n - 1) independent fluxes and concentrations n-1
H-l
N. - - E
N/,
and V c . = - ~ _ . VC/
/ 1
(6.217)
,/=1
The matrix from Eq. (6.213) becomes [B](N)
(6.218)
= -(Vc)
with the coefficients
Bii - xi
1 D' m
Di~
Din
k=l D:k k~i
(6.219)
Equation (6.218) becomes (N) - [B]- J(Vc) - [D] (Vc)
(6.220)
346
6.
Diffusion
where [D] is the matrix of Fickian diffusion coefficients, which are not symmetric. For species i, we have n-1
n-1
Ni = - - E D/j ( g c j ) = --c E D O.( g x j ) j=l j=l
(6.221)
Equation (6.221) shows that the flux of a component is dependent on the concentration gradients of all components in the mixture.
Example 6.7 Binary and ternary isothermal gas mixtures For a binary mixture of gases under isothermal and isobaric conditions and without shear forces, from Eq. (6.213) we have
By substituting
N 2 + N 1 = N, we
XlN 2 - x2N 1 = D12VCl
(6.222)
N 1 -- X l N - D12~rCl
(6.223)
find
For an equimolar diffusion (no net flow), Eq. (6.223) becomes
(6.224)
N 1 = -D12~7c1
For a ternary mixture under the same conditions, we have XlN2 - x2N1 + XlN3 - x3N1 = Vc1 D{2
(6.225)
D~3
x2N1 - XlN2 + x2N3 - x3N2 = ~7c2
D;2
(6.226)
D5
On the other hand, the Fickian-type relations are N 1 = -Dll~TC 1 - D12~7c 2
(6.227)
N2 -- -D21Vc1 - D22Vc2
(6.228)
From the inversion, we then have the diffusion coefficients defined by Dll =
O12 =
D13[XlO23 + ( l - x l ) O l 2 ]
XlO23 (D13 - D12)]
021 -- x2D13(D23 - D12)]
S 022 =
D23[x2D13+(1-
x2)D12 ]
with S -- XlD23 -Jr-x2D13 -Jr-x3D12
(6.229)
6.5
347
Diffusion without shear forces
Example 6.8 Diffusion in a dilute isothermal gas mixture Suppose that in a gas mixture, component n is in abundance and other components are in trace amounts. Then, we have Pn -~ P,C,,Vv,etZ,, ~O,c,,V,, ~ 1, and for a steady transport, Eq. (6.208) becomes the Navier-Stokes equation for the single component, xi ~ 0, i 4: n. m
0--p{v,.Vv,}i-VP+p,F,
+V'[2niS i + ¢ ( V ' v i ) I ]
(6.230)
This equation can be solved with appropriate boundary conditions. For a trace component i, the i-i momentum exchange will be negligible compared with the i - n exchange. This leads to a smaller shear effect than the diffusive friction effect. Therefore, for component i, we have 0 ~ V P i + PiE + P xix" (v,
- v i)
(6.231)
D;', When there is no external force, we get 0 .~. VP;. + P xix" (v, - v i)
(6.232)
D;.', or
1
0 -~ Vc; +
( x i N . - N i)
(6.233)
Di'Vc i
(6.234)
D/', N i ~. N n x i
6.5.1
-
-
Gas Diffusion in Meso-and Macroporous Media
Modeling of diffusion of gases in porous media involves averaging mass and momentum balances and considering the three-dimensional nature of the medium. In a practical engineering approach, we consider the counterdiffusion of gases through a porous medium, and assume that we can describe the geometry by means of a single effective pore radius, the porosity e, and a tortuoisity factor ~-. Following Kerkhof and Geboers (2005), the flux of a species with respect to a unit area of the medium is 8 t
- - -
N,,i,,v
2 Ni,av
m
(6.235)
T
where Nx.;,av = vx,;,avC; is the cross-section averaged molar fluxes based on cross-section averaged velocities, which depend on driving forces defined by
B,-
(6.236)
dP~
dx The averaged molar flux of a gas is ,
r~ Pay + DK
N;~'v - -~2 ~
AP RTL
(6.237)
where L is the length of the tube, 31is the dynamic viscosity, rp is the channel radius, and DK is the Knudsen coefficient, and is approximated by
[2 8TJ2]
DK ~ 0.89 ~r;)
where M is the molar mass.
(6.238)
348
6.
The driving forces may be defined by
B1
dP1
-RT
dx
B2
dP2 dx
Diffusion
[, [, gD
(x2Ntx,l,av - XlNrx,2,av ) + flmNx,l,av
-e,
(6.239)
(XlNtx,2,av -- x2Nx,l,av ) -+-f2mN.tr,2,av
--
(6.240)
D( 2
-RT
gD
D~ 2
8
where gD is the diffusion averaging factor, and fro is the wall friction factors. These equations show that the force on a component per unit volume is due to friction with the other components and due to shearing friction with the tube wall. These equations may be solved in an iterative manner. These driving forces can be extended to multicomponent mixtures.
,i
]2
dx
(6.241)
~ (xjN'x'i'av - xiN'x'j'av ) + fiimN'x'i'av m j=l D~ e
Here, fire is obtained from the binary friction model. For isobaric counterdiffusion, from the above equations, we have N tx,l,av
Am
~
N'x,2,av
(6.242)
Am
For the case of equimolar diffusion through a porous medium, we have a net total pressure gradient defined by dP
7.2
(6.243)
- -RTN'x 1 av(Am -- Am) - -
dx
''
8
For counter diffusion in large pores, the friction term dominates the wall-friction term.
6.5.2
Diffusion in Liquid Mixtures
A generalization of the Boltzmann equation for liquids and dense gases is / OV i __ __ n ~ [ [3i - - ~ --Pi {Vi "VV i } CiVT, PI,Zi -- ciViVP i @ p i g - c R r E XiXj L DT'i j=l D~. Pi
+V.[2r/iS i + ¢ ( V . v i ) l ] + c R T
~
DT'j ] V In T Pj
)
(6.244)
xixj
~ (vj-vi) j-1 D/~
where ¢ is the bulk viscosity and p, is the chemical potential. Equation (6.208) for monatomic gases differs from Eq. (6.244) for liquids. In Eq. (6.244), the total pressure is replaced by cRT; the partial pressure gradient has been replaced by the chemical potential gradient ci7/, i = ciVr,pt, i + ciV~VP, and the bulk viscosity is introduced. By disregarding convection and shear forces at steady state, we have
n .
.
.
.
.
.
(v;- vi)
(6.245)
We may remove the concentration and chemical potential of component n using the Gibbs-Duhem equation
• i=1
Ci V T,p ld6i --
0
(6.246)
6.5
349
Diffusion without shear forces
Equation (6.245) can be inverted to find the generalized Fickian formulation 1
1
(N) = N ( x ) - ~-f[B]- [ c R T [ F ] ( V x ) + ( c V ) V P - ( p F ) + ( c ~ ) R T V l n T ]
(6.247)
For many liquid mixtures, it is assumed that there is equivolumetric transport, and hence molecular volume contraction is negligible. For isothermal conditions and without external forces, the pressure gradient vanishes, and we have (N) = N(x)-[B]-Ic[F](Vx)
(6.248)
(N) = N(x)-[D]c[F](Vx)
(6.249)
Then, the Fickian formulation is
where [D] = - [B]-l[F].
6.5.3
Diffusion in Mixture of Electrolytes
Considering the force by an electric field on ion transport at isothermal conditions, we find from Eq. (6.245) ~ XiXj
0 --- --CiVT, pI,.Li -- ci~.VP i + c i z i F V ~ + cRT
~
(vj - vi)
j=~ D[/
(6.250)
where z i is the charge and F is Faraday's constant. Many electrolytes are in electroneutrality, given by 0 - Y__dn__lCiZi . The addition of Eq. (6.250) over all components yields VP = 0, and hence the system is isobaric. Then, Eq. (6.250) becomes n
0
--
--Cil~T,pl.tii + ciziFVO + cRT~_~ ~xixi j=l
(Vj - - V i )
D!}
(6.251)
Equation (6.251) can also be written as n
F,iVXi +ciziFVO+RT~_~ xiNj - x j S i
O=-ciRT i=l
j=l
!
(6.252)
D!/
The activity coefficients to be used in the thermodynamic factor F of ions are generally concentration dependent. For a very dilute mixture and taking the solvent as component n, and xn ~- 1, Eq. (6.252) becomes
0 = -cRTF~/Vx i + ciziFV~t- RT xnNi D~',
(6.253)
When the activity coefficients are equal to unity, we find the Nernst-Planck equation
N: -=-D,,
6.5.4
Vc i +ciz i --~V4~
(6.254)
Liquid Diffusion in Meso- and Macroporous Media
Modeling of diffusion of liquids becomes more complex when the steric effect of molecular exclusion inside the pores is accounted for. Following Kerkhof and Geboers (2005), a distribution coefficient between the pore and free liquid may be defined by K ; - cp f
(6.255)
Ci
Here, we assume a liquid mixture inside a pore, which is in equilibrium with the free liquid outside, leading to
350
6.
Diffusion
/~P =/,/f
(6.256)
In modeling diffusion, we assume that the local thermodynamic equilibrium holds and variables can be estimated from equilibrium relations. The concentration-dependent part of the gradient of chemical potential is
cp
RT
'
nl
(P/
j=l
-~j
= Ki Z L, jV
(6.257)
Here, we estimate the thermodynamic factors F at the hypothetical free liquid values, which would be in equilibrium with the actual pore liquid concentration. If Ki is not dependent on concentration, Eq. (6.257) becomes
cp ,,-1 Fc O VT PtAP = Ki £ ' V(c p) RT ' j=l -~j
(6.258)
For a binary liquid mixture, we have
cp dl'zp
P ? _RT[gD (xPNx,l,av XlNrx,2,av) flmNrx,l,av] 7"2
_
?
_
_
?
D~2
dx
_
_
e
cP dtZPdx - - R T [ gDD; 2 (xPNx'2'av' - x2Nx'la' v)P'
- f2mNx'2'av 1 ' "r2e,
(6.259)
(6.260)
From the definition
Bi
-- Ci ~ dx )r = Ci ~, dx
T,P + ciVi
~
(6.261)
we add Eqs. (6.259) and (6.260) to find
q T2 F dP --RTlflmN'x,l,av + f2mN'x,2,av ] dx t_ _~ e
(6.262)
For isobaric counter diffusion of liquids, we have the same relation as Eq. (6.242) NPx,l,av __ Nx,l,av __
flm
Nrx,2,av
f2m
N x,2,av
(6.263)
For equivolumetric liquid diffusion through a porous medium, we have 0 = V1Nrx,l,av +
(6.264)
V2Nrx,2,av
The total pressure gradient becomes
dP - - R
Vl) ,2
f lm -- f 2m -~2
8_.
(6.265)
and
dx
--RTN'x'l'avCPVl ( flm - f2m ~2 ) 'r28
(6.266)
6.6
351
Statistical rate theory
Using Eq. (6.259), we find equivolumetric diffusion
: -
N:,,,av
~m;i~) + C(.fim~2V2 +
(6.267) gD/D(21
7
where 05 is the volume fraction. For larger pores, the values o f fire become small and gD approaches unity, and we obtain a Fickian diffusion equation, with the thermodynamic correction estimated by
F c ij - 6ii + ci
'
6.6
.
.
.
.
OCj )ck P T
(6.268)
1-
C
-~n
STATISTICAL RATE THEORY
Onsager's reciprocal rules are valid for systems that are sufficiently close to global equilibrium. It is crucial to determine under what conditions the assumption of linearity will hold. Onsager's reciprocal rules hold if the flows and forces are independent of one another and are identified from the rate entropy production or dissipation function. Statistical rate theory may help in verifying Onsager's reciprocal rules and understanding the linearity criteria. Statistical rate theory is not based on the assumption of near equilibrium, and leads to rate equations consisting of experimental and thermodynamic variables that may be measured or controlled. Statistical rate theory is based on the local thermodynamic equilibrium. It is derived from the quantum mechanical probability that a single molecule will be transferred between phases or across an interface or that a forward chemical reaction will occur in a single reaction step. Therefore, it should be modified to apply to systems in which simultaneous multiple molecular phenomena would be significant. For a transport process or a chemical reaction process involving single molecular phenomena at some time scale, the statistical rate theory equation for the net rate of the flow J is
q[exp(/exp( 'b ASr
(6.269)
where Jeq is the equilibrium exchange rate of molecules between the phases, ASf and kSb are the entropy changes in the isolated systems as a result of a single molecule been transferred forward and backward, respectively, and k is the Boltzmann constant. In statistical rate theory, the microscopic transition rates between any two quantum mechanical states of molecular configurations that differ by a single molecule having been transferred between phases (or having undergone a chemical reaction) are equal. That means that the average of these rates does not change, and Jeq is a constant throughout the process and equal to the equilibrium exchange rate. As long as the entropy changes are large, Eq. (6.269) cannot be linearized. For example, chemical reactions and interfacial transport between two phases yield large entropy changes. Statistical rate theory leads to well-defined coefficients that can be measured or controlled, and hence the criteria for linearization may be explicitly expressed.
Example 6.9 Transport in biological cells: osmotic and pressure driven mass transport across a biological cell membrane After Elliott et al. (2000) consider a compartmental system shown in Figure 6.3. Here, a biological cell containing a dilute solute and water solution is immersed in the same solute and water solution. The cell is placed in a thermal reservoir with temperature TR. The cell exchanges the solute and water across the wall, and therefore, undergoes osmotic shrinkage or swelling. We assume that both the water and the solute are incompressible and the saturation concentration of the solute in water does not depend on pressure. The cell is in mechanical equilibrium, although the water concentration or pressure inside and outside the cell is different. The pressure difference inside and outside the cell causes and is balanced by a tension in the cell membrane. The cell and its surroundings are at constant temperature. The derivation of the transport equations starts with the formulation of the entropy production rate. A differential change of the entropy of the isolated system dSsy s is dSsy s -- dS ° + dS i nt- d S m + d S R
(6.270)
352
6.
Figure 6.3.
Diffusion
Schematic mass transport in a biological cell in a thermal reservoir.
where So, Si, Sm, and SR are the entropies of the fluid outside the cell, the fluid inside the cell, the cell membrane, and the reservoir, respectively. The differential entropy of the fluid outside the cell is 1 Po dVo dSo :FdUo +T
/Xw,od N w o _
r
~/Xs'° d N s o
T
(6.271)
'
where Uo, Po, and Vo are the internal energy, the pressure, and the volume, respectively, of the fluid outside the cell, /Xw,o and Nw,o are the chemical potential of the water and the number of moles of water outside the cell, and/Xs,o and Ns,o are the chemical potential of the solute and the number of moles of solute outside the cell. For the fluid inside the cell, we have 1 d U i + Pi d V i
dSi :--T
--T
dN w i -
~w,i
-
T
'
~s,i
T
(6.272)
dN s i
'
The subscript i indicates the properties for the fluid inside the cell. For the membrane, we have dSm : 7 d g m -
(6.273)
dam - Z tZm'k dNm,k k
T
where Urn, 7m, and Am are the internal energy, the tension, and the surface area, respectively, of the cell membrane. Here, the cell membrane is treated as a two-dimensional phase,/Xm,k is the chemical potential of the kth molecular species in the membrane, and Nm,k is the number of molecules of the kth species in the membrane. For a quasi-static heat transfer in the reservoir, we have dS R -- ---'~1 d U o _ F1d U i - - ~1d U m
(6.274)
After substituting Eqs. (6.271)-(6.274) into Eq. (6.270) and applying the following constraints d V o = - d V i, d N w ,o = - d N w , i, d N s ,o
=-dNs,i, d N m, k = 0
(6.275)
we have dS~y s _ P i - e o T
dVi
"Ym dA m +
~w,o --/£w,i
~w,o --/Zs, i dNw i + '
T
T
dNs T
'
i
(6.276)
By assuming that mechanical equilibrium holds for the membrane and that the cell is spherical with the radius r, we have Pi -
Po
-
2ym F
(6.277)
6.6
353
Statistical rate theory
Substituting Eq. (6.277) into Eq. (6.276), we obtain the rate of entropy production dSsys - / Z w ' ° -/'l'w'i '-'~"AA'lwi nt-/-/~w,o -- ~s,i ,.,l,A£fs dt
T
'
T
'
i
(6.278)
where Nw,i and Ns,i are the rates of change of the numbers of water and solute molecules inside the cell, respectively. The forces in Eq. (6.278) are related by the Gibbs-Duhem relation and are not independent. For a dilute solution, the difference in the chemical potentials of an incompressible solvent across the membrane is (6.279)
~ w , o - ]~w,i = V w ( P o - P i ) - k T ( x s , o - X s , i )
where Vwis the partial molecular volume of water, k is the Boltzmann constant, and Xs is the mole fraction of solute, which is approximately defined by x s = Cs/C w, where Cs is the concentration of the solute and cw is the concentration of pure water. For an incompressible solute with a pressure-independent saturation concentration, the difference in the chemical potentials of the solute across the membrane is ~s,o -- ~s,i -- Vs (/Do -- P i ) -
kT[ln(xs,o)-ln(xs,i)]
(6.280)
where Vs is the partial molecular volume of solute, and In (Xs, o ) - In (Xs, i ) = cs'°
_--Cs'i
(6.281)
Cs
where Cs,o m Cs,i CS z
In (Xs,o) - In (Xs,i) Substituting Eqs. (6.279)-(6.281) into Eq. (6.278), and rearranging, yields dSsys -[Vs]Vs,i--1- Vw~/w i](Po - Pi)--tdt
'
[ ]~rs,i
-
F
]Vwi] , kT(cs,o-Cs,i )
(6.282)
cw
We can identify the flows and forces from Eq. (6.282) and establish the following phenomenological equations Vs]Vs, i -4- Vw]Vw, i = L I I ( P o - Pi ) q- Ll2kT(cs,o - Cs,i)
[
/NTs i -' c
iN~w i ] ,' -- L21 (/Do -- Pi ) + L22 k T cw
(c s o
- Cs,i )
(6.283)
(6.284)
On the other hand, from statistical rate theory, we have ~rs~' - Js'eq [exp ( ASf-T]- exp ( ]]--~ASb
(6.285)
where Js,eq is the equilibrium exchange rate of solute molecules across the membrane. The forward entropy change is ASf - AS o + AS i --I-AS m -4- AS R
(6.286)
Each phase is a simple system, and we may write the appropriate Euler relations 1
PoA Vo -/Xw'---~°AN w o -/zs'° ZL/V T ' T s,o
ASo = rAC/o + T
(6.287)
354
6.
Diffusion
~si=F1 AUi + ?-Pi AVi
/'/'w,i _ i z~TVsi r ANwi, _ _/'Ls, T ' AAm - Z/zm'k ~Nm k k T
(6.289)
1AU i 1 -T-T ~um
(6.290)
AS m = -~- A Um -
ASR=
-T
1AU °
(6.288)
We formulate the ASb in a similar manner. Using Eqs. (6.286)-(6.290) and the following constraints A Vo = - A V i, ANw,o = ~/Ww,i - 0 ,
(6.291)
ANs, o - - - 1, ~Ws, i --1, L~/Wm,k - 0
in Eq. (6.285), we obtain
[ (so si/exp/ Si [/ wo wi) exp/Wiwo)]
]Vs,i - Js,eq exp
kT
]~rw,i = Jw,eq exp
kT
(6.292)
kT
(6.293)
kT
Equations (6.292) and (6.293) are the formulations of nonequilibrium thermodynamics and describe the osmotic transport of solute and water across the membrane. These equations can be linearized for small chemical potential differences, and we obtain ]~rsi
2Js'eq
]Vw,i
2Jw'eq
'-
kT
-
kT
(6.294)
(~s,o--~s,i)
(6.295)
(~w,o-~w,i)
Combining Eqs. (6.294) and (6.295) with Eqs. (6.283)-(6.286), we have 2 (Js eqVs Jw'eqVwcw) kT(cs'° -- Cs,i) Vsl~rs,i "k-VwJ~rw,i - K2 (Js,eqV? . k _ J w , e q V 2 ) ( p o _ P i ) + K Cs
Nsi c
cw
2(,seqs
kT
cs
i,+2('seq "weq)
cw
--k-f
~2Cs -+- Cw2
kT(cs,o
-- Cs,i)
(6.296)
(6.297)
Comparing these statistical rate theory equations, with Eqs. (6.283) and (6.284), we obtain the following phenomenological coefficients z~i = - f2f
(Js,eqW?+Jw,eqW2)
(6.298)
(
(6.299)
2 Vs /-'12 = k-T Js,eq c7
/
--Jw,eq-~ Cw, }
6.6
355
Statistical rate theory
L21 -- ~
Js,eq ~ -- Jw, eq Cs Cw )
2 (Js,eq Jw, eq) L22 = ~ k C2 -Jr- C; 2
(6.300)
(6.301)
Equations (6.299) and (6.300) show that Onsager's reciprocal rules hold. The Js,eq and Jw, eq have a microscopic definition represented by perturbation matrix elements and a macroscopic definition represented by the equilibrium exchange rate. As long as the criteria of linearization are satisfied, the statistical rate theory may be used to describe systems with temperature differences at an interface besides the driving forces of pressure and concentration differences.
6.6.1
Diffusion in Inhomogeneous and Anisotropic Media
Macroscopic diffusion model is based on underlying microscopic dynamics and should reflect the microscopic properties of the diffusion process. A single diffusion equation with a constant diffusion coefficient may not represent inhomogeneous and anisotropic diffusion in macro and micro scales. The diffusion equation from the continuity equation yields OP
Ot
- -v.a
(6.302)
where P and J are the density (probability or number) and diffusion flow of the particles. Following Christensen and Pedersen (2003), a definition for the diffusion flow J is a - -(P~2v v + fivP)
(6.303)
where V is an external potential,/2 is the mobility, and I) is the diffusion tensor given by the Einstein relation [zkT = l}
(6.304)
In Eq. (6.303), the first term represents the drift in the potential force field Vand the second is the diffusional drift given by Fick's law. Combining Eq. (6.302) with Eq. (6.303), we have OP
- V.{P/2V V + I}VP)
(6.305)
Since Eq. (6.305) cannot represent systems with inhomogeneous temperatures, we may have the following alternative equation OP Ot
- v . {Pt2v~ ~ + v. tiP) = v . ( P ( ~ 2 v v + v . f i ) + ISVP)
(6.306)
Equations (6.305) and (6.306) are different because of the drift term V.(PV. 1}), which is sometimes called a "spurious" drift term. These diffusion equations have different equilibrium distributions and are two special cases of a more general diffusion equation.
6.6.2
van Kampen's Hopping Model for Diffusion
The hopping model was originally introduced to discuss electron transport in solid materials, but it may be useful as a general model for diffusive motion~ In a one-dimensional diffusion equation based on hopping model, the diffusion medium is modeled by a large number of wells/traps in which the particles can get temporarily caught. The density of traps (o-) is the density times the cross-section of traps and may change throughout the media. In solvents, for example, the density represents the capability of the solvent molecules to form a cage around the suspended particle. The rate of escape of particles (a) is controiled by the local energy barrier • of the trap and the local temperature T
356
6.
Diffusion
a = a exp -~-T
(6.307)
Here, a defines the global time scale for escape out of the traps, and incorporates the spatial variation of the escape a into the potential barrier ~. Large values of o~ signifies shallow wells and hence fast diffusion, while large values of o- signifies small mean free paths and hence slow diffusion. Inhomogeneities in the medium may cause spatial dependencies of c~ and o-, such as in micelles, or by the interaction of two diffusing molecules. The isotropic diffusion equation based on van Kampen's one-dimensional hopping model may be extended to three dimensions using Cartesian coordinates in flat Euclidean space
OP_av.[exp(-~/kT ) (Vo- VV) (exp(-~/kT) 0.2 P o- + kT +V Ot o-2
P
)]
(6.308)
Equation (6.308) implies that the isotropic diffusive motion along the coordinate axes is independent. Here, 7V/KT is the drift due to an external potential force field V, while 7o-/o-represents an internal drift caused by a concentration gradient of the traps. The term PV(e-*/XT/O-2) is the "spurious" drift term. Equation (6.308) allows spatial variations of all parameters T, V, ~, and o- with inhomogeneous temperature. From Eq. (6.308), the diffusion coefficient becomes D= a
exp ( - • / kT)
(6.309)
o -2
The stationary solution of Eq. (6.308) for systems with a uniform temperature is
Ps = C o-exp(-V/kT) exp(-~/kT)
(6.310)
where C is the normalization constant. The stationary distribution depends on the local value of the macroscopic diffusion coefficient D and on the local value of one of the microscopic trap parameters o- or ~. Consider three special cases based on a simplification of Eq. (6.308): 1. o- oc exp (-~/KT). For this case, Eq. (6.308) becomes
OP Ot
(6.311)
This is the traditional diffusion model given in Eq. (6.305) with the diffusion coefficient D proportional to l/o-. For this case, the so-called "spurious drift" term vanishes because the effects of c~ and o- cancel each other out in the stationary state. The stationary distribution is proportional to the Boltzmann distribution exp(-V/kT) and independent of D. 2. o- = constant. Then, Eq. (6.308) becomes
OP Ot
+ ~
D
+ DVP
which is similar to the relation given in Eq. (6.306). The stationary solution is proportional to example, the particles would experience very slow diffusion in regions of low mobility. 3. • = constant. Internal drift does not vanish, and Eq. (6.308) becomes
OP - v'[DP ( VV kT + -~2FD ) +
(6.312)
exp(-V/kT)/D, for
(6.313)
which is different from both Eqs. (6.305) and (6.306). The stationary solution is V]x/-D Ps =exp -~--~
(6.314)
For isotropic systems, the diffusion equations for these three cases are mathematically equivalent since they can be transformed into each other by introducing effective potentials.
6.6
Statisttcal rate theory
357
Equation (6.305) has been used widely to model diffusion in liquids, but the above discussion shows that it is valid only where c~ ,x o-. Equation (6~306) is valid when the concentration of traps is constant, a situation that is more realistic. In all other cases, the diffusion equation is a combination of Eqs. (6.305) and (6.306).
6.6.3
Anisotropic Diffusion
The general diffusion equation, based on the hopping model, is
OP--g('cP)+av'!ex--p-P(-OP/kT)p{gCrat ,_ (re ---~- ~kTF}+ g . exp(-dP/kT)o 2 P~
(6.315)
where F is an external fbrce and ~ is the velocity held of the medium. If we assume that the parameter cr is isotropic while the trap potential is anisotropic and represented by the tensor D -- o
exp(-~/kT)
(6.316)
o"
The tensor + is required to be symmetric because of its relation with the diffusion tensor. Of course, the o- can also be anisotropic. The above equation may cover most phy~;ical systems, and it can even be used on curved manifolds. The anisotropy introduces two new f;eamres: (i) equations (6.305) and (6.306) cannot in general be transformed into each other, as the drift term g. 1) may not be a gradient field. Equation (6.306) can describe systems where the directions of the principal axes depend on the spatial position. (ii) Detailed balance implies that the diffusion flow a vanishes everywhere in the stationary state. However. this is not automatically satisfied for anisotropic systems and one needs to exercise extra care in the modeling of such systems. Inhomogeneity does not affect the detailed balance. (ii!) The diffusive part of the diffusion flow must be represented by a - - V l i ) P , while the drift is represented by (PV. D). In general, the difRlsion equation depends on all the microscopic parameters. The microscopic parameters of van Kampen's model are the local values of the effect!re trap density or, which is density times cross-section and work function • . The traditional dift~asion relation of Eqo (6.305) is valid only for isotropic diffusion and under the restrictive conditions that o-~cexp(-~/kT)~ It may be unsatisfactory even in a homogeneous system with nontrivial geometry. Equation (.6.306) is valid when the effective trap concentration is constant, which is more realistic fbr liquids.
6.6.4
Diffusion in Biological Solutes in Liquids
The diffusion of small molecules and macromolecules (e.g., proteins) in aqueous solutions plays an important role in microorganisms, plants, and animals. Diffusion is also a major part in food processing and in the drying of liquid mixtures and solutions, such as diflk~sing aroma constituents in fruit juice, coffee, and tea from solutions during evaporations. In fermentation, nutrients, oxygen, and sugar diffuse to the microorganisms, and products, waste, and sometimes enzymes diffuse away. The kidneys ~emove waste products like urea, creatinine, and excess fluid from the blood. Kidney dialysis removes ,,~aste products from the blood of patients with improperly working kidneys. During the hemodialysis process, the patient ~sblood is pumped through a dialyzer, and waste diffuses through a semipermeable membrane to the aqueous SOJtltlOl~l cleaning i!uid Macromolecules have large molecular weights and various random shapes that may be coil-like, rod-like, or globular (spheres or cllipsoids~ They l:b,~m t~t~e solutio>;. ]-~heir sizes and shapes affect their diffusion in solutions. Besides that, interactions of large molecules w|th the smal! solvent and/or solute molecules affect the diffusion of macromolecules and smaller molecules. Sometimes, reacts.on-diffusion systems may lead to facilitated and active transport of solutes and ions in biological systems. These type> ot'transport will be discussed in Chapter 9. Macromolecules often have a number ot sites i'or interactions and binding of the solute or ligand molecules. For example, hemoglobin in the blood binds oxygen at certain sites. Surface charges on the molecules also affect the diffusion. Therefore, the presence of macromolecules and small solute molecules in solutions may affect Fickian-type diffusion. Most of the experimental data on protein diffusivities have been extrapolated to very dilute or zero concentration since the diffusivity is often a function of concentration. Table 6.4 shows diffusivities of some proteins and small solutes in aqueous solutions. The diffusion coefficients for the macromolecules of proteins are on the order of magnitude of 5 × 10 l i me/s. For small solute molecuies, the diffusivities are around 1 × 10-9 m2/s. Thus, macromolecules difi\ise about 20 times slower then small moiec~ies. Small soi~f:cs such as urea and sodium caprytate ofien coexist with protein lnacromolecules in solutions. When these small molecules d~ffuse the i~rotein soitmc,n~ the diffusivity of the molecules decreases with increasing protein
358
6.
Diffusion
Table 6.4 Diffusivities of dilute biological solutes in aqueous solutions Solute
Molecular weight
Sucrose
Temperature (K)
342.3
Urea
310 293 293 298 293 298 310 298 298 298 293 293 293 293 293 293
60.1
Glycerol Glycine Creatinine Sodium caprylate Bovine serum albumin Urease
92.1 75.1 113.1 166.2 67500 482700
Soybean protein Lipoxidase Fibrinogen, human Human serum albumin y-Globulin, human
361800 97440 339700 72300 153100
Diffusivity (m2/s) 0.697 x 0.460 x 1.20 x 1.378 x 0.825 x 1.055 x 1.08 x 8.78 × 6.81 x 4.01 x 3.46 x 2.91 x 5.59 x 1.98 X 5.93 × 4.00 x
10- 9 10-9 10.-9 10 9 10-9 10-9 10-9 10-l° 10-11 10-ll 10-ll 10 ~1 10--11 10 1~ 10-al 10-11
Source: C.J. Geankoplis, Transport Processes and Separation Process Principles, 4th ed., Prentice Hall, Upper Saddle River (2003).
concentrations. This reduction is partly because of the binding of small molecules to proteins and is partly due to blockage by the large molecules.
6.6.5
Prediction of Diffusivities of Biological Solutes
For predicting the diffusivity of small molecules (with molecular weights less than about 1000 or molar volumes less than about 0.500 m3/kg) in aqueous solution, we may use the Wilke-Chang correlation to estimate the diffusivity in mZ/s 1.173 x 10-16 (~tMB)1/2 T DAB = ~B VO'6
(6.317)
where ~ is an association parameter of the solvent (~ is 2.6 for water, 1.9 for methanol, 1.5 for ethanol, 1.0 for benzene, ether, and n-heptane, and 1.0 for other unassociated solvents). MB and/ZB are the molecular weight and viscosity of solvent B in Pa s or kg/m s. VA is the molar volume of the solute at the boiling point, which may be obtained from Table 2.10 in Chapter 2. This relation should be used with caution outside temperature ranges of 278-313 K. For larger molecules, the equations for diffusivity estimations may not be too accurate. As an approximation, the Stokes-Einstein equation can be used 9.96 X 10-16T DAB =
~ V 1/3
(6.318)
where/x is the viscosity of the solution, and VA is the molar volume of the molecule. For a molecular weight above 1000, the following equation may be used 9.40 x 10-15T DAB----
~(MA)I/3
(6.319)
where M A is the molecular weight of the large molecule A. When the shape of the molecule deviates greatly from a regular spherical shape, this equation should be used with caution. During the diffusion of small molecules (with molecular weights less than about 1000 g/mol or molar volumes less than about 0.500 m3/kg) in protein solution, the diffusion may be blocked by the large molecules. In order to account for this effect, we need the diffusivity DAB of solute A in water alone, the water of hydration on the protein, and an obstruction factor. A semitheoretical relation to approximate the diffusivity of solute in a globular-type protein solution is DAp -- DAB (I -- I .8 I X l 0 -3 Cp)
(6.320)
6.6
359
Statistical rate theory
where Cp is concentration of P in kg/m 3. The approximate diffusion equation is
Nx=
OAp (CA1--CA2) z2
-
-
(6.321)
z1
When, however, a solute molecule A is bound to a protein, the diffusion flux of A is equal to the flux of unbound solute A and the flux of the bound protein-solute complex. This type of flux estimation requires data on binding. The equation used is
D A p - [DA~ (1 - 181 x l 0-3 CP)( free A%/100 + Dp (bound A%)]100
(6.322)
where Dp is the diffusivity of protein alone in the solution m m2/s. The percentage of free A can be determined from the experimental binding coefficient.
Example 6.10 Prediction of diffusion coefficients of macromolecules Predict the diffusivity of human serum albumin at 293 K in water as a dilute solution and compare with the experimental data in Table 6.4. Table 6.4 shows the molecular weight of human serum albumin A as MA = 72300 kg/kg tool. The viscosity of water at 298 K is 0.897 X 10.3 Pa s. Using Eq. (6.319)
DAB =
9.40 × 10-15 T /~(MA ) 1/3
=
9.40 × 10-15 (293) (0.897 × 10-3)(72300) 1/3
= 7.37×10 -11 m2/s
This value is 24% higher than the experimental value of 5.93 × 10-11 m2/s.
6.6.6
Diffusion in Biological Gels
Gels are semisolid and porous materials. Some typical gels are agarose, agar, and gelatin. Also, a number of organic polymers exist as gels in various types of solutions. They are composed ofmacromolecules. For example, the gel structure of agarose is loosely interwoven, and is composed extensively of hydrogen-bonded polysaccharide macromolecules. The pores that are open spaces in the gel structure are filled with water. The rates of diffusion of small molecules in the gels are usually less than in aqueous solutions. When there are no electrical effects, the gel structure mainly increases the path length for diffusion. Table 6.5 shows a few typical values of the diffusivity of some solutes in various gels. In some cases, the diffusivity of the solute molecule in pure water (wt% = 0) is given in Table 6.4. This shows how much the diffusivity decreases due to the gel structure. For example, at 293 K, Table 6.4 shows that the diffusivity of sucrose in water is 0.460 ;4 10.9 m2/s, while it is 0.107 × 10-9 m2/s in 5.1 wt% gelatin. This indicates a considerable decrease of 77%.
Example 6.11 Diffusion of solutes in biological gels A 0.02-m-long tube of a gel solution connects two chambers of agitated solutions of dextrose in water. The gel solution is 0.79 wt% agar in water and is at 278 K. The dextrose concentration in the first chamber is 0.4 g mol dextrose per liter solution and the other chamber concentration is 0.01 g mol dextrose per liter solution. Estimate the flow of dextrose in kg mol/s m 2 at steady state. Solution" Assume that the system undergoes steady state and one-dimensional diffusion. From Table 6.5, we read the diffusivity coefficient for solute dextrose at 278 K as DAB = 0.327 × 10.9 mZ/cm. The concentrations are
Cal--
0.4 1000
- 0.0004 g mol/cm 3 = 0.4 kg mol/m 3
360
6.
Diffusion
Table 6.5 Typical diffusivities of solutes in dilute gels in aqueous solutions Solute Sucrose
Urea
Methanol Urea
Gel
Wt% gel in solution
Gelatin Gelatin Gelatin Gelatin Gelatin Gelatin Gelatin Gelatin Gelatin Gelatin
Temperature (K)
Diffusivity (m2/s)
0
278
0.285 x 10 9
3.8
278
0.209 × 10 -9
5.1
293
0.107 x 10 -9
10.35
278
0.252 × 10 9
0 2.9 5.1 5.1 10.0
0.880 0.644 0.609 0.542 0.859
3.8
278
X X X X x
10 9 10 9 10 9 10-9 10 9
0.626 x I0 9
Agar
278
0.727 × 10 9
Agar Agar
278 278
0.591 × 10 -'~ 0.472 × 10 9
Agar
278
0.297 x 10-9
Agar
278
0.199 × 10
Agar
278
0.297 × 10
Dextrose Sucrose Ethanol
Agar
278
0.327 × 10
Agar
NaC1 (0.05 M)
Agarose Agarose
278 278 278 298
0.247 0.393 1.511 1.398
Glycerin
Agar
× x x x
10 10 10 -9 10 ...9
298
Source: C.J. Geankoplis, Transport Processes a n d Separation Process Principles, 4th ed., Prentice Hall, Upper Saddle River (2003).
(:12-
0.01
- 0.00001 g mol/cm 3 = 0.01 kg mol/m 3
1000
Since the urea concentration is very low, for dextrose diffusing through stagnant water in the gel, we may use NA~=--DAB
d f Az ..... _ D A B A C
dz
--
~
_
Az
_
DAB(CA1 --CA2) 2 2 -- Z 1
0.327×10-9(0.4-0.01) 0.02-0 NAz = 6.377 × 10-9 kg mol/s.
PROBLEMS 6.1
Derive modeling equations for diffusion through a stagnant phase.
6.2
Derive modeling equations for diffusion into a falling liquid film.
6.3
Two large vessels containing binary mixtures of gases A and B are connected by a truncated conical duct, which is 2 ft in length and has internal diameters of larger and smaller ends of 8 and 4 in, respectively. One vessel contains 80 mol% A, and the other 30 mol% A. The pressure is 1 atm and the temperature is 32°E The diffusivity under these conditions is 0.702 ft2/h. By disregarding the convection effects: (a) Calculate the rate of transfer of A. (b) Compare the results with those that would be obtained if the conical duct was replaced with a circular duct with a diameter of 6 in.
361
Problems
6.4
Oxygen in muscles is used for the oxidative removal of lactic acid. One theory suggests that a slab of muscle in contact with oxygen will possess a recovered oxygen zone at the muscle-oxygen interface, and an unrecovered lactic acid zone. The interfacial boundary between the two zones will advance with time into the lactic acid zone. Consider semi-infinite muscle tissue region x > 0 through which oxygen diffuses. At the external boundary x = 0, the oxygen concentration is Co - constant. The oxygen zone is 0 < x < L, where the boundary L between the oxygen zone and the lactic acid zone depends on time t, and L(0) = 0. The boundary conditions are C = C(O,t) = Co and C(L,t) = 0. The velocity of the advancing front dL/dt is assumed to be proportional to the oxygen flux at L
Ox
=L
dt
where A is a constant of proportionality. (a) When the front advances slowly, and in the oxygen zone, the concentration C satisfies the steady-state diffusion, find the velocity of the advancing f~ont. (b) Find the velocity if the oxygen diffusion is not assumed to be in a quasi-steady state. 6.5
Predict the diffusivity of lipoxidase at 293 K in water as a dilute solution and compare with the experimental data in Table 6.4.
6.6
Predict the diffusivity of soybean protein at 293 K in water as a dilute solution and compare with the experimental data in Table 6.4.
6.7
Estimate the diffusion coefficient of hemoglobin in water at 293 K. The globular hemoglobin molecule has a radius r ~ 30 A and the Boltzmann constant k = 1.38 × 10-16 erg/K. The viscosity is 0.01 E
6.8
A 0.02-m-long tube of a gel solution connects two chambers of agitated solutions of dextrose in water. The gel solution is 0.79 wt% agar in water and is at 278 K. The dextrose concentration in the first chamber is 0.4 g mol dextrose per liter solution and the other chamber concentration is 0.01 g tool dextrose per liter solution. Estimate the flow of dextrose in kg mol/s m 2 at steady state.
6.9
A 0.05-m-long tube of a gel solution connects two chambers of agitated solutions of urea in water. The gel solution is 1.05 wt% agar in water and is at 278 K. If the urea concentration in the first chamber is 0.25 g mol urea per liter solution and the other chamber concentration is zero. Estimate the flow of urea in kg mol/s m 2 at steady state.
6.10
Consider a thin rectangular region in a catalyst particle shown below. Component A diffuses across the top surface. After reaching one of the other three surfaces, component A undergoes an instantaneous reaction. Therefore, the concentration of A at the three surfaces of the region will be zero as shown in the figure below. There will be no net bulk motion within the region of catalyst. Derive the Laplace equation for a two-dimensional model of a catalyst passage. Use the separation of variables technique to obtain the exact solution.
CA= C(x)
CA =0
CA= 0
0
6.11
CA=O W
X
Consider the drying of a large sheet of wood with a uniform thickness of z. For a one-dimensional diffusion problem, the initial concentration profile through the wood will be a function of z. Develop an analytical method to describe the concentration profile of water moisture within the wood.
362 6.12
6.
Diffusion
Consider the absorption of oxygen from air in the aeration of a lake or the solid surface diffusion in the hardening of mild steel in a carburizing atmosphere. Both these processes involve diffusion in a semi-infinite medium. Assume that a semi-infinite medium has a uniform initial concentration of CAo and is subjected to a constant surface concentration of CAs. Derive the equation for the concentration profiles for a preheated piece of mild steel with an initial concentration of 0.02 wt% carbon. This mild steel is subjected to a carburizing atmosphere for 2 h, and the surface concentration of carbon is 0.7%. If the diffusivity of carbon through the steel is 1 × 10-11 m2/s at the process temperature and pressure, estimate the carbon composition at 0.05 cm below the surface.
REFERENCES R.B. Bird, W.E. Stewart and E.N. Lightfoot, Transport Phenomena, 2nd ed., Wiley, New York (2002). M. Christensen and J.B. Pedersen, J. Chem. Phys., 119 (2003) 5171. C.E Curtiss and R.B. Bird, Ind. Eng. Chem. Res., 38 (1999) 2515. E.L. Cussler, R. Aris andA. Bhown, J. Memb. Sci., 43 (1989) 149. J.A. Daoud, S.A. E1-Reefy and H.E Aly, Sep. Sci. Tech., 33 (1998) 537. Y. Demirel, Chim. Acta Turcica, 14 (1986) 114. Y. Demirel and S.I. Sandler, Int. J. Heat Mass Transfer, 44 (2001) 2439. EA.L. Dullien and A.EA. Asfour, Ind. Eng. Chem. Fundam., 24 (1985) 1. J.B. Duncan and H.L. Toor, AIChE J., 8 (1962) 38. J.A.W. Elliott, H.Y. Elmoazzen and L.E. McGann, J. Chem. Phys., 113 (2000) 6573. A. Katchalsky and EE Curran, Nonequilibrium Thermodynamics in Biophysics, Harvard University Press, Cambridge (1967). B.C. Eu, Kinetic Theory and Irreversible Thermodynamics, John Wiley, New York (1992). P.J.A.M. Kerkhof and M.A.M. Geboers, Chem. Eng. Sci., 60 (2005) 3129. D. Kondepudi and I. Prigogine, Modern Thermodynamics, From Heat Engines to Dissipative Structures, Wiley, New York, 1999. R. Krishna and G.L. Standard, AIChE J., 22 (1976) 383. J. Leffier and H.T. Cullinan, Ind. Eng. Chem. Fundam., 9 (1970) 84. R. Taylor, Ind. Eng. Chem. Fundam., 21 (1982) 407. R. Taylor and R. Krishna, Multicomponent Mass Transfer, Wiley, New York (1993). M.T. Tyn and W.E Calus, J. Chem. Eng. Data, 20 (1975) 310. A. Vignes, Ind. Eng. Chem. Fundam., 5 (1966) 189.
REFERENCES FOR FURTHER READING V. Alopaeus, Comp. Chem. Eng., 26 (2002) 461. A. Eftekhari, Chem. Phys. Lett., 374 (2003) 164. R. Krishna and J.A. Wesselingh, Chem. Eng. Sci., 52 (1997) 861. O.O. Medvedev and A.A. Shapiro, Fluid Phase Equilib., 208 (2003) 291. L. Minkin, Radiat. Prot. Dosimetry, 106 (2003) 267. G.L.J.A. Rikken and P. Wyder, Phys. Rev. Lett., 94 (2005) 016601. M.O. Vlad, J. Ross and EW. Scheneider, Phys. Rev. E, 62 (2000) 1743.
7 HEAT AND MASS TRANSFER 7.1
INTRODUCTION
Simultaneous heat and mass transfer plays an important role in various physical, chemical, and biological processes; hence, a vast amount of published research is available in the literature. Heat and mass transfer occurs in absorption, distillation extraction, drying, melting and crystallization, evaporation, and condensation. Mass flow due to the temperature gradient is known as the thermal diffusion or Soret effect. Heat flow due to the isothermal chemical potential gradient is known as the diffusion thermoeffect or the Dufour effect. The Dufour effect is characterized by the heat of transport, which represents the heat flow due to the diffusion of component i under isothermal conditions. Soret effect and Dufour effect represent the coupled phenomena between the vectorial flows of heat and mass. Since many chemical reactions within a biological cell produce or consume heat, local temperature gradients may contribute in the transport of materials across biomembranes. Various formulations and methodologies have been suggested for describing combined heat and mass transfer problems, such as the integral transform technique, in the development of general solutions. In this chapter, cross phenomena or coupled heat and mass transfer are discussed using the linear nonequilibrium thermodynamics theory. 7.2
COUPLED HEAT AND MASS T R A N S F E R
Using a dissipation function or entropy production equation, the conjugate flows and forces are identified and used in the phenomenological equations for simultaneous heat and mass transfer. Consider the heat and diffusion flows in a fluid at mechanical equilibrium not undergoing a chemical reaction. The dissipation function for such a system is
- -Jq' V In T -
~_~ Ji "aik
i.k=l
/=1
_>0
I Ow/
(7.1)
T./'.wi~ I
where aik = gik + w~/w,, 8~k is the unit tensor, and j; is the diffusion flow of component i. The heat flows are related through the internal energy flow J,,
Jr,-- J// q- £ hiJi - J; + £HiJi i=1
Jq and Jq !
tt
(7.2)
i--1
Similarly, the entropy flow is expressed by
J; £
J,,---T-+
sij i
(7.3)
i=1
where s; is the partial specific entropy and h; and u; are the partial specific enthalpy and partial specific internal energy, respectively. 7.2.1
Binary Systems
The independent forces and flows are identified by the dissipation function of Eq. (7.1). Therefore, the forces and heat and mass flows for a binary system are
364
7.
Heat and mass transfer
Xq = - V l n T
(7.4)
Xl = - 1--~ ~-------~1 0
VW1
(7.5)
w2 OWl T,P The linear phenomenological equations describe the flows (7.6)
-Jl = LlqVlnT +Lll w2 ~,Ow1)r,P
1(
0~1 ]
-Jq = LqqV lnT + Lql W2 OWl)T,P
VW1
(7.7)
By the Onsager reciprocal relations, the matrix ofphenomenological coefficients is symmetric, Llq -- Lql. Since the dissipation function is positive, the phenomenological coefficients must satisfy the inequalities
Lqq > O, Lll > 0,
LqqL11-L2ql > 0
(7.8)
Fourier's law describes heat conduction caused only by the temperature gradient
Jq
-kVT
(7.9)
where k is the thermal conductivity in the absence of a concentration gradient. Comparison of Eqs. (7.7) and (7.9) yields the relationship between the phenomenological coefficient Lqq and the thermal conductivity coefficient
Lqq = kT
(7.10)
Fick's law describes the diffusion flow caused only by the concentration gradient for an isothermal fluid Jl -- -pDVwl
(7.11)
which contains the diffusion coefficient D given by
D=Dll--Lll
1 ( 0/xl )
(7.12)
Pw2 ~,OWl T,P
The diffusion caused only by the temperature gradient is called the thermal diffusion (Soret effect). When the concentration gradient vanishes, Eq. (7.6) reduces to
Jl -- -Llq V In T
=
- p DT1VT
(7.13)
where DT1 is the thermal diffusion coefficient of component 1, and is related to the cross coefficient Llq by
Llq
DT1 --
(7.14)
P The Soret effect D~ may be defined by
D~ - DT1
-
-
Llq
(7.15)
WlW2T PWlW2T The thermal diffusion coefficient is usually smaller by a factor of 102-10 3 than the ordinary diffusion coefficient for nonelectrolytes and gases.
Z2
365
Coupled heat and mass transfer
The heat flow due to the Dufour effect arises only from a concentration gradient, and is expressed by
_, Jq - - L q l
( Otxl) TD('Vw 1
w,
Ow1 r,P VWl -- --11 ~,OW1 T,P
(7.16)
where D~ is the Dufour effect for component 1, and is related to the phenomenological coefficient Lql by Lql
D(, =
(7.17)
pwlw2 T
and Onsager's reciprocal relations yield (7.18)
D('-D(
So, Eqs. (7.6) and (7.7) can be expressed in terms of the transport coefficients of thermal conductivity k and diffusivity D -Jl
- J2 -
P(WlW2D(VT + D V W l )
(7.19)
p(W2WlD2 V T - DVwl)
(7.20)
or, as the sum of mass fractions is unity --J2 --
Jl
-
- J q - k V T + Pl ~, Ow 1 )
T,P
TD(' Vw I
(7.21)
Equations (7.19) and (7.20) suggest that D( - - D ;
(7.22)
The thermal diffusion ratio KT1 of component 1 is defined by
KT1-- D where ST1 is called the Soret coefficient for component 1, and is given by
D(]_ STI-
7
KT1 wlw2 T
L,, )
k aw, )j
(7.24)
Table 7.1 shows some experimental values of thermal diffusion ratios for liquids and gases at low density and pressure. IfKT1 is positive, component 1 diffuses to a cooler region; otherwise, it diffuses to a hotter region. The thermal diffusion .factor c~1 for component 1 is mainly independent of concentration for gases, and is given by
~l - T --5 - Ts T
(7.25)
The inequalities in Eq. (7.8) can now be written in terms of the transport coefficients of the thermal conductivity and mass diffusivity by using the thermodynamic stability condition (Otz~/Owl)r,p >- 0 k > 0 , D > O , (D{)2
0, the flow of a species may drag another species in the same direction; however, it may push the other species in the opposite direction if q < 0. For heat and mass flows, for example, the two limiting values of q are + 1 and - 1 . Since the degree of coupling is directly proportional to the product Q~(D/k) 1/2, the error level of the predictions of q is mainly related to the reported error levels of Q~ values. The polynomial fits to the thermal conductivity, mass diffusivity, and heat of transport for the alkanes in chloroform and in carbon tetrachloride are given in Tables C 1-C6 in Appendix C. The thermal conductivity for the hexane-carbon tetrachloride mixture has been predicted by the local composition model NRTL. The various activity coefficient models with the data given in DECHEMA series may be used to estimate the thermodynamic factors. However, it should be noted that the thermodynamic factors obtained from various molecular models as well as from two sets of parameters of the same model might be different.
374
7.
Heat and mass transfer
Table 7.3. Thermal conductivities of binary mixtures as a function of mole fraction of selected alkanes in chloroform at 30°C and 1 atm a Solute(I)
xl
k (mW/(m K))
n-Hexane
0.0 0.2572 0.4801 0.5808 0.6751 0.8471 1.0
110.79 104.28 104.90 105.51 107.25 112.06 115.19
n-Heptane
0.0 0.2295 0.4427 0.5437 0.6412 0.8266 1.0
110.79 105.50 106.68 108.75 110.62 115.55 120.41
n-Octane
0.0 0.2071 0.4106 0.5110 0.6105 0.8070 1.0
110.79 106.79 108.74 110.60 113.71 117.92 123.81
3-Methylpentane
0.0 0.2572 0.4801 0.5808 0.6751 0.8471 1.0
110.79 99.66 97.83 99.40 100.36 104.01 106.94
2,3-Dimethylpentane
0.0 0.2295 0.4427 0.5437 0.6412 0.8266 1.0
110.79 103.99 101.32 100.93 102.11 103.04 105.26
2,2,4-Trimethylpentane
0.0 0.2071 0.4106 0.5110 0.6105 0.8070 1.0
110.79 102.77 97.27 96.21 95.67 95.69 96.02
aRowley et al. (1988).
The degree of coupling and the thermal diffusion ratio KT1for the liquid mixtures are calculated from Eqs. (7.76) and (7.77), and shown in Figures 7.1 and 7.2. The liquid mixtures consist of six to eight carbon alkanes of n-hexane, n-heptane, n-octane, 3-methylpentane, 2,3-dimethylpentane, and 2,2,4-trimethylpentane in chloroform and in carbon tetrachloride. These systems represent straight and branched chains of the alkanes in two solvents. As the degree of coupling and the thermal diffusion ratio depend on the heat and mass transfer coefficients, the plots of q and KT1 versus the alkane compositions Xl show the combined effects of the transport coefficients on q and KT1 in various solvents. The figures reveal some important properties of coupling. The first is that the absolute values of q and KT1 reach peak values at a certain concentration of the alkane, and these peak values decrease as the molecular weight increases. Second, the solute concentrations at the peak values of coupling decrease gradually as the molecular weights increase. The third is that the behavior of alkanes is similar up to a certain concentration of solute depending
7.5
375
Coupling in liquid mixtures
in chloroform . . . . . . . . . . . .
0
= . . . .
.
. -
.
.
.
.
.
.
.
.
. . - _
0
-0.01
-0.01
-0.02
-0.02
q -0.03
-0.03
-0.04
-0.04
-0.05
\,..
-- , , " /
-0.05 0
0,2
0.4
0,6
(a)
0.8
!
0
0.2
0.4
0.6
0.8
1
(b)
XI
in carbon tetrlchloride 0
0 /
-0.01
-0.0I
k,
-0,02 q -0.03
/
-0.02
,:,: //
q -0.03
"- -_
~ S / / /
-0.04 -0.04
-0.05 0
0.2
0,4
(c)
0,6
0,8
i
0
0.2
0.4
0.6
(d)
X!
0.8
t
x,
Figure 7.1. Change of the degree of coupling q with the alkane concentrations xl at 30°C and ambient pressure: (a) and (c) straight chain alkanes, (--) n-hexane, (---) n-heptane, (---) n-octane; (b) and (d) branched-chain alkanes, (--) 3-methylpentane, (---) 2,2dimethylpentane, (---) 2,2,4-trimethylpentane. Reprinted with the permission from Elsevier, Y. Demirel and S.I. Sandier, Int. J. Heat Mass Transfer, 43 (2002) 75. in chloroform 0
0
-0,I
-0,1
-0.2
-0.2
-'--0,3
-0.4
-0.4
\%.
-0,5
,,'/
",,,L
-0,3
-'0.5
-0.6
-O.6 0
0.2
0,4
(a)
0,6
0.8
1
0
0,2
0.4
0.6
(b)
Xl
0.8
1
xj
in carbon t e t r a c h l o r i d e 0
0
-0.1
-0,I
-0,2
.
.
.
.
.
.
.
.
, .
.
.
.
.
: .
.
.
.
.
.
.
.
.
.
..-......._
_
_
.
.
.
_
-0,2
'\
~" -0,3
/2
i" -0,3
-0.4
-0.4
-0.5
-0.5 0
(a)
.
0.2
0.4
0.6 Xt
0.8
1
0
(b)
0.2
=
0.4
0.6
0.8
1
x,
Figure 7.2. Change of the thermal diffusion ratio KT1 with the alkane concentrations xl at 30°C and ambient pressure: (a) straight chain alkanes, (--) n-hexane, (---), n-heptane, (---) n-octane (b) branched-chain alkanes, (--) 3-methylpentane, (---) 2,2dimethylpentane, (---) 2,2,4-trimethylpentane. Reprinted with the permission from Elsevier, Y. Demirel and S.I. Sandier, Int. J. Heat Mass Transfer, 43 (2002) 75.
upon the combined effects of branching and the solvent on q and KTI (through approximately xl = 0.2), but at higher concentrations they behave differently. The fourth is that the absolute maximum extent of coupling is small, as expected, and the branching of alkanes has only marginal effects on the coupling phenomena. Tables 7.4 and 7.5 show the peak values of q and KT1 for both the straight and branched alkanes separately in both the solvents chloroform and carbon tetrachloride. The general trend is that the branching of the solute molecule has a minimal effect on the coupling for the considered alkanes. Tables 7.4 and 7.5 also show the effect of the solvent on q and KT1. The alkane concentrations where the peaks of q and KT1 occur are lower in chloroform than in carbon
376
7
Heat and mass transfer
Table 7.4. Degree of coupling q, and maximum ratio of dissipation ('r/,~)maxa
Straight-chain alkanes Solute
- qmax
Branched-chain alkanes
x~
(~TX)max
Solute
- qmax
x~
(XlO 4) (a) For alkanes in chloroform n-Hexane 0.056 n-Heptane 0.046 n-Octane 0.046
( 'r/)k)max
(XlO 4)
0.534 0.408 0.337
7.993 5.365 5.295
3-Methylpentane 2,3-Dimethylpentane 2,2,4- Trimethylpentane
0.053 0.048 0.046
0.475 0.392 0.340
7.085 5.815 5.226
(b) For alkanes in carbon tetrachloride n-Hexane 0.048 0.588 n-Heptane 0.045 0.527 n-Octane 0.041 0.341
5.718 5.203 4.185
3-Methylpentane 2,3-Dimethylpentane 2,2,4-Trimethylpentane
0.055 0.040 0.038
0.570 0.467 0.513
7.574 4.083 3.574
aDemirel and Sandler (2002).
Table 7.5. Thermal diffusion ratio for solute KT1a
Straight-chain alkanes Solute
Branched-chain alkanes
--KTj (l/K)
x~
(a) For alkanes in chloroform n-Hexane n-Heptane n-Octane
0.679 0.565 0.675
0.569 0.488 0.484
(b) For alkanes in carbon tetrachloride n-Hexane n-Heptane n-Octane
0.564 0.572 0.503
0.629 0.628 0.441
Solute
--KT1 (l/K)
x~
3-Methylpentane 2,3-Dimethylpentane 2,2,4-Trimethylpentane
0.596 0.591 0.622
0.527 0.449 0.402
3-Methylpentane 2,3-Dimethylpentane 2,2,4-Trimethylpentane
0.579 0.476 0.503
0.569 0.550 0.557
aDemirel and Sandler (2002).
tetrachloride. Generally, the peak values qmaxare also smaller in carbon tetrachloride than in chloroform. Consequently, it appears that concentration affects the degree of coupling in fluid mixtures. Concentration effects on the heats of transport and the thermal diffusion ratio of chloroform with various alkanes at 30°C and 1 atm are seen in Table 7.6. Table 7.7 shows the experimental heats of transport at various concentrations and at temperatures 298 and 308 K for binary mixtures of toluene (1), chlorobenzene (2), and bromobenzene (3) at 1 atm. The absolute values of heats of transport decrease gradually as the concentrations of the alkane increase. Table 7.7 also contains values of cross coefficients obtained from easily measurable quantities and the thermodynamic factor. Table 7.8 shows the thermal diffusion ratios and thermal diffusion coefficients obtained from Onsager's reciprocal rules for toluene, chlorobenzene, and bromobenzene at 1 atm and at 298 and 308 K. Thermal diffusion or heats of transport may be extremely sensitive to the molecular interactions in solutions (Rowley et al., 1988).
7'.5.3 Efficiency of Coupling Phenomenological stoichiometry z is defined by
ILqq
z--~
/ 1/2
(7.78)
With the definitions of the degree of coupling q and the phenomenological stoichiometry z, Eq. (7.66) can be written as
~7 -
zA + Q~/z 1/z
Q1A / z +
(7.79)
7.5
377
Coupling in fiquid mixtures
Table 7.6. Heats of transports and thermal diffusion ratio ET a of chloroform in binary mixtures with selected alkanes at 30°C and 1 atm b
Solute(l)
x,
- Q ~ (kJ/kg)
KT1 (l/K)
n-Hexane
0.1334 0.3725
60.0 49.5
2.10 2.29
0.5808
46.3
2.40
0.7637
40.0
2.26
0.9257
34.1
2.13
0.1169
56.5
2.30
0.3380 0.5437
45.8 41.1
2.33 2.27
0.7354
35.2
1.99
0.9147
31.9
1.80
0.1040
54.3
2.92
0.3093 0.5110 0.7092 0.9039
45.0 40.3 34.2 25.7
2.82 2.46 1.91 1.32
n-Heptane
n-Octane
3-Methylpentane
2,3-Dimethylpentane
2,2,4-Trimethylpentane
~'K~ -
D~ _ Dxtx2
0.1334
58.3
1.91
0.3725
48.5
2.26
0.5808 0.7637
44.3 40.1
2.43 2.40
0.9257
36.3
2.31
0.1169
55.0
2.34
0.3380
45.0
2.44
0.5437
39.9
2.31
0.7354 0.9147 0.1040
33.5 27.4 53.1
1.92 1.55 2.87
0.3093
42.6
2.69
0.5110
39.7
2.43
0.7092
31.0
1.73
0.9039
23.7
1.22
O('M,:MI RTM3 ( | + |~J i )
~Rowley et al. (1988).
Equation (7.79) shows that as the degree of coupling approaches zero, each flow becomes independent, and we have r / ~ z2A. If q approaches _+ 1, then the two flows are no longer associated with the forces, and the ratio of flows approaches a fixed value: A --. _+z. This case represents a complete coupling. Negative values of rt occur when the differentiation of chemical potential with respect to concentration is negative due to the nonideality of the mixture. The degree of coupling is not a unique characteristic of the system since there may be various ways of describing flows and forces consistent with a given equation for entropy production. For a complete coupling, q = _+ 1 for any choice of flows and forces, and z reaches a unique value. We can define the ratio of dissipations due to heat and mass flows in terms of the reduced force ratio and flow ratio
JqXq ~/A - - ~
(7.80)
jlY,
Equation (7.80) may be called the e f f i c i e n c y o f e n e r g y conversion. When jlX1 shows the input and J'q'Xqthe output power, then diffusion drives the heat flow. Since ~X is zero when either JqIt or Xq is zero, then it must pass through a maximum at an intermediate value. The values of ~/X are often small in regions of physical interest, and the maximum value depends on the degree of coupling only 2 ( T/Jt)ma x =
q
(1 + x/l_ q2 )2
(7.81)
378
7.
Heat and mass transfer
Table 7.7. Experimental heats of transports in binary mixtures of toluene (1), chlorobenzene (2), and bromobenzene (3) at 1 atm a i
j
Wi
T (K)
- Q ~ (kJ/kg)
1 + I'ii
-Loi (X 107) (kg/(m S)) b
1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3
0.3000 0.3996 0.5000 0.6025 0.3000 0.3000 0.4000 0.5500 0.6000 0.2010 0.7000 0.2000 0.3000 0.5000 0.6000 0.2000 0.3000 0.5000 0.6000 0.7000
298 298 298 298 308 308 308 308 308 308 308 298 298 298 298 308 308 308 308 308
6.21 12.08 10.50 14.98 16.19 15.38 16.59 20.81 24.82 23.35 19.33 14.20 9.41 8.55 9.31 23.71 24.84 12.30 15.08 16.57
1.020 1.020 1.020 1.026 1.025 1.020 1.024 1.026 1.025 1.019 1.018 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0
1.07 2.29 1.99 2.63 3.07 2.92 3.45 4.22 4.80 3.90 3.30 2.34 2.07 2.14 2.24 4.20 5.74 3.37 3.98 3.84
aRowley and Hall (1986).
bLoi = Pvo;gigjwiwj MavRT (1 - F ii)
Table 7.8. Thermal diffusion ratios and thermal diffusion coefficients from Onsager's reciprocal rules for toluene (1), chlorobenzene (2), and bromobenzene (3) at 1 atm a i
j
wi
T(K)
-KTi(1/K)
--DTi(×lolO)(m2/s)
1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3
0.3000 0.3996 0.5000 0.6025 0.3000 0.3000 0.4000 0.5500 0.6000 0.2010 0.7000 0.2000 0.3000 0.5000 0.6000 0.2000 0.3000 0.5000 0.6000 0.7000
298 298 298 298 308 308 308 308 308 308 308 298 298 298 298 308 308 308 308 308
0.0565 0.120 0.104 0.138 0.141 0.134 0.158 0.193 0.220 0.167 0.149 0.124 0.109 0.114 0.120 0.192 0.263 0.155 0.185 0.180
1.05 2.31 2.06 2.79 3.05 2.90 3.52 4.47 5.14 3.04 3.38 1.68 1.48 1.65 1.74 3.05 4.29 2.62 3.12 3.04
aRowley and Hall (1986).
7.5
7.5.4
379
Coupling in fiquid mixtures
Dissipation Function and Degree of Coupling
By substituting Eqs. (7.6) and (7.7) into the dissipation function of Eq. (7.1), we obtain the three contributions due to heat flow, mass flow, and the coupled transport, respectively
aJlt--Lqq(7 V1 T )
2
+Lll
~
l
0/x1
~
14,'. OW1
Vw 1
1
+2Llq
T
VT
~
1
0~1
~
w2 Ow1
Vw 1
(7.82)
We can also use the transport coefficients, given in Eqs. (7.73) to (7.77), and the degree of coupling in the dissipation function, and obtain 1
- kT T V T
1 [/ ) ] 1
)2
0/£1
+ pD --w, Owl~ (Vw 1
+
/ )E () 1
2pDQ 1 1
1
Z
~VT
0/x1
OWl~ Vw 1
(7.83)
By differentiating the chemical potential in terms F
(OIX---!-l) =MRT(I+Fll)OW1 WlM1M2
(7.84)
and using the definition ofq given in Eq. (7.78), we can express the dissipation function in terms of the degree of coupling q, the thermodynamic factor F, and the transport coefficients of thermal conductivity k and diffusion D
(1)2
* - kT ~ V T
+
p D M R T ( I + F l (I~7W l ) ) 21 WlW2MIM2
+2q
(pDkMR(I+F1)11/2 1 wlWzM1M2
(VT)(Vw 1)
(7.85)
The terms on the right of Eq. (7.85) show the dissipation due to heat flow and mass flow and coupling between the heat and mass flows, respectively.
7.5.5
Coupling in Multicomponent Mixtures
The heat of transport of component i is a measure of the local heat addition or removal required to maintain isothermal conditions as molecular diffusion of component i takes place from a higher chemical potential to a lower one. Since there are only n - 1 independent diffusion flows in an n-component mixture, there are only n - 1 independent heats of transport. Using the forces and flows identified in Eq. (7.1), and the Gibbs-Duhem equation for an n-component system at constant temperature and pressure, we obtain
VT]&n----Z Wk VTtZk
(7.86)
k=l
For isotropic, n-component, nonelectrolyte mixtures without external fields and pressure gradients, the phenomenological Eqs. (7.47) and (7.48) are expressed by
- k=l 1=1 -j; - L; v in r + Z Z Z L;j j=l k=l /=1
w,
+--
Wn
OWl T,P
Owl )r,e Vwt
(i = 1,2,...,n-I)
(7.87)
(7.88)
Coefficients Lqq and L~ are associated with the thermal conductivity k and the mutual diffusivity D, respectively, while the cross coefficients Liq and Lqi define the coupling. Thermal conductivity (k) is related t o Lqqby k = Lqq/T,while the thermal diffusion coefficient is related to Liq by L;q = pDri. Tables 7.9 and 7.10 show the values of the phenomenological cofficients Lij for the ternary mixture of toluene (1)-chlorobenzene (2)-bromobenzene (3) at 298.15 and 308.15 K. In ternary mixtures, there are two independent heats of transport related to the two independent crossphenomenological coefficients Lql and Lq2
380
7.
Heat and mass transfer
Table 7.9. Phenomenological coefficients in Eqs. (7.87) and (7.88) for ternary mixture of toluene (1)-chlorobenzene (2)-bromobenzene (3) at 1 atm a T(K)
wl
w2
Lll (X 109) (kgZ/(m s kJ))
298.15 298.15 298.15 298.15 298.15 298.15 298 15 298 15 308 15 308 15 308 15 308 15 308 15 308 15 308.15 308.15 308.15
0.100 0.200 0.200 0.200 0.362 0.397 0.450 0.600 0.200 0.200 0.200 0.325 0.326 0.397 0.400 0.450 0.600
0.500 0.200 0.400 0.600 0.317 0.303 0.174 0.200 0.200 0.400 0.600 0.175 0.274 0.303 0.100 0.450 0.250
6.85 13.4 13.0 12.5 19.9 20.8 22.4 21.9 14.9 14.4 13.9 21.5 21.1 24.2 23.1 22.5 23.6
L12 (Xl09) (kgZ/(m s kJ))
L21 (Xl09) (kgZ/(m s kJ))
-3.70 -3.23 -6.01 -8.78 -8.82 -9.28 -6.30 -9.51 -3.58 -6.68 -9.77 -5.15 -7.72 -3.76 - 10.3 - 17.2 -13.1
-3.44 -3.05 -5.79 -8.53 -8.76 -9.26 -6.29 9.63 -3.39 -6.43 -9.48 -4.99 -7.60 -3.65 - 10.3 - 17.0 -13.2
L22 (Xl09) (kgZ/(m s kJ)) 23.7 14.1 21.7 21.1 18.8 18.2 12.4 13.5 15.8 24.2 23.5 14.0 19.4 8.61 20.3 22.3 17.4
aplatt et al. (1983).
Table 7.10. Phenomenological coefficients in Eqs. (7.87) and (7.88) for ternary mixture of toluene (1)-chlorobenzene (2)-bromobenzene (3) at 1 atm a T (K)
wl
W2
Lqq (W/m)
Lql (Xl07) (kg/(m s))
Lq2 (X 10 7) (kg/(m s))
298.15 298.15 298.15 298.15 298.15 298.15 298.15 298.15 308.15 308.15 308.15 308.15 308.15 308.15 308.15 308.15 308.15
0.100 0.200 0.200 0.200 0.362 0.397 0.450 0.600 0.200 0.200 0.200 0.325 0.326 0.397 0.400 0.450 0.600
0.500 0.200 0.400 0.600 0.317 0.303 0.174 0.200 0.200 0.400 0.600 0.175 0.274 0.303 0.100 0.450 0.250
74.98 72.39 75.55 79.18 77.94 78.57 77.43 81.96 75.50 78.80 82.58 77.76 79.56 78.16 81.96 86.65 86.65
--0.937 --2.493 -- 1.543 --1.531 --2.620 --2.869 -3.154 -4.606 -3.430 -2.246 - 1.699 -3.037 -3.329 -4.553 -3.660 --4.025 --4.022
--1.510 --0.190 - 1.252 -0.647 -0.253 -0.081 -0.155 2.390 0.861 -- 1.013 -- 1.001 --0.510 -- 1.157 -0.627 -1.910 1.115 --0.024
aplatt et al. (1983).
Lql = l_qlQ1 + LzlQ2
(7.89)
Lq2 = I-q2Q~ + L22Q2
(7.90)
W e relate the t h e r m a l c o n d u c t i v i t y k a n d the t h e r m a l d i f f u s i o n c o e f f i c i e n t as
Lqq = kT
and
Liq = pOTi.
Dyi to
the p h e n o m e n o l o g i c a l c o e f f i c i e n t s
T h e r e f o r e , E q s . ( 7 . 8 7 ) a n d ( 7 . 8 8 ) c a n b e e x p r e s s e d in t e r m s o f t h e t r a n s p o r t c o e f f i c i e n t s
n-1 n-1 - J q = k V r + Z Z P Q ; D k ' Vw' k=l l=l
(7.91)
7.5
381
Coupling in liquid mixtures
Table 7.11. Thermal diffusion ratios from Onsager's reciprocal rules for toluene (1), chlorobenzene (2), and bromobenzene (3) at 1 atm and 35 °a u'l
u'2
-Kll
(I/K)
(I/K)
-KI2
0.300
0.700
0.141
0.300
0.700
0.134
-
0.400
0.600
0.158
-
0.550
0.450
0.193
-
0.600
0.400
0.220
-
0.201
0.799
0.167
-
0.700
0.300
0.149
-
0.0
0.200
-
0.192
0.0
0.300
-
0.263
0.0
0.500
-
0.155
0.0
0.600
-
0.185
0.0
0. 700
-
0.200
0.200
0.145
-0. 040
0.180
0.200
0.400
0.095
0.048
0.200
0.600
0.074
0.049
0.325
0.175
0.128
0.024
0.326
0.274
0.141
0.055
0.400
0.100
0.194
0.029
0.400
0.300
0.156
0.091
0.450
0.450
0.175
- 0.055
0.600
0.250
0.177
0.001
~Rowley and ttall ( 1 9 8 6 ) .
n-1
-Ji - pDTiV l n T + ~ , p D i t V w l
(7.92)
1=1
Equations (7.91) and (7.92) are valid for mixtures at mechanical equilibrium, containing no external body forces, and with negligible surface effects. Also, mass-average velocity is small even under an initially large concentration gradient. For a ternary mixture, Eqs. (17.91 ) and (7.92) become -J
- k V T + p(Q1Dl l + Q2D21 )VWl + p(Q1D12 + QzDzz)VW2 -Jl
-J2
(7.93)
--
PDTI V
T + pDl l VW 1 -t- PD12 g w 2
(7.94)
-
P D T 2 V l n T + pD21Vwl + pD22Vw2
(7.95)
In
where Dit is the diffusion coefficient, and related to the phenomenological coefficients as 1 Dil -- -- Z /9 i =1 k=l
i
wk Lq.i (~i/, + ~ Wn
/{ ) O~k -OWl
(7.96)
T,P
Table 7.11 shows the thermal diffusion ratios obtained from Onsager's reciprocal rules for toluene (1), chlorobenzene (2), and bromobenzene (3) at 1 atm and 35°C. The heats of transport for the ternary mixtures are shown in Tables 7.12 and 7.13. For the ternary mixture of toluene ( 1 )-chlorobenzene (2)-bromobenzene (3), the heats of transport are tabulated at 298 and 308 K. The temperature- and composition-dependent heats of transport values are fitted by the following equations by Platt et al. (1982) with a deviation below 25%: Q1 - (MavWlW2)[-29.687 + 0 . 0 7 0 6 T + 0 . 0 8 7 1 5 T w l - ( 9 8 . 7 0 - 0 . 4 1 8 T ) w 2 + ( 7 8 7 . 3 - 2.765) % w 2 - 21.59w~ - 0.1071Tw~ ]
Q2 - (Mavwl w2 ) [ - 39.370 + 0 . 1 3 0 2 T + 0.01679Tw~ - (120.6 - 0 . 4 1 9 9 T ) w 2 +(993.7-
3 . 3 0 8 T ) w , u' 2 + 7.006Wl 2 - 0.04601Tw22 ]
382
7.
Heat and mass transfer
Tables 7.14 shows the fitted parameters ai for the phenomenological coefficients in Eqs. (7.87) and (7.88) to the following equation: L i k = a o + a 1w 1 nt- a 2 w 2 -k- a 3 w 2 nt- a 4 w 1w 2 nt- a 5 w 2
for the ternary mixture of toluene (1)-chlorobenzene (2)-bromobenzene (3) at 1 atm, and at 298.15 and 308.15 K.
Table 7.12. Heats of transport of ternary mixture of toluene (1)-chlorobenzene (2)-bromobenzene (3) at 1 atm a
T (K) 298 298 298 298 298 298 298 298 308 308 308 308 308 308 308 308 308
W1A
W2A
-Q~ (kJ/kg)
-Q2 (kJ/kg)
0.100 0.200 0.200 0.200 0.362 0.397 0.450 0.600 0.200 0.200 0.200 0.325 0.326 0.397 0.400 0.450 0.600
0.500 0.200 0.400 0.600 0.317 0.303 0.174 0.200 0.200 0.400 0.600 0.175 0.274 0.303 0.100 0.450 0.250
18.2 20.0 16.5 20.1 15.9 18.1 16.8 19.3 23.0 19.8 21.1 16.4 20.9 25.9 21.3 33.9 29.6
9.3 5.9 10.3 11.4 6.1 9.7 9.8 -4.1 -0.2 9.7 13.0 9.7 14.3 22.6 16.6 21.2 22.4
aG. Platt, T. Vongvanich, G. Fowler and R.L. Rowley, J. Chem. Phys., 77 (1982) 2121.
Table 7.13. Heats of transport of ternary mixture of carbon tetrachloride (1)-benzene (2)-acetone (3) at 1 atm and 298.15 Ka W1
W2
Q] (kJ/kg)
Q2 (kJ/kg)
0.1 0.4 0.6 0.8
0.3 0.1 0.1 0.1
17.3 23.9 18.3 28.0
3.6 13.4 18.7 30.2
aS.C. Yi and R.L. Rowley, J. N o n - E q u i l i b . T h e r m o d y n . , 14 (1989).
Table 7.14. Parameters in the fitted equation for phenomenological coefficients for the ternary mixture of toluene (1)-chlorobenzene (2)-bromobenzene (3) a Ljk = ao + al wl + a2 w 2 + a3 w2 4- a4 w~ w2 + a5 w2, where w; is the weight fraction Lik
Lqq
Lll L22
T
(K)
298.15 308.15 298.15 308.15 298.15 308.15
a0
a1
a2
a3
a4
67.0426 69.9331 -3.2126 -3.2166 -3.37541 -3.19829
10.8125 11.2839 101.41 111.835 7.13475 7.1013
11.9207 12.451 93.8366 -104.64 -1.68174 -1.69981
8.81046 9.11118 6.23984 5.10796 108.765 118.181
17.9015 18.7197 -25.0849 -24.6334 -36.1574 -36.1694
aDemirel and Sandler (2002).
a5
5.73382 6.05464 -4.27267 -3.37569 -104.986 -114.52
7.5
7.5.6
383
Coupling in liquid mixtures
Degrees of Coupling in Multicomponent Mixtures
For a ternary mixture, there are two independent degrees of coupling between the heat and mass flows, and are given by Lql qql
Lq2
( LqqLl l
)1/2 '
qq2
(LqqL22
)1/2
(7.97)
These equations show the relationships between the degrees of coupling and the cross coefficients Lqi. The sign of heat of transport is an artifact of numbering the substances since -Q~ - Q2 in a binary mixture of substances 1 and 2. The negative sign in the numbering system used here indicates that the heat flows toward the more concentrated substance. The absolute values of the degree of coupling decrease with increasing temperature for binary mixtures, while the effect of the composition on the degree of coupling is more complex. The degree of coupling decreases gradually with increasing concentration of toluene for toulene (1) and chlorobenzene (2), while it increases with increasing chlorobenzene concentration at 35°C, and it remains almost the same at 25°C for chlorobenzene (1) and bromobenzene (2). The heats of transport have a complex composition dependence, and are sensitive to the composition of the heavy component bromobenzene. For the ternary mixture, the parameters of the fitted equations for the phenomenological coefficients computed from the diffusion coefficients are given in Table 7.14. The fitted values of Lql and Lq2 in kg/(m s) are estimated from Eqs. (7.89) and (7.90) as functions of composition and temperature as follows Lql - 10-7 ( 1 5 . 6 1 - O . 0 5 9 T - O . O 5 0 1 T w l
+ 2.687Tw2)
(7.98)
Lq2 - 10-7wlw3 [ - 5 3 3 . 0 + 2.185T-441.5w 1 + ( 7 1 8 . 1 - 4.025T)w 2 + 1011 w,w 2 + (4096-12.53T) w 2 + 300.3w~ ]
(7.99)
Besides the cross coefficients, the straight coefficients Lqq, L l l, and L22 should also be calculated. The values of Lqi, Lqq, and Lii, are used in Eqs. (7.125) and (7.126) to calculate the degrees of coupling in the ternary mixture. Figure 7.3 shows the degrees of coupling for 0.1 < wl < 0.6 (i = l, 2) and w3 > 0.1 at 25°C and 35°C and ambient pressure; the degree of coupling qq2 changes direction with changing bromobenzene composition. At high concentration of bromobenzene, qq2 is positive and the flows of the components are in the same direction; at lower concentration, however, qq2 becomes negative and hence the components flow in the opposite directions. The cross-phenomenological coefficient Lq2 changes its sign as a function of the mass fraction of the heavy component bromobenzene. This means that the direction of coupling due to the heat transported by the flow of chlorobenzene relative to the mass-average velocity in the toluene-chlorobenzene mixture can be reversed by controlling the mass fraction ofbromobenzene in the mixture. From the standpoint of thermal diffusion, the addition of bromobenzene to the toluene-chlorobenzene mixture can change the magnitude and direction of the separation. The effect of temperature on qql and qq2 is mostly marginal. As the number of components increase, the relative compositions of each component may play an important role in the coupling between two-flow systems.
o,3/ o,ooo°. 0 . 2 /
r
A 0,000
0.0002 qq2
qq2
V
O.l
•
.... j /
?
_o_
0.3
Figure7.3. Change of degree of coupling %1 and qq2 with weight fraction of toluene wl and chlorobenzene w2 at (a) 25°C, (b) 35°C. Reprinted with the permission from Elsevier, Y Demirel and S.i. Sandier, Int. J. Heat Mass Transfer, 43 (2002) 75.
7.
384 7.6
Heat and mass transfer
COUPLED MASS AND ENERGY BALANCES
We consider the general balance equations of mass and energy in the absence of chemical reactions, and electrical, magnetic and viscous effects. The partial differential equations of these general balance equations represent the mathematically and thermodynamically coupled phenomena, which may describe some complex behavior due to interactions among various forces and flows within a system. 7.6.1
Binary Mixtures
For a binary mixture under mechanical equilibrium and without chemical reaction, the general balance equations are P ~ at
= - V'nl
(7.100)
OT --V.q Ot
pCp
(7.101)
where nl and q denote the total flow of species 1 and total heat flow defined by nl = Jl +Pl v
(7.102)
Ju = q = Jq + EJihi
(7.103)
i where v is the mass-average velocity and h i is the partial molar enthalpy of species i, and Ji and Jq' are the diffusion flow and conduction heat flow, respectively. These equations show that the changes in concentrations and temperature are due to diffusion, conduction, and bulk motion. Using the molecular transport only in these differential equations, we have
Owl 3= - V.jl p .--~pep
OT = - V . J q
(7.104)
"
(7.105)
• hi). By using the Fick and Fourier laws in where Jq,, is also called the vector of reduced heat flow (Jq,, = q _ ~ni=1]i one-dimensional uncoupled molecular transport, Eqs. (7.104) and (7.105) become
Owl)
P ~-~)
--
02 Wl
pD11 0),2
OT)
02T pCp --~ = k~oy 2
(7.106)
(7107)
Without the external mass and heat transfer resistances, the initial and boundary conditions with the y-coordinate oriented from the centerline (y = 0) to the surface (y = L) are w(O,+_L) = Wls,
dw 1(t, 0) = 0 t > 0, dy
and
dT(t O) T(0,_L) = Ts, ~ = 0 t> 0 dy
(7.108)
By substituting Eqs. (7.64) and (7.65) into Eqs. (7.104) and (7.105), we have
0Wl] _ V. /-qlQl*VlnT + / - 1 1 -
P Ot )
W2 ~OW1)T,P
(7.109)
Z6
385
Coupled mass and energy balances
•
pCp ~ t - V. LqqVlnT+Lq,Q, 1_~(Ol~l
VW 1
/
(7.110)
W2 ~ OWl T,P Here, the cross coefficients are eliminated by using the heats of transport. These equations may be solved by using appropriate initial and boundary conditions in Eq. (7.105). If we use Eq. (7.101) instead of Eq. (7.105) and a thermal diffusion coefficient for one-dimensional heat and diffusion flows for a binary mixture, we have the following coupled balance equations:
(o.:,)_o(
or +pD-~yow, ])
(7.111)
pC p - ~ - --~y ,k -~v. + --~v. PD Q; oy ) (M1 M2 )2
Ox2
03y )
(7.112)
where Mav is the average molecular weight of the mixture, M~ is the molecular weight of component i, x 1 is the mole fraction of component 1, and hE is the molar excess enthalpy. At steady state and considering molecular transport, coupled mass and energy balances in Eqs. (7.109) and (7.110) become (7.113)
O - V . LllQl VlnT+Lll-w2 _Owl
(7.114)
7.6.2
Multicomponent Mixtures
By substituting Eqs. (7.87) and (7.88) into the following balance equations:
/
p --07
--V'ji , pCp .--07 - - V ' J q
"
we have the thermodynamically and mathematically coupled mass and heat balance equations
P k,--07-} V" PDTiVIn T +
E
Li/ 6/k + -
Wn
j=l k=l l=1 '
OWl ) T,P
0% -;7 -v. krVlnr+Z Z Z L. a:+-/=I k=l
/=I
Wn
Vw/
(i = 1,2,...,n-I)
~ OWl ) T,P
w,
(7.115)
(7.116)
Here, the phenomenological coefficients Lqq and L;,/are related to thermal conductivity k and thermal diffusion coefficient DT by Lqq = kT and Liq = pDvi. For a ternary mixture under mechanical equilibrium and without chemical reaction, mass and heat balance equations are
WI )--
p --~
-V'n 1
(7.117)
Ow2 ) Pk, Ot - - V'n2
(7.118)
386
7.
Heat and mass transfer
OT pCp at - - V . q
(7.119)
where nl and n 2 denote the total flows of species 1 and 2 defined by n i = Ji +Piv. Using the diffusion flows in Eqs. (7.117) and (7.118), we have
Owl ) P ~.-~-- = - V ' J l
w2 )=
P -~
(7.120)
-V'j2
(7.121)
OT pCp Ot - - V . q
(7 122)
By substituting Eqs. (7.87) and (7.88) into Eqs. (7.120) to (7.122), we have
r )
2
P ~. O~ = V" ~-~(PDTiVlnT +
(7.123)
(i=1,2)
pCp OT = 17. kVT + p ~ Z QkDkiVwi + p Z Z hkkDkiVwiVWk E * i=1 k=l
(7.124)
i=1 k=l
where hE is the excess specific enthalpy defined in terms of excess enthalpy he
(2h)
OW2 T,P
If we consider Jq' instead of q in Eq. (7.122) then Eqs. (7.123) and (7.124) become
P ~. Ot = V'(pDT1VlnT + PDllVW1 + PD12Vw2)
OW2 )
p --~
(7.125)
= V'(pDT2V In T + pD21Vw1+ pD22Vw2 )
pCp OT = V.[kVT+p(Q~D11+Q2D21)Vw1+P(Q1D12 +Q2D22)Vw2 ]
-05-)
*
*
Under steady-state conditions, equations above would be
0 = V'(pDT1VlnT + POllVW1+ POl2VW2) 0 = V •(pDy2 V In T + pD21VW1 + PD22Vw2 )
(7.126)
0 = V'[kVT + p(Q1D11 + Q£D21)Vw1 + p(Q~D12 + Q2D22)Vw21 For a ternary mixture, equations above can describe thermodynamically and mathematically coupled mass and energy conservation equations without chemical reaction, and electrical, magnetic and viscous effects. To solve these equations, we need the data on heats of transport, thermal diffusion coefficient, diffusion coefficients and thermal conductivity, and the accuracy of solutions depend on the accuracy of the data.
7. 7
7.7 7.7.1
Separation by thermal diffusion
387
SEPARATION BY THERMAL DIFFUSION Thermal Field-Flow Fractionation
Thermal diffusion plays an important role in achieving the purification of macromolecules, isomeric substances, and isotopic elements. It is also important in models for predicting the composition profiles of oil fields. Thermal field-flow fractionation separates macromolecules and colloids. The separation is based on the Soret effect. A relatively large temperature gradient is applied across a solution flowing through the narrow gap of a concentric tube. The thermal gradient drives solute molecules with larger thermal diffusion ratio from the solutes with smaller thermal diffusion ratio to the concentric tube walls where friction slows down the flow. Therefore, solutes with large Soret coefficients are retained and separated from the solutes with smaller Soret coefficients.
7.7.2
Soret Coefficients for Aqueous Polyethylene Glycol Solutions
The flow of a solute in a nonisothermal solution is caused by the solute concentration gradient and by the temperature gradient J - -DVc--cDrVT
(7.127)
where ,I is the molar flow of the solute in the volume-fixed frame of reference and DT is the thermal diffusion coefficient. When the ordinary diffusion (-DVc) counterbalances the thermal diffusion flow (-CDTVT) at steady state, we have DT _ _ I ( V @ , ) _ KT-- D c
(7.128)
where KT is the thermal diffusion ratio and represents the relative change in solute molarity per degree of temperature. Equations (7.127) and (7.128) are applied to molecular transport in a binary and dilute multicomponent solution of noninteracting solutes. When the values of KT and DT are positive, the solute diffuses from a warmer to a cooler region of a nonisothermal solution, and the solvent simultaneously diffuses toward the warmer region. The separation of polymers due to thermal diffusion may be quite large. For example, the thermal diffusion ratio for dilute solutions of polystyrene in tetrahydrofuran is around 0.6 K-1. This indicates that the change of polystyrene concentration per degree is 60%. The type of solvent and polymer pair may have a considerable effect on both the thermal diffusion ratio and the thermal diffusion coefficient. The segmental model predicts the DT independent of the polymer molecular weight, and is given by
UsDseg
DT = - -
RT 2
(7.129)
where Dsegis the polymer segmental diffusion coefficient, Us is the activation energy for solvent viscous flow, and R is the gas constant. Schimpf and Semenov (2004) developed the following model for the thermal diffusion coefficient for dilute solutions of flexible polymers D~ =
16c~rZAps 27r/Vs
(7.130)
where r/, Vs, and c~ are the viscosity, molar volume, and thermal expansivity of the solvent, respectively, rp is the effective radius of the polymer segment, and Aps is the Hamaker constant for polymer-solvent interactions, which can be estimated from the Hamaker constants of the pure polymer App and pure solvent Ass using Aps = (AppAss) °5. Schimpf and Giddings's (2003) correlation is DT
_
1.19X10
-4
k p _ k s 0.374
U0.623
(7.131)
where kp - ks is the difference of thermal conductivities of the polymer and solvent, in W/(m s), Us is in J/mol, and DT is estimated in cm2/(s K).
388
7.
Heat and mass transfer
Another model based on Emery and Drickamer's theory (Schimpf and Semenov, 2004) yields
D T = ~ s -s
14---
Us -
tiP
~PP
Up
2RT 2
(7.132)
where Us and Up are the activation energies for the liquid polymer and solvent, Mp and Ms are the molar weights for the polymer and solvent, pp and Ps are the densities of the polymer and solvent, and Vp and Vs are the molar volumes of the polymer and solvent. Equation (7.132) is capable of predicting negative DT values, and shows that DT is proportional to the product MpD for a given polymer solution. The molality-based thermal diffusion ratio tr is related to molarity-based thermal diffusion ratio K x = CoVotr+ c~
(7.133)
where Co and V0 are the solvent molarity and molar volume, respectively, and a is the thermal expansivity. Table 7.2a shows the mutual diffusion coefficients and thermodynamic factors and Table 7.2b shows the experimental values of thermal diffusion ratios, thermal diffusivities, and heats of transport for aqueous ethylene glycol and PEG at 25°C.
Example 7.2 Separation by thermal diffusion Consider two vessels connected by a thermally insulated conduit. The system is filled with a solution of hydrogen and nitrogen. The hydrogen mass fraction is Wn2 = 0.2. Estimate the difference between the mass fraction of the components in the two vessels at stationary state when one of the vessels is at 200 K and the other at 370 K. At stationary state JH2 -"--JN2 = 0, and from Eq. (7.19), we find K~
VWH2 = -- --'H~2 VT
(7.134)
T
The integration of this equation for a one-dimensional system between the temperature limits T1 and T2 yields
WH2,II
-- WH 2
,I
=--f TIIKT JT~
T
dr
Since the thermal diffusion ratio is temperature dependent, the integral above can be integrated by assuming a constant thermal diffusion ratio for a reference temperature obtained from Tr =
Tn-----~TIIn Tn - ( 3 7 0 ) ( 2 0 0 ) In 370 = 268 K TI 370- 200 200
TII - TI
TII WH2,II -- WH2,I -- - K T l n ~ ri
Table 7.1 shows the values of the thermal diffusion ratio in the form: which is very close to reference temperature 268 K. Therefore, we have K T = 0.0548 MH:MN2 = 0.0548
KT(MZv/M1M2)=0.0548
at T = 264 K,
(MN:WH2+Mn2WN:)2
Ma2v
MN 2 MH 2
K T = 0.0548 [28(0.2) + 2(0.8)] 2 = 0.0507 28(2) As the thermal diffusion ratio is positive, the hydrogen diffuses into the vessel at a lower temperature. We estimate the difference between the hydrogen mass fractions in two vessels by WH2,II --WH2,I - - - K T In TI-LI= - 0 . 0 5 0 7 1 n
TI
370 = -0.0312
200
7. 7
389
Separation by thermal diffusion
Example 7.3 Total energy flow and phenomenological equations For mixtures, the energy flow contains the conductive flow qc, and the contributions resulting from the interdiffusion qd of various substances and the Dufour effect qD; we therefore express the total energy flow relative to the mass-average velocity q = qc + qd + qD
(7.13 5)
where qc = -kVT, and qd = £ h i J i , and h i is the partial molal enthalpy. When we express the energy flow e with respect to fixed stationary coordinates by disregarding the Dufour effect, the viscous effect, and kinetic energy, we have e = -kgT
+ Z
hiNi
(7.136)
Equation (7.136) is the usual starting point for simultaneous heat and mass transfer. Mass flow is associated with the mechanical driving forces and thermal driving force (7.137)
Ji = Ji, x + Ji, P + Ji, g q- Ji, r
where Ji,x is the ordinary diffusion, c 2 ~_, ai, x
-
pRT
MiMiDi/ i
"
"
,, OGj xj Z k = 1 " k~j
Vx k
(1 4= j , k )
(7.138)
C)Xk T,P,x:
Ji, P is the pressure diffusion,
Ji, e -
c pRT
i
M i M jD!/ x j V j
Jig = - c - - - pRT
i
V/ Mj
1 VP p
(7.139)
--gk
(7.140)
Ji, g is the forced diffusion
'
MiMiDii . .
x j Mj gj . .
k=l
1)
where g is the body force, and Ji, r is the thermal diffusion, J i,r - -DTi V ln T
(7.141)
where Gj and ~ are the partial Gibbs free energy and partial volume, respectively, D# are multicomponent diffusion coefficients, and Dr/- are multicomponent thermal diffusion coefficients; these coefficients show the following properties: Dii - O
and
~ DTi -- O
(7.142)
i
and
•
(MiM/Dil - MiMkDik ) - 0
(7.143)
i=1
Ordinary diffusion depends on the partial Gibbs free energy and the concentration gradient. The pressure diffusion is considerable only for a high-pressure gradient, such as centrifuge separation. The forced diffusion is mainly important in electrolytes and the local electric field strength. Each ionic substance may be under the influence of
390
7.
Heat and mass transfer
a different force. If the external force is gravity, then 'Ji,g vanishes since all gi are the same. Thermal diffusion leads to the separation of mixtures under very steep temperature gradients. For a binary system, ordinary, pressure, forced, and thermal mass flow is expressed as follows:
OlnxA
RT
1) VP - MApSxA ] p pR----------~(gA-g~)+K~'VlnT
MA
(7.144)
where (dGA)T,p= RTd In aA, and KT is the thermal diffusion ratio defined by P DTA K y = C2MAMB DAB
(7.145)
When KT is positive, component A moves to the colder region; otherwise, it moves to the warmer region. Some typical values of thermal diffusion ratios for binary fluid systems are given in Table 7.1.
Example 7.4 Modified Graetz problem with coupled heat and mass flows The Graetz problem originally addressed heat transfer to a pure fluid without the axial conduction with various boundary conditions. However, later the Graetz problem was transformed to describe various heat and mass transfer problems, where mostly heat and mass flows are uncoupled. In drying processes, however, some researchers have considered the thermal diffusion flow of moisture caused by a temperature gradient. Consider a fully developed flow of a Newtonian fluid between parallel plates with a parabolic velocity distribution (Coelho and Telles, 2002) v = Vmax(1-- ~'/2)
(7.146)
where ~ = y/H, and H is the distance from the wall to the center line. The fluid consists of a solvent and n number of solutes. The flow enters the channel with uniform concentrations Cio and uniform temperature To. At the inlet, the confining walls and the fluid are in thermodynamic equilibrium. The wall temperatures vary. Steady-state mass and energy balances are
pv
OCi =-V.ji Ox
(i=l,2,...,n-1)
OT = Ox
pCpv ~
-V'Jq
(7.147)
(7.148)
where Ji and Jq are the mass and heat flows, respectively. Assuming that the local equilibrium holds, we have the following linear phenomenological relations for n - 1 independent concentrations: n-1
Ji = -DsiqVT- ~ DikVCk
(7.149)
k=l n-1 Jq =
-kVT- ~
DDqkVC k
(7.150)
k=l
Here, Ds and Do are the coefficients representing the Soret and Dufour effects, respectively, Dii is the self-diffusion coefficient, and Dik is the diffusion coefficient between components i and k. Equations (7.149) and (7.150) may be nonlinear because of, for example, reference frame differences, an anisotropic medium for heat and mass transfer, and temperature- and concentration-dependent thermal conductivity and diffusion coefficients.
7. 7
391
Separation by thermal diffusion
Substitution of Eqs. (7.149) and (7.150) into mass and heat balance equations yield n-1
E ~ikZ~2@k + ~iq A20
=
(1-- r/2) O@i
(7.151)
OZ
k=l
n-1
Z ~qkA2q)l< nt- A 2 0
:
(l--
00 T/2) --
(7.152)
OZ
k=l
where
T-T o Ci =--, O - AT ' qPi Ci 0
Z=
x DD@Cko , ~)qk = ~ , HPe kAT
_ DsiqATCio , P e = ~H, v m a x DikCpCk° d/~iq~ik : kCio , kCio ol
a--
k
(7.153)
oG
and the dimensionless Laplacian operator in Cartesian coordinates is
A2
02
1
02
= ~ + ~ -3T/2 pe 2 Oz2
(7.154)
The boundary conditions for temperature of the upper (U) and lower (L) plates are
O(Z,--1)=OL(Z), O(Z,+I)--Ou(Z ) lim:_,~ 0L(z) = Ou(z ) = 0
(7.155)
Both plates are held at specified, variable temperatures. For asymmetric wall temperature boundary conditions, the lower plate may be held at the input temperature, while the top wall temperature would be at a specified variable value. The boundary conditions for concentration reflect the asymptotic approach to the fluid composition at the inlet and the permeability properties of both walls lim:_+~ ~ i ( Z ,
1'1) --- 1
ji(z,--
1).n
ji(z,+
1).n - Kui (Ci(z,+ 1)- Cui ) = 0
-
KLi(Ci(z,-1)-CLi ) = 0 (7.156)
Here, Ki are the mass transfer coefficients (permeabilities) for each wall, and CLi and Cui a r e the ambient concentrations of each component i outside the lower and upper walls, respectively. Sometimes, selective membranes may be used as the walls. These membranes may be permeable to selected components only. For example, in a purification process, the membrane would be permeable to one of the solutes only. In a concentration process, both walls can be impermeable to the selected solute. Equations (7.151) and (7.152) describe the thermodynamically and mathematically coupled heat and mass flows at stationary conditions and may be solved with boundary conditions and with some simplifications (Coelho and Telles, 2002).
Example 7.5 Cooling nuclear pellets Spherical nuclear fuel pellets generate heat at a rate per unit volume, q, and being cooled at the boundary by convection heat transfer. For a single pellet on start up, we have
at
-ol
~ -~ Or
-I-~ pCp
392
7.
Heat and mass transfer
subject to initial and boundary conditions
~ = R , - k av~= h(v- vf) Or ~0T = 0 Or T(0,r) = To
r=0, t=0,
where a = k/(pCp), h is the heat transfer coefficient, and Tf is the flowing coolant fluid temperature. Derive an expression to predict transient response T(r, t). Solution: Assume constant physical properties, constant heat source. Temperature with steady T(r) (where t approaches infinity) and unsteady T'(r,t) parts as follows T(r,t) = T(r) + T'(r,t)
(7.157)
For steady state case 0 = a~-~r
q +0%
r2
(7.158)
m
Integrating directly with the symmetry condition
OT
= 0 at r = 0, we obtain
Or
T(r) . . . . q r2 + I k6
(7.159)
From the first boundary condition describing the cooling at the surface we obtain the constant/, and temperature becomes T(r) = Tf + q_fi_R+ qR 2 1 3h 6k
(7.160)
For unsteady state heat transfer, we have
OT' Ot
1 0 (r2 OT' ~ -~-&r Or )
-a
OT'
r=R,
-k
r=0,
OT' ~=0 Or
t = o,
T ' ( o , r) - ro - ~'(r)
Or
(7.161)
= h ( T ' - rf)
m
Equation (7.161) becomes OT' Oz
E2 82 OS
08 J
(7.162)
with r R'
at R2
s=0,
~=0T 0 0e
e = l,
~OT = BiT 0s
Bi-
hR
k
(7.163)
7. 7
393
Separation by thermal diffusion
Apply the separation of variables method:
T'(/3,r) = X(/3)Y(r)
(7.164)
Substituting Eq. (7.164) into Eq. (7.163), we get
l d(/3 2dx ) A2 e2d/3 ~ + X = 0 dY+A2y dr With X =
u(/3)//3,Eq.
(7.165)
0
(7.166)
(7.165) becomes d2bl ~ - + - A2u = 0 d/32
(7.167)
The solutions for u, X, and Y are u(/3 } - ,4 sin(A/3) + B cos(A/3}
X(/3)- A sin(hf_)_) + B cos(h/3_________~)) /3
(7.168)
/3
y - C exp(-A2r) The combined solution becomes"
T ' - ( A sin(A~3)~3+BC°S(A/3~))exp(-A2r)+C+D/3 --/3 T' is finite at the center
dT'/dr- O, and B -
(7.169)
C - D - 0 (r --+ ~c)
T'( /3,r) - A sin(a/3) exp(-A2r) /3
at/3 = 1 - dT'/d/3 = BiT and sin(A) - acos(A ) - BiSin(A). There are infinite numbers of values of A, which satisfy this solution, and the general solution is a superposition of all possible solutions 7~
T'(/3,r)- E A,, sin(An/3~)e x p ( - a , 2 r ) /3
tl-- 1
Finally, from the orthogonality condition and the initial condition, we have Y'(/3, r ) = 2Bi~[~ ~} - rr
{N / AT, - 1)
,,=, ( B i - 1 + cos2 An)
cos a, sin(an/3 ) ~ exp(-A2z) A,
e
where
qR2 N
With specified values of Bi and
N, the temperature
__
profiles may be constructed.
(7.170)
7.
394
7.8
Heat and mass transfer
NONLINEAR APPROACH
Nonlinear systems of transport and chemical kinetics analyzed by the generalized Marcelin-de Donder equations consider two competing forward and backward directions of an elementary process. These equations characterize the flow of matter and energy through the energy barrier and contain potentials F = (-ix~T, 1/T) in exponential forms
Jr = Jrf - J r b = Jr0
exp -
l)fi
-exp
(
bi)]
Pbi~
-~ i
where l)ik is the stoichiometric coefficients that are positive for products and negative for a chemical reaction and satisfies the mass balance EviMi = 0. In Eq. (7.171),/x = 0 for i = 0, which corresponds to the energy transfer, while i = 1, 2, 3,..., n refers to species transfer. For elementary transport processes of heat and mass, stoichiometric coefficients in both directions are equal vfi = Pbi l)i" The term Jr0 denotes the exchange current. The Jr0 and Pik are common for both directions. The ratio of absolute flows is -
-
Jrf Jrb
exp
(
/
R
(7.172)
Based on Eq. (7.171), generalization of the isothermal kinetics of the Marcelin-de Donder yields Jr = Jr0 exp ~ . v fi - ~ f
exp
(7.173)
-I)bi ~ b
The generalized form can represent slow transport processes and nonisothermal effects, and satisfies the detailed balance at thermodynamic equilibrium. The exchange current Jr0 is
Jr0=kfexp
/~i -l"fi~Tf)=kbexp
.i0/
--Pbi--~b
and assures vanishing affinities at equilibrium. For the isothermal chemical kinetic system of S = P, using Eq. (7.173), we have
r: fexp ) (~so
(exp
~s
exp /xe (7.175)
For the cross symmetry property, the partial derivatives of flows with respect to potentials are
OFk
--Lik = - J i o ~
exp -
R (7.176)
OFi
-L~i - --Jko --~ exp -
(7.177)
R
Lik=
L~. Not being confined to linear rate relations, the general symmetry yields Equation (7.171) in terms of the generalized forward and backward potentials of 1-Ifand li b is
= (exp( / exp/ )/ (7.178) where
1--If-
and l i b
--
i=1
f
f
=
-
i=1
b
b
7.8
395
Nonlinear approach
Both the generalized potentials are state functions. If the chemical kinetics represented by the chemical potentials is ignored in Eq. (7.178), heat effects are described by the generalized potentials as follows (Hfx. - l-Ibk )heat -- P0,bk
P0,fk _ PO,bk -- P0,fk _+_ PO,fk (Tf - T b )
Tb
Tf
Tb
(7.179)
TfT b
Based on Eq. (7.171), kinetic equations for mass and heat flows are
Js-Js0
r +Vqs
exp Vss ~
-U
r - e x p Vss ~--~
+Vqs b
Jq-Jqo
exp
VSq - ~
+vqq - - ~ f
-exp
VSq - ~
f
(7.180) b
+Vqq --R-T b
(7.181) b
The Onsager coefficients of Eqs. (7.180) and (7.181) are Lik, eq -- Jio
~
exp --Vki Fk'eqR ) = Ji, eq PkiR
(7.182)
For a symmetric matrix uki,both absolute equilibrium flows Js,eq and Jq, eq m u s t be identical and replaced by a universal constant Jeq. However, if the matrix ukiis not symmetric, which is usual, the equilibrium flows are related to each other so that the Onsager symmetry is achieved PSq Lsq, eq - LqS, eq - J s , eq R
PqS - Jq, eq R
(7.183)
Therefore, the generalized kinetic equations for exchange (transport) processes and chemical reactions are of similar structure. During a diffusion-controlled reaction, matter is transported around an interface, which separates the reactants and the product. The progress of the reaction is strongly determined by the morphology of the interface with a complicated structure, which controls the boundary conditions for the transport problem. The morphological stability of interfaces with nonequilibrium systems may undergo self-organization or pattern formation arising in biology, physics, chemistry, and geology.
Example 7.6 Fokker-Planck equation for Brownian motion in a temperature gradient: short-term behavior of the Brownian particles The following is from Perez-Madrid et al. (1994). By applying the nonequilibrium thermodynamics of internal degrees of freedom for the Brownian motion in a temperature gradient, the FokkerPlanck equation may be obtained. The Brownian gas has an integral degree of freedom, which is the velocity v of a Brownian particle. The probability density for the Brownian particles in velocity-coordinate space is
f ( v , r , t ) = p(v,r,t) m
(7.184)
where r is the position, t is the time, and p is the mass density of the Brownian particles. The mass density of a system consisting of Brownian gas and a heat bath is P -- PH + PB -- PH +
mlf(v,r,t)dv
(7.185)
For a constant PH, the Gibbs equation is
6ps - ~1S p e - - ~m S /x(v,r,t) 6f (v,r,t)dv (7.186)
396
7.
Heat and mass transfer
where 6 represents the total differential of a quantity,/z(v, r, t) is the chemical potential gradient of the Brownian gas component with internal coordinate v, and T(r) is the temperature of the bath at position r. The chemical potential is related to energy (e) and entropy (s) per unit mass
pe- Tps + P = f/.L(v,r,t) p(v,r,t)dV + PH~H
(7.187)
Here,/~H is the chemical potential of the heat bath and P is the hydrostatic pressure. The mass energy and entropy balance equations are needed. The rate of change of probability density with time is
O f = - r . Of-O-~-.J v = - v . O f - O-~.J v Ot Or Ov Or Ov
(7.188)
The conservation of mass for the Brownian particles (B) is obtained from integrating Eq. (7.188) OPB
at
--
--
V"
(7.189)
PBPB
where VB is the average velocity of the Brownian particles obtained from 1, VB(r,t ) = --:-J p(v,r,t)vdv PB
(7.190)
Ope _ -- - V ' J q Ot
(7.191)
The energy conservation is
where Jq is a heat flow in the reference frame in which the heat bath is at rest. The entropy balance equation is derived assuming that the gas is at local equilibrium. We also assume that the suspension of Brownian particles in the heat bath may be a multicomponent ideal solution, and the thermodynamic potential is expressed by
Iz(v,r,t) = kT ln f(v,r,t)+C(v,r,t)
(7.192)
m
where the potential function C(v, r, t) can be a function of the local thermodynamic state variables T(r) and pB(r, t). Then, the chemical potential must satisfy the following conditions: 1. Entropy at constant energy and density of Brownian particles has a maximum, is uniform in velocity space, and equal to the thermodynamic potential of a Brownian ideal gas, so we have
tZB=tZl'eq(V'r't)=kT( -m - lnpB -m - - - 23In (2~rkT)/ m
(7193)
2. The distribution functionf(v) is Maxwellian at local equilibrium, and is defined by
fl, eq(V,r,t) -- exp ( l/zvB2-)~m
kT
(7.194)
Using Eqs. (7.193) and (7.194) in Eq. (7.192), we find C(v) =(1/2)v 2, and Eq. (7.192) becomes /z(v,r,t) = k--T-Tlnf(v,r,t)+ lu2 m 2
(7.195)
From Eq. (7.186), conservation of mass and energy, and the chemical potential, the rate of change of entropy per unit volume is obtained as
Ops_ - - V . j s +~ Ot
(7.196)
7.8
397
Nonlinear approach
where the entropy flow as, the entropy source strength ~, and the modified heat flow (J'q) are obtained from the second law of thermodynamics J~ - a'q - k I f(v, r, t)(ln f ( v , r , t ) T
~=
T2 V T -
0
1)vdv
(7.197)
(7.198)
f
1
(7.199)
J'q - Jq - mI-~v2vf (v,r,t)dv
One of the contributions to the modified heat flow is the motion of the Brownian particles. The entropy source strength is due to heat flow and due to diffusion in velocity space (internal degree of freedom), which is the contribution of the motion of the Brownian particles in the heat bath. Equation (7.197) is based on the identity 0
df k l V.-d~r l n f d v = k -~r . ~ v f ln f dv -
kJ v . -df t'
(7.200)
while Eq. (7.197) results from a partial integration over velocity space by assuming Jv ~ 0 as v --, _+oc.
7.8.1 Phenomenological Equations Since the system is isotropic and assuming locality in velocity space, and using the linear nonequilibrium formulations based on the entropy production relation in Eq. (7.198), we have the linear phenomenological equations
j:,
-j'kL v 7vv
(7.201)
dv
(7.202)
Jv--L 7 r -
The Onsager relations yield Lvq - -Lqv. The heat conduction is expressed by A Lqq/T 2 and the friction coefficients are y - Lvq/fT, ~ - mLv~/fT. With these relations, Eqs. (7.201) and (7.202) become =
(7.203)
VT 1
J, - -~f T -
~
(
k T O- f- )
fv +--
m
(7.204)
0v
Using Eq. (7.204) in Eq. (7.188), the Fokker-Planck equation for the Brownian motion in a heat bath with a temperature gradient is obtained Oj O [ kT Of ) 3' 0 0 T Of _ _ v . _ _ + ~ • fv + ~ B + .f __ at Or -~v ~ m Ov T -~v Or
(7.205)
Using Eq. (7.203) in the energy conservation. Equation (7.191), a differential equation is obtained
02T Ope _ A ~ T _ Ot c)r~
0~(kTOf) 02T my--" fv+~ -- dv- A--T-y Or , m Ov Or
c)
-~r
"PBVB
(7.206)
398
7.
Heat and mass transfer
Equation (7.206) disregards the small contribution to the heat flow arising from the kinetic energy of the Brownian particles. Equation (7.206) is mathematically and thermodynamically coupled and describes specifically the coupled evolutions of the temperature field and the velocity-coordinate probability distribution of the Brownian particles. However, for larger times than the characteristic time/3-1, the system is in the diffusion and thermal diffusion regime.
7.8.2
The Thermal Diffusion Regime
Conservation of momentum may be used to simplify the equation of motion for the Brownian gas for long time behavior: t >>,8 -~. At this regime, the Brownian gas will reach an internal equilibrium with the heat bath. From Eq. (7.188) and the mean velocity in Eq. (7.190), the equation of motion for the mean velocity becomes dVB _ _ V.PB (r, t) + mf Jvdv
(7.207)
PB ( r , t ) = m f f ( v - v a ) ( V - V a ) d v
(7.208)
PB dt where PB is the pressure tensor given by
And the substantial derivation is d/dt = O/Ot+ Va" (O/Or). By substituting Eq. (7.205) into Eq. (7.207), the equation of motion becomes VT dva+ 1 V.Pa(r,t)+y__f__=_[~v a dt PB /
(7.209)
For the Brownian gas at internal equilibrium, the distribution function is approximated by
{ II~B--1/2(V--VB)2]} kT
f ( v , r , t ) ~ f, eq(V,r,t) = exp m
(7.210)
and the pressure tensor is reduced to gas pressure PB: PB = PBU, PB = pBkT/m, where U is the unit tensor. The inertia term on the left side of Eq. (7.209) can be neglected, and we have
JD = PBVB = --DVPB -- DT
VT
(7.211)
where the diffusion coefficient D and the thermal diffusion coefficient (DT) are defined by
kT (Tm) D - m[~' D T = pBD l+-k-~-
(7.212)
With Eqs. (7.194) and (7.210), and mf Jvdv = VPB, the entropy production equation becomes VT VPB (I) -- - J q " - ~ - J D " ~
pT
(7.213)
Using the relation PB = pBkT/m, Eq. (7.213) becomes ~ : _j,q VT (k/m)VpB " U - - JD" PB
(7.214)
k VPB j'q = Jq + P B J D =-A'VT-DTT--~ m PB PB
(7.215)
where the modified heat flux is
399
7.8 Nonlinearapproach
where the heat conduction coefficient A' is defined by A' = A + (k/m)(D2T/DpB). Both Eqs. (7.213) and (7.214) can identify the conjugate forces and flows in ordinary space for which the Onsager relations will hold, and the linear phenomenological equations become J) - - A ' V T -
DTT
k VPB
-- - L q q V T - LqDVPB
m PB
(7.216)
JD -- --DT V__TT_ DIFp B = _LDqlF T T
_
(7.217)
LDDIFpB
Example 7.7 Absorption of ammonia vapor by lithium nitrate--ammonia solution The following modeling is from Venegas et al. (2004). For simultaneous heat and mass transfer during the absorption of ammonia vapor by lithium nitrate-ammonia (A) solution droplets, the ammonia concentration profile in the liquid phase can be estimated from the continuity equation without a source term OPA ~ Ot
(7.218)
V ' ( p d D V x ) + V" (PAV) = 0
where Pd is the density of the dispersed phase and PA is the density of ammonia. As there is no source term, the ammonia production and sink do not exist. In an adiabatic chamber, when the solution is dispersed in droplets, we may assume that the density and diffusion coefficient are constant. Therefore, Eq. (7.218) becomes Ox _ DV 2x + v" Vx = 0 Ot
(7.219)
In spherical coordinates and considering symmetry in & direction, Eq. (7.219) reduces to OZx 2 0 x cot00x Ox Ox + V r ' m + v o Ox ~ + ~ 2~ +O0 Ot Or r O0 - D ~ o r 2 + r- Or
02X)
1
m
)
-7 j
(7.220)
Energy conservation equation for performing the same analysis in spherical coordinates yields
~+v o,
r'~+
,0
cot0
- D / ~ + - ~ + ~ 0o t, or r Or
~+ 00
1
-
- |
(7.221)
77 00
The second and third terms on the left of Eq. (7.221) represent the temperature variation in the radial and angular directions, respectively, due to the convection effect. The terms on the fight correspond to the temperature variation in the radial and angular directions due to conduction. Initial and boundary conditions for the absorption of refrigerant vapor by solution droplets are as follows: Att=0x=x
0T=T o forallr
OT - 0 for t > 0 Or Or Atr=Rx=xeqT=Teq fort>0
Atr=0
At0=0
Ox
and
0=Tr
Ox
00
-
OT
00
-0
(7.222) fort>0
For a low-pressure absorber at constant pressure, it is common to relate the temperature and concentration at saturated equilibrium by a linear function Xeq = -0.00372Teq + 1.58226. Another boundary condition suggests that the heat flux at the droplet surface is proportional to the absorbed mass of refrigerant vapor, and we have (7.223)
400
Z
Heat and mass transfer
where M-Iv is the latent heat. Equation (7.223) is based on the assumption that absorption heat release occurs only at the droplet surface and that no heat is exchanged between the liquid and vapor phases. Relative movement between the droplet and the surrounding fluid can induce circulatory motion inside the droplet and affect the mass and heat transfer. These circulatory velocities depend on the Reynolds number, and start to occur for Reynolds numbers higher than 20. For Re < 1, the stream function q~ in spherical coordinates with the origin on the sphere center is
v ( R2r2-r4 ) q~= ~-2
4(1+3,)
sin 2 0
(7.224)
where v is the droplet velocity and 3' is the ratio of liquid and vapor viscosity. The velocity components are
1 ]2 r
---
Pe
0q~
r 2 sin 0 00
1 VO - -
/ /2/cos0
(7.225)
Pe
0q~
r sin 0 Or
(7.226)
where Pe' is the modified Peclet number defined by Pe
Pe ' =
(7.227)
4(1+~,)
and used to correct the effects of the liquid and vapor viscosity ratio. To simplify the modeling, the following dimensionless variables are introduced _ X--
x
T -
,
T ,
xo
_ r r = ~
TO
Dt ,
(7.228)
7"-
R
R2
Using Eqs. (7.225) and (7.226) in Eqs. (7.220) and (7.221), we have
2 0 Y cot 0 0 g OE - -02y _ _ + _ ~ + ~ ~ OF 2 F OF F 2 O0 Or
~
1~ 02y F 2 002
(7.229)
The energy conservation equation for performing a similar analysis in spherical coordinates yields _--z+---+ F OF
cotoo 102 / ( t
F2
00
+Pe'
(1-F2)cosOOf
F 2 002
0F
+/
/ sn°OT/
272-1 F
(7.230)
O0
The initial and boundary conditions are Forr-0x=T=l
for allr m
Atr=0
OY
aT
- 0 for r > 0 OF 0g Atr=lY=xeqT=Teq forz>0
At0=0
-
and
0=rr
OY 00
-
OT 00
-0
(7.231)
fort>0
7.9
401
Heat and mass transfer in discontinuous system
Another boundary condition suggests that the heat flux at the droplet surface is proportional to the absorbed mass of refrigerant vapor, and we have
Xeq
0"00372T° )Teq -~- 1.58226 -
-
x0
AHvD
Cpc~
{
(7.232)
x0
/
xo
02 _
0.00372To O~=1
aT
(7.233)
O~=1
m
At:0=0
and
0-rr
02
O0
-
aT
-0
fort
(7.234)
O0
The simultaneous solutions of Eqs. (7.229) to (7.234) describe the mass and heat transport processes between refrigerant vapor and solution droplets at constant pressure and Reynolds numbers smaller than 1. Typical lithium nitrateammonia solution properties and ammonia vapor properties are available in Venegas et al. (2004).
7.9
HEAT AND MASS TRANSFER IN DISCONTINUOUS SYSTEM
Transport problems in discontinuous (heterogeneous) system discuss the flows of the substance, heat, and electrical energy between two parts of the same system. These parts or phases are uniform and homogeneous. The two parts make up a closed system, although each individual part is an open system, and a substance can be transported from one part to another. There is no chemical reaction taking place in any part. Each part may contain n number of substances. For example, thermal diffusion in a discontinuous system is usually called thermal osmosis. If the parts are in different states of matter, there will be a natural interface. However, if both parts are in liquid or gas phases, then the parts are separated by a porous wall or a semi-permeable membrane. The postulate of local thermodynamic equilibrium in a discontinuous system is replaced by the requirement that the intensive properties change very slowly in each part, so that the parts are in thermodynamic equilibrium at every instant. The intensive properties are a function of time only, and they are discontinuous at the interface and may change by jumps. In the following sections, thermomechanical effects and thermoelectricity are summarized. Considering the dissipation function below B , - Jil -AT ~ + £ JiArl.ti i 1
(7.235)
where
Artz i -- V;.~P + E
/--1
AW/
Ow i T , P , w / :¢=14'i
We can choose the following thermodynamic forces Xq -
AT
(7.236)
T
X i - Ar/Xi
(7.237)
Based on these thermodynamic forces, the linear phenomenological equations become
Aw/ /=1
/=!
i=1
Ow i )
T,P,w/~w i
(7.238)
z. Heatandmasstransfer
402
Jm=LmqXq+£LmiXi=Lmq--~-+£Lmi i--1
ViAP+ E
i=1
Otzi
j--1
~OWj ) T,P,wj =/=Wi
/
(7.239)
where m = 1,2,...,n. By Onsager's reciprocal rules of Lqi-- Liq and Lmi-- Lira, and the dissipation function is positive
xI~--ZqqX2 +£(Ziq-+-Zqi)XiXq+ £ LmiXmXi >0 i=1
(7.240)
i,m=l
Therefore, the phenomenological coefficients satisfy the following conditions"
Lqq > O, Lii > O, LmmLii-LZi > 0
(i,m = 1,2,...,n)
(7.241)
By introducing heats of transport of the components (when AT = 0)
Zmq = £ LmiQ;
(7.242)
i=1
into the phenomenological equations, we have
Jm = £ Lmi(Xi + Q~.Xq)
(7.243)
i=1
Jq = LqqXq +
Lim)(iQm= Lqq i,m=1
LmqQm Xq + m=l
QmJm
(7.244)
m=l
For an isothermal fluid, the thermodynamic force Xq vanishes, and the heat of transport for component m becomes
(7.245)
Qm = Ji~m =0,AT=0
This represents the heat transported per unit flow of component m without the flow of other components and without the temperature difference. With the explicit thermodynamic forces identified in Eqs. (7.236) and (7.237), the mass flow in Eq. (7.243) becomes
i=1
k=l ~, OWk)
T,P,wj~k
+ aT]
(7.246)
For a binary fluid, Eq. (7.246) yields the flows
(7.247)
(7.248)
where the heats of transport are related to the phenomenological coefficients by taking into account the Onsager rules
Llq = LllQ1 +/-qzQ2
(7.249)
7.9
Heat and mass transfer in discontinuous system
403
L2q - L21Q1 + L22Q2 - LI2Q1 + L22Q2
(7.250)
The heats of transport may be estimated from Eqs. (7.247) and (7.248)
Q1-
LIv
LI2
L2q
L22
, ]I~21Llq Q2 =
(7.251)
L2q
D
where D is the determinant D = L 1 IL22 - LI2L21. Assume that, at the beginning, the two parts are mixed homogeneously, and zXP = 0 and ture difference is the only remaining thermodynamic force, then Eq. (7.240) becomes j"
q-Lqq
Awj =
0. If the tempera-
AT = k A AT
T
(7.252)
(~
The thermal conductivity in this case becomes
k-
• Lqq
(7.253)
A T At stationary state, when the flows of substances vanish (Ji = 0), Eq. (7.244) yields
4 ' - - . Lqq - i=1 LiqQi
T -- ks --6AT
(7.254)
where ks is the thermal conductivity at stationary state, and related to the phenomenological coefficients by
(~
ks ~ ~-~
i=1
Liq Qi
- -~
i,m=l
LimQi m
(7.255)
Since the thermal conductivity is always positive, the heat flow is directed from the warmer part to the colder part. At stationary state, Eq. (7.246) becomes
0 - rAP+ k=l
OWk T,P,w/~x
Awk + Q__LAT T
(7.256)
Using the relations 2~=1Awk = 0 and E"k--t ~Xwk - 1, we can determine the pressure difference and (n - 1) mass fraction differences. 7.9.1
Thermal Effusion
For a binary fluid, we have
O-V1AP+(OtIx---L
Awt + Q__LAT T
(7.257)
O-V~AP+(°tx---L)
Aw~ + Q__LAT T
(7.258)
OWl T,P
"-
OWl T,P
Using the Gibbs-Duhem equation
OWl
T,P
-I-W2( 0~---~2 )OW 1 T,P
404
z.
Heat and mass transfer
Equation (7.258) becomes
O -- V2Ap _ W_____(~l°312'2] Aw 1 + ~ AT T w2 t OW1) r,p
(7.259)
After eliminating the pressure difference from Eqs. (7.259) and (7.257) and using the specific volume of the binary solution V = w l V 1 - w z V 2, we have
VT
(7.260)
t OWl )T,P
This equation describes the change of the difference of the mass fraction of component 1 with respect to a change in temperature at stationary states. This effect is called thermal effusion.
7.9.2
Thermomolecular Pressure
If we eliminate the mass fraction difference AWl from Eqs. (7.259) and (7.257), we obtain 'Sat' - - w2Q~ + w2Q2 AT VT
(7.261)
This equation describes the change of pressure difference with the temperature difference. This pressure difference is called the thermomolecular pressure.
7.9.3
Thermoosmosis
In a special membrane system, we may use a membrane, that is permeable only to component 1 so that J2 = 0. At a stationary state, we also have J1 = 0. If the membrane is movable and the pressure difference is zero, then from Eq. (7.257), we estimate the mass-fraction difference Aw 1
=
_
Ql*
AT
(7.262)
(0[D1/0W1)T,P T This equation describes a difference in the mass fraction arising because of a temperature difference. This phenomenon is called thermoosmosis, which is thermal diffusion in a discontinuous system.
7.9.4
Osmotic Pressure and Temperature
If the membrane is motionless and if there is no temperature difference, then from Eq. (7.257) we obtain the pressure difference kip = - (0~1 / Owl)v'e Aw 1
(7.263)
vl This equation describes the pressure difference because of the mass fraction difference when there is no temperature difference. This is called the osmotic pressure. This effect is reversible because AT = 0, J2 = 0, and at stationary state J1 = 0. Therefore, Eq. (7.244) yields Jq' = 0, and the rate of entropy production is zero. The stationary state under these conditions represents an equilibrium state. Equation (7.263) does not contain heats of transport, which is a characteristic quantity for describing nonequilibrium phenomena. The temperature difference arising from a mass fraction difference is called the osmotic temperature (7.264)
tOWl
Z9
7.9.5
Heat and mass transfer in discontinuous system
405
Thermomechanical Effects: One-Component System
Consider a system with two parts. The parts are separated by a permeable membrane. The two parts may have different temperatures and pressures. Therefore, two generalized flows of substance and heat occur, while the temperature difference and pressure difference are the two thermodynamic forces. In terms of entropy flow Js, the dissipation function is
qt - _ a , . V T - ~ Ji • V/x; i=1
(7.265)
This local equation should be integrated across the membrane to find an expression for a discontinuous system. For a steady state system with a single component, the integrated form is xp = J A T + J1A~I
(7.266)
After identifying the conjugate forces and flows, for small forces of AT and A/x, the heat flow and mass flows may be represented by the following linear phenomenological equations -J1 = L11A/Xl +LI2 A T
(7.267)
- J s = L21Atxl + L22AT
(7.268)
Here, the reciprocal rules hold, and we have L12 - L2~. The introduction of the explicit form of chemical potential for a single component Atx ~ = - S A T
(7.269)
+ VAP
into the phenomenological equations yields
-J1 = LI 1VAP1 + (L12 - L11S) AT
(7.270)
- J s = L21VAP1 +(L22 - L21S) AT
(7.271)
where S and V are the partial molar entropy and the partial molar volume of the component, respectively. One important case would be at steady state when a constant temperature difference is applied to the discontinuous system. Under these conditions, a pressure difference develops across the membrane that leads to dl = 0. The magnitude of this stationary pressure is obtained from Eq. (7.269)
z2LR) -~
-(LI2 - L11S)
-L12
S
VL I1
VL I1
V
.I,-:o
(7.272)
Equation (7.271) shows that the thermoosmotic effect is dependent on two factors. One is proportional to the ratio L12/Lll, and represents a coupling between the flow of the substance and the flow of entropy (heat). The other is proportional to the partial molar entropy S, since the difference in temperature causes a difference in chemical potential, as Eq. (7.269) shows. The coupling phenomenon may be described in terms of the entropy of transfer S*, which is the entropy transferred by a unit flow of substance under conditions of uniform temperature, and defined by
(7.273) Lll
~r=o
Combining of Eqs. (7.272) and (7.273) yields AP j •~
=-(S*-S) I~=0
V
(7.274)
406
7.
Heat and mass transfer
This equation describes the steady state pressure difference induced by a temperature difference. For a system containing a single component, the total heat flow Jq and the reduced heat flow Jq' are TJ~ = Jq - lXlJ 1
(7.275)
Jq' = Jq - H 1 J 1
(7.276)
where H1 is the partial molar enthalpy. Equations (7.275) and (7.276) help to derive the heat of transfer and energy of transfer for the discontinuous system, and relate them to S*. Introducing Eq. (7.275) into Eq. (7.276) yields
J~
1
Jq
(7.277) where (Jq/J1)Ar= o may be called the energy of transfer U* so that we have TS* = U* - tx
(7.278)
Using Eq. (7.276), we find
0-.1)-u--1
(7.279)
and the heat of transfer is (Jq'/J1)~V= 0 =Q*. Therefore, Q* = U* _ H. From the relation H =/z TS and the heat of transfer, we have Q* = T(S* - S). Using the heat of transfer, we may describe the steady state thermoosmotic effect by AP/J - -Q* ST- l=0 VT
(7.280)
A general example by Denbigh may clarify the concept of quantities of transfer. Consider two compartments separated by a barrier that is permeable only to molecules of a relatively high energy. If the molecules with higher energy penetrate from compartment 1 to compartment 2, then the energy of the transported molecules in compartment 2 will be greater than the average energy of the molecules in compartment 1. Compartment 1 will lose energy because of transfer, while compartment 2 will gain energy. In order to maintain a uniform temperature, heat equal to the transferred amount should be added to compartment 1 and removed from compartment 2. For the significance of thermal osmosis, consider the transfer of water in biological systems. An estimated heat of transfer of water across plant cell membranes is approximately 17,000 cal/mol. If this value is used in Eq. (7.280) along with appropriate values of V and T, we estimate that a temperature difference of 0.01 °C would cause a stationary pressure difference of 1.32 atm. However, the maintenance of a temperature difference of 0.01°C leads to a rather large temperature gradient of 10,000°C/cm as the membranes are about 100 A thick. Unless the membrane has a very low thermal conductivity, such a high temperature gradient may be difficult to maintain. In contrast, many chemical reactions taking place within a cell produce or consume heat, and therefore some local temperature gradients may exist and contribute in the transport of substances across biological membranes.
7.10
THERMOELECTRIC EFFECTS
Thermoelectric effects demonstrate the existence of coupling between electrical and thermal phenomena.
7.10.1
Seebeck Effect
In a thermocouple, heating one junction of a bimetallic couple and cooling the other produces electromotive force in the circuit. This observation was originally was made by Seebeck in 1821. Besides the use of thermocouples, transistor electronics and semiconductors are important areas of interest for thermoelectric phenomena. Thermocouples made of semiconductors can develop relatively large electromotive potentials and are used to convert heat into electricity.
7.10
7.10.2
407
Thermoelectric effects
Peltier Heat
In 1834, Peltier observed that the passage of electric current I through a bimetallic circuit caused the absorption of heat at one junction and rejection of heat at the other junction. The heat flow per unit current at constant temperature was called the Peltier heat qpe, and defined by heat added or removed
(7.281)
qPe =
I
With the use of semiconductors, it is possible to achieve rapid heating or cooling by using the Peltier effect. Relatively large temperature differences, as high as 70°C, can be maintained between hot and cold junctions.
7.10.3
Thomson Heat
Figure 7.4 shows the Thomson heat system. Consider a homogeneous wire heated to 373 K with two endpoints cooled to 273 K. If there is no current passing through the wire, the temperatures at points a and b would be the same. However, after passing a current (I), the temperatures at points a and b are different. Therefore, the electric current disturbs the temperature gradient, and the original gradient can be maintained only by adding or removing heat. The heat necessary per unit current and per unit temperature gradient is called the Thomson heat qTh, which is dependent on the nature of the wire. The Thomson heat for a metal wire a is defined by
qTh,.
-
1 6q
(7.282)
I dT
Figure 7.5 shows a composed of a bimetallic couple metal wires "a" and "b" with one junction maintained at temperature T and the other maintained at T+ dT. An electromotive force E causes a current I to pass through the wires. A Peltier heat qpe(T+dT) per unit current will be absorbed at the warm junction and an amount of heat qpe(T) will be given off at the cool junction. To maintain a temperature gradient, Thomson heat (qrh,a)(dT) must be supplied to the metal a, and an amount of heat (qTh,b)(dT) must be removed from b, since the current is in the opposite direction in metal wire b. In a closed work cycle, the electric energy is fully converted to heat. Therefore, the energy balance per unit current by the first law of thermodynamics is (7.283)
dE - qpe(T + d T ) - qpe(T) + qTh,adT-- qXh,bdT If we expand qpe(T + d T ) in a Taylor series and retain the first two terms, we have
qPe ( T + d T ) - qPe (T) + qPe dT dT 273 K CD
Ta a
373 K ~
Tb b
(7.284)
273 K ~ I
Figure 7.4. System for the Thomson heat demonstration. The uniform wire is at 373 K at the middle point. At the end points, temperatures are held at 273 K. After passing a current (/), the temperature at points a and b are measured.
qTh,b
qPe(T) T
Figure 7.5. A bimetallic couple of metals a and b, the two junctions (points 2 and 3) are held at different temperatures Tand T + dT. qPe and qmh show the Peltier and Thomson heats respectively, while E is the electromotive force.
408
7.
Heat and mass transfer
Using this relation in Eq. (7.283), we obtain dE
dqPe + qTh,a -- qTh,b
dT
(7.285)
dT
This is the first equation of Thomson for thermoelectricity. 7.10.4
Flows and Forces in a Bimetallic Circuit
The metallic circuit we consider has only electrons flowing, and the dissipation function in terms of entropy flow as is -- --Js " V r - J e
"V/&e
(7.286)
With the identified conjugate forces and flows, the linear phenomenological equations are - J s = L l l V T + L12V/&e -ae -
L21VT+
(7.287) (7.288)
L22V/&e
Consider the Seebeck effect resulting from two junctions maintained at two different temperatures as shown in Figure 7.5. Assume that points 1 and 4 are at the same temperature To. These points are connected to a potentiometer so that the electromotive force E can be measured with zero current Je = 0. Under these conditions and using the reciprocal rules, Eq. (7.287) yields li7/&e - - L 2 1
VT---S*VT
(7.289)
L22
where S* is the entropy of transfer and represents the entropy transferred per unit flow of electrons at uniform temperature
V/Xe
(7.290)
/Je) VT=0
L22
To estimate the total electromotive force of the circuit in Figure 7.5, Eq. (7.289) must be integrated between points 1 and 4. Assuming one-dimensional gradients, the result is obtained by summing the following integrals /&e2 -- ~&el -- -/&e4 --/&e3 -- --
I;a
* S a dT,
I;
* +dT Sa d T ,
/&e3 -/&e2
__
f T+dT , --a r Sb dT
rT+dT /&e4 - / & e l z AlL e z .IT
(7.291) *
($2 - Sb )dT
Since points 1 and 4 are at the same temperature, and due to electroneutrality in the circuit, there is no concentration gradient for the electrons, and we have A/&e = - F E , where F is the Faraday constant. Therefore, Eq. (7.291) becomes E = - - - 1 FT+dT(s a _ S b ) d T FaT
(7.292)
After differentiating Eq. (7.292) with respect to T, we find d E = _ Sa* - S b dT F
(7.293)
This equation is called the relative thermoelectric power of the metal "a" against "b". Since the transfer of entropy depends on the cross coefficients L12 or L21, this derivation represents coupling between the electrical and thermal phenomena.
7.10
Thermoelectric effects
409
junction a
b 1
J ,/,%,.
Jq, b ,~
qPe Figure 7.6. Schematic of a junction between metals "a" and "b";/is the electric current, and Jq,a and and qPe is the Peltier heat absorbed at the junction.
Jq,b are
heat flows in two metals,
Consider the Peltier effect where a heat flow accompanies a current under isothermal conditions. Figure 7.6 shows the junction between metals "a" and "b" at which the Peltier heat is absorbed. After applying Eq. (7.288) to both metals "a" and "b," we have (7.294)
Jc -- FL22,aVE = FL22,bVE
After introducing the current I, defined by I
(7.295)
- -FJ~
Equation (7.294) may be expressed in the form of Ohm's law - I = F 2L22,.VE
and
- I = F2L22,bVE
(7.296)
The heat flows passing the two metals a r e TJs, a and TJs, b and are not equal, since the Peltier heat must be absorbed at the junction to maintain constant temperature. Therefore, (7.297)
TJs, b - TJs, a - qPe
Using Eq. (7.287) at uniform temperature, we find TJ,,, - - TL~2,, V~c,
TJ~,,b = -TL12,b Vtxc
(7.298)
After dividing Eq. (7.298) by Eq. (7.296) side by side and using Eq. (7.290), we get [ TJ,,, ) I
TL12,a _ L22,a
VT=0
TL12,b _ L22,b
VT=O
TS; F
(7.299)
TS;
F
(7.300)
By inserting these expressions into Eq. (7.297), we have
qPe n
r(Sb --Sa F
(7.301)
Comparing Eq. (7.297) with Eq. (7.293), we find qPe-
T dE dT
(7.302)
This is known as the second equation of Thomson. This relation is based on the Onsager reciprocal rules, and the experimental verification of Eq. (302) would be additional confirmation of Onsager's rules.
z
410
Heat and mass transfer
Differentiation of Eq. (7.302) with respect to temperature yields dE + T d2 E dT dT 2
d q p e __
dT
(7.303)
Using the first equation of Thomson and Eq. (7.293), we have
qTh,a--qTh,a - - - - - -
dT 2
F
dT
dT
(7.304)
This relation represents the Thomson heat with specific entropies of transfer of individual metals "a" and "b"
qTh,a =
T dS a F d T ' qTh,b
T dS b F dT
(7.305)
Therefore, using Eqs. (7.302), through (7.304), the Peltier heat, its variation with temperature, and the Thomson specific heats may be estimated.
PROBLEMS 7.1
Derive equations that describe the temperature profiles for a plane wall, long hollow cylinder, and hollow sphere. Assume constant thermal conductivity, and temperature at the walls as T1 and T2.
7.2
Derive modeling equations for (a) Heat conduction with an electrical heat source. (b) Heat conduction with a nuclear heat source.
7.3
Derive modeling equations for (a) Heat conduction with a viscous heat source. (b) Heat conduction with a chemical heat source.
7.4
(a) Consider a gas mixture of species A and B confined between walls, which are apart from each other with a distance L, and maintained at different temperatures T(L) >> T(0). The walls are impermeable. The gas mixture is free of chemical reactions and convection flows. Derive a general equation to estimate the mole fraction difference of species A, XA(L) -- XA(0), induced by the temperature gradient between the walls. (b) Consider the same gas mixture. The species A is consumed by a fast, irreversible reaction 2A --+ B occurring on the surface of one of the walls at L = 0. The solid walls are impermeable, and there is no heat flow through the walls. Temperature and composition at L = 0 are To and CA0. Estimate the heat flow within the walls because of the Dufour effect.
7.5
Assume that the temperature distribution in a circular rod with internal heat source q may be represented by the following ordinary differential equation (ODE) d2T + 1 dT dr 2
r
dr
+q=0
The nondimensional radius changes over the range 0 < r < 1. The boundary conditions are
r ( r = 1)= 1,
dT )r = 0 -a-; -o
For the heat source values of q = 100 and 1000 J/m 2, plot the temperature profiles.
Problems
411
7.6
Seawater is pumped into a well-mixed tank at a rate of 0.5 m3/h. At the same time, water is evaporating at a rate of 0.02 m3/h. The salty seawater flows out at a rate of 0.5 m3/h. Seawater concentration is 6 g/L. The tank initially contains 1.5 m 3 of the input seawater. Determine: (a) The change of volume as a function of the time. (b) The salt concentration as a function of the time.
7.7
A sphere of ice with a 10 cm radius is initially at 273.15 K. This ice sphere is placed on a sponge, which absorbs the melted water. Determine the diameter of the ice sphere as a function of time. Assume that the heat transfer coefficient is 5 W/(m 2 K). The latent heat of melting is 333 kJ/kg, and the density of ice is approximately 0.917 kg/m 3.
7.8
Consider a tapered conical cooling fin. Assume that the temperature distribution of the cone can be described by the nondimensional differential equation should be d2Y+ dx 2
-aT
0
dx
where a is a nondimensional parameter defined by
a=~
hL/ 1 + ~ 4 / k
2m 2
where h is the heat transfer coefficient, k is the thermal conductivity, L is the length of the cone, and m is the slope of the cone wall. The boundary conditions are x - 0 T = 0 and x = 1 T = 1. Plot the temperature profile for oL- 30. T(x=O)=O
X
T(x=
7.9
Consider a 10-cm long thin rod. Solve the heat conduction equation OT 3t
-k~
02T dx 2
using the boundary conditions ofx - 0 T = 100°C, and x = 10 T = 25°C, and the initial condition of at t = 0 T = 0. The thermal conductivity k - 2.5 W/(cm K).
7.10
Consider a 10-cm long thin rod. Solve the heat conduction equation o30
oq20
37"
dy 2
using the boundary conditions of at v - 0 0 - 100°C, and y - 10 0 = 25°C, and the initial condition of at ~- = 0 0 - 0 (0 = < y - y>--0
Determine the temperature change in time and distance.
REFERENCES R.B. Bird, W.E. Stewart and E.N. Lightfoot, Transport Phenomena, 2nd ed., Wiley, New York (2002). J. Chan, J.J. Popov, S.K. Kehl and D.G. Leaist, J. Sol. Chem., 32 (2003) 197. R.M.L. Coelho and A.S. Telles, Int. J. Heat Mass Transfer, 45 (2002) 3101. Y. Demirel and S.I. Sandler, Int. J. Heat Mass Transjer, 44 (2001) 2439. Y. Demirel and S.I. Sandler, Int. J. Heat Mass Transjer, 45 (2002) 75. D. Kondepudi and I. Prigogine, Modern Thermodynamics, Wiley, New York (1999). A. Perez-Madrid, J.M. Rubi and R Mazur, Physica A, 211 (1994) 231. G. Platt, T. Vongvanich, G. Fowler and R.L. Rowley, J. Chem. Phys., 77 (1982) 2121. G. Platt, T. Vongvanich and R.L. Rowley, J. Non-Equilib. Thermodyn., 8 (1983) 1. R.L. Rowley and M.D. Hall, J Chem. Phys., 85 (1986) 3550. R.L. Rowley, S.C. Yi, V. Gubler and J.M. Stoker, Fluid Phase Equilib., 36 (1987) 219. R.L. Rowley, S.C. Yi, V. Gubler and J. M. Stoker, J. Chem. Eng. Data, 33 (1988) 362. M.E. Schimpf and J.C. Giddings, J. Polym. Sci. Part B, 27 (2003) 1317. M.E. Schimpf and S.N. Semenov, Phys. Rev. E, 70 (2004) 031202. D. Tondeur and E. Kvaalen, Ind. Eng. Chem. Res., 26 (1987) 50. M. Venegas, M. Izquierdo, R Rodriguez and A. Lecuona, lnt. d Heat Mass Transfer, 47 (2004) 2653. S. Wisniewski, B. Staniszewski and R. Szymanik, Thermodynamics of Nonequilibrium Processes, D. Reidel Publishing Company, Dordrecht (1976). S.C. Yi and R.L. Rowley, J. Chem. Phys., 87 (1987) 7214.
REFERENCES FOR FURTHER READING A. Barletta and E. Zanchini, Int J Heat Mass Transjer, 40 (1997) 1007. J.M.O. de Zarate, J.A. Fornes and J.V. Sengers, Phys. Rev. E, 74 (2006) 046305. R.V. Devireddy, D.J. Smith and J.C. Bischof, J Heat I'ransfer, 124 (2002) 365. E Guillemet, J.P. Bardon and C. Rauch, hit. J Heat Mass Transfer, 40 (1997) 4043. J. guettmer-Strathmann, J Chem. Phys., 119 (2003) 2892. J.M. Simon, D. Bedeaux, S. Kjelstrup, J. Xu and E. Johannessen, J. Phvs. Chem. B, 110 (2006) 18528. S.E. Wright, D.S. Scott, J.B. Haddow and M.A. Rosen, Int..1 Eng. Sci., 39 (2001) 1961. J. Xu, S. Kjelstrup, D. Bedeaux, A. Rosjorde and k. Rekvig, ,l Colloid Intelf Sci., 299 (2006) 452. M. Zhang and F. Muller-Plahte, J. Chem. Phys., 125 (2{}{}61 124903.
8 CHEMICAL REACTIONS 8.1
INTRODUCTION
A chemical reaction is an irreversible process that produces entropy. The general criterion of irreversibility is diS > O. Criteria applicable under particular conditions are readily obtained from the Gibbs equation. The changes in thermodynamic potentials for chemical reactions yield the affinity A. All four potentials U, H, A, and G decrease as a chemical reaction proceeds. The rate of reaction, which is the change of the extent of the reaction with time, has the same sign as the affinity. The reaction system is in equilibrium state when the affinity is zero. This chapter, after introducing the equilibrium constant, discusses briefly the rate of entropy production in chemical reactions and coupling aspects of multiple reactions. Enzyme kinetics is also summarized.
8.2
CHEMICAL REACTION EQUILIBRIUM CONSTANT
A single reaction can be represented by
(8.1)
piBi i--1
where B i is species i, n is the number of different species, and u/are the stoichiometric coefficients for species i, which are positive for products and negative for reactants. The amount of Ni of species i at any time is given by N i -
Nio
(8.2)
+ l,, i E
where N/0 is the initial amount of species i, and e is the extent of the reaction, which is an extensive property. Differentiation of this equation is dN
(8.3)
i -- pid¢ "
For a single reaction in a closed system, the differential of the Gibbs energy is
dG - - fdt + VdP + { £ vi txi l
(8.4)
where/x; is the chemical potential of species i. At constant T and P, we have the Gibbs energy of reaction AG r
--
T,P
Pi[Li
= AG r
(8.5)
i=1
At chemical equilibrium, the Gibbs energy of the system is minimum, and from Eq. (8.5) we have
/
Pi[d,i,eq-- 0
--
T,P
i=1
(8.6)
416
8.
Chemical reactions
The chemical potential in terms of activity a is o
Id'i "-" [£i + R T In a i = t x ° + R T In c i')/i
(8.7)
where/x ° is the standard chemical potential (/xi =/z ° at a = 1), and activity is defined by ai = ciYi, where Yi is the activity coefficient of species i, which is a strong function of concentration. Assuming that the activity coefficient is either unity or close to unity, the chemical potential becomes ~ i -- t"L° nt- R T
In c i
(8.8)
Substituting Eq. (8.8) into Eq. (8.6), we have 1/
//
1,itx ° = - R T ~ _ ~ 1,i i=1 i=1
In Ci,eq
= -RT
In I-I (Ci,eq)ui
(8.9)
i=1
Using Eqs. (8.5) and (8.9), we relate the Gibbs energy of reaction and equilibrium constant K zXG ° = - R T
(8.10)
ln K
and
(8.11)
K = r I (¢i,eq ) vi i=1
Example 8.1 Equffibrium constant of a reaction Consider a homogeneous reaction between species B and P: B = P. An ideal solution of 1 L containing 1 mol of B initially has the concentration 1 M. Define the equilibrium constant K. The Gibbs energy of the reaction is G = N B (tx~3 + R T
ln[B] + Np (/x~, + RT[P])
Using the extent of reaction E, we find NB = 1 - e and Np = E. With these amounts, the relation above becomes G = ( 1 - e)/x~3 + e/X~ + RT[(1- E)ln(1- E) + e In E] At equilibrium (OG/Oe)p = 0, and the equilibrium extent of reaction is 8eq then we find
o
o
tXp - tXB = - R T
In
{eq / ...... 1 - Eeq
RT
In K
Example 8.2 Equilibrium compositions Consider the following chemical reaction 1,1A 1 + 1,2A 2 ~ 1,3A 3 + 1,4A 4
Determine the equilibrium compositions in terms of the extent of the reaction. The reaction coordinate 8 is: dE = dni/vi dE=
d/'tl _ d//2 _ d//3 _ d//4 1"1
d n i = 1,id8
1'2
dl/l i = 1,i t
1,'3
de
1,'4
n i = nio + 1"iE
8.2
Chemical reaction equilibrium constant
417
n = E ni = E nio + e , E Pi = no + e,P i
i
i ni _ nio + Pie,
Yi-
n
n o + t.,e,
For a multireaction, we have nio + ~ , j l"i,e,.j Yi z
n o + ~.,,j P j e , j
For the following reaction, we have 2 mol
CH4, 1 mol H20, 1 mol CO, and 4 mol H2 CH 4 + H 2 0 ~ C O + 3H 2
The summations of stoichiometric coefficients and the initial compositions are" u= ~u~ =-1-1+1+3=
2
i
no = ~ n i 0 = 2 + 1 + 1 + 4 = 8 i
We have the compositions in terms of the extent of reaction
YCtt4
8.2.1
2--e" 8 + 2e
1--e, 8 + 2e" YH2
YH,O
4+3e, 8 + 2e" YCO
l+e, 8+2e,
Changes of Thermodynamic Properties in Chemical Reactions
From Eqs. (8.4) and (8.5), the Gibbs energy becomes d G - - S d T + VdP + ~ @de,
(8.12)
The reaction entropy and volume change by (OS)
Oe
=
,e
=_AS r
OAG r
/
]r,~
= -AVr
OP
OS
(8.13)
(8.14)
(8.15)
where ASr is the reaction entropy and A Vr is the reaction volume. The Gibbs energy of reaction in terms of the enthalpy of reaction M-/r is (8.16)
AG r - AH r - TAS r
The enthalpy of reaction AHr can be calculated from the Gibbs-Helmholtz equation
•
-T 2 (O(AGr/T) OT
)
= p
~I
r
(8.17)
418
8.
Chemical reactions
The standard enthalpy of reaction &/_/oand the standard Gibbs energy of reaction AG ° can also be obtained from the standard enthalpy of formation of species i ~/~i and the standard Gibbs energy of formation of species i AG'~i by the following equations
A H ° -- ~ P i A H ~ i
and
AG ° - ~
i=1
pima~i
(8.18)
i=1
From the definition H = G + TS, we have ~ H r -~ A G r -+"T A S r
(8.19)
From Eqs. (8.10) and (8.17), we can relate the equilibrium constant with the standard enthalpy of reaction 0 In K ) _ M-/° OT p RT 2
(8.20)
This is the van't Hoff equation. If M-/r° is assumed to be independent of temperature, then the integration of this equation between T1 and T2 yields
1!
(8.21)
The derivative of the enthalpy of reaction with respect to temperature yields the heat capacity of the reaction at constant pressure (8.22)
ACpr -- ~ Pifpi i=1
If Cpi does not change significantly in the experimental temperature range, the enthalpy of the reaction will change linearly with T, while entropy changes logarithmically AH o = M-/°(298.15K) + A C por ( T - 2 9 8 . 1 5 )
(8.23)
A S ° ( T ) - AS°(298.15 K) + ACpln ~ 298.15K
(8.24)
Substituting these relations in the following relation AG ° = M-/° - TAS ° = - R T In K yields lnK =-AH°(298"15K) RT
+ '5S°(298"15 K) R
o(
ACpr 1R
298"15K~- In ~ T T 298.15K
(8.25)
Using this equation, we can estimate the effect of temperature on K.
Example 8.3 Temperature effect on equilibrium conversion Consider the elementary reversible reaction B ~- P with no initial product P, while the initial concentration of B is B0. The standard Gibbs energy and standard enthalpy of the reaction are AG ° (298.15K) - - 1 4 . 1 kJ/mol and AH ° (298.15K) -83.6kJ/mol. Assume that the specific heats of solutions are equal to that of water. Estimate the equilibrium conversion of B between 25°C and 120°C. For an ideal solution K = Cp,eq/CB,eq, AGr = - R T in K, and we have 14100 / K(298.15K) = exp 8.314(298.15) = 296.2
8.3
1
'
-
0.8
!
~
~
i
~
-~ . . . . .
I
!
i
0.7
----/
OOo,
:5" W
419
The principle of detailed balance
......
.....
.................
±
.......... i
04
.
.
.
.
.
.
.
ii
.
i
0.2
3()0 310
320
Figure 8.1.
330
340
350 T,K
360
370
380
390
400
Change of conversion with temperature.
Since the specific heats of the reactant and product are the same, AC~, = 0, and
AH°(T)
= AH°(298.15 K ) +
From Eq. (8.21) with Ta = 298.15 K and
lnK =
2~H°(Tz - T l)
RT~T2 lnK
=
T2 =
AC°p(T -
298.15) = - 8 3 . 6 kJ/mol
393.15 K, we find the value of K in terms of T2
+ In K(298.15) -
AH°(T2-298"15) RT2 (298.15)
AH°r (T2 - 298.15)
RT2(298.15)
10055.3 + ln(298.15) = ~ T2
+ ln(298.15)
28.0
On the other hand, the equilibrium constant in terms of conversion becomes
Cp,eq _ CB0 -- CB,eq _
CBoXB,eq
K
CB,eq where the conversion is X B , e q
-- (CB0 --
CB,eq
CB0 (1 -- XB,eq )
CB,eq)/CBo.From
_m
XB,eq 1 - XB,eq
the above equation, we have
XB,eq K
_.._
1 - XB,eq
or K
XB,eq (T2) -
I+K
_
exp[(10055.3/T2)-28.0 ] 1 - exp[(10055.3/T2) - 28.0]
Figure 8.1 shows the change of conversion with the temperature. In order to secure 80% conversion, the temperature should be below 340 K.
8.3
THE
PRINCIPLE
OF
DETAILED
BALANCE
In chemical kinetics, the reaction rates are proportional to concentrations or to some power of the concentrations. Phenomenological equations, however, require that the reaction velocities are proportional to the thermodynamic force or affinity. Affinity, in turn, is proportional to the logarithms of concentrations. Consider a monomolecular
420
8.
Chemical reactions
reaction between a reactant R and a product P in the vicinity of global equilibrium R = R The rates of reaction by the kinetic theory are d C R -- - k f C R nt- kb C P
(8.26)
dt
d C p _ NfCR _ kbCp
(8.27)
dt
where kf and kb are the forward and backward reaction rate constants. Chemical kinetics distinguishes two reaction flows with two constants. However, there is only one macroscopic flow of reaction from the left to the right, and is given by Jr =
dCR - dCp
dt
(8.28)
dt
At equilibrium, the reaction flow vanishes and we have kfCR,eq = k bcP,eq or K-
kb - CR'eq kf Cp,eq
(8.29)
where CR,eq and Cp,eq are the equilibrium concentrations of R and P, while K is the equilibrium constant of the reaction. If we define the following deviations, xi, of the concentrations from their equilibrium values when the reaction system is close to equilibrium, we have ~R --" CR -- CR,eq, ~P -- Cp - Cp,eq
(8.30)
with
~R
~P
CR,eq
~P,eq
-0
(8.70)
j=l
The phenomenological equations are t
t
Jrj = E LJk Ak = Z LJk k=l
k=l
vik Mi tzi
(8.71)
i=1
and the Onsager reciprocal relations link the coefficients Ljk- Lkj. The dissipation function is a quadratic function of the chemical affinities l
- T ~ = __E LjkAiAk~ >- 0 j,k=l
(8.72)
8.5
425
Reaction velocity (flow)
The positive values of • imply the following inequalities
Lkx. > O, Lx.kLi i -L~./> 0
8.5
(8.73)
REACTION VELOCITY (FLOW)
The velocity of a chemical reaction is proportional to the product of the concentration of the species. For a reaction kf
A< the reaction rates for the forward Jrf and backward
/%
>B
(8.74)
Jrb directions
are
P-f
J~f - kr l-I ci,',: i
(8.75)
Pi.b
Jrb -- kb U Ci,b i
(8.76)
For a reaction j, the rate of reaction is given by
,i]
U i Ci Jr/ -- Jrf - Jrb -- Jrf i 1- -----~~
(8.77)
where K c - kf/kb is a constant at equilibrium conditions. For an ideal gas, the chemical affinity is given by iv[ A~
-
- - ~ uii tz~ /=-[
H i Ci
-
-RT
I n -
(8.78)
KC
where
HiCi
i
( A/ )
- exp -
(8.79)
so that the reaction rate becomes
.4/
(8.80)
If we expand the expression in brackets, and consider the case of a near-equilibrium state, which may be specified by the inequality IA//RT 0 in general, the right side of Eq. (8.102) is a quadratic form, and the matrix L is positive semi-definite. Each 2(,. can be regarded as a driving force, and each rate of change of a reaction coordinate Jj as a flow that depends on all the forces; this indicates a coupled phenomenon. The reciprocal relations are valid both for the states that are in chemical equilibrium and for those that are not.
Example 8.5 Conservation of mass in chemical reactions Describe the mass conservation in closed and open systems with n number of components and the number of moles changing because of chemical reactions. (a) Closed system: Consider the following single reaction
N 2 +3H 2 ~ 2 N H 3
(8.103)
The change of mass with time is d m i "-
uiMide, or in terms of the number of moles n i d n
i -- pide.
(8.104)
where vi is the stoichiometric coefficient of species i in the reaction. The values of Pi are positive for reactants (N2, H2) , and negative for products (NH3), and Mi is the molar mass of component i. The term e is the degree of advancement or extent of the reaction. For the reaction in Eq. (8.103), we have
dmN----~2= dmH-------=-~2dmNH------=--~3de or dtIN2
-- d/'/H2
-1
-3
--MN2
--3MH 2
2MNH3
-- d/'/NH3
2
--
de
(8.1o5)
The total mass is m
=
Zmi
(8.106)
i
From Eqs. (8.106) and (8.104), we have (8.107)
dm = (E ~'iMi) de" =
8.6
429
Multiple chemical reactions
where ~ i P i m i = 0 is called the stoichiometric equation. From Eq. (8.104), we define the chemical reaction rate or reaction flow as follows d~3
Jr -
(8.108)
dt
The time change of the number of moles is (8.109)
piYr - dfli dt
For r number of simultaneous chemical reactions, we have
= ~ Uikdek
dn i
(8.110)
k=l
Consider the following simultaneous reactions: 2C+O 2~2CO C +0
(k-- 1)
(8.111)
(k = 2)
2 ---~C O 2
(8.112)
The changes of the number of moles from Eq. (8.110) are dn c = - 2 d e
1 -de2,
dno2 = - d e
I
-de2, dnco = 2de1,
dF/co 2 = d e 2
(8.113)
(b) Open systems" The change of mass of component i has two contributions (8.114)
dm i = d e m i - d i m i
where d e m i represents the mass exchanged between the system and its surroundings, and dimi represents the change within the system. Using Eq. (8.110), we have
dmi :demi ff-Mi~PikdF.k ordn i :deni+~PikdF, k k=l
(8.115)
k=l
and with the stoichiometric equations Ev/kM/= 0, we obtain (8.116)
dm - dem
Equation (8.116) shows that the change of the total mass is equal to the mass exchanged with the surroundings.
Example 8.6 Calculation of entropy production for a reversible reaction Consider the following reaction between a substrate S and a product P: S
P
(8.117)
At time t, we have the concentrations of the species ~3
~3
C s - Cso - - - and Cp = Cpo + V V
(8.118)
The reaction velocity is
1 de J r -- J r f - J r b
-
V dt
(8.119)
430
8.
Chemical reactions
where the forward and the backward rates of reaction are Jrf = kf (Cso
-e) and Jrb
= kb (Cpo + s) for unit volume of V = 1
(8.120)
Combining Eqs. (8.119) and (8.120), we have d~ dt
- -~(kf
+ kb)+ kfCs0 -
(8.121)
kbCpo
~(o)=o. The solution is e = ~b[1- exp(-bt)]
(8.122)
where a
b = kf + kb, a = k f C s o - k bCpo, and 05 = -
b
After using these relations in Eq. (8.120), we obtain the affinity of the reaction considered A = R T ln Jrf - k f ( C s ° - s ) =
Jrb
kb (Cp0 +e)
kf[Cs0- 4)(1- exp(-bt))] kb[Cp0 + ~b(1- exp(-bt))]
(8.123)
The entropy production becomes
(i) = A 1 ( R T l n Jrf] T jr = ~(Jrf--Jrb) Jrb = R{kf[Cso
8.7
-
= R[kf(Cso-8)-kb(Cpo
+e)Jln kf(Cs° -e) kb (Cp0 -4-8)
(8.124)
~b(1- exp(-bt))]- kb [Cp0 + ~b(1- exp(-bt))]} In kf [Cs° - ~b(1- exp(-bt))] kb [Cp0 -[-~b(1- exp(-bt))]
STATIONARY STATES
At stationary state, all the properties of a system are independent of temperature. Stationary states resemble equilibrium states in their invariance with time; however, they differ in that flows still continue to occur and entropy is produced in the system. If a property is conservative, then the divergence of the corresponding flow must vanish; for example, Op/Ot = - d i v J. Therefore, the steady flow of a conservative quantity must be source-free and in stationary states; the flows of conservative properties are constant. If we consider the change of local entropy of a system at steady state Os/Ot = 0, the local entropy density must remain constant because external and internal parameters do not change with time. However, the divergence of entropy flow does not vanish: div ,Is = ~. Therefore, the entropy produced at any point of a system must be removed or transferred by a flow of entropy taking place at that point. A steady state cannot be maintained in an adiabatic system, since the entropy produced by irreversible processes cannot be removed because no entropy flow is exchanged with the environment. For an adiabatic system, equilibrium state is the only time-invariant state. Consider a monomolecular elementary chemical reaction S( klf ) X ( k2f > P klb
(8.125)
k2b
occurring under nonequilibrium conditions. In an open system, the substrate S is constantly supplied and the product P is constantly removed. The concentration of X is maintained at a nonequilibrium value. The evolution of X is defined by dCx
dt
-
k l f C S - k l b C X - k 2 f C X nt- k 2 b C P
(8.126)
8.7
431
Stationary states
Entropy production per unit volume in terms of reaction affinities is q~- A1 J~l + Az
T
(8.127)
T J~e
Flows of S and P keep the chemical potentials of/Xs and/Xp fixed leading to a fixed total affinity "41 -+ A2 - ( ~ s -/'Zx ) q- (/'Lx - ~P ) - ~ s - ~P -- A
(8.128)
Substituting this equation in Eq. (8.127), we get 1 diS
__A1Jrl
A-A1
Jr2
(8.129)
Jr1 - L l l -A-1 and Jr2 = L22 A - A 1 T T
(8.130)
VdtV
P-(I)-
T
+
T
The reaction velocities in the vicinity of global equilibrium are
Combining these relations with Eq. (8.129), the entropy production as a function of A1 becomes
Al2 ~ ( A1 ) - L11 _TT + L22
(A- 4 ) 2
(8.131)
T2
When the differentiation of • with respect to A~ is zero, we have a minimum O ~ ( A 1 ) _ LI 2 A 1 2 ( A - A1) OA 1 -17 + L22 T2 = 0
(8.132)
This equation shows that
A1
A2
(8.133)
Lll T + L22 --T-- - Jrl - Jr2 -- 0
The entropy production is minimized at the stationary state. The entropy production can also be expressed in terms of the concentrations. The value of Cx that minimizes the entropy production is the concentration of X at stationary state. The entropy production in terms of the reaction rates is
~ - R [ ( J r l f - - ' ] r i b )In( JrlbJrl----f-f) + (Jr2f - Jr2b) ln( Jr2j72~)]
(8.134)
The forward and backward reaction rates can be expressed in terms of concentrations Jrlf = klfCS, Jrlb - klbCx, Jr2f = k 2 f C x ,
Jr2b = k2bCp
(8.135)
At equilibrium, forward and backward rates of each reaction become equal to each other. The equilibrium concentrations are obtained from the principle of detailed balance
Cx'eq
-
klf k2b CS,eq Cp,eq kl b k2f
(8.136)
We may define small deviations in concentrations from the equilibrium 6 S = C S - Cs,eq, 6p -- Cp - Cp,eq , 6 X = C X -- Cx,eq
(8.137)
432
8.
Chemical reactions
These small deviations occur because of small fluctuations of the flows of S and R and 6s and 6p are fixed by these flows, while 6x is determined by the chemical reaction. Combining and rearranging Eqs. (8.134), (8.135), and (8.137), we have the entropy production in terms of the deviations (I)(aX) = R
[
(klfaS - klbaX )2 -Jr-(k2fax _ k2ba P )2 klfCs,eq
k2fCx,eq
J
(8.138)
If we set O~/06x = 0, the value of deviation 6x that minimizes the entropy production becomes 6x =
klfa S -Jr-k2ba P
(8.139)
klb q- k2f
Using Eq. (8.137) in Eq. (8.126), we have the stationary state
dCx dt
-
(8.140)
klfa S - klba x -- k2fa x -Jr"k2baP -- 0
This equation yields the stationary value of 6x given in Eq. (8.139). The stationary value of X minimizes the entropy production. At stationary state, the total entropy of the system is constant
dS - - 4S - +
4S
dt
dt
dt
(8.141)
=0
Therefore, the entropy change with the environment becomes negative
deS -
diS
This equation shows that the phenomenological coefficients are related to the reaction rates at equilibrium. The principle of detailed balance or microscopic reversibility is incorporated into Jr3f = J r 3 b - - J r 3 f , eq, and hence the Onsager reciprocal relations are valid. The second law requires the total entropy production resulting from all the simultaneous reactions to be positive. This has been verified experimentally. Sometimes, a system has two simultaneous coupled reactions, such that 4Jr1 < 0, A2Jr2 > 0
(8.166)
although the sum of the entropy production is always positive A1Jrl + A2Jr2 > 0 VT
(8.167)
8.7
435
Stationary states
Example 8.8 Entropy production in a homogeneous chemical system The oxidation of sulfur dioxide to trioxide over a vanadium pentoxide catalyst is 1 02 ---+SO 3 SO 2 + ~"
(8 168)
Sulfur trioxide is used to produce sulfuric acid, one of the most common chemicals used in industry. The reaction is strongly exothermic. Here, a tubular reactor is considered. The following relations are from Kjelstrup et al. (1999). The entropy production rate per unit volume of a chemical reactor is given by
/ A)
d~ - J~ - ~ =Jr X where Jr is the reaction rate, energy of reaction
(8.169)
(-A/T) is the thermodynamic driving force X, and A is the affinity, which is the Gibbs A = -AG r
(8.170)
The minimization problem is formulated as a Lagrange optimization
6
(8.171)
aX f (-JrX + AJr)dz= O where A is the Lagrange multiplier for the constraint of constant output J
J = Ac~Jr dz
(8.172)
where Ac is the cross-sectional area of the reactor. Since X can be related to concentration c, we can differentiate with respect to c instead of X, and obtain from Eq. (8.171) 2~ t"
6c
J(-Jr x + aJr)
dz -
(8.173)
0
The solution of this equation is possible for a particular expression for Jr. Considering a nonlinear chemical reaction, a practical solution may be obtained by the integration of the Gibbs-Helmholtz equation from equilibrium to optimal state
AHrd
Xopt = - f
(8.174)
eq
At equilibrium, Aeq = 0. The equilibrium temperature for a given conversion is found by using the condition Jr = 0. When we know the enthalpy of the reaction as a function of temperature, we can carry out the integral in Eq. (8.174). Since Yopt is known from Eq. (8.173), we can then find Topt. For a reactor operating with constant output, the criterion for optimal performance is for the cooling medium to have the highest possible temperature in the heat removal system. For a working example of the nonadiabatic reactor, there are 4631 cylindrical tubes with inner diameters of 7 mm packed with a catalyst and surrounded by a constantly boiling liquid at 703 K. Sulfur dioxide and air are fed into the reactor at a total pressure PT, in volume fractions of Yso2 = 0.11 and Yo2 - 0.10. The empirical expression of Jr takes into account diffusion and reaction kinetics, and we have
Jr = - k
Pso,
]
PSO3 )2
[Po2-{
]
(8.175)
The rate constant k has the dimension of mol/g of the catalyst, while the partial pressures are given in bar. The equilibrium constant is based on partial pressures, Kp (bar-°5). The rate of the reaction is negligible below 670 K,
436
8.
Chemical reactions
and the temperature of the materials in the reactor should not exceed 880 K; therefore, the temperatures of 670 and 880 K are practical boundaries for real operation. An expression for k is given by k = exp I - ~-97800 8 T 110 ) +ln(l"T
913]
(8.176)
where T is in K. Similarly, Kp in terms of temperature is given by - e x p -11800 --F-
_ 11.2]
(8.177)
The volume flow into the reactor is 3590 kmol/h. By introducing the degree of conversion x and the pressure P as variables, we obtain Jso3 =
[
kp 1- x Pso2 0.91-- 0.5X x
P
_
1 - 0.0055x
)2]
x
(1-x)Kp
(8.178)
where p is the mass density of the gas mixture. The affinity of the reaction is 1
A =/Xso 3 -/Xso 2 - ~/x02
(8.179)
and the driving force is
A REin[/ / x -
T
Pso2 P
~
-
In
0.91-0.5x
]
Kp
(8.180)
Equations (8.178) and (8.180) show that Jr is a complicated function of P, T, and composition, and it cannot be expressed in the form of linear phenomenological equations with constant coefficients. Estimations indicate that 80% of the conversion is accomplished within the first 20% of the reactor length. We obtain the optimum force Yopt as a function of A by solving Eq. (8.173) with the help of Eqs. (8.179) and (8.180) R F[(1/x(1- x))+(O.225/(1-O.O55x)(O.91-O.5x))] Xopt = -A - R [(F/Zx(1 - x)) + [(0.45Pso 3/(100.055x) 2)(P/Pr )] + (2x/(1 - x) 3K 3)] where
( F = P SO2 1 - 0.055x
(8.181)
)2
(1-x)Kp
The results for Yop t can be combined with Eq. (8.171) for a simultaneous determination of the conversion of x and P. The optimal driving force is small, around 10 J/(molK), and is almost constant through the reactor. However, the actual force is large at the inlet (about 30 J/(mol K)), and it passes a minimum (about 10 J/(mol K)) in the central region before it flattens out (about 16 J/(mol K)). This minimum is due to high temperature produced by the exothermic reaction. The large deviation between the actual and the optimal forces indicates that the lost work can be reduced without changing the rate of production. For example, maintaining stepwise control of temperature by using heat exchangers at the right positions may bring the temperature profile or the actual driving force close to the theoretical optimum, and reduce entropy production. When the optimal driving force is found for a given production rate, one may consider further optimization and, for example, further trade-offs can be made between the production rate and reactor size. Generally, a chemical reaction has a nonlinear relation between the rate Jr and the driving force -A/T. Chemical reactors are often designed to operate at the maximum rate of reaction. An alternative is a reactor operated with
8. 7
Stationary states
437
minimum useful work lost. The lost work per unit time in a chemical reactor is given by the Gouy-Stodola theorem, and is obtained by integrating the entropy production rate over the reactor volume W~ost - To~%IC~dx
(8.182)
where • is the entropy production per unit volume, Ac is the cross-sectional area that is assumed to be constant, and x is the length. Equation (8.182) implies a plug-flow reactor, since the integration is over the volume; in a batch reactor, integration is over time. The reaction takes place at temperature T, while To shows the temperature of the surroundings. Any reduction in lost work can result in economic gains, but these are not uniquely dependent on the reduction in lost work, since the reduction varies with energy and labor cost. So, it may be useful to undertake energy optimization studies separately from economic analyses. The minimization of Wlost(=Elost) entails the minimization of the integral in Eq. (4.182) with the constraint of constant production rate J (mol/h) and use of the Euler-Lagrange method
J = AcIJr dx
(8.183)
where Jr is the reaction rate (flow). When a chemical reaction is the only process in a system, the entropy production is the products of flows Jr and conjugate forces X
dp -- Jr ( - - ~ ) = JrX
(8.184)
Here, A is the affinity (the Gibbs energy of reaction) and T is the temperature. The flow for reaction is given by
Jr = L ( X ) X
(8.185)
where L is a phenomenological coefficient, which may be dependent on the force. An Euler-Lagrange minimization problem is formulated as
a f(JrX+AJr)dx=O 6X
(8.186)
where A is the Lagrange multiplier to be obtained in terms of Jr. For a required production rate, minimum lost work can be obtained by creating operating conditions compatible with the design parameters. This may be achieved when dcI)/dJr is constant. Using Eq. (8.184), the minimum lost work is obtained when
dX dJr
X + Jr --
=
constant
(8.187)
The solution to Eq. (8.186) is obtained with Eq. (8.187) when A = constant. Equation (8.187) means that the operating temperature should be parallel to the equilibrium temperature, when the enthalpy of the reaction is constant.
Example 8.9 Chemical reactions far from global equilibrium Consider the following chemical reaction far from equilibrium: A~B
(8.188)
The reaction rate is
Jr = klcA = k2CB
(8.189)
438
8.
Chemical reactions
where kl and k2 are the constants, and c A and CB show the concentrations of A and B, respectively. The following relations are from Kjelstrup and Island (1999). The normalized concentration c, due to mass conservation, is c = CA = 1 -- CB. The output of B per unit time J is given as
J = Act~[klC- kz(1- c)]dx
(8.190)
The driving force of the reaction is
- T (txA
tXB)
- R In
- In Keq
(8.191)
where Keq = kl/k2 is the equilibrium constant and R the gas constant, and the mixture is assumed to be ideal. The rate Jr, given in Eq. (8.189), is related to the force, Eq. (8.191), in a nonlinear way, as is the case for chemical reactions far from equilibrium. The total entropy production is determined by using the local value of the reaction rate and conjugate driving force in Eq. (8.184). Since the force X can be expressed in terms of concentration c, Eq. (8.186) becomes 0
- - ( J r X + AJ r) = 0 Oc
(8.192)
The result is given by
lneq] '1+'2(1+ lc The first term on the left of this equation is the thermodynamic force. The force is not necessarily constant when we have minimum lost work. The optimum force that gives the minimum total entropy production rate is obtained from
X°pt
(1 k1 + k2 1---~
To find the value ofh as a function of Jr and the rate constants of the reaction, we solve c(x) from Eq. (8.193) and substitute into Eq. (8.190). This value can be used with Eq. (8.194) to determine the value of the optimal force along the reactor. The optimum value of the Lagrange multiplier h is obtained from Eq. (8.186) (8.195)
0 In Jr By substituting Eq. (8.165) into Eq. (8.195), we have
A 2
1 2
Xopt = - - + - ( X + A ) ~
OlnL OlnX
(8.196)
The solution for A is obtained as - 2 J + JrAc ~ L*( - X ) 2 dx A=
I (L + (-X)L*
where L*=
OL OX
(8.197)
8.7
439
Stationary states
L being independent o f - X means that L can depend only on the state variables. The coefficient may depend on the temperature, pressure, and some of the concentrations. We should be able to vary - X independent of L. The optimal force is constant when the phenomenological coefficients are independent of the force. For reactions close to equilibrium, L is assumed to be independent of the forces, although it may vary along the path because the composition, temperature, and pressure vary along the reactor. By assuming L* = 0, Eq. (8.165) yields - (O In - l aJrX ]
-- - --TA
(8 .198)
From Eq. (8.196), we obtain [X(x)]Eo F = - -
A
(8.199)
2
Here, the subscript EoF means the equipartition of the force, and the force is constant. From Eq. (8.197), the Lagrange multiplier A is obtained as
(8.2oo)
t4 The specified production rate in terms of A becomes A J , - - -2 tAc I L ( x ) d x
(8.201)
Z
and the constant force is obtained as
JA
isL l
1
(8.202)
The force distribution that gives minimum lost work is uniform over the length of the steady-state reactor or over time in a batch reactor. Since the affinity A < 0 for a spontaneous reaction, A is a negative constant. Equation (8.195) indicates that the force is close to constant when we have L L* < < - - X
(8.203)
Near equilibrium, the local entropy production rate per volume is given by
(I)min ( X ) -
l ) 2[I L(x) dx] -2
LX2oF - L -JA ~
(8.204)
If the phenomenological coefficients are constant, then • is constant through the reactor, and (I)mi n becomes the summation of local entropy productions 2
• - 4]¢;.mm
'-gc
(8.205)
This equation shows that • becomes smaller for larger values of L(x) for a given reactor, and it can be used to determine the minimum lost work for a process. Equation (8.194) reduces to a constant force when the reaction is close to equilibrium. Consider the perturbation 8c on concentration at equilibrium Ceq c - Gq + 8c
(8.206)
440
8.
Chemical reactions
The concentration at equilibrium is obtained for Jr = 0
k2 Ceq
kl + k2
(8.207)
Close to equilibrium, we have Jr = (kl + k2)6c
(8.208)
X = R¢~c(klff-k2)2
(8.209)
and
k, k2
x
-
r(1 +1 3
~ k1 + k2 1 - Ceq
~ >0 Ceq
(8.210)
Equation (8.209) shows a constant force, while Eq. (8.210) shows that even i f - X i s constant the Jr need not be so since Ceq varies along the path. Therefore, the path of minimum lost work may not be a path of constant entropy production rate per unit volume. We can expect minimum lost work by the equipartition of forces when the reaction enthalpy varies little with composition, and if the reaction rate is a linear function of the conjugate force. The construction of particular operating path for a reactor with a specific output requires knowledge of reaction kinetics. The path that gives the minimum lost work entails a trade-off between the energy efficiency and the entropy production rate for a given reactor. It may be profitable to increase reactor investments and lower the energy cost for the same product, or to lower the production rate for a given energy input. For example, a 30% reduction in the lost work can be achieved by lowering the production by only 5%.
Example 8.10 Time variation of affinity We wish to derive the time variation of affinity in an open system at constant temperature and pressure. The affinity for reaction i is
4
----~lYi['Zk
(8.211)
k=l
The time variation o f A i at constant temperature and pressure is
dt
k=l
(8.212)
T,P dt
We use the change of concentration represented by (8.213)
dc k = dec k + dic k
where deck is the concentration exchanged through the surroundings, and dick is the change within the system. Using the following relations
l
Jri
i:1
V
(8.214)
d i M k = Mdic k = ~,,va. ~ d t
we have
d4_ dt
04 k= 1
deck+ T,P dt
Zl,,#m k=l m
V
(8.215)
8. 7
441
Stationary states
This equation shows that there are two contributions to the time variation of affinity. The first represents the supply of matter from the surroundings. The second is due to a chemical reaction within the system. Therefore, the two terms may counterbalance each other at a certain supply of matter. This leads to maintaining at a constant value. Such a phenomenon is only possible in an open system. In time, the system approaches a stationary state where all the reaction flows are zero except Jrm.
Ai
Example 8.11 Time variation of entropy production in simultaneous chemical reactions Consider two simultaneous reactions and derive relations for the time variation of the entropy production. Assuming that linear laws hold for a two-reaction system, we have (8.216)
Jr2-
L2~/~-)+L22(--~)
(8.217)
If Onsager's reciprocal rules hold, the rate of entropy production for two simultaneous chemical reactions is
P - diSdt - L~l
+ 2L12
T2
+ L22
(8.218)
D0
We also assume that the phenomenological coefficients are constant. The time variation of the entropy production is ldP
1 d
2 dt
2dt
dt }
L11T-k-L12T 7
-k- L12T+L22 T 7
>0
(8.219)
Using the linear laws of Eqs. (8.216) and (8.217) in the equation above, we have
2 dt
d-t
,,220,
-T + Jr2 7
For a closed system, the affinities A1 and A2 are expressed in terms of P and T, which are kept constant, and the extent of reactions el and e2, and we have 1 dip 2 dt
Jrl --[( OA1] Jrl "Jr-(OA1] Jr2 ] +---~---~ Jr2 [(OA21L 081 Jrl + ~.~08 (om2] 7"--)p,T 2)p T Jr2 ] T
(8.221)
From the definition of affinity, we write
08"2) P,T
C]el P,T
OelO~32 P,T
(8.222)
Using this equation in Eq. (8.161), we have
ldP
l{aA1 2 0A1 aA2 ) 0e I Jrl +2 JrlJr2 +~Jr22
2 dt = T
Oe2
Oe2
P,T < 0
(8.223)
Because terms OAilcgei belong to a definite quadratic form, the time variation of entropy production can only decrease with time.
442
8.
Chemical reactions
For an open system, affinities are functions of the mole numbers. From
dm= dem + dim, we have
dA1 (OA1]d#'li ~lOA1]deni (OAlldini:E(OA1) d131..p. (OA1]de#'li d' ~i t~ni)--~-:iT"~tOyli.JT"sr-~i COni)--~ i toni) pil--'~" ~i toni) 'dr
(8.224)
nt-E( OA1] dE2 fOAl)Jr1 .qt_ (OA1)det'li (OA1)dlli i t Oni)Pi2 -'~ = t 081) ~i t Oni) "-~ + t ~2 ) ~ Substituting this equation in Eq. (8,220), we have 1 dP
1 (OA1j21+20A 1 JrlJr2 +OA2jr2)+
2dr
]" ~,081
082
0~:2
1 l~{ OA1 7. Jrl ~-n/-t- Jr2
OA2] deni
Oni ) dt
(8.225)
In open systems, the time variation of the rate of entropy production has two contributions. The first contribution is an internal term, which is always negative 1 diP_ 1
(OA
T 081 Jr1 -k-20e2 JrlJr2 -k-~Jr20/32 "~ 0
2dt
(8.226)
The second contribution is an external term representing the exchange with the surroundings with no definite sign
l deP- l ~i (Jrl OA1 OA2]deni 2 dt T . -~n/-+-Jr2Oni ) dt
(8.227)
For transport processes, the entropy production always decreases in time.
Example 8.12 Minimum entropy production Consider the following synthesis reaction H 2 + Br2 = 2HBr
(8.228)
This results from the following intermediate reactions Br2 < 1 >2Br
(8.229)
H 2 +Br< 2 " H B r + H
(8.230)
H + B r e.. 3 " H B r + B r
(8.231)
The affinity of the net reaction is maintained at a constant value by the flows H 2 and Br2. The affinity of the first reaction is unconstrained. We wish to show that the stationary state leads to minimal entropy production. Assume that the system is homogeneous with a volume V = 1. The net reaction is obtained from the summation of reactions (8.230) and (8.231), and the affinity of the net reaction is A = A2 + A3
(8.232)
The value of A is maintained at a nonzero value, and hence at nonequilibrium conditions. The entropy production per unit volume is d~-
1 diS -
A1Jrl -[- A2 Jr2 + A3 Jr3 > 0
v dt
r
7-
-/-
(8.233)
-
The linear phenomenological reaction flows with vanishing cross-coefficients are
Jri = Lii
T
(8.234)
8.8
Michaelis-Menten kinetics
443
From Eqs. (8.228) and (8.229), we have (I)-- A1 Jr1 if- & 4 2 q- A - A2 Jr3 ~ 0
T
(8.235)
T
Using the linear reaction flows given in Eq. (8.234), and minimizing the entropy production by setting O~/OAi = O, for the independent affinities of A1 and A2, we find Jrl - 0 and Jr2 = Jr3
(8.236)
These equations support the kinetic equations at stationary states d C H - Jr2 -- Jr3 = 0
(8.237)
dt
dCBr -
2J~l - Jr2 if- Jr3 = 0
(8.238)
dt
8.8
MICHAELIS-MENTEN KINETICS
All chemical reactions in a biological cell take place with the direct participation of enzymes as catalysts. Enzymes are proteins, which are macromolecules composed of a combination of the 20 amino acids. Enzymes, as catalysts, are highly efficient and selective in binding small molecular species called ligands. A ligand that is acted upon by an enzyme to cause a chemical reaction is called a substrate. Only a certain, small portion of the amino acids that comprise an enzyme is involved in the catalytic reaction. This region is called the active site, and is directly involved in the formation of product(s). For example, the amino acid residues of proteins are greatly influenced by their local pH values, and the activity of proton acceptors and donors occurs in the active site. In 1913, Michaelis and Menten assumed that the enzyme and substrate react reversibly to form a substrate-enzyme complex. Later, the complex dissociates to form the free enzyme and product(s). The reactions are as follows k1
Fo
(8.239) k2
F1
~ >Fo + p
(8.240)
where Fo and F1 are the concentrations of the enzyme and substrate-enzyme complex, respectively. The terms kl, k2, and k3 represent the rate constants in the forward and backward directions. Michaelis and Menten also assumed that the first reaction is rapid and equilibrium is established quickly. The kinetic equations in accordance with the law of mass action are dS dt
- -lqFoS + k2F1
dE1 - k l F o S - ( k
2 + k3)F 1
(8.241)
(8.242)
dt dP dt
- k3F~
(8.243)
As the total concentration of the enzyme is constant, E - F0 + F1 - constant, Eq. (8.242) becomes - k
dt
ES-(k2 +
+ k S)Y
(8.244)
444
8.
Chemical reactions
Jr
Zero order behavior when S is large
Jr, max = k3E First order behavior when S is small .... ~
Jr=
L
\
_~ -
~ / ' / -
~t " Rate-concentration curve for Michaelis-Menten kinetics
max
2
l
~P~'atKM=S
KM=S
Figure 8.2. Some properties of the Michaelis-Menten kinetics. If S >> E in the stationary state, we assume that dFo/dt = dF1/dt = 0. From Eq. (8.244), we express F1 by
klES k2 + k3 + klS
F1
(8.245)
Since P = - S coincides with the reaction rate Jr, we obtain Jr -
aP at
-
k3klES k2 +k3 + klS
k3F1 =
(8.246)
Dividing the numerator and denominator of Eq. (8.246) by kl, and defining the Michaelis constant KM by
+ k3 kl
KM -
(8.247)
leads to the Michaelis-Menten equation
k3ES Jr =
(8.248)
KM +S
This equation shows that the reaction rate is directly proportional to E, and the relation between the reaction rate and the substrate concentration is not linear. However, if we study the reaction in certain regions, it can be approximated by linear kinetics. A plot of Jr versus S yields a rectangular hyperbola, and approaches Jr,max asymptotically
k3E
Jr,max =
(8.249)
at large values of S leading to a concentration-independent rate of reaction as seen from Figure 8.2. So, Eq. (8.248) can also be written as J r , max S Jr = ~
(8.250)
k +S
When the substrate concentration is large, the reaction rate is dependent on the substrate concentration. This represents zero-order kinetic behavior. When the concentration is very low, then the kinetics may be represented by first-order kinetic behavior. At Jr = Jr,max/2, the value of the Michaelis constant KM is obtained as S. The MichaelisMenten equation can be linearized, and the Lineweaver-Burk plot (Figure 8.3) is obtained from the following form 1 _
Jr
1
Jr,max
t
K M
1
(8.251)
Jr,max S
From the Lineweaver-Burk plot, we can obtain -1/KM from the abscissa intercept, and intercept.
1/Jr, ma x
from the ordinate
8.8
445
Michaelis-Menten kinetics
1
Jr
f J
f
/ /
"/,.max
1 KM
1 S
Figure 8.3. Lineweaver-Burk plot of the Michaelis-Menten equation.
The Monod equation describing the initial rates for microbial growth using the steady-state concentrations
j_j
{ cs j ,max
Ks + Cs
(8.252)
or dual Monod equation
rmax! CS l/ CP / Ks + Cs
"
(8.253)
K p + Cp
are well-known relations, although neither of them is general. Here, the concentrations Cs and Cp are the substrate and product concentrations, respectively, and Jr,ma×is the maximal growth rate. The terms Ks and Kp are the half saturation constants for growth, which are phenomenological constants. The Monod equation considers one irreversible step, although more steps are involved in the growth rate. Under relatively stable conditions, some chemical species remain uniform in concentration, or the overall reaction remains close to or far from global equilibrium. Michaelis-Menten kinetics leads to another rate equation to be considered using the Michaelis-Menten constants Ks and Kp, which are the ratio of reaction rate constants exp(A)- 1
Jrs - JrS,max O[exp(A) + (Jrs,max/JrP,max)]
(8.254)
where
(b
/ KS / ~ + 1
C S + Cp
Using Cs + Cp = constant, this equation relates the reaction velocity to affinity only. A plot of JrS/Jrs,max v e r s u s affinity shows an inflection point, representing a maximum slope. Therefore, the quasi-linear approximation between the reaction velocity and the affinity will be valid with a deviation of less than 15%.
E x a m p l e 8.13 Growth of a pathogenic bacterium Brucella abortus When the B. abortus is grown on an
agar medium, the colony has a smooth outline, and the constituents are called S-cells. During growth, mutations occur and form colonies having a wrinkled appearance. These cells are called R-cells. A flask of nutrient broth that is initially inoculated with S-cells contains a majority of R-cells by the time a stationary population size is attained. When the broth is analyzed, we see a decline of S-cells coinciding with the appearance of alanine,
446
8.
Chemical reactions
produced by S-cells, which kills S-cells but does not affect the growth of R-cells. Therefore, the growth curve exhibits a decline starting about 4 days after inoculation. Describe the growth kinetics for the S- and R-cell interaction. Discuss the equilibrium states of the system. Assume that A is the amount of alanine present in the broth at time t, and S and R denote the number of smooth and rough colony-forming cells. With very little or no interactions, S and R cells obey logistic growth laws. Meanwhile, because of alanine production, S cells die at a rate of AS. Since R cells arise because of mutation of S cells, there is an additional production of R cells at a rate proportional to S, if we disregard the effect of backmutations on the growth of S cells. We also assume that alanine is produced at a rate that is proportional to S. Therefore, we would have the following growth relations (8.255)
- - = k l S - fll $2 - aSA dt dR dt
dA dt
- k z R - f12R2 - AS
(8.256)
- yS
(8.257)
where kl, k2, ill,/~2, '~, and y are constants. The initial condition is (S, R, A) = (So, 0, 0). We can estimate the concentration from Eq. (8.257) and use it in Eq. (8.255), and we find t
(8.258)
dS _ k l S _ fllS2 _ a y S ~ S ( t ) d t dt o
So Eqs. (8.258) and (8.256) represent the growth kinetics. At stationary states, they become
(
o = s k~ - I ~ S - ~
)
S(t) dt
(8.259)
0
(8.260)
0 = R ( k 2 - t2 R) - AS
To satisfy Eq. (8.259), either S = 0, or the contents of the bracket equals zero. As the amount of S in the integral in Eq. (8.259) will increase with time, there is no fixed value of S that makes the contents of the bracket zero. Therefore, the only solution for Eq. (8.259) would be S - 0. Now, Eq. (8.260) becomes 0 = R ( k 2 -/32R )
(8.261)
and has two solutions with the roots, R = 0 a n d R - k2/f12. Hence, there are two stationary points: which is unstable, and (S, R) = (0, k2/~2), which is stable (Rubinov, 1975).
8.8.1
(S, R ) -
(0, 0),
Flow Force Relations for Microbial Growth
Consider a microbe coupling catabolism, which consists of processes that liberate free energy, and anabolism, which is biosynthesis. In the vicinity of global equilibrium (and for some far-from equilibrium systems) linear relations exist between thermodynamic forces, which are the corresponding Gibbs free energy differences A G, and flows. Therefore, a microbial growth may be represented by J=-LAG=-L(AG°+RTln
S)
(8.262)
where AG ° is the standard free energy difference, L is a phenomenological constant, which may depend on the enzyme, and R is the gas constant. Equation (8.262) can be used for catabolism C and anabolism A, and we have Jc = - L c A G c
(8.263)
8.9
447
Coupled chemical reactions
(8.264)
JA = --LAAGA
The dissipation function • is always positive, as the dissipation of the Gibbs free energy should occur because of the irreversibility of the system ~{r = _ J c A G C _ J A A G A > 0
(8.265)
Overall, the thermodynamic efficiency of growth rt is the ratio of the Gibbs free energy output and the Gibbs free energy input rl = -
JAAGA
~ < 1 JcAGc
(8.266)
Here, conventionally, biosynthesis is represented by a negative flow of anabolism. Thermodynamic efficiency can be estimated when the yield and actual concentrations of the substrates and products are known. The ratio JA/Jc is determined from the yield of the carbon and energy substrate, while AGA/AG C is determined by the standard free energy differences and concentrations of substrates and products. The theoretical efficiency of energy conversion may be optimized by varying AG A and the degree of coupling between catabolism and anabolism. Catabolism is the conversion of the growth-supporting energy source with the concurrent generation of ATE This ATP is utilized in anabolism for converting low-molecular-weight substrates via monomers into biomass. On the other hand, the leakage reactions are involved in all those processes that consume ATP without coupling to anabolism.
8.9
COUPLED CHEMICAL REACTIONS
Besides the transport, the most important processes in biological systems are those related to chemical reactions of metabolism. One of the typical aspects of such reactions is the requirement regarding the apparent stoichiometry of two partially coupled reactions, and the study of the efficiency of such reactions as limited by the constraints of the second law of thermodynamics. When we have the number of moles of species i and reactions j, we can express the Gibbs equation in terms of the extent of reaction ej and the affinity A/ P
dS - dU7/, + --T d V - ~_~. ~
de /
(8.267)
.!
where dej = dNij/vi/, and Aj = - Xiv(j~j In an isolated system in which U and V do not vary, d U = dV = 0, the dissipation function is given by the second law of thermodynamics q ~ - r dS dt
.
Ai "
>--0 dt )
(8.268)
According to the law of mass action of chemical kinetics, the rate of reaction is proportional to the product of the concentrations. The reaction velocity is d~3j
Jr; -
dt
(8.269)
With the respective affinities as the thermodynamic forces, Eq. (8.268) becomes
~ - E A /Jri >__0 / Therefore, A and Jr must have the same sign. In the chemical equilibrium state, we have A = 0.
(8.270)
448
8.
Chemical reactions
8.9.1 Two-reaction Coupling Despite its limitations for chemical reactions, the linear net theory has a useful conceptual base. Consider the linear phenomenological equations for two chemical reactions with flows Of Jrl and Jr2 Jrl =/-q 1A1 + L12A2
(8.271)
Jr2 = L214 + Lz2A2
(8.272)
These relations are based on the dissipation function
'~ = A1Jrl + A2Jr2 -> 0
(8.273)
For this sum to be positive, either both contributions can be positive, or one of them (A 1Jrl) can be negative while the other is positive and large enough to compensate for the negative effect of the first term. When, for example, A1Jrl is negative, the first reaction is carried out in the direction opposite to the direction imposed by its affinity. This is only possible if a coupling occurs between these reactions, and a sufficiently large entropy production is attained by the other reaction. This effect may cause one process to drive another process. Coupled processes are of great interest in biological systems, since in many situations the synthesis of reactions or the transport of substrates takes place in the direction opposite to that predicted by its thermodynamic force. The degree of coupling is defined by the ratio
q12
L12 (L11L22)1/2
(8.274)
The second law imposes L llL22 ~ (L 12)2, and therefore the degree of coupling is limited between - 1 and + 1. When q = + 1, the system is completely coupled and the two processes become a single process. When q = 0, the two processes are completely uncoupled and do not undergo any energy-conversion interactions. Let us consider Jrl as output and Jr2 as input flows, and the ratio Jrl/Jr2 is the stoichiometric ratio, which indicates the number of moles reacting in reaction 2 in order to produce a certain rate of reaction 1. From Eqs. (8.271) and (8.272), we have
Jr1 = LI1A1 + LlzA2 Jr2 L21A1+L22A2
(8.275)
Dividing the numerator and the denominator of this equation by (LllL22)l/2A2, we obtain Jr1 _
Jr2
q + ZX
Z~ 1+ qzx
(8.276)
where z andx are z = (Lll/L22) 1/2 andx = AlIA 2. This relation shows that the ratio of flows (rate of reactions) depends on the ratio of forces and the degree of coupling. When the degree of coupling q goes to zero, J r l / J r 2 - - Z2X, SO that z is a stoichiometric parameter. In the case of two coupled reactions, one spontaneous and the other forced, it is customary to define efficiency as the ratio of the dissipation in reaction 1 to the dissipation in reaction 2
JrlA1 r/= - ~ Jr:A2
(8.277)
From Eq. (8.276), we can write
r/= -
zx(q + zx) 1 + qzx
(8.278)
449
Problems
For the maximum efficiency of the energy transfer from reaction 2 to 1, we differentiate ~ with respect to x and equate to zero, and then we have -l+x/1-q 2 Xmax =
(8.279)
qz
After rearranging terms, the maximum efficiency is obtained as
~ma×
z
q2 [1 + X/1- q2 ]2
(8.280)
The value of TImaxdepends only on the extent of coupling. When two reactions are completely uncoupled, the efficiency must be zero. When q = 1, we have the maximum efficiency of unity.
PROBLEMS 8.1
Calculate the entropy production for racemization reaction S = E and plot the change of entropy production with time. Use kr = 2.0, kb = 0.01, Cs0 = 2.0 tool/L, Cp0 = 0.01 tool/L, V = 1 L, T = 300 K, and R - 8.314 J/(mol K).
8.2
Estimate the change of concentrations with time for the following elementary reaction: X + Y = 2 Z Use k~ = 0.5, kb = 0.01, C~0 = 1.0, C,,0 = 2.0, and (7_-o= 0.
8.3
Estimate the change of entropy production • with time for the following elementary reaction: X + Y = 2 Z Use kr = 0.5, kb = 0.01, C~0 = 1.0, C,,0 = 2.0, and Go = 0.
8.4
In the gas phase, isopropyl alcohol is dehydrogenated to produce propionaldehyde in the following reaction (CH3)2CHOH(g) = CH3CH2CHO(g ) ÷ H2(g) The standard heat and the Gibbs free energy for isopropyl alcohol are AH ° - 55.48 kJ/mol, AG ° - 17.74kJ/mol,
ACp- 16.736 J/(molK)
Determine the equilibrium composition of isopropyl alcohol at 500 K and 1 atm if ACp = 16.736 J/mol. 8.5
The following reaction shows the production of ethylene dichloride using ethylene dibromide as a catalyst C2H4 + C12 - C2H4C12 The normal boiling point of ethylene dichloride is 83.47°C. Estimate the equilibrium conversion of ethylene if the reaction takes place at 50°C and Iatm.
8.6
We heat a component A. The pure component A, originally at 298.15 K and 1 atm, decomposes upon heating and the following reactions occur in the gas phase: A=B+C A=D+E (a) Estimate the equilibrium composition of component A at a pressure of 1 atm over a temperature range of 1000-1500 K. (b) Estimate the equilibrium composition of A over a temperature range of 1000-1500K if pure A at 298.15 K and 1 atm is heated in a constant-volume reactor.
450
8.
Chemical reactions
For this temperature range, the following K values may be used: T(K)
Kal Ka2 8.7
1000 2.907 534.3
1200 38.88 2581
1400 246.0 7754
1500 512.6 11950
Ethyl benzene is produced in a constant-volume reactor by the following reaction: C6H6(g) + C2H4(g ) = C6H5C2H5(g ) For this purpose, 1 mol of ethylene and 1 mol of benzene are heated to 600 K. The initial pressure within the reactor is 1 atm. At 600 K, Ka = 345.0 and AHr = 103940 kJ/mol for the reaction. Determine: (a) The equilibrium compositions of ethyl benzene, benzene, and ethylene. (b) The heat to be removed to maintain the isothermal reaction above at 600 K.
8.8 8.9
Determine the entropy and enthalpy change of an electrochemical cell reaction. The kidneys transport glucose from the urine to the blood against a concentration gradient, in which the transport occurs from a low concentration to a high concentration. This is called active transport, and can occur only if the transport is coupled with a spontaneous chemical reaction. Assume that the initial concentration of glucose in the urine is 5 × 10-5 mol/kg. The glucose concentration becomes 5 × 10-6 mol/kg after leaving the kidneys. The blood contains an almost constant concentration of glucose of 5 x 10-3 mol/kg. Estimate the minimum work or Gibbs free energy supplied per mole of glucose transported across the kidneys.
8.10
Consider a nonisothermal reactor. The concentration C and temperature T change with time within the reactor may be described by the following equations:
dC - - c exp ( -10) dT - - lOOOCexp(- ~ ) - lO(Twhere the initial conditions are T(0) = 288.15 K and C(0) = 1.0 gmol/L. Plot the concentration and temperature of the reactor as a function of time. 8.11
Consider the following chemical reaction kinetics equations for the components x, y, and z:
dx --O.031x-lOOOxz dt dt dz dt
- -2500xy - -0.03 lx - 1000xz
-
2500yz
where the initial conditions are x(0) = y(0) = 1 and z(0) = 0. Plot the concentration as a function of time. 8.12
k3 If we consider the back reaction F1 < k4 >F° + P' where k4 is not zero in the reactions of the enzymesubstrate system, modify the Michaelis-Menten kinetics. Show that when equilibrium is established, after a very long time, equilibrium concentrations of substrate and product are related by the following Haldane's relation
eeq kxk3 Seq k2k4 _
References
8.13
451
In an autocatalysis system, the product P of a reaction catalyses its formation from a substrate S. If the initial concentration values of substrate and products are So and P0, respectively, define the rate equations for the concentrations and find their dependence on time. Assume that an intermediate complex is not formed in the reaction.
REFERENCES Zs. Ablonczy, A. Lukacs and E. Papp, Biophys. Chem., 104 (2003) 240. S.R. Caplan and A. Essig, Bioenergetics and Linear Nonequilibrium Thermodynamics." The Steady State, Harvard University Press, Cambridge (1983). E.P. Gyftopoulos and G.P. Beretta, Thermodynamics. Foundations and Applications, Macmillan, New York ( 1991 ). S. Kjelstrup and T.V. Island, Ind. Eng. Chem. Res., 38 (1999) 3051. S. Kjelstrup, E. Sauar, D. Bedeaux and H. van der Kooi, Ind. Eng. Chem. Res., 38 (1999) 3046. D. Kondepudi and I. Prigogine, Modern Thermodynamics. From Heat Engines to Dissipative Structures, Wiley, New York (1999). G. Nicolis and I. Prigogine, Exploring Complexly, Freeman & Company, New York (1989). S.I. Rubinow, Introduction to Mathematical Biology, Wiley, New York (1975). M.T. Suchiya and J. Ross, Proc. Natl. Acad. Sci., 100 (2003) 9691.
REFERENCES FOR FURTHER READING R.A. Alberty, Biophys. Chem., 124 (2006) 11. H. Qian and E.L. Elson, Biophys. Chem., 101-102 (2002) 565. J. Ross and M.O. Vlad,Annu. Rev. Phys. Chem., 50 (1999) 51. S. Sieniutycz, Int. J Eng. Sci., 36 (1998) 577.
9 COUPLED SYSTEMS OF CHEMICAL REACTIONS AND TRANSPORT PROCESSES 9.1
INTRODUCTION
Nonisothermal reaction-diffusion systems control the behavior of many transport and rate processes in physical, chemical, and biological systems. A reaction-diffusion system with appropriate nonlinear kinetics can cause instability in a homogeneous, steady-state system and generate stable concentration patterns. Some of the chemical reactions coupled with the transport of species can lead to pumps and chemical cycles in biological systems, such as a sodiumpotassium pump. A considerable work has been published on reaction-diffusion systems. This chapter discusses mathematically and thermodynamically coupled differential equations of nonisothermal reaction-diffusion systems. Here, the thermodynamic coupling refers that a flow occurs without its primary thermodynamic driving force, or against the direction imposed by its thermodynamic force. The principles of thermodynamics allow the progress of a process without or against its primary driving force only if it is coupled with another process. This is consistent with the second law, which states that a finite amount of organization may be obtained at the expense of a greater amount of disorganization in a series of coupled spontaneous processes. Modeling of spatio-temporal evolution may serve as a powerful complementary tool for studying experimental nonisothermal reaction diffusion systems within a porous catalyst particle and a membrane. The linear nonequilibrium thermodynamics approach may be used in modeling coupled nonisothermal reaction-diffusion systems by assuming that the system is in the vicinity of global equilibrium. In the modeling, the information on coupling mechanisms among transport processes and chemical reactions is not needed.
9.2
NONISOTHERMAL REACTION-DIFFUSION SYSTEMS
The basic equations for an unsteady-state process of one-dimensional (in the y-direction) heat and mass transport with a simultaneous chemical reaction in a porous catalyst pellet are
OCA, o? --D"-~ OvO---(ybOC----~A OV ) (pCp)-~-
ke
--
--
~, O v
Ov
-k-(-AHr)Jr(CA,T)
(9.1)
(9.2)
where ~L/r is the heat of reaction, De and ke the effective diffusivity and thermal conductivity, respectively, and v is stoichiometric coefficient, which is negative for reactants. Here, b describes the shape: b = 0, slab; b = 1, cylinder; and b - 2, sphere. These partial differential equations are mathematically coupled. The initial and boundary conditions with internal and external and resistances across the boundary are Att=0 At v - 0
(surface conditions)
CA=CA~, a n d T = -c ) C- A
--
0 and
~c)T= 0
Ov - - D e OLT----~A" - - k ( C A - -
&, CAO )
Ov
-k~ -°r- = h ( r - r0) 0v
(symmetry conditions) (external mass transfer effect)
(external heat transfer effect)
(9.3)
454
9.
Coupled systems of chemical reactions and transport processes
The linear nonequilibrium thermodynamics approach can provide a quantified description of the fully coupled phenomena for systems in the vicinity of global equilibrium.
9.2.1
Effective Diffusivity
In a multicomponent fluid, a species can be driven not only by its own thermodynamic force (its own concentration gradient) but also by concentration gradients of all the other species. Flow of speciesj in an n multicomponent fluid system is n-1
Nj = - ~ , CDjkVY k + yj k=l
Nk j = 1,2,...,n- 1 k=l
where C is the total concentration. The last term represents the bulk flow of the fluid. To simplify this equation, a common approach is to introduce a mean effective binary diffusivity for species j diffusing through the fluid mixture
Nj = -CDjm Vyj + yj ~ N k
(9.4)
k=l
For ideal fluid mixtures, the Maxwell-Stefan equation yields
-CVyj = ~
1---~-(ykNj -- yjNk) Djk k--/=j
(9.5)
k=l
For a binary mixture, this equation becomes !
-CVy 1 = -__:__.[N 1 - Yl (N1 + N2)] D12
For a multicomponent gas mixture, the effective binary diffusion coefficient for species j diffusing through the mixture may be found by equating the driving forces Ayj in Eqs. (9.4) and (9.5) //
1
~.k=l(1/Djk)(Yk -yj(Nk/Nj)) -
(9.6)
n
Djm
1-- yj Z k=I(Nk/Nj )
This equation reduces to Wilke equation, when it is used with the zero flow rates for k = 2, 3. 1
Dim
_
1
'(-" Yk
1 - Yl k=2Z-"D1k
This equation represents the diffusion of species 1 through stagnant species 2,3..., n in the reacting system, and is mainly suitable for very dilute solutions. When the other species are not stagnant, the steady-state flow ratios are determined by the reaction stoichiometry. For a reaction, N/vj = constant, and Eq. (9.6) becomes n
1 _ ~-~ke=j(1/Djk)(yk - Yj (vk/vj)) n Djm 1- yj Z k = l (l~k/PJ ) or
1
Djm
/
1+ 6jyj k~j Dj---k Yk - Yj ~j [
(9.7)
(9.8)
where j is the reactant species, and 6j = (Evi)/vj. Usually, some average composition Yj,av is used to determine an average value of Dim.
9.2
4.55
Nonisothermalreaction-diffusion systems
For certain applications, we define an effective binary diffusivity with the flow relative to the fixed solid and include only bulk flow in the values of D~m Nj -- -CD)mVy j
Then, the same procedure yields
--
(9.9)
Yk--Yj
k~ i Djk
The effective diffusion coefficient may be obtained from the molecular diffusion coefficient Dj, the catalyst porosity 8p and tortuoisty %. _ ~p
Dej
mDj ?p
(9.10)
Tortuoisty describes the deviation of the pores from an ideal structure.
Example 9.1 Effective diffusivity Consider the following chemical reaction aA + bB ~ rR + sD Equation (9.8) yields
1
DAm
where
-
' [2(
1 + 6 AYA
6 A = (r + s - a - b)/a.
YB--YA
+
l(
DAR
YR +YA
a) l(
')1
-t-1 Ys + Y A - DAS a
The flow of species A is dYA NA =-CDAm---~-y +YA (NA +NB +NR + N s )
or N A = _ C D Am dY AT _F YA N A ( + ~b a- r - s
Here the counter diffusion is used. The flow of species A is N~
z
m
CD Am dY A
l + 6 A y A dy
After integrating this equation between y - 0 and y - L with NA = constant and an average constant value of DAm, we have NA _ CDAm In 1+ 8AYA0 L8 A
1 + 8 A YAL
_
CDAm YA0 - - YAL L
where YfA is the film factor defined by
YfA --
When
•A
--
(1 + 6AYA0)-- (1 + 6AYAL) (1 + 6A YA0 )/(1 + 6AYAL)
0 that represents equimolar counterdiffusion, then YfA = 1.
YfA
456
9.2.2
9.
Coupled systems of chemical reactions and transport processes
Effective Thermal Conductivity
For a heterogeneous solid where one solid phase is dispersed in a second solid phase, or one solid phase contains pores, we introduce an effective thermal conductivity to describe steady-state conduction. The geometry of the dispersed solid or pores affect the thermal conductivity. If we have a material made of spheres with thermal conductivity kl dispersed in a continuous solid phase with thermal conductivity ks, then the effective thermal conductivity ke is ke - 1 + 305 ks (kl + 2ks )/(kl - ks ) - 4'
(9.11)
where 4' is the volume fraction of the embedded material. This equation is called Maxwell's Derivation and assumes that spheres do not interact thermally and the volume fraction 4~ is small. For solids containing gas pockets with thermal conductivity kl, thermal radiation may be important, and an effective thermal conductivity may be approximated by ke_
1
ks
1 + (k 1/ks6 + 4oT3L/ks ) - 4,
(9.12)
where o- is the Stefan-Boltzmann constant and L the total thickness of the material in the direction of the heat conduction. For gas-filled granular beds, the thermal conductivity of the gas may be very low. Since gas phase heat conduction mainly occurs near the points of contact between adjacent solid particles, the distance for heat conduction over the gas phase may approach the mean free path of the gas molecules. This reduces the thermal conductivity of the gas further, since the whole system may become rarefied for evacuated beds of fine powders. In separation processes and chemical reactors, flow through cylindrical ducts filled with granular materials is important. In such systems conduction, convection, and radiation all contribute to the heat flow, and thermal conduction in axial ke,x and radial ke,r directions may be quite different, leading to highly anisotropic thermal conductivity. For a bed of uniform spheres, the axial and radial elements are approximated by ke, x = 0 . 5 p C p v o D p
ke, r = O. l p C p v o O p
(9.13)
where v0 is the superficial velocity, and Dp the diameter of the particles. These equations are used for highly turbulent flow.
9.2.3
Balance Equations
Assuming a steady state, for first-order reaction-diffusion system A ~ B under nonisothermal catalyst pellet conditions, the mass and energy balances are 0 =-V.J-kvC A O -- - V . J q -+-( - , ~ r
(9.14) )kvC A
(9.15)
By using Fick's and Fourier's laws in one-dimensional transport in a slab catalyst pellet (Figure 9.1) with, equimolar counter-diffusion under mechanical equilibrium, Eqs. (9.14) and (9.15) become d2CA O - D e dy 2 - kvC A
(9.16)
O-- ke d2T dy---5- + ( -
(9.17)
AH r)kvCA
9.2
457
Nonisothermal reaction-diffusion systems
I I
y=0
y=L
I
CAs I
II cA
I I
\\,
,, I ] Stagnant A I film I I I
Catalytic surface B
Figure 9.1. Schematic heterogeneous reaction-diffusion system.
Without the external mass and heat transfer resistances, the boundary conditions with the x-coordinate oriented from the centerline (y - 0) to the surface ( y - L) are dC A (0) = 0 and T ( L ) - ~ , c(L) - CAs, ~
d~'
dT(O)
dy
- 0
The values of De tend to be smaller than those of ordinary gas diffusivity, while the values of ke are smaller than those of the thermal conductivity for a similar nonporous solid. The effective reaction rate kvCa is based on the total rate of reaction within any small, representative volume. Eliminating the reaction terms from Eqs. (9.16) and (9.17), and integrating twice with the boundary conditions above, the temperature is related to concentration by q~-1-/3(0-1)
(9.18)
where 0
CA = c-2,'
T
(-AHr)DeCAs '
Equation (9.18) is valid for any particle geometry under steady-state conditions, and can be used to eliminate the 0 or q~ from one of the differential Eqs. (9.16) and (9.17). The nondimensional parameter/3 (positive for exothermic reactions) is a measure of nonisothermal effects and is called the heat generation function. It represents the ratio between the rate of heat generation due to the chemical reaction and the heat flow by thermal conduction. Nonisothermal effects may become important for increasing values of/3, while the l i m i t / 3 - , 0 represents an isothermal pellet. Table 9.1 shows the values of/3 and some other parameters for exothermic catalytic reactions. For any interior points within the pore where the reactant is largely consumed, the maximum temperature difference for an exothermic reaction becomes ( - A H r )OeCAs
ke
= fi~
(9.19)
458
9.
Coupled systems of chemical reactions and transport processes
Table 9.1. Parameters for some exothermic reactions used in the thermodynamically coupled model Reaction systems Synthesis of vinyl chloride from acetylene and HC1 Dissociation of N20 Hydrogenation of benzene Oxidation of SO2 NH3 synthesis Oxidation of CH3OH to CH20 Hydrogenation of ethylene Oxidation of ethylene to ethylene oxide Oxidation of H2
~b
/3~/3'
y
Le
•a
(.oa
0.27 5.0 0.05-19 0.9 1.2 1.1 0.2-2.8 0.08 0.8-2.0
0.25 0.64 0.12 0.012 0.000061 0.0109 0.066 0.13 0.1
6.5 22.0 14-16 14.8 29.4 16.0 23-27 13.4 6.75-7.52
0.1 0.01 a 0.006 0.0415 0.00026 0.0015 0.11 0.065 0.036
0.001 0.001
0.001 0.001
0.001
0.001
aAssumed values.
Source." Hlavecek et al. (1969). Transient forms of Eqs. (9.16) and (9.17) become 00 OT 1 0q~
Le 0~-
-
0expE,/' +/]
020
OZ2
(9.20)
02@
(9.21)
OZ2
after using the following dimensionless parameters
y Z----"
L'
E
Det (!/)2 L2ko exp(E/RTs) T--
--~-'
; Y = ~ s s ' Le .
De
ke/pCp _ a e . De. .
De
(9.22)
where Le is the modified Lewis number and ae the effective thermal diffusivity. The nondimensional group 3' is called the Arrhenius group, and represents a nondimensional activation energy for the chemical reaction. The initial and boundary conditions are 0(0, z) - 1, 0(~-,1) - 1,
dO(r,O) = 0
dz dq~(~-,O) =0 ~(0, z ) - 1, q~(~-,1) = 1, dz After substituting Eq. (9.18) into Eq. (9.20), steady and nonisothermal concentration profiles become
d20 dz 2 - ch2Oexp( - 1yfl(O-1) -/3(0-1) )
(9.23)
Diffusion may reduce average rates relative to those obtained if the concentration everywhere was the surface concentration CAs (Froment and Bischoff, 1979). This limitation is quantified as the effectiveness factor rl defined by 77 - 1 ] Jr(C/)dV V
(9.24)
Jr(Cis)
where V is the volume. For exothermic reactions (/3 > 0) a sufficient temperature rise due to heat transfer limitations may increase the rate constant kv, and this increase may offset the diffusion limitation on the rate of reaction (the decrease in reactant concentrations CA), leading to a larger internal rate of reaction than at surface conditions CAs. This, eventually, leads to rl > 1. As the heat of reaction is a strong function of temperature, Eq. (9.24) may lead to multiple solutions and three possible values of the effectiveness factor may be obtained for very large values of/3 and a narrow range of ¢h values (approximately 0.47-0.49). In common catalytic reactions,/3 is usually 1
(9.30)
9.2
9.2.4
461
Nonisothermal reaction-diffusion systems
Effectiveness Factor for Various Geometries
For a spherical pellet, from the steady-state mass balance
D e l d F 2 dCA -kvC A -0 r- dr
(9.31)
dr
and using the similar boundary conditions, the effectiveness factor becomes 3 b coth b - 1 b
(9.32)
b
where b = Rx/kv/De, and R is the sphere radius. For a cylinder and other geometries, a general modulus for a first-order reaction is
_Vp /kv cb-~
(9.33)
ap ~ De where lip and ap are the volume and external surface area of the pellet. A general modulus for an nth-order irreversible reaction is
__ _Vp //7 + 1 (--V A)kv(CAs) n-I _
lJ) --
.t
ap ~
2
n --> 0
(9.34)
De
The effectiveness factor in terms of/3 and y for an nth-order reaction may be calculated from a finite series for a region of low/3 (Tavera, 2005)
rt - ~ ' 0 5
exp 1+/3
1 - exp(-A)
k=0
k' •
/3 -< 0.1
(9.35)
where
A
z
(1+~ 2)
For a first-order reaction, this equation reduces to
oxp l+t
i
,-
+
/3 --< O. 1
(9.36)
With Eqs. (9.35) and (9.36), overestimation of the nonisothermal effects does not exceed 5% (Tavera, 2005).
9.2.5
External Diffusion Resistance
To consider external diffusion resistance for a first-order reaction, we need to determine the surface concentration of reactant Cas from the boundary condition
kg(Cb-CAs)-De[dCA I dy
s
(9.37)
The solution of this equation in terms of the bulk concentration Cb is
CAs --Cb
( y/L ) cosh ch cosh 05 + (Dedp/Lkg)sinh 05
(9.38)
462
9.
Coupledsystems of chemical reactions and transport processes
This equation, when is used in the definition of the effectiveness factor based on the fluid bulk concentration Cb, leads to the combined resistance of fluid and particle 1 _1+
rig
_
+m
rt kgL/De tanh4~
(9.39)
kg
Example 9.4 Effectiveness for a first-order reversible reaction Consider a first-order reversible reaction
R < kf >P. The rate of reaction is kb Jrv (CR) = ~7Y7[(1 + where K =
kf/kb and C = Cas + Cps = (1 +K)
K)C R -C]
CR, eq. The mass balance equation is
De d2CR = Jrv(CR) dy 2 After the first integration of this equation with constant De, we have
CRs __ kvDe ~s I DeJrv(CR)dCR [ ( I + K ) C R -C]dC R K
CR,eq
CR,eq
K
2
R'eq) -- C(CRs - CR'eq)
]
This equation can be used in the generalized effectiveness factor obtained from Eq. (9.24)
De dCR (L)
7-y
]1/2
,5 ICc
LJrv(CRs ) - L[Jrv(CRs) ] ~ DeJrv(CR)dCR
{ [
L[Jrv (CRs)]
kvD e I + K ( C _ C 2 K
]}12
(9.40)
R,eq ) -- C(CRs - CR,eq )
2
which yields an asymptotic effectiveness factor for Cc = CR, eq. Here, Cc is the reactant concentration at the centerline and may be assumed as the concentration of the reactant at equilibrium
r / ~ ~-7, where
9.2.6
= n
ap
External T e m p e r a t u r e Gradients
The balance Eqs. (9.16) and (9.17) for simple slab geometry are
0--De
d2CA 2 -kvCA dy
d2r + (- a G )kvCA 0=ke dy--T
K
(9.41)
9.2
Nonisothermal reaction-diffusion systems
463
As before these balance equations can be combined as follows
~2/ DeCA + ~keT / - 0
dv 2
( - AH r )
and integrated twice from the pellet center to the surface using the following boundary conditions
De
dCa dT - kg(¢ b - GAs), ke ~ - - hf(Tb -Ts) dv av •
where Cb and Tb are the fluid bulk concentration and temperature. The integration yields the following sum of external and internal temperature differences
kg De T - Tb - (-AH r ) ht----f(Cb--CAs)+-(--~Hr)-f--(CAstQ --CA )
(9.42)
The maximum temperature difference occurs when CR = 0, and after rearranging terms, the equation above becomes
~axTb ~b:~bESh l1 ~CAsi -+-~~ ----7 Nu Cb J The ratio
Cas/Cb is obtained in terms of the observable
rate Jrv, obs
1L De(dCaj kg Jrv,obs = T f 0 JrvdY- T dy L --T(Cb-CAs) where Sh' is the modified Sherwood number Sh' = at bulk fluid conditions is defined by
(9.43)
(9.44)
kgL/De, Nu' the modified Nusselt number Nu' = hfL/ke, and/~b
D e
/~b -- ( - / ~ H r ) ~ C b t%l b
(9.45)
After substituting Eq. (9.44) into Eq. (9.43), we have
1)]
Tmax--]'b=/~b I1 +~b ( 1 Tb
Nu'
Sh'
~946,
where (I) b is the observable Weisz modulus and is defined by
-
with 4~-
L2~DcCb Jrv'°bs
= Sh' ( G 1 -A st~)b
-
~2
L~Jkv/D~ and 1/r/g - l/r/+ 052/Sh'
The internal and external temperature differences can be obtained from Eq. (9.46)
[[
-)
,
Tmax Tb __ /~b I1 -- (I)b(~-7) 1 ~ int
!~axTb ~b)~b,i 1/ ext NUT/"
(9.47)
464
9.
Coupled systems of chemical reactions and transport processes
Example 9.5 Maximum overall temperature difference in the hydrogenation of benzene Consider the hydrogenation of benzene, which is exothermic with a heat of reaction 50 kcal/mol. For a catalyst pellet containing 25% Ni-0104P, 25% graphite, 50% 7-A1203 (Harshaw A1-0104T), the effective thermal conductivity and diffusivity are 3.5 × 10 -3 cal/(cm sK) and 0.035 cm2/s, respectively. The fluid bulk concentration of benzene is 5.655 × 10-6mol/ cm 3, and the fluid bulk temperature is 412 K. The characteristic length of the pellet is 0.31 cm. The observed rate for the reaction is 22.4 × 10 -6 mol/(gcat s) and the density of the catalyst is 1.57 g/cm 3. The modified Sherwood and Nusselt numbers are 401 and 1.35, respectively. From Eqs. (9.6) and (9.7), we find the values of/3b and ~bb /3 = (50,000)(0.035)(5.655 × 10 -6) = 0.007 (3.5×10-3)(412) The reaction rate is
Jrv,obs = 22.4 × 10 -6 mol/(gcat s) × 1.57 gcat/cm 3 = 35.168 × 10 -6 mol/(cm 3 s) L2Jrv obs
~b = ~ '
DeCb
=
(0.31)2(35.168×10 -6 ) (0.035)(5.655×10 -6 )
=16.7
The external temperature difference can be obtained from Eq. (9.46)
( /
,'"(T-ax - Tb ,]ext -- Tb/~b(I)b 1
Nu'
007167(1 /
=
= 36 K
The experimental value is 11 °C. This lower temperature may indicate that the internal pellet concentration may not be exactly zero, contrary to expectations. The internal temperature difference is
[ (ill
(Tmax --Tb)in t = Tb/~b 1--~0b ~
/
=(412)(0.007) 1 - - ~ }
2.8K
This result shows close agreement with the experimental value of 2 K and the low internal temperature gradient, as the value of/3 is small.
9.2.7
Criteria for Negligible Transport Effects
After Dekker et al. (1995), criteria for negligible transport effects in steady-state kinetics are as follows. The criterion for negligible external mass transport resistance in steady-state kinetics is
Jr,obs
g(Sp/Vp)Cb
< 0.05
(9.48)
For negligible intraparticle mass transport resistance, the criterion is rt4~2 =
Jr obsL2 n + 1 ' DeCA
< 0.1
(9.49)
2
For negligible external heat transport resistance, the criterion is
/~y(Cbcb-CA ) < 0.05
(9.50)
The criterion for negligible internal heat transport resistance is (9.51)
9.3
465
Chemical reaction with coupled heat and mass flows
The criterion for external mass transport resistance is
kga
--t
>_ 2.9
(9.52)
where a is the particle surface area per unit volume of packed bed, and c the void fraction of the catalyst bed. For negligible external and internal mass transport resistance, the criterion is
De 8R 2
9.3
t -> 0.25
for Bi m -
kgR De
--> 20
(9.53)
CHEMICAL REACTION WITH COUPLED HEAT AND MASS FLOWS
Nonisothermal reaction-diffusion systems represent open, nonequilibrium systems with thermodynamic forces of temperature gradient, chemical potential gradient, and affinity. The dissipation function or the rate of entropy production can be used to identify the conjugate forces and flows to establish linear phenomenological equations. For a multicomponent fluid system under mechanical equilibrium with n species and Nr number of chemical reactions, the dissipation function q~ is * -- Tq) -
--~-JqVTi=l
J i ( V # i ) + Z JrjA/ j=l
>- 0
(9.54)
where, Jq is the vector representing heat flow, d; the vector representing mass flow, kc; the chemical potential of species i, A the affinity A = -EVil, i, v the stoichiometric coefficients, and Jr the reaction velocity. Equation (9.54) consists of scalar processes of chemical reactions and vectorial processes of heat and mass flows, and excludes viscous flow, electrical, and magnetic effects. Equation (9.54) and the rate of entropy production may be used to identify the conjugate forces and flows to establish linear phenomenological equations. Excluding the coupling of chemical reactions with the heat and mass flows, and using the relationship/X/x a = (OtJ6A/OCA)z~iCA,Eq. (9.54) reduces to 1
- -'--Jq~T-JAAT,pVC
T
A ~ 0
(9.55)
where AT,P --(1--t-(VACA/VBCB))(0[,.I,A/0CA )T,P' and V, is the partial molar volume of component i. Assuming that the reaction-diffusion system is not far from global equilibrium, the linear phenomenological equations based on Eq. (9.55) are 1 - J A - LAAAT.p~'CA + --~ L A q V T - DeVCA + DS,eVT 1
(9.56)
1 - J q - LqAAT,pVC 4 - + - T L q q V T - DD,eVC A -+-keVT
(9.57)
where DD,e is a coefficient in m 2 J/(mol s) related to the effective Dufour effect, and Ds, e a coefficient in mol/(m s K) related to the effective Soret effect (thermal diffusion). When there is no volume flow, the mass flow JA is JA = --LAAAT,p~7CA
(9.58)
and comparing Eq. (9.58) with Fick's law J = -DeVCA, the coefficient LAA is related to the effective diffusion coefficient by LAA --
De AT,P
(9.59)
Using Fourier's law Jq -- -keVT in Eq. {9.57), the primary coefficient Lqq is related to the effective thermal conductivity ke by
Lqq = k~T
(9.60)
466
9. Coupledsystems of chemical reactions and transport processes
The thermal diffusion coefficient for species A is
DT - -
LAq
(9.61)
CAT
For liquids, the diffusion coefficient D is of the order of 10 -5 cm2/s, and the thermal diffusion coefficient D T is of the order of 10-8-10 -1° cm2/(s K). For gases, the order of magnitude for D and D T is 10 -1 cm2/s, and 10-4-10 -6 cm2/(s K), respectively. By using the flows JA and Jq from Eqs. (9.56) and (9.57), respectively, in Eqs. (9.14) and (9.15), we have 0 = V'(DeVC ~ + DS,eVT' ) - kvC'A
(9.62)
0 : ~'(DD,e~C~k nt-kel~T')-+-(-ASr) kvCrA
(9.63)
where T' and C), represent the temperature and concentration in the coupled system. As before, the elimination of the reaction terms from Eqs. (9.62) and (9.63) yields q~'= 1-/3'(0-1)
(9.64)
where
O'
C'A" q9' r' = Cs' = Tss' and
~ , _ (Oe(-AHr)-kDD,e)Cs - (ke + Os'e(-~r-/r) ) ~
(9.65)
and the maximum temperature difference becomes AT' /3' ~ (qgcenter- 1) . . . . .
(9.66)
rs
The modified dimensionless group/3' represents the ratio between the chemical reaction's rate of heat and rate of heat conduction when only the heat and mass flows are coupled (Demirel, 2006). By disregarding the coupling effects, we would have/3'=/3. Using Eqs. (9.62) and (9.63) with the Arrhenius equation kv = koe-E/Rr, we have the transient forms of the coupled heat and mass flows for a single component
OC'A - V.(DeVC~ +Ds,eVT')-(koe-U/Rr')C' a Ot
(9.67)
pCpOT'ot ~'( DD'e~7C~ q-kegT'+[-(AHr)] ) (k°e-E/"r')
C'A'
(9.68)
The initial and boundary conditions are the same as those given for Eqs. (9.20) and (9.21). For a simple plane geometry and one-dimensional unsteady state, Eqs. (9.67) and (9.68) become
00' 0T
D
020' -+E02¢-----~P' 02 Z ~20 exp [T (1 - ~ , ) ] 0Z2
1)]
1 Oq~' O2q)' ~+ ~+13'q520'exp T 12 Le Or OZ2 (.00Z 2 where
Z ~y
L'
Ds,e Ts
DD,eCs
DeC s
keTs
L2ko exp(E/RT s) E k~/pCp _ Ole 7"-- Det 492_ De ' Y - ~ s s ' Le --~-' De De
(9.69)
(9.70)
9.3
467
Chemical reaction with coupled heat and mass flows
Equations (9.69) and (9.70) represent the modeling of reaction-diffusion systems with the thermodynamically coupled heat and mass flows excluding the coupling effects due to reaction. After combining Eqs. (9.64), (9.69), and (9.70) steady-state balance equations with the coupled heat and mass transfer become
de0 ' (y/3'(0'-l) ) ( 1 - fi'e) dT~ = 0520' exp - 1 - / 3 ' ( 0 ' - 1)
(9.71)
(9.72)
Since the dynamic behavior of a reaction-diffusion system may be more apparent with state-space diagrams, the temperature and concentration profiles are replaced with the spatial integral averages obtained from I
1
()
0
(9.73)
Example 9.6 Coupled heat and mass flows in oxidation of CH3OH to CH20 The modeling Eqs. (9.69) and (9.70) are used with parameters for the exothermic catalytic oxidation of CH3OH to CH20 00'
020 ' --+s~-q~20'exp c)z2 Oz2
Or 10q~' Le Or
02~ '
+o)~+ Oz2 Oz2
t, 1 -
[( 1)]
exp y 1 -
-~
where Le is the modified Lewis number. The initial and boundary conditions are
o(o,z)- 1, O ( r , 1 ) =
1,
dO(.~,O)
= 0
dz d p ( r , O) ¢ ( 0 , z ) - 1, ¢ ( r , 1 ) - 1, ~ = 0 & The other parameters are defined by
DDxCs
Ds,~T~ g - - ~ ,
DeCs v L
Z =:z---,
to-
keys
E D~t 052 - - L2k o exp(E/RT~) De ' Y ~ s ' L2 '
T - - ~ -
ke/pCp L e - ~De-
ole D~
Here % is the effective thermal diffusivity. The nondimensional group y is called the Arrhenius group, and represents a nondimensional activation energy for the chemical reaction. MATHEMATICA is used to solve the partial differential simultaneous equations of the mathematically and thermodynamically coupled systems given in Eqs. (9.69) and (9.70). g = 16.0; le = 0.0015; f - 1.1; e = 0.001; w = 0.001; b = 0.0109; da =f'f; Print[" ~b= ",f,"; y = ",g,"; [3 = ",b, ". le = ", le, "; e = ",e,"; m = ",w, "; oxidation of CH3OH to CH20"]; eq 1 = D[c 1 [t,x], t] = = D[c 1 [t,x] ,x,x] + e* D [q[t,x] ,x,x] - d a * c 1 [t,x] *Exp [g* ( 1.0 - 1. O/q[t,x])]; eq2 - D [q[t,x] ,t] --= le*D [q[t,x] ,x,x] + le*weD [c 1 [t,x] ,x,x] + le*b*da*c 1 [t,x]*Exp [g* ( 1.0- 1.0/q[t,x])]; soll -NDSolve[{eql, eq2, cl [0,x]--- 1.0,cl[t,1] 1.0, Derivative[0,1][cl][t,0]-- 0.0, q[0,x] == 1.0,q[t, 1] - - 1.0,Derivative[0,1] [q] [t,0] - - {0.0},{cl,q},{t,0,1 },{x, 1,0},PrecisionGoal --, 0.08,MaxStepSize --, 0.0005, MaxSteps --, Infinity]:
468
9.
Coupled systems of chemical reactions and transport processes
Plot3D [Evaluate [c 1 [t,x]/.soll [[1 ]]] ,{t,0,1 },{x, 1,0},PlotPoints --, 40, A x e s L a b e l --, {"'r"," z "," 0' "}, DefaultFont --, {"Times-Roman", 13}]' Plot3D [Evaluate [q[t,x]/. sol 1 [[ 1]]], {t, 0,1 }, {x, 1,0},PlotPoints --, 40, AxesLabel --, {'"r"," z "," cp' "},DefaultFont --, {"Times-Roman", 13}]' ParametricPlot [{NIntegrate [ Evaluate [c 1 [t,x]/.sol 1 [[ 1]]], {x,0,1 }], NIntegrate [ Evaluate [q[t,x]/.sol 1 [[ 1]]], {x,0,1 }] }, {t,0,1 },PlotRange --, All, Frame --, True, GridLines --, Automatic, PlotStyle --, {PointSize[0.007],Thickness[0.009]}, FrameStyle --, Thickness[0.007], FrameLabel --, { "0' ", "~ '"}, RotateLabel --, True, DefaultFont -, {"Times-Roman", 13}]' Plot[NIntegrate[ Evaluate[c 1 [t,x]/.soll [[1 ]]], {x,0,1 }],{t,0,1 }, Frame --, True, AxesStyle --, Thickness[0.007], FrameStyle --, Thickness[0.007], DefaultFont --, {"Times-Roman", 13}, PlotStyle --, {PointSize[0.007],Thickness[0.009]}, GridLines --, Automatic, F r a m e L a b e l --, {"~-.... 0 '(,r)"}]' Plot[NIntegrate[ Evaluate[q[t,x]/.soll [[1]]], {x,0,1 }],{t,0,1 }, Frame -, True, AxesStyle --, Thickness[0.007], FrameStyle --, Thickness[0.007],DefaultFont --, {"Times-Roman", 13}, PlotStyle --, {PointSize[0.007],Thickness[0.009]},GridLines -, Automatic, F r a m e L a b e l --, {"~-","~ '(~-)"}]' Table 9.1 shows some of the experimental and assumed values of the parameters considered for catalytic oxidation of CH3OH to C H 2 0 with/3 = 0.0109 and hence display relatively fewer nonisothermal effects. The thermal diffusion coefficient is usually smaller by a factor of 102-103 than the ordinary diffusion coefficient for nonelectrolytes and gases. Therefore, for the present analysis the values for e and co are assumed to be 0.001. y = 16.0, Le = 0.0015, ~h = 1.1, /3 = 0.0109, e = 0.001, co = 0.001
1
0.9 0' 0.~ 0.'
(a)
1.00002 1.00002 rp' 1.00001
1
1.00001
0.8
0.6
1
0
"--... 0.2 ~
3.4 0.4
(b)
I:
z
0.2
0.6 0.8 1
0
Figure 9.4. Dynamic behavior of thermodynamically coupled nonisothermal reaction-diffusion system of catalytic oxidation of CH3OH to CH20: (a) concentration surface, (b) temperature surface. The parameters used are in Table 9.1.
9.3
469
Chemical reaction with coupled heat and mass flows
The numerical solutions from Mathematica are obtained with precision goal = 0.08, maximum step size = 0.0005, and maximum steps = infinity. Figure 9.4 shows the dynamic behavior of coupled values of concentrations and temperatures for the catalytic oxidation of CH3OH to CH20. The surface of constant temperature closely follows the change in concentrations. Figure 9.5 shows the changes of the spatial integral averages of concentration and temperatures with time. Figure 9.5c shows the state-space representation of temperature versus concentration when the time changes from 0 to 1. For catalytic oxidation of CH3OH to CH20, the temperature reaches its maximum value when the dimensionless concentration is approximately 0.97, as seen in Figure 9.5c.
1.00002 1.00001 1.00001 1.00001 1 .oooo I
0.8
0.75
0.85
0.9
0.95
1
O' (a)
0.95 0.9 0.85 0.8 i
0.75
,
!. . . . . .
0
0.2
0.4
0.6
0.8
(b) .,
.
.
.
1.00002 1.00001 1.00001
i
1.ooooI 1.00001
S 0
.............
0.2
0.4
0.6
0.8
!
(el Figure 9.5. Spatial integral averages for the catalytic oxidation of CH3OH to CH20; (a) change of concentration with time, (b) change of temperature with time, (c) change of temperature with concentration when the time varies between 0 and 1. The parameters used are in Table 9.1.
470
9.
Coupled systems of chemical reactions and transport processes
This analysis considers the thermodynamic coupling between heat and mass flows in an industrial reactiondiffusion system with a low value of/3. Modeling with the coupling effects of Soret and Dufour opens the path to describing more complex reaction-diffusion systems by adding the two new controlling parameters e and w.
9.4
COUPLED SYSTEM OF CHEMICAL REACTION AND TRANSPORT PROCESSES
Adenosine triphosphate (ATP) as a universal free energy transmitter undergoes the following turnover reaction ADP + Pi = ATP + H20. This reaction represents a simplified synthesis of ATP and hydrolysis of ATE which releases energy utilized in the transport processes. The quoted standard Gibbs free energy AG ° of this reaction is 30.5 kJ/mol at physiological conditions of 37°C, 1 atm, pH 7, pMg = 3, and 0.2 M ionic strength. However, the reported values of AG ° vary between 28 and 37 kJ/mol. The relations AG ° = -RTlnK' with the apparent equilibrium constant K' = {[ATP]c°/[ADP][Pi]}eq with c o representing the standard concentration may not be enough to determine AG ° directly. By introducing a H+-translocating ATP synthase, the coupled reaction cycle above becomes ADP + Pi + nHi+n = ATP + H 20 + nHo+ut where "in" and "out" denote two phases separated by a membrane, and n is the ratio H+/ATP which is the level of transmembrane proton transport for each ATP molecule to be synthesized. The apparent equilibrium constant K ' - {[ATP]c°(H+)out/([ADP][Pi](H+)in}eq is related to the pH level by -AG ° log(K') = ~ + nAPHeq 2.303RT where ApH is the transmembrane difference between outside and inside pH. The H+/ATP coupling ratio of 4 is a widely accepted value and determined from energy balance and proton flow measurements. The respiratory electron transport chain in the inner membrane of mitochondria creates a proton motive force across the membrane, which is used in synthesizing ATE Consequently, the hydrolysis of ATP is coupled in transporting substrates, leading to osmotic work of active transport and other mechanical work. Moreover, the ATP synthesis, in turn, is matched and synchronized to cellular ATP utilization according to the chemiosmotic theory. Enzyme-catalyzed reactions, including the electron transport chain and proton translocation, are composed of series of elementary reactions that proceed forward and backward. One of the methods in describing this thermodynamically and mathematically coupled complex chemical reaction-transport system is the nonequilibrium thermodynamic model, which does not require the detailed knowledge of the system. Lateral gradients of ions, molecules, and macromolecules may occur in an anisotropic medium, such as between mitochondrial compartments. An anisotropic medium or a compartmental structure may support the coupling between a chemical reaction and the transport processes of heat and mass according to the Curie-Prigogine principle. Such a coupling requires interactions between the scalar process of a chemical reaction and the vectorial process of heat and mass transport. In the vicinity of global equilibrium, coupled mass and heat flows and the reaction velocity can be derived by the linear nonequilibrium thermodynamic approach without the need for the detailed mechanisms of coupling phenomena. The modeling equations may be helpful in describing and controlling the evolution of some complex systems, such as reaction-diffusion phenomena with heat effects and active transport in biological systems.
9.4.1
Balance Equations
Consider a reversible homogeneous elementary reaction between a substrate S and a product P S
P
where kf and kb are the forward and backward reaction rate constants, respectively. This type of reaction system is highly common and plays an important role in chemical and biological systems, such as unimolecular isomerization,
9.4
Coupled system of chemical reaction and transport processes
471
enzyme kinetics, or racemization of molecules with mirror-image structures. For this reaction-transport system, general balance equations of mass and energy under mechanical equilibrium are _
'3Cs
- -V. N s + vsd r
- -
'3Cp _
(9.74)
(9.75)
4- vpd r
-V'Np
ht
pCp
where
~-/r is the heat of reaction.
OT
- -V.q +(-AHr)J r
'3t
(9.76)
The symbols Ns and Np denote the total flows of species S and P defined by Ns - Js + Csv
(9.77)
Np - Jp + Cpv
(9.78)
where v is the molar average velocity. The total heat flow q is
q-Jq+~_~JiHi
(9.79)
i
where H; is the partial molar enthalpy of species i. The reaction velocity is (9.80)
d C s _ dCp _ Jr ~'sdt vpdt
where the parameters Vs and Vp are the stoichiometric coefficients, which are negative for reactants (Vs- - 1 ) . Equations (9.74)-(9.76) show that the changes in concentrations and temperature are due to diffusion and convection, and chemical reaction. Using only the molecular transport in these differential equations, we have '3Cs '3t
(9.81)
- - V. a s + vsd r
'3Cp _
(9.82)
-- --V" Jp + vpd r
- -
'3t OT
p u p (J[ - - - V ' J q
+(-AHr)J
r
(9.83)
where Jq is the vector of the reduced heat flow Jq - q - ~ J;H;. By using Fick's and Fourier's laws in the onei=1
dimensional y-direction, Eqs. (9.81)-(9.83 ) become '3Cs - Ds,e O2Cs + V s J r
,31
0Cp
'3t ~T
0v 2
-
Dp,e 02 Cp
'3v2
+ VpJ r
(9.84)
(9.85)
O:T
p C p -~t -- ke - - + ( - A H r ) J r Ov 2
(9.86)
472
9.
Coupled systems of chemical reactions and transport processes
I I I I I I I
, I I I I
T
"5 Css
Cs
I y=+L
y=-L
I
I y=0
I I I
Figure
9.6. Schematic temperature and concentration profiles in a thin film.
where Di, e is the effective diffusivity for component i, and ke the effective thermal conductivity. Assuming a simple slab geometry (Figure 9.6), the initial and boundary conditions are t = 0
y = ±L
y = 0
C S -- CS0 Cp = Cpo
(9.87)
T = TO
C s -- CSs Cp -- Cps T = Ts (surface conditions)
OC s _ OCp _ OT _ 0 Oy Oy Oy
(symmetry conditions)
(9.88)
(9.89)
where L is the half-thickness of the slab. At stationary state, eliminating the reaction terms from Eqs. (9.84) and (9.85), and integrating twice with the boundary conditions given in Eqs. (9.88) and (9.89), concentrations of the species are related to each other by 0p = a 1 + a2(1- 0s)
(9.90)
with _ Cs 0s - < ,
_ Op
Cp
_ Cps
KCss , a 1
KCss , a 2 -
Ds,e KDp,e
Cps , K-
Css
(9.91)
The value of al determines the direction of the reaction; the net reaction is toward the P if al < 1. Similarly, eliminating the reaction terms from Eqs. (9.84) and (9.86), the temperature is related to the concentration by q~: 1+/3(1 - 0s) where T
(-AHr)Ds,eCss
The value of/3 is a measure of nonisothermal effects. As/3 approaches zero, the system becomes isothermal.
(9.92)
9.4
Coupled system of chemical reaction and transport processes
473
Example 9.7 Diffusion in a liquid film with a reversible homogeneous reaction Consider the reaction
S
P. After relating the two concentrations of species S and P by Eq. (9.91), at stationary state, Eq. (9.84) kb becomes
d20s
(9.93)
(Da s + Dap)0 s - (Dasa 1 + Dap)
-
dz 2
with the boundary conditions
dOs
(9.94)
0s(+_L) = 1, - - 7 ( 0 ) = 0 dz where
krE
Y
z = - - " Da s L'
Dap-
Ds,e
kbE Dp,e
Das and Dap are the Damk6hler numbers, and represent the ratios of the forward and backward reaction rates to the diffusion velocities Di,e/L; they measure the intrinsic rates of the reactions relative to those of the diffusions, and represent an interaction between reaction and diffusion. For the product, P, an expression similar to Eq. (9.94) can also be derived. Figure 9.7 displays the concentration profiles obtained from MATHEMATICA for reactant S and product P for two different sets of Damkohler numbers when 3' = 0 Da s - 5 0 . 0 ,
Dap-40.0,
Da s - l . 0 ,
Dap=0.5,
y=0.0 y=0.0
When the chemical reaction is fast (with large Damkohler numbers or with very low diffusivities) the reactant and product reach their equilibrium concentration throughout most of the film. The concentration gradients are very steep at the nonequilibrium region. The set of parameters Das = 1.0, Dap = 0.5, 3' = 0.0 represent slow reaction and nonequilibrium film.
0.8
"
m -...
m
0.6
0.4
0.2 ~.
•" .
"
"',
,
i
0.2
,
,
,
i
0.4
~
........
.
.
m
.
""
.
.
-
.
0.6
0.8
Figure 9.7. Concentration profiles of reactants and products for diffusion in a stagnant film with reversible homogeneous chemical reaction: Das = 1.0, Dap = 0.5, y = 0; bold dashed line is for 0s; gray dashed line is for 0p; Das = 50.0, Dap = 40.0, y = 0; bold solid line is for es, and gray solid line is for 0p.
474
9.
Coupled systems of chemical reactions and transport processes
MATHEMATICA code: (*a for S and b for P *) al : 50.0;bl = 40.0;g: 0.0; a2 : 1.0;b2 : 0.5; soil = NDSolve[{ ta 1"In] == (a 1+b 1)*tal [n]-(a l*g+b 1),tal [0]== 1.0,ta 1'[1 ] ==0.0, tb 1"[n] ==(a i +b 1)*tb 1 [n]-(a 1*g+b 1),tb 1 [0] ==0.0,tb 1'[1 ] ==0.0, ta2" [n] = = (a2 +b2)*ta2 [n]-(a2*g+b2),ta2 [0] = = 1.0,ta2' [ 1] ==0.0, tb2" [n] ==(a2 + b2)*tb2 [n]-(a2*g+b2),tb2 [0] = =0.0,tb2' [ 1] = =0.0}, {tal,tbl, ta2,tb2},{n,0,1}] Plot [Evaluate [{ta 1 In] ,tb 1 [n], ta2 [n],tb2 [n]}/.sol 1],{n,0,1 }, PlotRange ~ {{0,1 },{0,1 }}, Frarne ~True, GridLines ~Automatic, GridLines--,Automatic, PlotStyle~{{Thickness[0.007], Dashing [{0.0, 0.0}]}, {Thickness[0.005], Dashing [{0.0, 0.0}]}, {Thickness [0.007], Dashing [{0.025, 0.02}]}, {Thickness[0.004], Dashing [{0.025, 0.02}]}}, FrameStyle ~ Thickness [0,005], FrameLabel ~{"z", "0s, 0p"}, RotateLabel ~ True, DefaultFont ~ {"Times-Roman", 14}];
9.4.2 Linear Phenomenological Equations For a multicomponent fluid system under mechanical equilibrium with n species and Nr number of chemical reactions, the entropy production function is
o
(V/Ai)T'P -Jr ~ Aj T j=l --f- Jrj > 0
,q
(9.95)
i=1 where Ji is the vector of mass flows, }tZ i the chemical potential of species i, and A the affinity A = -Evil.Li; ifA > 0, the reaction proceeds toward the right. Equation (9.95) is derived from the general balance equations including the entropy balance and the Gibbs relation, and identifies a set of independent conjugate flows Ji and forces Xk to be used in the linear phenomenological equations when the system is in the vicinity of global equilibrium
Ji - Z LikYk
(9.96)
Equation (9.95) excludes viscous flow, electrical, and magnetic effects. For the chemical system considered in Eq. (9.72), Eq. (9.95) yields ~ = Jq "~7(T1--)- Js "(~7/zS)T'~P T - JP "(V/zP)T'~P T + Jrs A T -> 0
(9.97)
where
n-1 OILi (V/LLi)T,P = 2 - ~ / V C i i=1 By using the Gibbs-Duhem equation at constant temperature and pressure Cs~TIAS -k-Cp~7/.Zp = 0
(9.98)
JsVs + JpVp = 0
(9.99)
and no volume flow condition
9.4
Coupled system of chemical reaction and transport processes
475
where Vi is the partial molar volume of species i, Eq. (9.97) becomes
Op
-
1 -J
1
A
V T - 'Is ~- asVCs + J,s --T >- 0
T2
q
(9.100)
where
"(' +
CS
for(Vs =Vp)
/T,p
Using the net flows and net forces based on Eq. (9.100), we obtain the linear phenomenological equations as follows 1
1
A
1
A
Lsq --~VT + Lsr -~-
'Is - -Lss y AsVCs 1
'I,t = -Lq s -T AsVCs - Lqq 7 VT + Lqr --T 1
Jrs
-
1
-Lrs -~-AsVCs - - Lrq - ~ V T
A
(9.102)
(9.103)
Lrr ~-
+
(9.101)
The above phenomenological equations obey Onsager's reciprocal rules, and hence there would be six instead of nine coefficients to be determined.
9.4.3
Degree of Coupling
The linear phenomenological equations help determine the degree of coupling between a pair of flows; the degree of coupling between heat and mass flows qsq and between the chemical reaction and the transport process of heat and mass flows, qsr and qrq are
-
qsq
-
Lsq (LssLqq)12, qsr ,
-
-
Lsr (LssLrr)l/2 , qrq
_
_
Lrq
(LrrLqq)l/2
(9.104)
Equations (9.101)-(9.103) assume coupling between the vectorial flows of transport processes and the scalar chemical reaction velocity. This type of coupling is possible in an anisotropic medium only according to the Curie-Prigogine principle. Consequently, the nonvanishing values of cross coefficients, Lsr, LrS, Lqr, and Lrq, must have vectorial characteristics. For example, during the active transport of sodium ions, in which the hydrolysis of ATP is coupled with the flow of sodium ions, the direction of flow is determined by the property of the membrane in the mitochondria. The medium may be locally isotropic, although it is not spatially homogenous. In this case, the coupling coefficients are associated with the whole system.
9.4.4
Efficiency of Energy Conversion of a Reaction-Diffusion System
The following entropy production function from Eq. (9.100) shows the input and output energies when there is no heat effect 1
A
1
1
- - ' I s -~ AsVCs + Jrs -~ - output + input -> 0
(9.105)
Applying the entropy production function above to the active transport in a biological cell, the chemical reaction term (Jrs(A/T)) represents the hydrolysis of ATE which can pump an ion in a direction opposite to the direction imposed by its thermodynamic force, and hence we have
,
)
-3 s-~hsvC s < 0
(9.106)
476
9.
Coupled systems of chemical reactions and transport processes
However, the hydrolysis of ATP can pump the ions only if some degrees of coupling exist between the reaction velocity and the mass flow. The efficiency of energy conversion for pumping a substrate with the help of a chemical reaction may be related to the degree of coupling by using Eq. (9.105) -
output input
-
-Js(1/T)AsVCs _ x((Lss/Lrr) 1/2x + qSr) -
Jrs(A/T)
[qsr
+(Lrr/Lss) 1/2 ]
(9.107)
where x is the ratio of thermodynamic forces defined by
(1/T)AsVCs (A/T)
x=
(9.108)
The optimal efficiency would be a function of the degree of coupling.
9.4.5
Phenomenological Coefficients
The diagonal elements of the coefficient matrix [L] may be identified by using Fick's, Fourier's, and the mass action laws. Comparing the first term on the right of Eq. (9.101) with Fick's law, J = -Ds,~VCs, yields Lss -
Ds,eT
As
(9.109)
Similarly, comparing the second term in Eq. (9.102) with Fourier's law, Jq = -keVT , yields
Lqq = keT2
(9.110)
The cross coefficients Lsq and Lqs may be represented by the Soret coefficient ST or the thermal diffusion coefficient DT, which are related to each other by
Lsq = STDs,eT2Cs = DTT2Cs
(9.111)
The Soret coefficient is the ratio of the thermal diffusion coefficient DT to the ordinary diffusion coefficient D
DT ST --
Ds,e
(9.112)
at steady state, and has the dimension of T-1. It changes in the range 10-2-10 -3 1/K for gases, nonelectrolytes, and electrolytes. The term Lqs(= DTT2Cs) is expressed by the Dufour coefficient DD
Lqs
DDCsT =
(9.113)
As
For Lqs = Lsq, we have D D -- DTTAs, which is proved experimentally. We may define two new effective diffusion coefficients, DT,e and DD,e, which are related to the thermal diffusion and the Dufour effect, respectively 1
DT, e --
Lsq --~ = STDs,eC s = DTC s
%
DD,e = L qs--f-= D DCs
(9.114)
(9.115)
With these newly defined primary and cross coefficients, Eqs. (9.101)-(9.103) become A Js - -Ds,eVCs - DT,eVT + Lsr "~-
(9.116)
9.4
Coupledsystem of chemical reaction and transport processes
A
Jq - - D D ~VC s - keV T + Lqr
-
'
Jrs 9.4.6
1
= -Lrs ~ AsV C s - Lrq ~ 1
1
V T nt-
477
(9.117)
-
T
kf Cs,eq A
R
(9.118)
T
Reaction Velocity
For the reaction S
P, the affinity is A -/Xs-/Xp. The local rate of entropy production due to chemical reaction is ~_A
~-Jr
(9.119)
where Jr is the reaction velocity. The phenomenological form of Jr is A Jr - L~r -~
(9.120)
where Lrr is the phenomenological coefficient. The reaction velocity is also defined in terms of the forward (f) and backward (b) reaction rates Jr = J r f - J r b
(9.121)
= kfC S - k bCP
The affinity is also related to the forward and backward reaction rates as follows
A - RTln Jrf / or
Jrf _Cskofexp(-Ef/RT)(A)
(9.122)
Cpkobexp(-Eb/RT) = exp
Jrb
where Ef and Eb are the activation energies for the forward and backward reactions, respectively. Using Eqs. (9.12 l) and (9.122), the reaction velocity Jr in terms of affinity A is given by
xpl ¢/)
(9.123)
Far from global equilibrium, the reaction velocity is not only related to affinity but also depends on the concentration of species. If we expand Eq. (9.123) and consider the near global equilibrium state ( A~ Rrl P2, was placed in the two compartments, and the steady-state flow of oxygen across the membrane was measured. The presence of hemoglobin enhanced the flow of oxygen at low oxygen pressure, however, the facilitation of oxygen transfer disappeared at a higher pressure of oxygen. Katchalsky and Curran (1967) used the linear nonequilibrium thermodynamics theory for the facilitated oxygen transport by the hemoglobin based on the linear flow-force relations. The following reaction is used for the facilitated oxygen transport by the carrier hemoglobin nO + H b ~ HbO •
2t~
(9.206)
A representative dissipation function takes the form XF -- J1 ~7(- ~1 )-+- J2V( - ~ 2 ) -Jr-J3V( - / z 3 )
(9.207)
A - n#, + / x - / x
(9.208)
The affinity A is given by
where the subscripts 1, 2, and 3 refer to oxygen, hemoglobin (Hb), and oxyhemoglobin (HbO2n), respectively. We assume that the rate of reactions is more rapid than that of diffusion, so that the reaction is at equilibrium, and hence A- 0
(9.209)
Applying the gradient operator to Eq. (9.208) at equilibrium, we obtain ,V#l + V/~2 - V/~3
(9.210)
This equation relates forces that lead to a coupling of flows. Passing at any point in the membrane are the flows free oxygen J1, hemoglobin J2, and oxyhemoglobin J3. The externally measured flow of oxygen Jl equals to the flows of free oxygen and oxygen carried by hemoglobin
JJ - J1 + n J3
(9.211)
Since no external flows of hemoglobin J2 take place, we have •12 - J2 + J3 - 0
(9.212)
492
9.
Coupled systems of chemical reactions and transport processes
Using Eq. (9.210), we transform the dissipation equation as follows
xt* = J ; V ( - / x 1) + J2V(-/x2 )
(9.213)
From Eq. (9.213) the following linear phenomenological equations are obtained
J( = -L~lVa~- L~2Va2
(9.214)
J2 --0 = - L 2 l V ~ I -
(9.215)
L22V~2
Equation (9.214) can also be expressed in terms of the diffusivity coefficients J1 - -D1 dcl + nD2 dc2 dx dx
(9.216)
By integrating Eq. (9.216) between x = 0 and x = h, and assuming that J(, D1, and D 2 are constant, we get j~ = D1 c o - c~ + n D 2 c~ - c~ h h
(9.217)
The external oxygen pressure determines the first term on the right side of Eq. (9.217). If P1 > P2 and co > c~, then the contribution of the carrier is considerable. The transport of oxygen increases with increasing hemoglobin concentration. When P2 - 0 and c~ = 0 the facilitated oxygen transfer decreases with increasing P1.
Example 9.14 Nonisothermal facilitated transport An approximate analysis of facilitated transport based on the nonequilibrium thermodynamics approach is reported (Selegny et al., 1997) for the nonisothermal facilitated transport of boric acid by borate ions as carriers in anion exchange membranes within a reasonable range of chemical potential and temperature differences. A simple arrangement consists of a two-compartment system separated by a membrane. The compartments are maintained at different temperatures T1 and T2, and the solutions in these compartments contain equal substrate concentrations. The resulting temperature gradient may induce the flow of the substrate besides the heat flow across the membrane. The direction of mass flow is controlled by the temperature gradient. A set of example chemical reactions for boric acid that take place in an anion exchange membrane is HB + OH- ~ B- + H 20 ( a - b)B- + bHB ~- HbBa -(a-b) where HB denotes boric acid, B- the borate, which is the cartier ion, and a and b are the stoichiometric coefficients. Assuming that the system is in mechanical equilibrium and at steady state, the entropy production is
AT A t.tB _+_Z jr i ~. OP=Jq~av -+-JB Tav i ' Tav where A~B =/*B,I- ~B,II, A T - TI - TII, and Tav = (T[ + TII)/2, Jq is the heat flow, which takes into account the enthalpy of mixing and the heat of reactions, Jri the flow for reaction i, A~B the chemical potential difference of H3BO3, AT the temperature difference across the membrane, and JB the absolute flow of boric acid directed from compartment I to compartment II when the absolute flow of water through the membrane is negligible. Equation above shows the three contributions to the rate of entropy production due to heat flow, mass flow, and the chemical reaction, respectively, and excludes the viscous and electrical effects. As the membrane is assumed to be an isotropic medium, there will be no coupling between the vectorial heat and mass flows and scalar chemical reaction, according to the Curie-Prigogine principle. Under these conditions, entropy production equation identifies the conjugate forces and flows, and linear relations for coupled heat and mass flows become AT
Jq = Lqq ~
Tav
+ LqB
A/xB
Tav
(a)
9.6
493
Facilitated transport
AT
JB -- LBq ~
Tav
A/.zB
+ LBB ~
(b)
Tav
According to the Onsager reciprocal rules, we have LqB = LBq. If we replace the chemical potential in terms of the concentration (instead of activity), we transform Eqs. (a) and (b)
AT (CBI ] Jq - Lqq ~ q- LqBRln Tav k CBII )
(c)
AT (CBI ] JB -- LBq ~ + LBBR In Tav ~,CBII )
(d)
To estimate the flow of borate, and assuming that the phenomenological coefficients are dependent on the average concentration (CBI + CBn)/2 linearly, we have
LBB -- aBB if- ]~BBCav LBq -- OLBq-ff ~ BqCav Substituting these relations into JB, we get
JB
_
{
(
1
OlBq'-~ff-~Bq%]~T'+" Tav Ta~)
1
Cav ]
O/BBT--~vq-ffiBB Tav ) ~/J'B
The coefficients ffBB,/~BB may be determined by the isothermal mass transport
J
1
,,so =
Cav)
Therefore,
LBB --
JB iso ' Tav - C~BB-I- ~BBCav '~P-.~B
(e)
On the other hand, the difference between nonisothermal and isothermal mass transport AJB yields a relation to estimate the coefficients aBq, ~Bq
AJB - JB --JB,iso -
C~Bq~
1
Ev
Cav / if-/~Bq ~ AT
ray
After rearranging this relation, we find AJB 2 LBq -- A T Tar - °lBq + [~3qCav
(t)
The linear Eqs. (e) and (f) help in determining the coefficients aBB,/3BB as well as ~Bq,/~Bq. Equation (e) produces a linear plot between (JB,iso/Al~B)Tav and Car, while from Eq. (f) a straight line results when plotting (AJB/AT)T~v versus Cav.The coefficients obtained from the slopes and intercepts of these straight lines are tabulated in Table 9.2 (Selegny et al., 1997).
494
9.
Coupledsystems of chemical reactions and transport processes Table 9.2. The phenomenological coefficients in the facilitated transport of boric acid by borate ions
Coefficient 2.8 × 10-7 K mol2/(m2 s J) 7.8 × 10 -9 K mol m/(s J) -9.15 × 10-3K mol/(m 2 s) --1.0 X 10-14K m/s
O/BB /~BB O/Bq
/~Bq
Source: Selegny et al., 1997.
Effect of heat of reaction From phenomenological equations AT A]..£B Js = Lsq ~ + LBB
AT A/xB Jq = Lqq ~ + LqB
rav
rav
7av
Vav
we have the isothermal facilitated heat and mass flows Ax/_______~B
Jq,is°=Lq B h~"-""~B Tav '
JB,iso = LBB Tav
Under these assumptions, the stationary heat of transport for formation of the complex per mole of B transported becomes Jq,iso = AH(JB,iso )
This equation shows that heat and mass flow in the same direction if &/-/> 0 (symport) and in opposite directions (antiport) if AH < 0. The two equations above and Onsager's rules yield
LqB -- LBq = L B B Z ~ The AJB is obtained from
A JB = JB -- JB,iso -- LBB ~
-L5-
The direction of AJB is determined by the sign of the product (~-/. AT). The difference of the heat of transport is related to
AJq = Jq -Jq,iso
= ~J-/(AJB)
This relation yields
Lqq = LBB~ 2
Level flow and static head Level flow occurs at zero load, which is at A/z = 0. At level flow, the mass flow is induced by AT and the phenomenological equation becomes
J ,max =
LBq
AT /av
9. 7
495
Active transport
With LqB = LBq = LBBZ~r--/, this equation is transformed into the facilitated form
JB,max
---
L B B A H AT
r:2v
At static head (JB -- Jq = 0), we have the maximum mass load A~max. From equation
AT T,7,,
JB -- LBq -X-V-+ LBBRIn
{CBI ] CBa
we have AT_ LBq To;,v
LBq AT
A/xB
--LBB ~,,, A~B'max . . .LBB . Tar
By using LqB = LBq -- LBBAH, this equation can be transformed into
~/d,B,ma x -- -- AH
9.7
AT
LV
ACTIVE TRANSPORT
Diffusion occurs spontaneously from a region of higher chemical potential/.L 1 to a region of lower chemical potential /x2, and the direction of flow is the same as the direction of decreasing chemical potential. The total Gibbs energy change for such a system is expressed by
- d G - - d N [&l + dN tx 2 - dN (ix 2 - ~ )
(9.218)
However, if a compensating process with a coupling mechanism, is added into the isolated system, causing the total free energy of the complete system to decrease, then it is possible to envisage diffusion against a potential gradient or an electrochemical potential gradient. This type of mass diffusion is called active transport for which the product of the flow d~ being acted upon by the generalized force X, is negative JiXi < 0. This inequality implies that the flow is occurring in the direction opposite to the direction of the force. If the compensating process is a chemical reaction, then the coupling will only be allowed in an anisotropic medium, according to the Curie-Prigogine principle. Since the chemical potential gradient is a vector quantity grad/~ - i ~0/~i + j Ol~i +k O/zi
&r
Oy
Oz
(9.219)
and the chemical reaction rate is a scalar quantity, no chemical reaction can impose directional properties onto the flow of substances unless the spatial gradient of a thermodynamic potential is altered. In biological systems, active transport involves the use of proteins that require the use of cellular energy (usually the energy released by the hydrolysis of ATP) to actively pump substances into or out of the cell (Figure 9.10). For example, protein found in the cell membrane of neurons pumps sodium ions from the inside to the outside of the neuron and pumps potassium ions in the opposite direction. This process sets up a high concentration of sodium ions outside the cell and a high concentration of potassium ions inside the cell. This concentration difference across the membrane is important for the generation of nerve impulses for transmitting information from one end of the neuron to the other. The sodium-potassium pump in red blood cells is operated by an oligomeric protein called Na+-K ÷ ATPase, which is embedded within and transverses the plasma membrane. Almost all cells have an active transport system to maintain nonequilibrium concentration levels of substrates. For example, in the mitochondrion, hydrogen ions are pumped into the intermembrane space of the organelle as part of producing ATR Active transport concentrates ions, minerals, and nutrients inside the cell that are in low concentration
496
9.
Coupled systems of chemical reactions and transport processes
3Na + ~i, Na+_K + ATPase Outside cell
aN~
Membrane
Jr
ATP Inside cell
ADP +
JK
Figure 9.10. Schematic active transport coupled to hydrolysis of ATE outside. Active transport also keeps unwanted ions or other molecules that are able to diffuse through the cell membrane out of the cell. Another common active transport system class pumps molecular nutrients such as glucose and amino acids into the cell at a much faster rate than can be achieved by passive or facilitated transport. For example, in the cells of higher animals, glucose active transport is dependent on the cotransport of Na + into the cell, in which Na + is pumped back out with simultaneous ATP hydrolysis by the Na+-K + ATPase. If an uncharged substrate is pumped from a region where its concentration is 0.001 M to the another region where this substrates concentration is 0.1 M, then we need the following amount of energy at 298 K to drive the transport
AG°=[8"314J/(m°lK)](298K)ln{O'l) =ll'41kJ/m°lO.O01 Free energy for active transport is supplied by a driving system from a high potential state to a low potential state. In primary active transport, the driving system is a chemical reaction away from equilibrium, and in secondary active transport, it is a concentration gradient. A diffusional flow against its conjugate thermodynamic force driven by dissipation of another diffusional process is called an incongruent diffusion, not active transport. Conventional methods for establishing the existence of active transport are to analyze the effects of metabolic inhibitors, to correlate the rate of metabolism with the extent of ion flow or the concentration ratio between the inside concentration and the outside concentration of the cells, and to measure the current needed in a short-circuited system having identical compositions solution on each side. Measurements indicate that the flow contributing to the short-circuited current, and any net flow detected are due to active transport, since the electrochemical gradients of all ions are zero (A~b- 0, Co ci). -
Example 9.15 Long-term asymptotic solution of reversible reaction diffusion systems The formation of dynamic reaction-diffusion fronts occurs when two species A and B are uniformly distributed on opposite sides of an impenetrable barrier, which is removed at time t = 0 at isothermal conditions. The species A and B start to diffuse and react upon mixing, and produce species C. This creates a dynamic reaction front, and the spatio-temporal evolution of this front may exhibit some unique features, which may be valuable in understanding many phenomena in physical, chemical, biological, and geological systems. The long-term behavior of this reversible reaction-diffusion system was studied by Koza (2003). For the reaction system A + B < - - ~ C , kb differential equations
we have the following mathematically coupled nonlinear partial
OA Ot
02A Ox2
DA ~ -
Jr
(9.220)
9. 7
497'
Active transport
DB __02 B _ Jr Ox2
OR _
3t OC
O2C Dc ~ + Jr
-
at
(9.221)
(9.222)
3x 2
where JR is the effective reaction rate with forward kf and backward kb reaction rate constants
kf AB
Jr -
-
k bC
(9.223)
Here, the concentration of species is a function of distance x and time t; A(x,t), B(x,t), C(x,t), and D; show the diffusion coefficient of species i. The initial conditions are
A ( x , O ) - ~ , B ( x , O ) - B o, C ( x , 0 ) - 0
(9.224)
We measure distance, time, and concentration in units o f x/DA/kf A o ,1/kf A 0 ,and A0, respectively. The problem will have four independent control parameters: DB, Dc, B0, and kb by assuming D A - l, A 0 - 1 and kf = 1. Long-term solutions of Eqs. (9.220)-(9.223) are series with ~-= l/t, and coefficients being some functions of r-s = x/qt
A ( x , t ) - ~ ~"a,(e)
(9.225)
n=()
-/
B(x,,;) -- ~ tl
-::
r'b,(e)
(9.226)
(i
C(x,t) - ~ "r'c, (e)
(9.227)
n=0
and
.1~- ~ ~-"r,,(e)
(9.228)
n=O
where r,(e) = Ej=oaj(e)b,,_i(e) - kt, c, (e), and a, b, c, and r are the scaling functions. When collecting coefficients with ~ + l where n - 0, 1, 2 .... , we have c__t2 a ,, ( e___~~-~- s
de 2
DB
Dc
2
d 2b, (e) de 2
da,,(e) +-ha, -- r,+ 1(e) = 0 de e db, (e) 2
de
(9.229)
+nb, - r,,+] (e) - 0
(9.230)
+n G + r , + l ( e ) - O
(9.231)
d2c,,(e) , s dc,(e) de 2
,:.~ de
Therefore, the lowest order terms (at ~.0) satisfy d:a,)(e)
e dao(e) t
d~ 2
2
de
~i(e)- 0
(9.232)
498
9.
Coupled systems of chemical reactions and transport processes
d 2 b 0 (e) e db 0 (e) DB~ -+ de 2 2 de
d 2c O(e)
DC
de 2
t
e d c O(e) 2
de
r1(e) = 0
(9.233)
+-r1(e) = 0
(9.234)
ro(e ) = 0 = ao(e)bo(e ) - kbCo(e)(k f = 1)
(9.235)
lim ao ( e ) = Ao, lima 0 (e) = 0, limbo (s) = 0, and limb o (e) = B o
(9.236)
with the boundary functions
Equations (9.232)-(9.235) have four unknowns functions, a0(e), b0(e), Co(e), and rl(e), which control the asymptotic (long-term) properties. Equation (9.235) shows that in the long term the chemical reaction tends toward a local chemical equilibrium at which the forward and backward rates become asymptotically equal. Because of the nonlinear of form Eq. (9.235), the solutions to these functions can be found for specific cases. One of these specific cases would be to consider a system with equal diffusion coefficients D A = D B = D c = 1, k1 = 1, and Ao = 1 The remaining unknown parameters are Bo and kb. By adding Eq. (9.222) to Eq. (9.221) and (9.220), we have two diffusion relations A ( x , t ) + C ( x , t ) = -~ erfc
(9.237)
B ( x , t ) + C ( x , t ) = ~ Boerfc
(9.238)
These equations have well-known solutions a0(e ) = bo(~) =
*(,)-
kb + J±(~) 2
-*(~)-
kb +
,/±(~)
2
(9.239)
erfc(e/2) + B o e r f c ( - e / 2 ) + 2k b - 2 ~x/-~ Co(~) =
4
~(~) = kbB0 (1 + k b + B o ) e x p ( - e 2 / 2 ) 27r[A(e)] 3/2
where ~(~)
= Aoerfc(e/2 ) - B0erfc(-e/2 ) 2
A ( e ) = [ ( I ) ( e ) - k b ] 2 + 2kberfc(2 /
Since ~(e) decreases monotonically from Ao to B 0, equation ~(e) = 0 has a unique solution denoted by ef. In the case of an irreversible reaction, where kb ~ 0, we have
~¢(s), lim a 0 (e) = [0, kb~0
s < sf} e ~ ef
(9.240)
9. 7
499
Active transport
lira bo(e) -
-0,~ ( e ) ,
f} e8 ~< gef
(9.241)
k b --' 0
lim c o (e) -
~ B° erfc
(9.242)
erfc
lim q (e) = B° exp(-e2/4) 6(e - el) kb~0 x/~erfc(ef/2)
(9.243)
where 6 is the Dirac's delta distribution. Both species A and B will be segregated at el, which may be identified as the position of the reaction front. Beyond this front, the reaction rate approaches zero and the species concentrations satisfy the diffusion equations. The reaction must be restricted to a region narrower than ~7, which is in line with the fact that for irreversible reactions, the width of the reaction front grows as t~ with c~ < 1/2. If we consider the opposite limit where kb--+ % the concentration of species C approaches 0, while the concentrations of species A and B evolve as if there was no reaction, and the scaling functions become
1
l i m a 0 (e) = -;- erfc(e/2) z
k b --,
1
lim bo (e) = -g Boerfc(- e/2) z, kb--+~f_ lira c o (e) = 0
(9.244)
kb-, ~,_
Example 9.16 Nonisothermal heterogeneous autocatalytic reactions-diffusion system Consider an autocatalytic reaction A
k, >B
A+(n-1)B
k2 >nB
where n is the order of the autocatalytic system, and kl and k2 the rate constants of the first- and second-reaction steps, respectively. Here, the product of the reaction acts as a catalyst and hence affects the conversion of reactant to product. The first step is slower than the second autocatalytic step. Effectiveness factors for autocatalytic reactions can be much larger than unity for exothermic and endothermic reactions. One example of such reactions is the catalytic cracking of paraffins on a zeolite catalyst to produce olefins. A mathematical model was suggested by Neylon and Savage (1996) with the assumptions: (i) the effective diffusivities of species A and B are the same, (ii) the catalyst pellet is symmetrical, (iii) effective diffusivities and thermal conductivities are constant, and (iv) external resistances to heat and mass transfer are negligible. It is also assumed that the diffusion is equimolar and heats of reaction for both reaction steps are equivalent to M/r, which is assumed to be constant. With power-law kinetics and the Arrhenius relationships for the reaction rate constants kl and k2, the rate of reaction becomes
- - J r A -- k01 e x p
--~-~ C A -Jr-k02 e x p --~-~- CAC~n-')
(9.245)
Under the conditions assumed and at steady state, the rate of diffusion of A within the pellet is equal to the rate of consumption due to reaction, and we have DA
d 1,, d/ a/
ko exp ' 77
CA
E2 )
~(n-l)
+ ko2 exp -~-~- C A,~B
(9.246)
500
9.
Coupledsystems of chemical reactions and transport processes
where 1 is the spatial coordinate and s the shape factor of the catalyst particle; s = 0 for a slab, s = 1 for a cylindrical shape, and s = 2 for a spherical shape. Similarly, the rate of heat conduction is equal to the rate of heat generation due to the reaction, and we have
lS dl
-~
/
=(-AHr)
exp/
E
k01exp --~-f CA
]
(9.247)
The boundary conditions are Atl=L
CA=CA0
At 1 = 0 dCa = 0
and and
dl
T=T o ~=dT 0
dl
where L is the characteristic length of the particle and CA0 and To the concentration and temperature of the fluid at the surface of the particle, respectively. Equations (9.246) and (9.247) represent mathematically coupled system. Temperature is related to concentration at any internal point by Prater's relation
T_To = _ (~Hr)DA (CA0-C A) ke
which may be used in Eq. (9.246) to reduce the number of variables from two to one by eliminating temperature within the Arrhenius relationships. Once the concentration profile is obtained, then the effectiveness factor is determined by
DAap z
m
--JrA0 where ap is the external surface area and Vp the volume of the particle. The reaction rate at the surface conditions is represented by-JrA0. With n = 2, we have a quadratic expression for the rate of reaction, which can display multiple steady states with high values of effectiveness factor for exothermic autocatalytic reactions.
9.8
NONLINEAR MACROKINETICS IN A REACTION-DIFFUSION SYSTEM
For a reaction S --, R the affinity is A =/Zs - ~p. After substituting the chemical potentials of the substrate and product in an ideal system,/x =/x ° + In C, where/x ° is any reference state, or the component compositions in the form (Sieniutycz, 2004)
Ck=exp
/zk exp -
(9.248)
we have the Marcelin-de Donder form Jr=Jrf-Jrb=kfCs-kbCp-Jro(T)
exp vs ~
- e x p Vp
(9.249)
where Jro is the exchange current and is expressed in terms of forward or backward reaction rate constant
(
(
Jro ( T ) = kf exp - v s ~-f = kb exp -Vp RT)
(9.250)
Here, the stoichiometric coefficients Vs = - 1 and Vp 1 are used. The exchange current Jro satisfies the microscopic reversibility at the state of thermodynamic equilibrium. These relations can be applied to chemical reactions with ionic substances by replacing the chemical potentials with electrochemical potentials. =
501
Problems
9.8.1
Generalized Chemical Kinetics
Consider an elementary generalized chemical reaction step
VofBor + ~ ....
pkt.Bkf--~V0bB0b+ k=l £ VkbBkb
(9.251)
1
where u is the stoichiometric coefficient matrix. Equation (9.251) involves the energy B0 (for k = 0) as well as the chemical species B~. (for k = 1,2,...,s, where s is the number of state coordinates in the generalized system) and therefore includes the heat effects. Here, the forward f and the backward b processes are considered. The net reaction rate is defined by
/ /
Jr--J0 exp --~v/dk=l
--exp --~Vbf--~ k=l
Where F is the potential defined by F-
1 /x)
(9.253)
7'7
The 1/T and tx/T appear in the thermodynamic conjugates of the extensive variables in the Gibbs equation for the system entropy dS = 1 d E T
(9.254)
Ixk dc k T
When the transport process is fast, potentials are equal at forward and backward states. With the total number of system coordinates s comprising the energy ( s - 0) and n components, we have the extended affinity defined by
A
--
(Pkbfkb-- U/'t"F1i")--
k=0 =--1If --[|b
£ Pkf[~/,T /"Of Tf k=l rf
I£Pkb[&kb P0b / k=l
rb
rb
(9.255)
where 1-If and H b are the generalized potentials of forward and backward states, respectively. The second terms within the generalized potentials describe heat effects through the energy barrier. These potentials are state functions, and the reaction proceeds from a higher to a lower potential, which is similar to heat flowing from higher to lower temperature.
PROBLEMS 9.1
Consider the hydrogenation of benzene, which is exothermic with a heat of reaction 50 kcal/mol. For a catalyst pellet containing 25% Ni-0104R 25% graphite, 50% T-AI203 (Harshaw A1-0104T), the effective thermal conductivity and diffusivity are 3.5 × 10 .....3 cal/(cm s K) and 0.035 cm2/s, respectively. For a benzene surface concentration of 4.718 x 10 -6 mol/cm 3, and a surface temperature of 340 K. (a) Estimate the maximum internal temperature difference. The observed rate for the reaction is 22.4 × 10-6mol/(g cat s) and the density of the catalyst is 1.57 g/cm 3. (b) Using the surface concentration as fluid bulk concentration, and the modified Sherwood number 401, estimate the internal temperature difference and compare with the one obtained from part (a).
9.2
Consider the hydrogenation of benzene, which is exothermic with a heat of reaction 50 kcal/mol. For a catalyst pellet containing 58% Ni on Kieselguhr (Harshaw A1-0104T), the effective thermal conductivity and diffusivity are 3.6 × 10 -.4 cal/(cm s Ki and 0.052 cm2/s, respectively. The fluid bulk concentration of benzene is 5.655 × 10-6 mol/cm s, and the fluid bulk temperature is 412 K. The characteristic length of the pellet is 0.296 cm. The observed rate for the reaction is 2.258 x 10 --~ mol/(g cat s) and the density of the catalyst is 1.88 g/cm 3. The modified Sherwood and Nusselt numbers are 215 and 10.8, respectively. Estimate the internal and external
502
9.
Coupled systems of chemical reactions and transport processes
temperature differences. (Note: The experimental values of the internal and external temperatures are 35 and 40 K, respectively. The difference between the surface and bulk fluid temperature is 6-7K, Froment and Bischoff, 1979). 9.3
Derive modeling equations for diffusion with a heterogeneous chemical reaction A ~ B. (a) The reaction is instantaneous. (b) The reaction has a rate NA = kCA. (c) Diffusion with a homogeneous chemical reaction A + B ~ AB, which is a first-order reaction with respect to component A.
9.4
Model a process of diffusion and chemical reaction A ~ B inside a porous catalyst. Assume that the reaction takes place on the catalytic surface within the porous medium.
9.5
Chemical vapor deposition is an important technique in the production of all kinds of solid-state devices. In the process, the active metal organic vapor is swept into a two-dimensional slit reactor by a carrier gas, and deposition occurs at the hot top and bottom plates. The reaction at the plate surfaces can be written MOoM+O Assuming a laminar operation, develop an expression to compute the rate of loss of MO for diffusioncontrolled process.
9.6
A hollow tubular reactor has inside walls that are coated with a catalyst. A feed of reactant A and an inert fluid pass through the reactor. At the tube wall, the irreversible catalytic reaction takes place A~P Assuming a plug-shaped velocity profile, estimate the concentration profile.
9.7
In a nuclear fuel rod, the rate of production of neutrons is proportional to the neutron concentration. Derive the differential equation to describe the flow of neutrons in the fuel rod.
9.8
A compound diffuses in a 4-cm long tube while it reacts. The compound starts from a source with a concentration of 0.2 M at the beginning of the tube. The reaction is a first-order irreversible reaction. Compound A is absorbed completely by an adsorbent at the other end of the tube, so that the concentration is assumed to be zero. Determine the concentration as a function of distance in the tube. Assume that the diffusivity is D = 1.5 × 10 -5 cmZ/s and the reaction rate is constant k = 5 × 10 -61/s.
9.9
A heterogeneous reactionA ~ 2B with nth order kinetics JrA = -kC~ ( n > 0) takes place on a catalyst surface. The component A with initial concentration CA0 diffusses through a stagnant film on the catalyst surface at isothermal and isobaric conditions. Assume one-dimensional diffusion, and determine the concentration profile of component A within the film of thickness 6 if the k is constant.
9.10
Derive a finite difference formulation for a steady-state reaction-diffusion system.
9.11
Consider the antibiotic production of Cycloheximide from streptomyces (Ablonczy et. al., 2003). For the first 24 h, streptomyces grows quickly and produces, a small amount of cycloheximide. After that the mass of streptomyces remains relatively constant and cycloheximide accumulates. Once the cycloheximide reaches a certain level, extracellular cycloheximide is degraded (feedback is initiated). One way to alleviate this problem and to maximize cycloheximide production is to continuously remove extracellular cycloheximide. The rate of growth of streptomyces is
dt
-
~max 1 -
x(O) = 1
Xmax
x
503
References
where x is the mass concentration,/Xma~ the maximum specific growth rate, and Xmax the maximum mass concentration. Experiments show that ]£max 0.3 l/h, and Xmax 10 g/L. (a) For 0 < t 0, Lp > 0, and LoLp-LopLpo > 0. L o shows the influence of substrate availability on oxygen consumption (flow), and
Lp is the feedback of the phosphate potential on ATP production (flow). The cross-coupling coefficient Lop shows the phosphate influence on oxygen flow, while Lpo shows the substrate dependency of ATP production. Experiments show that Onsagers's reciprocal relations hold for oxidative phosphorylation, and we have Lop = Lpo. By dividing Eq. (11.150) by Eq. (11.151), and further dividing the numerator and denominator by Xo(LoLp)1/2, we obtain
"q-ix=-
x+q q+ l/x
(11.152)
where j = (Jp/JoZ), x - (XpZ/Xo). Here, Z is called the phenomenological stoichiometry defined by (11.153) Equation (11.152) shows the exergetic efficiency r/in terms of the force ratio x and the degree of coupling q. The ratio Jp/Jo is the conventional phosphate to oxygen consumption ratio: P/O. Figure 11.5 shows the change of
11.7
1
,
583
Exergy use in bioenergetics
,
,
,
,
,
'
0.9 0.8 0.7
qopt= 0.6195
~~...
~- 0.6
~_q~C
0
" 0.5
/
.m 0
0.4
0.3 0.2 0.1
/o,:/ -0.9
Figure 11.5.
-0.8
. . . . . . -0.7
-0.6 -0.5 -0.4 force ratio, x
-0.3
-0.2
-0.1
0
The change of efficiencies 7, given in Eq. (11.153), in terms of flow ratio x and for the degrees of coupling qf, qp, q~C, and q~C.
efficiencies r/in terms of flow ratio x between-I and 0, and for the particular degrees of coupling qf, q p , q~C, and q~C.As Eq. (11.148) shows, the optimum efficiency values are dependent only on the degrees of coupling, and increase with increasing values of q.
11.7.3
Exergy Losses
For the oxidative phosphorylation described by Eqs. (l 1.151) and (l 1.152), the exergy loss can be obtained from Eq. (11.146) in terms of the force ratio x and the degree of coupling q, and is given by - (x 2 + 2qx + 1)LoXo2
(11 154)
Minimum exergy loss or minimum entropy production at stationary state provides a general stability criterion. There are two important steady states identified in the cell: static head (sh) and level flow (lf). At the static head, where ATP production is zero since Jp =--0, the coupling between the respiratory chain and oxidative phosphorylation maintains a phosphate potential Xp, which can be obtained from Eq. (11.151) as (Xp)sh = --qXo/Z, and the static head force ratio Xsh becomes X~h= --q. The oxygen flow Jo at the static head is obtained from Eqs. (11.151) and (11.152) (Jo)sh
=
LoXo(1-q 2)
(11.155)
where Lo may be interpreted as the phenomenological conductance coefficient of the respiratory chain. If an uncoupling agent, such as dinitrophenol, is used, the ATP production vanishes and hence Xp = 0; then, Eq. (11.151) becomes (Jo)unc -- LoXo
(11.156)
(do)sh -- (do)unc ( 1 - q2)
(11.157)
Combining Eqs. (11.155) and (11.156), we obtain
Using the experimentally attainable static head condition (state 4 in mitochondria) and the uncoupled oxygen flow (Jo)unc, we can determine the degree of coupling q
1/2 (Jo)sh q--
1-- (jo)un
(11.158)
c
At constant Xo, Eq. (11.154) yields the minimum value of exergy loss at x (q~)~h - (q~)mi. - (1- q2 )LoX2o
-q
(11.159)
11.
584
Thermodynamicsand biological systems
1
0.8
0.6 470.4
0.2 ..._...
.
-().9
-0.
8
-().7
-().6
-0'.5 -0'.4 -0'.3 -().2
- .O' 1
0
force ratio,x Figure
11.6. The change of ratio 4'1, given in Eq. (11.160), in terms of flow ratio x and for the degrees of coupling qf, qp, q~C, and q~C.
The ratio of dissipations expressed in Eqs. (11.159) and (11.154) depends only on the force ratio and the degree of coupling, and becomes an exergy distribution ratio over the values ofx ¢hl -
1- q2
a'Itsh __
(11.160)
x 2 + 2qx + 1
Figure 11.6 shows that the values of ¢hl reach unity at various values of x, and the economical degrees of coupling q~C and q~C yield lower values of 4'1 than those obtained for qf and qp. The exergy loss at the static head is relatively lower at the degrees of coupling corresponding to economical ATP production and power output. At level flow, the phosphate potential vanishes, and no net work is performed by the mitochondria, and the flow ratio becomes Jlf -
Lpo Lo
- qZ
(11.161)
Combining Eqs. (11.160) and (11.161) yields an expression for estimating the phenomenological stoichiometry Z from measured Jp/Jo = P/O ratios at level flow Z=
P/O
(11.162)
41-(Jo)sh/(Jo)unc The efficiency expressed in Eq. (11.153) is zero at both the static head and level flow, due to vanishing power at these states. Between the static head and level flow, efficiency passes through an optimum, which is given in Eq. (11.149). The force ratio at optimal efficiency is expressed by _
q
Xopt----Cpt--l+41--q
2
(11.163)
The rate of optimal efficiency of oxidative phosphorylation is not characterized by the exergy decrease, and the exergy loss at optimal efficiency is given by
Xi/,opt __ x 2 )2 _ (1l+x-_------------~LoX2
(11.164)
17.7
585
Exergy use in bioenergetics
1.8
/,///~
1.6 1.4 1.2 O4
e-
1 0.8 0.6 0.4 0.2 0
t._
-0.9
-0.8
i
-0.7
I
I
-0.6 -0.5 -0.4 force ratio, x
I
I
I
-0.3
-0.2
-0.1
Figure 11.7. The change of ratio ~h2,given in Eq. (11.165), in terms of flow ratio x and for the degrees of coupling qf, %, q~C,and q~C.
The ratio of dissipations ~opt and ~ , given in Eqs. ( 1 1.164) and (11.154), respectively, shows the effect of optimal operation on the exergy loss in terms of the force ratio x and the degree of coupling q ~)2 --
opt _
qf
(I-- v2) 2 " (1 + x 2 )(x 2 + 2qx + 1)
(11.165)
Figure 11.7 shows that the values of 052 reach peak values higher than unity, and the exergy loss is not minimized at optimal efficiency of oxidative phosphorylation. The exergy loss is the lowest at the degree of coupling corresponding to the economical power output. For the optimal efficiency to occur at steady state, oxidative phosphorylation progresses with a load. Such a load JL is an ATP-utilizing process in the cell, such as the transport of substrates. A load, which will make the steady state the optimal efficiency state, can be identified through the total exergy loss q~c q/c -- J p X p -Jr-J o X o
+
JLX
(ii.166)
Here, Jk is the net rate of ATP consumed and X is the driving force. If we assume that the ATP-utilizing process is driven by the phosphate potential Xp, and Jc is linearly related to Xp, then we have JL - LXp. Here, L is a phenomenological conductance coefficient. The dissipation caused by the load is JkXp = LX2p, and the total exergy loss becomes
(11.167)
Using Eq. (11.163), and from the extremum of Eq. (11.167), Stucki (1980, 1984) found the condition L _ x/l_q2
(11.168)
Lp
which is called the conductance matching of oxidative phosphorylation. After combining Eqs. (11.23) and (11.168) when the conductance matching is satisfied with a percentage/3, we obtain qZc _ ix211 + fiX/1_ q2 )+ 2qx + 1)LoX,~
(11.169)
The exergy lost at the static head with conductance matching, (Xlrc)sh , and at the state of optimal efficiency, (Xlrc)opt, are expressed by Stucki (1984) as follows
11. Thermodynamicsand biological systems
586
(attc)sh = 4 I - - x 2 (X2 + 4 I - - X 2 )LoX2o
(11.170)
x2 i + x 2 LoX2o
1 -
(atrc)op t
u
I t is now possible to estimate the ratios of exergies 4~3 and q by using Eqs. (11.169)-(11.171)
4'3-
(11.171)
~4 in terms of the force ratio x and the degree of coupling
,,/1- x2 (x2 + x/1- x2 )
(XIYc)sh _ _
(11.172)
x2 (l + ~x/l- q2 )+ 2qx + l
"~II"c
(XItc)opt __
I--X 2
/)4 ~ - ~
4
i
i
I
I
I
I
I
i
2.5 g
2
0.5 0
g
I
-1
I
-0.9-0.8-0.7-0.6-0'.5-o'.4-d.a-d.e
-.1 o'
0
force ratio, x Figure 11.8. The change of ratio (b3, given in Eq. (11.172), in terms of flow ratio x and for the degree of couplings qf, qp, q~C, and q~C with load, and/3 - 1 and/3 = 0.9 conductance matching.
i 1.7
Exergy use in bioenergetics
587
Equations (11.172) and (11.173 ) are analogous to Eqs. (11.160) and (11.165) and show the ratios of exergy with and without the load. Figure 11.8 shows the change of ~3 when a load is attached with the values/3 = 1 and/3 = 0.9 at the static head. With the load, the exergy destruction increases considerably. This increase is larger with decreasing values of/3. Figure 11.9 shows the values of &4 with/3 = 1 and/3 = 0.9. At optimal efficiency, exergy destruction is the lowest at qf, corresponding to the maximum production of ATR while the exergy destruction is relatively higher at the economical power output with a minimal effect of/3. The comparison of energy conversions in linear and nonlinear regions requires a combination of thermodynamic and kinetic considerations to express the exergetic efficiencies of nonlinear TinI and linear r h modes
/ AoX
tin'--
--v-' and TII----
-7-
1
(11.174)
"p o
Denoting y as a measure of the relative binding affinity of a substrate H + on the either side of the membrane, the following inequalities are obtained for y >- 1
i
i
i
!
i
I
i
i
i
i
9=1 _
2.5
e- 1.5
qf 0.5
-1
I
I
i
I
I
I
I
I
I
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0
force ratio, x 3 p=0.9 2.5 P
2-
1.5
0.5
0
-1
-0.9
1
1
-0.8
-0.7
I
I
I
-0.6 -0.5 -0.4 force ratio, x
I
I
I
-0.3
-0.2
-0.1
Figure 11.9. The change of ratio ~h4,given in Eq. (11.173), in terms of flow ratio x and for the degrees of coupling qf, q> qp, and q~C with load and/3 = 1 and/3 = 0.9 conductance matching.
588
....
11.
Thermodynamicsand biological systems
~n] > ~7~ for ~'-
T/n1 < T]I
for ~"-
1
(11.175)
q 1
(11.176)
q
where ~"is the reduced stoichiometry Z/n, which is subjected to kinetic limitation (l/q) > ( >_q, and n is the mechanistic stoichiometry. Inequalities (11.175) and (11.176) suggest that at a specified value of (, the efficiencies of nonlinear and linear modes become equal to each other, and there exist values for ~"where the energy converter operates more efficiently. The ratio of efficiencies (called as gain ratio) at linear and nonlinear modes is 'r/1
r/r -
(11.177)
Tlnl
This shows a measure for the efficiency gain in linear mode operation. The efficiency in linear modes depends on only q (Eq. (11.149)), while the efficiency in nonlinear modes depends on input force Xo besides q. In nonlinear regions, the efficiency decreases at high values of input force, and the force ratio at optimum operation Xopt.nl is shifted towards the level flow where x = 0. In oxidative phosphorylation, the input force is the redox potential of the oxidizable substrates and the output force is the phosphate potential. If these two forces are balanced, the system operates close to reversible equilibrium. Experiments show that in mitochondria, q < 1, and the input force is well above 50RT. For a fully coupled system in the nonlinear region of a single force, the phosphate potential Xp would be very small. However, a dissipative structure can only be maintained with a considerable Xp. On the other hand, in the linear mode of operation, optimum force ratio Xop t does not depend on the input force (Eq. (11.163)). Gain ratio T/r can be calculated at a reference force ratio, such as Xopt, which is a natural steady-state force ratio of oxidative phosphorylation. This is seen as a result of the adaptation of oxidative phosphorylation to various metabolic conditions and also as a result of the thermodynamic buffeting of reactions catalyzed by enzymes. The experimentally observed linearity of several energy converters operating far from equilibrium may be due to enzymatic feedback regulations with an evolutionary drive towards higher efficiency. A living organism requires q] < 1. The particular value of the degree of coupling depends on the nature of the output required from the energy converter. For example, fatty acids decrease the degree of coupling and act as uncoupler. Uncoupling is not restricted to thermoregulation; some uncoupling activity is favorable for the performance of the metabolic and even the energy-conserving functions of cellular respiration. Mitochondria can regulate their degree of coupling of oxidative phosphorylation depending on the energy demand of the cell. For example, for fed rats, oxidative phosphorylation operates very close to the conductance matching, i.e., at the state of optimal efficiency with an economical degree of coupling. The load in a living cell fluctuates and compromises the optimal efficiency of oxidative phosphorylation. Some enzymes operate as sensitive thermodynamic buffering to decrease deviations from optimal efficiency. ATP-utilizing reactions act as a load as well as thermodynamic buffers. This regulatory mechanism allows oxidative phosphorylation to operate with an optimal use of the exergy contained in the nutrients. Every reversible ATP-utilizing reaction can, in principle, act as a thermodynamic buffer. For example, adenylate kinase can buffer the phosphate potential Xp to the value permitting optimal efficiency of oxidative phosphorylation in the presence of too high loads. The adenylate kinase reaction is reversible, and acts as a buffer. If this buffer is treated as another load with a conductance La, the overall load conductance L* becomes E =LaO+L
(11.178)
where
0=
c~+ RT ln(1 + e x p - (6 + Xp/RT))
Xp
-1
11.7
Exergy use in bioenergetics
m
589
[AMP] [ATP] + [ADP] + [AMP]
z
6 = kGp - RT ln[Pi ]
Stucki (1984) expressed the dissipation function with buffering from Eq. (11.167) in terms of L*
('qY'c)b-- X2 1%"
+2qx+l
Lp
Lo X2
(11.179)
Dividing Eq. (11.179) with Eq. (11.167) shows the effect of thermodynamic buffering on the exergy loss
!
!
'
'--1--
'
'
,
i
i
!
i
|
1 L=0.9 0.95 0.9 0.85 0.8 -C- 0.75 0.7
qo
0.65 0.6
0 (a2S)eq < 0
S, P
(6H)cq = 0
~ H :> 0
(Z~)e q > 0
(62/-/)eq ~> 0
T, V
(aA)e q = 0
6A > 0
(z~l)e q > 0
(a2A)eq > 0
T, P
(6G)~q = 0
6G > 0
(AG)e q > 0
(32G)eq > 0
For a closed system at uniform temperature and pressure, we have TdiS = T 6 s - d U - p d V >- 0
(12.16)
Equation (12.16) results from the entropy balance di S = d S - deS >__O
with deS = 6q/T and 6q = dU + PdV, and leads to a stability criterion for thermodynamic equilibrium; the inequality shows that equilibrium state is stable. For a small and arbitrary increment 6, Eq. (12.16) provides the stability criterion 6 U + ['6 V - T6S >- O. At constant S and V, the stability condition becomes 6 U >- 0, indicating that the internal energy assumes its minimum value for stable equilibrium state: (6U)e q = 0. Stability after a perturbation with finite amplitude is (A g)e q > 0, or after an infinitesimal perturbation is (6 2 U)e q > 0. Table 12.1 shows the equilibrium and stability criteria for various boundary conditions; the last column is not always satisfied for metastable systems, although we often describe both stable and metastable systems as stable states.
12.2.6
Configurational Heat Capacity
Heat absorbed by a system existing in two isomeric forms changes pressure, temperature, and extent of transformation between isomers 6q = - V d P + dH = (hr, ~ - V ) d P + Cp,~dT + hr,pde
(12.17)
where
where Cp,~ is the heat capacity at constant composition and with very slow relaxation of the transformation. The heat capacity at constant pressure is
/
/
The second term on the right is called the configurational heat capacity, and is due to the relaxation of system to the equilibrium configuration (Kondepudi and Prigogine, 1999)
C t,~, = ( OH
(12.18)
For a transformation at equilibrium (A = 0), we have OH
(12.19)
604
12.
Stability analysis
The configurational heat capacity for a transformation at equilibrium and constant pressure is defined by combining Eqs. (12.18) and (12.19) OA
(12.20) , ~.OT}p,A=O
Since the stability condition for a chemical reaction is (OA/Oe)< 0, the heat capacity at constant composition is always less than the heat capacity of a system that remains in equilibrium as it receives heat. Certain fluid molecules, such as supercooled liquid glycerin, can vibrate but not rotate freely, which is called libration. As the temperature increases, more molecules rotate, and the variable e becomes the extent of libration-rotation transformation. If the transformation equilibrium is reached rather slowly, the heat capacity (Cp,~) will be lower than the heat capacity measured in slow heating. 12.2.7
Phase Stability
If a system is in the thermodynamic equilibrium (dS/dt >- 0), instabilities can occur only at phase transition points, and the new phase may be in a more ordered state (vapor ~ liquid), which is a self-sustaining structure. Phase splitting due to thermodynamic instability and hence symmetry breaking in equilibrium in a feed mixture affect problems associated with the simulation and design of distillation and extraction. Knowing the exact number of phases contributes considerably toward successfully predicting phase equilibria. For ternary mixtures, for example, feed points located within the binodal curve split only into two liquid phases; therefore, knowing the position of the feed leads to the exact number of phases. For multicomponent mixtures, the distance between the tangent plane and the Gibbs energy of mixing surface is used for phase stability analysis. When the distance D for a composition x is negative, a phase with feed mole fractions z is unstable, and the molar Gibbs energy of the mixing surface Gm = AGmix/RT falls below a plane tangent to the surface at z. The distance D is obtained from
D(x)=Gm(x)-Gm(z)-~( OGm] ( x i - z i =1 ~',
i)
(12.21)
OXi Jz
The subscript z indicates the evaluation of partial differentials at x = z, and n is the number of components. The Gibbs energy of mixing Gin, and the reduced excess Gibbs energy g~ are n
GE
Gm (x) = ~_, x i In x i + gE (x) and gE _
(12.22)
RT
i=1
The tangent plane distance analysis minimizes the D subject to the mole fractions by solving the following system of nonlinear equations, which provide the stationary points (Gecegormez and Demirel, 2005)
with
Lk ax,
k - .Jj
=
ox.
0 i = 1,...,n-1
(12.23)
z
i=l Xi -- 1.
12.3
STABILITY AND ENTROPY PRODUCTION
The Gibbs stability theory condition may be restrictive for nonequilibrium systems. For example, the differential form of Fourier's law together with the boundary conditions describe the evolution of heat conduction, and the stability theory at equilibrium refers to the asymptotic state reached after a sufficiently long time; however, there exists no thermodynamic potential with a minimum at steady state. Therefore, a stability theory based on the entropy production is more general. The change of total entropy is: dS = deS + diS. The term deS is the entropy exchange through the boundary, which can be positive, zero, or negative, while the term diS is the rate of entropy production, which is always positive for irreversible processes and zero for reversible ones. The rate of entropy production is diS/dt = EJeXk. A near equilibrium system is stable to fluctuations if the change of entropy production is negative z~iS < 0. For isolated systems,
I2.3
605
Stability and entropy production
dS/dt >- 0 shows the tendency toward disorder as deS/dt = 0, and dS = diS >- O. For nonisolated systems, diS/dt > 0
shows irreversible processes, such as chemical reactions, heat conduction, diffusion, or viscous dissipation. For states near global equilibrium, diS is a bilinear form of flows and forces that are related in linear form. The second law for isolated systems shows that the excess entropy, AS = S - Seq -< 0, increases monotonically in time, d(AS)/dt >_ O. Therefore, it plays the role of a Lyapunov function, and defines a global stability. So, diS/dt is a Lyapunov function that guarantees the global stability of stationary states that are close to global equilibrium. For nonequilibrium systems far from global equilibrium, the second law does not impose the sign of entropy variation due to the terms deS and diS, as illustrated in Figure 12.2. Therefore, there is no universal Lyapunov function. For a multicomponent fluid system with n components, entropy production in terms of conjugate forces X~, flows J,., and I number of chemical reactions is
+ =
J;x, = a . . v
-
a;. v v
i
- F, +
(W)
i=1 1
(12.24)
l
+ - - E A/Jr/>- 0
T j=l
'
where F i is the force per unit mass of component i. Here, the rate of entropy production is the sum of contributions from heat, mass, momentum transfer, and chemical reactions, excluding electrical, magnetic, and other effects. Equation (12.24) identifies a set of the conjugate flows and forces to be used in the phenomenological equations with the coefficients satisfying Onsager's reciprocal rules in the linear regime of the thermodynamic branch. For a chemical reaction, entropy production is diS _ A de dt
(12.25)
T dt
An approximation of A for a given small fluctuation in the extent of reaction e, c~ = ( e - geq), is expressed by eeq), and used in Eq. (12.25) to obtain the following stability condition
A = (OA/Oe)(e
-
~ A d e = [ a~ OA
AiS =
aO
7
oldol -
eq
< 0
eq
(12.26)
where de = dc~. For I number of chemical reactions, Eq. (12.26) becomes
AiS--
~
1 (04.)a O, ~
dt
< 0
(12.54)
A function L satisfying Eq. (12.54) is called a Lyapunovfunction. The second variation of entropy L = - 62S may be used as a Lyapunovfunctional if the stationary state satisfies E 3XiaJi > 0; hence, a nonequilibrium stationary state is stable if
d 62S dt 2
- ~ aXiaJ i
(12.5 5)
A functional is a set of functions that are mapped to a real or complex value. Equation (12.55) indicates that the quantity d(62S)/dt has the same form for the perturbations from the equilibrium state as well as the nonequilibrium state. In the vicinity of equilibrium, the quantity ~i,6Xi6Ji is called the excess entropy production, and it shows the increase in entropy production. The quantities 6Ji and 6Xi denote the deviations of Ji and X~.from the values at the nonequilibrium steady state. The increase in entropy production for a perturbation from a nonequilibrium state is
6P=6xP+6~P
(12.56)
Since 3S < 0 under both the equilibrium and nonequilibrium conditions, the stability of a stationary state is accomplished if
d 62S dt 2
-- E ~Yi~Ji > 0
(12.57)
Example 12.4 Stability of an autocatalytic reaction For a simple example, we may consider the following autocatalytic reaction, which appears in the reaction scheme of the Brusselator 2X+Y, We have the forward Jrf of reaction Jr are
=
kf "3X ku
kfc2Cyand backward Jrb = kbC3Xreaction rates, respectively. The affinity A and the flow A = RT In Jrf , Jrb
Jr = Jrf - Jrb
12.5 Stability in nonequilibrium systems
611
The excess entropy production is written in terms of 6X = 6A/T and 6Jr
1 da2S _ _R(2kfcxsCys _ 3kbC~) (6Cx)_____~ 2 2 dt Cxs The excess entropy production can become negative if kf >> kb, and hence the stationary state may become unstable. Coupling between chemical kinetics and transport may also lead to dissipative structures, which are caused by auto- and cross-catalytic processes with positive and negative feedback, influencing their own rates of reaction. For example, the Belousov-Zhabotinsky reaction exhibits a wide variety of characteristic nonlinear phenomena. In the nonlinear region, the possible instabilities in chemical and biological systems are (i) multisteady states, (ii) homogeneous chemical oscillations, and (iii) complex oscillatory phenomena. The thermodynamic buffer enzymes may represent a bioenergetics regulatory principle for the maintenance of a far from equilibrium state.
12.5.1
Stationary States
The study of the behavior of the stationary state ofjth order after a disturbance may be helpful for stability considerations. For this purpose, we designate the index s for all forces and flows in the stationary state. Assuming that the force Xk has been disturbed by 6Xk, and keeping the remaining forces constant, we have
XI,. = X,.t,. +tXk, n >-k >- j + l
(12.58)
dk = J,,k + LkktXk
(12.59)
The new value of the flow Jk becomes
According to the Prigogine theorem, d,..k = 0, and we have
Yk = LkI,-tXI,-
(12.60)
After the disturbance, the rate of entropy production is
• = ~ _ J i Y i - ~ L i / X , . i Y , . / + [ ~ _ ~ ( L i k + L l , q)YitYk]-t-Lkk(tYk)2>O i=1
i,j=l
(12.61)
i=1
The first summation term on the right is the minimum entropy production corresponding to the stationary state. The second sum on the right is zero, according to the Onsager reciprocal relations and the Prigogine principle. Therefore, we have Lkk(~Xk) 2 > 0
(12.62)
or
JktXk > 0 This inequality is identical to the Le Chatelier principle for nonequilibrium stationary states, since the disturbance
6Xk has the same sign as the flow dk, indicating a decrease in the disturbance. For example, an increase in the gradient of the chemical potential will cause the mass flow and diminish the gradient. Hence, the stationary state will return to its original status. The change of entropy production can be measured by a sensitive calorimeter during the growth of bacterial colonies. The heat dissipated by the system is directly related to the entropy production. The high-precision calorimetric measurements indicate that the entropy increases sharply and reaches a maximum, and finally decreases to a minimum. This is a stationary state, in which the colony no longer grows. The colony is then limited to maintaining itself. If no growth occurs, the system can be described by means of linear phenomenological laws, and tends toward the minimum entropy production. Similarly, the resting states of biological cells can often be interpreted as stationary states described by linear laws; the active phases or growth are mainly examples of nonlinear phenomena.
612
12.5.2
12. Stability analysis
Stability of Stationary States
In the linear nonequilibrium thermodynamics theory, the stability of stationary states is associated with Prigogine's principle of minimum entropy production. Prigogine's principle is restricted to stationary states close to global thermodynamic equilibrium where the entropy production serves as a Lyapunov function. The principle is not applicable to the stability of continuous reaction systems involving stable and unstable steady states far from global equilibrium. For chemical reactions, the change of entropy production with time P is
dt
T dt
--f- Jrk
(12.63)
The affinity and the reaction velocity are expressed by A = A(P,T,e) and Jrk = £kLki(Ai/T). At constant pressure and temperature, we have
dP dt
2T~i/l~)pTJriJrj
(12.64)
Equations (12.27) and (12.64) show the stability of the nonequilibrium stationary states in light of the fluctuations
6ei. The linear regime requires P > 0 and dP/dt < 0, which are Lyapunov conditions, as the matrix (OAi/Oej)is negative definite at near equilibrium. Dissipative structures can sustain long-range correlations. The temperature and chemical potential are well defined with the assumption of local equilibrium, and the stationary probability distribution is obtained in the eikonal approximation; so the fluctuation-dissipation relation for a chemical system with one variable is
Jr(x) =
2D(x)tanh -2kBT
( 1
= 2D(x)tanh -2kBT VOx
/
(12.65)
where Jr(x) is the net reaction rate representing the flux, D(x) is a probability diffusion coefficient and shows the strength of chemical fluctuations, A(x) is a species-specific affinity representing the thermodynamic force, and q~(x) is the stochastic potential. Equation (12.65) shows a nonlinear relationship between the flux and force, and due to the hyperbolic tangent, the reaction rate approaches toward a constant value for large values of the specific affinity.
12.5.3
Evolution Criterion
An evolution criterion can be obtained from the rate of change of volumetric entropy production P = fEJXdV> 0 as follows:
dt - I v
J k - - ~ j d V + ~.v
/
X k--~ dV= dt ' dt
(12.66)
This equation is independent of the type of phenomenological relations between fluxes and forces. In contrast, linear phenomenological equations and the Onsager reciprocal relations yield
E Jk dXk = E LjkXj dXk = 2 Xj d (LjkX k ) = E Xj dJ k
(12.67)
So, in the linear region, we have
dxP - djP _ 1 dP dt dt 2dt
(12.68)
For time-independent boundary conditions, we have the general conditions for the stability of a state
dP - 2 dxP 0 and dxP < 0 are called the Lyapunov conditions. The fundamental quantity, which determines the stability, is the sign of excess volumetric entropy production (12.70)
P(6S)=IE6JaXdV>-O
where 6S is the perturbation in entropy. The stability condition for chemical reactions is (Glansdorff and Prigogine, 1971) P(6S)
=
Ei(~'Jric~Ai -->0 T
(12.71)
Consider the following rate of entropy production:
P= ~
T dt
(12.72)
. T dt
and
d P= ZJkdXk7d k
-
(12.73)
7d
with the following approximately defined differentials:
@_d -- - - ( hA(~CA + hB6cB + Cp(~T) dt dt d--~=R~-
ci~,
RE,,
# )
= _(d(6T)
--,
...........
T,
Substituting these into Eq. (12.64), we have
d×P = -R(d(6CA) d(6ca ) f d(6CB) d(6cB) dt cA,, dt CB,,~
Cp d(6T) d(T)) dt L
(12.74)
Equation (12.74) can be used in Eq. (12.69) for stability analysis at near global equilibrium. The Lyapunov function resembles the thermodynamic entropy production function and the asymptotic stability principle. If the eigenvalues of the coefficient matrix of the quadratic form of the entropy production are very large, then the convergence to equilibrium state will be rapid.
Example 12.5 Macroscopic behavior in systems far from equilibrium Consider the nonequilibrium chemical
system R---~ X --~ p X--"-
X----
Concentrations of both R and P are maintained at constant values, while the concentration of the intermediate component X may vary with time. Assume that Xs denotes a steady state (stable or not). The behavior of such a system may be controlled by the position of the steady state: (i)
If the steady state is close to equilibrium, then the system is stable; linear nonequiiibrium thermodynamics can be used. Considering the entropy production below p=l 7EJr/~ i
"
614
12. Stability analysis
The evolution criterion becomes 1 1 11 d x P = --_ '~, Jri ( dAi ) = ~_ d, P = - -- d P < 0 i 1'-2T
This equation is the result of minimum entropy production at stable steady state and shows that if the system is disturbed by a small perturbation, it will return to the steady state. (ii) However, if the steady state is far away from equilibrium, the system may be stable or unstable. A perturbation may lead to multiple states, since the system may enhance the fluctuations instead of damping them, and the system may choose one of the states according to the hydrodynamic and kinetic conditions the system is in. Even if the system is stable, the behavior of the system may vary; the path to the steady state may be spiral or the system may rotate around the steady state. A larger variety of possibilities may exist for the unstable steady-state case. For far from equilibrium conditions, the overall stability is no longer a consequence of the stability with respect to the diffusion, as is the case for conditions in the vicinity of equilibrium.
12.6
LINEAR STABILITY ANALYSIS
Hamiltonian dynamics show that classical mechanics is invariant to ( - t) and (t). In a macroscopic description of dissipative systems, we use collective variables of temperature, pressure, concentration, and convection velocity to define an instantaneous state. The evolution equations of the collective variables are not invariant under time reversal Chemical reaction: dcA -- --kCACB, A + B
k >D
(12.75)
dt
Heat conduction:
OT
= c~V2T, ol > 0
(12.76)
Ot
Diffusion: v--7-~= D V 2 c , D > 0
(12.77)
Ot
Here, T and c are called the even variables whose signs do not change upon time reversal, while convection velocity, and momentum of a particle are called the odd variables whose signs change with time reversal. The general form of a dissipative system with macroscopic variables X1, ..., Am, space r, and time t, may be defined by OX i
- fi(X1,...,Xn,r,t,
tx)
Ot
(12.78)
The evolution of the variables X/is influenced by the variation of some control parameters represented by/x that can be modified by the environment. The control parameters may be the diffusion coefficient, thermal conductivity, chemical rate constants, or initial and final concentrations of reactants and products. Stability analysis has to consider a variety of variables characterizing problems of transport and rate processes. The variables often are functions of time and space. The functionf has the following properties:
f/([Yj, eq ],/Zeq ) = 0, at equilibrium
(12.79)
fi ([Xj, s ],/x s) = 0, at nonequilibrium steady state
(12.80)
The components [X~s] represent the stationary and spatially uniform solution. These relations are associated with certain restrictions, such as T > 0 and c > 0; detailed balance is achieved, and hence physical systems are highly unique (Nicolis and Prigogine, 1989).
615
12.6 Linear stability analysis
The state variables X1, ..., if,., which are continuously subjected to either internal fluctuations or external perturbations, are represented by a column vector X 0X
Ot
- F(X,/z)
(12.81)
Here, F is an operator, and acts on the space in which X is defined. Stability analysis determines if the stationary solutions will remain stable to small perturbations of x(t) X(t) = X~ + x(t)
(12.82)
The stationary state Xs is a particular solution of Eq. (12.81 ): F(X,~) = 0
(12.83)
Using Eq. (12.82) in Eq. (12.81), and by retaining the linear terms only in the Taylor expansion ofF, we obtain 0X
at
- F([X, + x],/x)- F(X,,/x) = Jx
(12.84)
where J is the Jacobian matrix with the elements (Ofi/O~.)~ calculated at stationary state. Equation (12.84) presents an eigenvalue problem: Jq~ = hq~. The solution in terms of each eigenvector (q~) and its eigenvalue (h) becomes x - y_~c k exp(A~t) q~k
(12.85)
k
The coefficients cx. are determined by the initial conditions. Stability depends on whether the perturbation x grows or decays with time. A perturbation may be due to the interference of the environment with the intrinsic dynamics of the system or intrinsic internal deviations called fluctuations that the system generates spontaneously. The property of stability refers to several responses of systems to various types of perturbations (Nicolis and Prigogine, 1989): (i) Perturbations remain smaller than a critical value for all times, and the state X~ is stable in the sense of Lyapunov. Then we can define the notion of orbital stability as the distance between the reference and perturbed trajectories as the whole sequence of possible states. (ii) Perturbations decay in time, and % is asymptotically stable, which implies irreversibility. (iii) State X(t) does not remain in the vicinity of %, and x(t) cannot remain less than a critical value for all times. Then the reference state X,. is unstable: the system experiences the rapid growth of perturbations leading to orbital instability. (iv) State X(t) remains in some vicinity of X, for x(t) - critical. This represents a locally stable but globally unstable state %. How the perturbations affect the state of the system depends on the eigenvalues hk. If any eigenvalue has a positive real part, then the solution x grows exponentially, and the corresponding eigenvectors are known as unstable modes. If, on the other hand, all the eigenvalues have negative real parts then a perturbation around the stationary state exponentially decays and the system returns back to its stable state. The linear stability analysis is valid for small perturbations (1x I/1%1 - k4 --~--k3(klA]
2
This kind of instability is space independent and leads to order in time.
(12.94)
618
12. Stabilityanalysis
Example 12.8 Linear stability analysis with two variables Consider the following reaction scheme previously used: A
kl >X,
B+X
k2 >Y+E,
k2 = 1.0
k3 >3X,
k3 =1.1
2X+Y X
k1 =1.0
k4 >F,
k4 =1.1
The initial values of A and B are maintained at CA = 1 and CB = 1.6, while the products E and F are removed. Calculate: (a) The particular (stationary) solution (b) The Jacobian matrix at steady state (c) The homogeneous solution Solution: (a) the particular solution can be obtained by starting with the rate of change of the intermediate components, Cx and Cy dCx - k,c,~ - k ~ c ~ c ~ + L c ~ c ~ dt dCy k2C~Cx - k3C2Cy dt
- qCx
(a)
-
At steady-state condition and with CA = 1 and CB = 1.6, Eq. (a) becomes 0 = (1)(1)- (1)(1.6)Cx + (1.1)C~Cy - 1.1C× 0 = (1)(1.6)Cx - (1.1) C2Cy
(b)
These simultaneous equations yield the particular solutions or steady-state solutions
X,
~, 1.600 )
(b) The Jacobian matrix is
J
df~(Cx,Cy) dl(cx,Cy) dCy dCx d~ (Cx,Cy ) df2(Cx,Cy) dCy dCx
= [-2.7 + 2.2CxCy k 1.6 - 2.2CXCv
1"1C2 ] -1.1CZx
By substituting the steady state into the above matrix, the Jacobian matrix at steady state becomes
J=
0.5 -1.6
0.9091 1 -0.9091
(c) With MATLAB we have the following eigenvalues matrix and eigenvectors matrix:
A=
[-0.2045 + 0.9789i 0
0 ] - 0 . 2 0 4 5 - 0.9789i
[-- 0°3516 - 0o4885i w0.3516 + 0o4885i1 ('P-0,7985 0°7985
(c)
12.7
619
Oscillatingsystems
Using the eigenvalues and eigenvectors the homogeneous solutions becomes
-
-
cos (0.9789t) [
sin (0.9789t)
x2(t )
0 e -0"20450t
+12
sin(0.9789t)l
(d)
cos(0.9789t) -0.48850
+ e -020450t
Cx (0) - 1 and Cr (0) - 1.
The constants 11 and 12 are obtained from the initial conditions:
I l - -0.7514, [2 - 0.3547 The real parts of the eigenvalues are negative, and the perturbations will decay in time as Figure 12.3 illustrates. When the value of B is 2.4 then the oscillations are sustainable. Figure 12.3b and 12.3d show the state-space plot of concentrations Cx and Cy for the different values of B. Regardless of whether the eigenvalues are real or complex, the steady state is stable to small perturbations if the two conditions tr[J] < 0 and det[J] > 0 are satisfied simultaneously. Here, tr is the trace and det is the determinant of the square matrix J.
12.7.2
The Brusselator Model with Diffusion
The Brusselator reaction-diffusion system is capable of sustaining spatial and temporal structure. When the concentrations of A and B are controlled, the one-dimensional approach to complex reaction-diffusion systems under isothermal conditions yields the kinetic equations for X and Y. Consider the following set of equations representing space-independent evolution: 0Z
1.6
- F(Z, ~)
(12.95)
N,..j--~__
1 >. 1.4 r.j
......
1~
!
G'I ---...
..... .
0.8 6 ....
l0
~0
(a)
3o . . . .
0.7
;0
t
.
.
.
2
2
. .
0 (c)
1
0.9 Cx
.
0
0.8
(b)
.
.
.
.
.
.
10
.
.
.
.
.
.
20 t
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
30
0.5
40 (d)
0.75
1
1.25 Cx
1.5
1.75
2
Figure 12.3. (a) Oscillations of concentrations of Cx and Cy with kl - 1.0, k2 = 1.0, k 3 - 1.1, k4 = 1.1, A = 1.0, B = 1.6. Bold line indicates the concentration of Cy. (b) State-space plot, (c) oscillations of concentrations of Cx and Cy with kl - 1.0, k2 = 1.0, k3 = 1.1, k4 - 1.1, A - 1.0, B = 2.4. (d) State-space plot in Example 12.8.
620
12. Stability analysis
where/x represents some controlling parameters. The perturbations z around a steady state Zs are defined by Z = Zs + z with Izl/Iz, I True, GridLines->Automatic, PlotStyle->{Thickness [0,008]}, FrameStyle->Thickness [0,004], FrameLabel->{"S", "X"}, RotateLabel->True, DefaultFont- > {' 'Times-Roman", t 2 }]:
12.7.4
Belousov-Zhabotinsky Reaction Scheme
Sometimes, even spatially homogeneous chemical systems can cause bistability and show complex behavior in time. For example, autocatalysis may occur due to the particular molecular structure and reactivity of certain constituents, and reactions may evolve to new states by amplifying or repressing the effect of a slight concentration perturbation.
3.5
X 2.5
1.5
,
8.3
,
,
i
.
8.35
.
.
.
~
.
8.4
.
.
.
,
.
.
.
.
8.45
,
.
8.5
.
.
.
,
.
8.55
.
.
.
.
.
.
.
.
.
8.6
S Figure 12.4. Multiple steady states in Example 12.10.
8.65
626
12. Stability analysis 0.0002 0.0002 0.00015
0.00015
×
0.0001
0.0001
0.00005
0.00005
500 (a)
1000 t
1500
2000
500
1000
(b)
1500
2000
t
F i g u r e 12.5. C h a n g e of composition of X with time, kl = 1.28, k2 = 8.0, k3 = 8 × 105, k4 = 2 × 103, k 5 -- 1.0, H = 0.06, B = 0.02.
(a) f - 0.51319711709999 and (b) f= 0.5131971170999999.
The Belousov-Zhabotinsky reaction system is one example leading to such chemical oscillations. One of the interesting phenomena is the effect of the very narrow range of controlling parameter ~ on the stability of the BelousovZhabotinsky reaction system. The following reactions represent the Belousov-Zhabotinsky reaction scheme: H+Y
k, >X+P
H+X X +Y
k~ >2X+2Z 'k3 >2P
2X
(12.118)
k4 >H+P
The evolution equations for the Belousov-Zhabotinsky system are dX
- klHY+k2HX-k3XY-2k4
X2
dt
dY--klHY-dt k3XY + ( f ) kSBZ dZ dt
(12.119)
(12.120)
- 2kzHX- ksBZ
(12.121)
Figure 12.5 shows the effect of the kinetic and controlling parameterfon the evolution of concentration of X estimated from Eqs. (12.119) to (12.121).
Example 12.11 Reaction-diffusion model The linear stability analysis (Zhu and Li, 2002) may be used to investigate the evolution of a reaction-diffusion model of solid-phase combustion (Feng et al., 1996). The diffusion coefficients of the oxygen and magnesium (g) are the two controlling parameters besides kinetics Mg(s) + 02 bMg(g) + 0 2
kl >MgO(s) + aMg(g)
k2 >MgO(s) + cMg(g)
Mg(g) + 02env
k3 >MgO(s)
The flows of Mg and 02 are"
Mg(g) k4
> MgO(g)env
and 0
2 ( k5
02env
12.7
by assuming that X denotes
02, and Y
627
Oscillatingsystems
denotes Mg(g), the mass-action law yields
dX
k2XY b + ks(X 0 - X)
-- -klX-
dt
dY
- a k l X + ( c - b) k2XY b - k3Y + k 4 (Yo - Y)
dt
For the combustion, the linearized equations become dx - - w x dr
x y b -~ 1 - x to
dy _ a w x + ( c _ b ) x y b - uy-~ Yo - Y t1 dr
where X --
X
Y
k2xbt,
w--
k,
Xo ' y = ~-0-0, r =
k2 Xb
k3
k2xb, u-k2xb
, t o --
ks
k2Xob , t,-
With the following numerical values: a = 1, b = 2, c - 3, w = 1/650, v = 1/20, and Y0 = 0.006 (Feng et al., 1996), the linearized equations reduce to ----
x-- xy 2 + ~ to
dr
+ -- (6+o)x + x y 2 --
y+
0006,
dr
t~
From the eigenvalue problem for these equations, the two controlling parameters to and t~ are obtained. As the parameters to and t 1 contain kinetics and transport coefficients, they represent a combined effect, and make the study more interesting and complex.
Example 12.12 Adiabatic stirred flow reactor Consider the following reaction: A.
k, "B k, _
The reaction occurs in an adiabatic stirred flow reactor with feed flow rate F, transient compositions CA, and CB and reaction rate Jr, and total mass of reacting mixtures M. For small perturbations around the stationary state(s), the following expansions are used: CA (t) -" CA,s -Jr-~CA (t), CB (t) = CB,s Jr- ~CB (t), T ( t ) = T, + 6 T ( t ) Jr (CA, CB, T) -- Jrs (CA,s,CB,s,Ts) + ~Jr (CA,CB, T ) to find the differential model equations of heat and mass balances d(~CA) . . . . dt
d(,~CB) dt
F M
6c A - 6 J r
F - - - ~ a C B + a Jr M
628
12.
Stability analysis
d ( 6 T ) - - F - - - 6 T + Q ~Jr dt M Cp In stable systems, such disturbances vanish in time and the stationary values are restored. For very small perturbations, the reaction rate disturbance may be expanded with negligible second order and higher terms as follows:
~Jr : ( OJr ] ~CA"k-(OJr ) OJr) OCA)s ~~CB)s aCBq-( OT .)s ~r This expansion may lead to the linearization of differential equations above.
PROBLEMS 12.1
Using the truncated virial equation of state with the second virial coefficient B(T)
P V - 1 + ~(TB,________~ RT V Obtain the thermodynamic stability condition based on the constraint on B(T). 12.2
If we have a fluid in a closed system at constant entropy and pressure, prove that the stability condition of the fluid is Cp > O.
12.3
Using the truncated virial equation of state
PV
B C -1-+----+-~ RT V V2
and the constant-volume heat capacity C v = a + b T. For what values of B and C this fluid undergo a vapor-liquid phase transition? 12.4
Solve the following initial value problem as an eigenvalue problem, and prepare a state-space plot.
dXl__ - 9 x 1 + 4x 2 dt
dx2_
dt X 1 (0)
12.5
_ _ 2x 1 + 2x 2
--
1; x 2 (0) = - 1
Solve the following initial value problem as an eigenvalue problem, and prepare a state-space plot of xl versus x2 and x2 versus x3.
dXl -
8x 1 - 5x 2 + 10x 3
dt
dx2 _ 2 x 1 + x 2 + 2 x 3 dt
dx3_
dt
_ - 4 x 1 + 4x 2 - 6x 3
The initial conditions are x 1( 0 ) = x 2 ( 0 ) = 2; x 3 ( 0 ) - - 3
References
12.6
629
We have a first-order homogeneous reaction, taking place in an ideal stirred tank reactor. The volume of the reactor is 20 × 10-3 m 3. The reaction takes place in the liquid phase. The concentration of the reactant in the feed flow is 3.1 kmol/m 3 and the volumetric flow rate of the feed is 58 × 10-6 m3/s. The density and specific heat of the reaction mixture are constant at 1000 kg/m 3 and 4.184 kJ/(kg K). The reactor operates at adiabatic conditions. If the feed flow is at 298 K, investigate the possibility of multiple solutions for conversion at various temperatures in the product stream. The heat of reaction and the rate of reaction are 2U-/,. - -2.1 × 108 J/kmol Jr = k C -
4.5 × 106 C (kmol/m 2) exp[(-62800/(RT)] kmol/(m 3 s)
REFERENCES G. Captained, lnt. J. Heat Mass Transfer, 46 (2003) 3927. Y. Demirel, Int. J. Thermodynamics, 8 (2005) 1. C.-G. Feng, Q.-X. Zeng, L.-Q. Wang and X. Fang, J. Chem. Soc. Faraday Trans., 92 (1996) 2971. H. Gecegormez and Y. Demirel, Fluid Phase Equilibria, 237 (2005) 48. E Glansdorff and I. Prigogine, Thermodvnamic Theoo, oj'Structure, Stability and Fluctuations, Wiley, New York ( 1971 ). W Horsthemke and EK. Moore, J. Phys. Chem. A., 108 (2004) 2225. G. Izfis, R. Deza, O. Ramirez, H.S. Wio, D.H. Zanette and C. Borzi, Phys. Rev. E, 52 (1995) 129. D. Kondepudi and I. Prigogine, Modern Thermodynamics, Firm Heat Engines to Dissipative Structures, Wiley, New York (1999). W.B. Li, K.J. Zhang, J.V. Sengers and R.W. Gammon, J. Chem. Phys., 112 (2000) 9139. G. Nicolis and I. Prigogine, Self-Organization in Nonequilibrium Systems, Wiley, New York (1977). G. Nicolis and I. Prigogine, Exploring Complexity, Freeman & Company, New York (1989). J.M. Ortiz de Zarate, R.P. Cordon and J.V. Sengers, Phys. A. 291 (2001 ) 113. M. Tsuchiya and J. Ross, Proc. Natl. Acad. ScL, 100 (2003) 9691. A. Turing, Phil. Trans. R. Soc. B, 237 (1952) 37. M.O. Vlad, A. Arkin and J. Ross, Proc. Natl. Acad. Sci., 101 (2004) 7223. L. Yang and I.R. Epstein, Phys. Re~: Lett., 90 (2003) 178303-1. R. Zhu and Q.S. Li, Theot: Chem. Acc., 107 (2002) 357.
REFERENCES FOR FURTHER READING I.D. Epstein, An Introduction to Nonlinear Chemical Dynamics: Oscillations, Waves, Patterns, and Chaos, Oxford University Press, Oxford, 1998. I.R. Epstein and V.K. Vanag, Chaos, 15 (2005) 047510-1. A. Goldbeater, Biochemical Oscillations and Cellular Rhythms: The Molecular Bases of Periodic and Chaotic Behavior, Cambridge University Press, Cambridge 1996. I.A. Halatchew and J.P. Denier, Int. J. Heat Mass Transfer, 46 (2003) 3881. T. Kolokolnikov, T. Erneux and J. Wei, Phys. D, 214 (2006) 63. I.S. Kovacs, Nonlinear Anal., 59 (2004) 567. M. Mincheva and D. Siegel, Nonlinear Anal., 56 (2004) 1105. M.G. Newbert, H.Caswell and J.D. Murray, Math. Biosci.. 175 (2002) 1. J.M. Ortiz de Zarate and J.V. Sengers, J Stat. Phys., 115 (2004) 1341. S. Petrovski, B.-L. Li and H. Malchow, Bull. Math. Biol., 65 (2003) 425. I. Schreiber, E Hasal and M. Marek, Chaos, 9 (1999) 43. R. Sureshkumar, J. Non Newtonian Fhtid Mech., 97 (2001) 125. M. Vendruscolo, Trends Biotechnol., 20 (2002) 1. R. Wittenberg and P. Holmes, Phvs. D, 100 ( 1. >/2
.
'
I l lt_ 1 It~ 17 Itlltlh [ II IX IX II
b
L Y ' \,)' Yx' YW"~J' \ .,
0
....
,
....
10
(c)
i
20
t
....
i
30
. . . . .
.
0.5
40
(d)
1
1.5
.
.
.
.
.
2
.
.
.
.
.
2.5
.
.
.
.
.
.
.
3
.
3.5
X
Figure 13.5. The oscillations and the limit cycle obtained from the Brusselator scheme in Example 13.3: (a) X and Yversus time, where the bold line displays the concentration of Y, and (b)limit cycle with kl = 1.3, k2 = 1.0, k3 = 1.0, k4 = 1.0, A = 1.1, B = 3 . 0 (c) X and Y versus time, where the bold line displays the concentration of Y, and (d)limit cycle with kl = 1.0, k2 = 1.0, k3 = 1.0, k4-- 1.0, A = 1.0, B = 3.0.
With the parameters kl = 1.0, k2 = 1.0, k3 - 1.0, k4 = 1.0, A = 1.0, B = 3.0, we have the Hopfbifurcation point B H = 2.0 < B = 3.0 Since B > BH, the oscillations are sustainable. Figure 13.5c and d shows the oscillations and the limit cycle. Starting from the neighborhood of steady state as an initial condition, the system asymptotically attains a limit cycle in (X,Y) space. Therefore, for long times, X(t) and Y(t) exhibit periodic undamped oscillations, and the system always approaches the same asymptotic trajectory regardless of the initial conditions. The M A T H E M A T I C A code is: kl=l.3;k2=l.O; k3=l,O; k4=l,O; a=l.1; b=3.0; sol 1-NDSolve [{x' [t] - = k 1* a-k2*b*x[x] + k3* (x[t] ~ 2) *y [t]-k4*x[t] ,x[0] = - 1.0, y' It] = =k2*b*x[t]-k3*(x [t] A2)*y[t] ,y[0] - - 1.0},{x,y},{t,0,40}, MaxSteps-> 1000] Plot [Evaluate [{x It] }/.sol 1], {t, 0,40},Frame->True, GridLines->Automatic, PlotStyle-> {PointSize[0.03],Thickness[0.02]}, FrameStyle->Thickness[0.004], FrameLabel->{"t", "X"}, RotateLabel->True, DefaultFont-> {"Times-Roman", 12}]; Plot [Evaluate [{y[t] }/.sol 1], {t,0,40},Frarne->True, GridLines->Automatic, PlotStyle->{PointSize [0.03] ,Thickness [0.02] }, FrameStyle->Thickness[0.004] ,FrameLabel->{"t", "Y"}, RotateLabel->True, DefaultFont-> {"Times-Roman", 12}]; Plot [Evaluate [{x [t] ,y[t] }/.sol 1], {t, 0,40},Frame- >True, GridLines->Automatic, PlotStyle->{PointSize[0.02],Thickness[0.01 ]}, FrameStyle->Thickness[0.004],FrameLabel->{"t", "X,Y"}, RotateLabel->True, DefaultFont-> {"Times-Roman", 12}]; ParametricPlot [Evaluate [{x[t] ,y[t] }/.sol 1],{t,0,40}, PlotRange->All,Frame->True, GridLines- >Automatic, PlotStyle->{PointSize[0.03],Thickness[0.02] }, FrameStyle->Thickness[0.004],FrameLabel->{"X", "Y"}, RotateLabel->True, DefaultFont-> {"Times-Roman", 12 }];
640 13.6.1
13.
Organized structures
Limit Cycle in the Brusselator Model
In dissipative systems, we frequently encounter behavior called the limit cycle, in which, the system tends toward a certain orbit or an oscillation regardless of the initial conditions. If we assume the following perturbations X = X s + ceexp(At)
(13.13)
Y = Ys +/3exp(At)
(13.14)
and introduce these perturbations into Eq. (13.12), we obtain (13.15)
(13.16)
Here, we have disregarded the terms quadratic in ce and/3. For nonzero, initial perturbations, the determinant of Eqs. (13.15) and (13.16) should be equal to zero, which leads to the relation A 2 nt- (a 2 + 1 - b) A + a 2 = 0
(13.17)
where a = A(kl/k4)(k3/k4) 1/2 and b = B(k2/k4) If the solutions for Eq. (13.17) have a real and positive part, then the perturbations grow and make the system unstable. If the real parts of A1 and A2 are negative, then the perturbations decrease, and the system becomes stable. The solutions o r a l and A2 are given by A1,2 = _ ( 2 1 _ . ) ( a 2 + l _ b ) _ + _ [ ( a 2 + l _ b ) 2 4 a 2
]1/2
(13.18)
The system will be stable if b < (a 2 -+- 1), and it will be unstable if b > (a 2 + 1). When b - a 2 + 1, and A 1 and A2 are purely imaginary, we have an undamped oscillation similar to the Lotka-Volterra model. I f (a2+ 2a + 1 ) > b > (a 2 -- 2a + 1), and A1 and A2 have a nonzero imaginary part, then we have oscillatory behavior. These oscillations damp out in the stable z o n e : (a 2 -+- 2a + 1) > b > (a 2 + 1). When b > (a 2 -+- 2a + 1), limit cycle oscillations occur, which are independent of the initial values of the perturbations in X and Y. Figures 13.5b and d display the limit cycles obtained in Example 13.3 under the two set of parameters.
Example 13.4 Order in time and space with the Brusselator system The Brusselator model with unequal diffusion may produce order in time and space. When the concentrations of A and B are controlled, the one-dimensional approach to complex reaction-diffusion systems with the spatial coordinate r under isothermal conditions yields the kinetic equations for Xand Y(Eqs. (12.98) and (12.99)) OX
02X
Ot - k l A - kzBX + k 3 x Z Y - k4X + Dx Or2
OY Ot
- k 2 B X - k3xZY + Dy
02y Or2
where Dx and Dy are the respective diffusion coefficients. The boundary conditions are
( r3 r=-L
r=+L
=0
13.6
Order in chemical systems
641
For a specified value for A and some variation of B if one of the eigenvalues becomes positive, then Det < 0, and the system becomes unstable, and the propagating wave or the Turing structure occur
B >_ 1-L(k4 +o-2Dx) (l k3(klA/ Yk4)2 The following is a slightly modified version of the MATLAB demo program the BRUSSODE, and displays order in time and space (see Figure 13.6). The parameter N -> 2 is used to specify the number of grid points; the resulting system consists of 2N equations. The problem becomes increasingly stiff and increasingly sparse as Nis increased. The Jacobian for this problem is a sparse constant matrix (banded with bandwidth 5). The property 'Jpattern' is used to provide the solver with a sparse matrix of l's and O's showing the locations ofnonzeros in the Jacobian. By default, the stiff solvers of the ODE Suite generate Jacobians numerically as full matrices. However, when a sparsity pattern is provided, the solver uses it to generate the Jacobian numerically as a sparse matrix. Providing a sparsity pattern can significantly reduce the number of function evaluations required to generate the Jacobian and can accelerate integration. [Source: E. Hairer and G. Wanner,
Solving OrdinaryDifferentialEquations lI, Stiff and Differential-AlgebraicProblems, Springer-Verlag, Berlin, 1991 .] The Brusselator for N = 6 0
The Brusselator for N = 6 0
I
--"g
3d¢. " "
2.1.-" "
.-rI
".
-- ~
,
I
,
i
,--
~l ~ ~
......
- - , _
,
i
i
I i
,
1
~'1-~
I
.0
'"
6
i
. .~'~ "
.t
1
0.2 0
time
0
time
0
The Brussdator for N = 60 I"
/
~
i
0
space
space
The Brusselatcr ...~
~~.~
,
,...-- ""
,
i
for N = 60
~ ~ ....
I
i ~ "" ~
-
~
7
t
0
time
0
space
o
time
The Brusselator for N = 8 0 ~ -i" ..- ,-- " •-- ~-
t
T h e Brusselator for N = 6 0
_
i i
o
.~ "T" ~ .
'"i ~ ~ ~ i ~ ~ ~
"-
._8
""
6
"5 •-6
== -~
co
1
time
0
0
space
1
time
0
0
space
Figure 13.6. Brusselator reaction scheme and order in time and space produced by Brussode demo of matlab.
642
13.
Organized structures
function b r u s s o d e ( N ) if nargin< 1 N = 50; end tspan = [0; 30]; y0 = [1 +sin((2*pi/(N+ 1))*(I:N)); repmat(3,1,N)]; options = odeset('Vectorized','on','JPattern',jpattern(N)); [t,y] = ode 15s(@f, tspan,y0,options,N); u = y(:,l:2:end); x = (1 :N)/(N+ 1); figure; surf(x,t,u); view(-40,50); xlabel('space'); ylabel('time'); zlabel('solution u'); title(['The Brusselator for N = ' num2str(N)]); /0 .
.
.
.
.
.
function dydt - f(t,y,N) c - 0.000005 * (N+ 1)"2; dydt - zeros(2*N, size(y,2)); % preallocate dy/dt i=l; dydt(i,:) = 1 + y(i+ 1,:).*y(i,:)."2- 4*y(i,:) + c*(1-2*y(i,:)+y(i+2,:)); dydt(i+ 1,:) = 3*y(i,:) - y(i+ 1,:).*y(i,:)."2 + c*(3-2*y(i+ 1,:)+y(i+3,:)); % Evaluate the 2 c o m p o n e n t s of the function at all interior grid points. i = 3:2:2"N-3; dydt(i,:) = 1 + y(i+ 1,:).*y(i,:).A2- 4*y(i,:) + ' c* (y(i- 2, :)- 2*y(i, :) +y(i+ 2,:)); dydt(i+ 1,:) = 3*y(i,:) - y(i+ 1,:).*y(i,:).A2 + '.. c*(y(i- 1, :)-2*y(i+ 1,:)+y(i+3,:)); i - 2*N-l; dydt(i,:) = 1 + y(i+ 1,:).*y(i,:)."2 - 4*y(i,:) + c*(y(i-2,:)-2*y(i,:)+ 1); dydt(i+ 1,:) = 3*y(i,:) - y(i+ 1,:).*y(i,:)."2 + c*(y(i-l,:)-2*y(i+ 1,:)+3); % function S = jpattern(N) B = ones (2 *N, 5) ; B(2:2:2*N,2) = zeros(N, 1); B(l:2:2*N-1,4) = zeros(N, 1); S = spdiags(B,-2:2,2*N,2*N);
13.6.2
The Belousov-Zhabotinsky Reaction Scheme
One of the best-known oscillatory reactions is the Belousov-Zhabotinsky reaction scheme, which contains a set of oxidation-reduction steps; the oxidizing agent is bromate (BROW), the reducing agent is malonic acid [H2C(COOH)2], and cesium ions are used as the catalyst. The concentrations of Ce+3 and Ce +4 vary periodically with a frequency on the order of 0.01 Hz. The Belousov-Zhabotinsky reaction also organizes itself into bands. The Belousov-Zhabotinsky reaction is a chemical oscillatory reaction, and originally consisted of a one-electron redox catalyst, an organic substance that is easily brominated and oxidized, and a bromate ion is dissolved in acid. The typical catalyst ferroin, in its oxidized state, has a blue color, while in its reduced state, ferroin is red. As the Belousov-Zhabotinsky reaction alternates between the oxidized state, and the reduced state, the solution changes its color. The overall reaction is complex; however, its oscillatory effects can be understood by studying the following reaction steps (Field et al., 1972) Ce+3
BrO3' H+ >Ce+4
(oxidation)
(13 19)
13.6 Order in chemical systems
Ce+4
malonicacid
>Ce+3 (reduction)
643 (13.20)
At the start, the cycle begins with a certain amount of Ce+4 ions. The second reaction provides Br- ions, which inhibit the first reaction. This leads to an increase in concentration of Ce +3. After reaching a certain amount of Ce +3, the oxidation reaction starts, since little Ce+4 remains. The system can no longer produce sufficient Br- to inhibit the reaction, and Ce +3 decreases rapidly, producing Ce +4 until the cycle is completed. It is possible to maintain indefinite oscillations with constant frequency in a continuous flow stirred reactor into which bromate, malonic acid, and cerium catalyst are being supplied at a uniform rate. The Belousov-Zhabotinsky reaction scheme can also produce moving spatial inhomogeneties in unstirred solutions. Spatial waves develop as an oxidizing region advances into a region of low but finite bromide ion concentration that falls below a critical value. The autocatalytic production ofbromous acid at the interface advances the wave faster than the diffusion of any other molecules proceeds (Field et al., 1972). Nagy-Ungvarai and Hess (1991) used the electrochemical method to produce experimental data on the two-dimensional concentration profile of three variables in distributed Belousov-Zhabotinsky solutions. Biological systems might be oscillatory, which may be the direct result of biological evolution on earth. The earth moves around the sun and rotates on its own axis. These periodicities induce rhythms in the changes of temperature, light, humidity, and their effects are reflected in the physiology of living systems. The periodicity of day and night is reflected in the characteristics of living systems. Related to these rhythmic changes are the concepts of biological clocks and circadian rhythms for oscillations with a time span of approximately 24 h. The circadian rhythms are generated internally, since their periods are practically independent of environmental factors. Periodic self-oscillatory processes are characteristics of the processes of glycolsis (anaerobic catabolism of glucose) and in the conversion of ADP to ATP. Some reactions display oscillatory behavior in the transport of substrates across membranes, such as facilitated transport.
Example 13.5 The Belousov-Zhabotinsky reaction scheme Field et al. (1972) explained the qualitative behavior of the Belousov-Zhabotinsky reaction, using the principles of kinetics and thermodynamics. A simplified model with three variable concentrations producing all the essential features of the Belousov-Zhabotinsky reaction was published by Field and Noyes (1974). Some new models of Belousov-Zhabotinsky reaction scheme consist of as many as 22 reaction steps. With the defined symbols X = HBrO2, Y = Br-, Z = Ce 4+, B = organic, A = BrO3 (the rate constant contains H+), FKN Model (Field et al., 1972) consists of the following steps summarized by Kondepudi and Priogogine (1999): (i) Production of HBrO2 BrO 3 + BrA+Y
2H +
>HBrO 2 + HBrO k, > X + P
(13.21)
(ii) Autocatalytic production of HBrO2 BrO3 + HBrO2 A+X
H + ' Ce 3+
>2HBrO2 + 2Ce4+ k2 >2X+2Z
(13.22)
(iii) Consumption of HBrO2 HBrO 2 + Br-
H+
>2HBrO
2HBrO 2 -, BrO 3 + HBrO + H + X +Y k3 >2P 2X
xr4 > A + P
(13.23)
(13.24)
(iv) Oxidation of the organic reactants Ce4+ + ( 2 ) C H 2 ( C O O H ) 2 + BrCH(COOH)2--+(@) B r - + Ce3+ + Products
(13.25)
644
13.
Organized structures
The oxidation step is approximated with the reaction above. Concentration of the organic compounds (B) is assumed constant. Effective stoichiometry (f) is a variable, and the oscillations occur whenfvaries in the range 0.5-2.4. Representative kinetic equations of the Belousov-Zhabotinsky reaction scheme based on Eqs. (13.21)(13.25) are
dX - klAY + k z A X - k 3 X Y - 2k4 X2 dt dY - - k , A Y - k 3 X Y + ( f ) ksBZ dt dZ - 2 k z A X - ksBZ dt
(13.26)
By using the following data and maintaining the concentrations of A and B constant, the oscillatory solutions of concentrations can be obtained: kl = 1.28, k2 = 8.0, k3 = 8.0 × 105, k4 = 2 x 103, ks = 1.0 L(mol s); A = 0.06 M, B = 0.02M, f = 1.55. Figure 13.7 displays the oscillations in the concentrations obtained by the following MATHEMATICA code (* The Belousov-Zhabotinsky reaction*) (*FKN Model*) (*X-HBrO2, Y=Br-, Z=Ce4+, B=org, A=BrO3-*) Clear kl = 1.28;k2-8.0;k3=8.0* 10 ^ 5;k4=2" 10 ^ 3;k5= 1.0; A=0.06;B=0.02 ;f= 1.55; Soln3=NDSolve[{X' [t] ==kl *A*Y[t] +k2*h*X[t]-k3*X[t]*Y[t]-2*k4*X[t] ^2, Y' [t]==-kl *A*Y[t]-k3*X [t]Y[t] + (f/2)*k5*B*Z [t], Z' [t] = = 2*k2*A*X [t]-k5*B*Z [t], X[0] = =2" 10 ^-7,Y[0] = =0.00002,Z[0] ==0.0001 },{X,Y,Z}, {t,0,800},MaxSteps-> 10000] Plot[Evaluate[{X[t]}/.Soln3],{t,0,800},Frame->True, FrameLabel->{"t", "X"}, DefaultFont->{"TimesRoman", 14}, PlotStyle->{PointSize[0.03],Thickness[0.02]}, FrameStyle->Thickness[0.0075], PlotRange->{0.0,10^-4}] Plot[Evaluate[{Y[t]}/.Soln3],{t,0,800}, Frame->True, FrameLabel->{"t", "Y"}, DefaultFont->{"TimesRoman", 14},PlotStyle->{PointSize[0.03],Thickness[0.02]}, Frame Style- > Thickness [0.0075]] Plot[Evaluate[{Z[t]}/.Soln3],{t,0,800},Frame->True, FrameLabel->{"t", "Z"}, DefaultFont->{"TimesRoman", 14},PlotStyle->{PointSize[0.03],Thickness[0.01 ]}, Frame Style-> Thickness [0.0075]] Plot[Evaluate[{X[t],Y[t]}/.Soln3],{t,0,800}, Frame->True, FrameLabel->{"t", "X,Y"}, DefaultFont- > {"TimesRoman", 14},PlotStyle-> {PointSize [0.03] ,Thickness [0.01 ]}, FrameStyle->Thickness[0.0075], PlotRange->{0.0,0.0001 }] ParametricPlot[Evaluate[{X[t],Y[t]}/.Soln3],{t,0,800}, Frame->True, FrameLabel->{"Y", "Z"}, DefaultFont->{"TimesRoman", 14},PlotStyle->{PointSize[0.03],Thickness[0.01]}, FrameStyle->Thickness[0.005], PlotRange->{0.0,10"-4}]
Example 13.6 Order in time: Thermodynamic conditions for chemical oscillations Consider the following set of reactions S + X < klf ,,') 2 X X+Y < Y2Y >D
(13.27)
645
13.6 Order in chemical systems
0.0001
l
0.00008
]
0.00006
t
t
1 0.00004 0.00002
26o
4oo -~
(a)
800
600
t
o.oool 0.00008 0.00006 0.00004 0.00002 ,
,
,
,
,
0
200
400
600
800
(b)
t
0.0012 0.001 0.0008 N
0.0006 0.0004 0.0002 0 0
2()0'
(c)
'
'4()0'
'600'
'
'800
t
0.0001 0.00008
>.. X-
0.00006 0.00004 0.00002
0 (d)
200
400
600
80(
t
Figure 13.7. Concentration diagrams for Belousov-Zhabotinsky reaction with parameters: k1=1.28, k2 = 8.0, k3 = 8.0 × 105, k4 = 2 × 103 , k5-- 1.0, A - O . O 6 M , B--O.O2M, f - 1.55; (a) X - H B r 0 2 versus time, (b) Y - - B r - v e r s u s time, (c) Z - C e 4÷, (d) X, Y versus time, Y is displayed with the bold line.
The concentrations of initial and final products S and D are maintained at constant values. So, there are two independent variables Xand Y. The kif and k,~ denote the forward and backward chemical reaction rate constants, respectively. The overall affinity characterizes the thermodynamic state of the chemical system, and is found form
A_ tx, +ljc2+tJc3_ RTln( klfk2fk3fklbk2bk3b S
(13.28)
646
13.
Organized structures
At chemical equilibrium, we have
F
eq
klfk2fk3f
'
-- klf S, Xeq - "klb
Yeq -- klf k2f S klbk2b
(13.29)
Within the vicinity of a nonequilibrium (A 4: 0), the magnitude of the affinity may determine the time behavior of the system: (i) If (S/D) is close to its equilibrium value, then affinity and the reaction velocity are related linearly. Using the subscript s for steady-state values, we have
Jrl,s = klf SXeq AR 1T '
Jr2,s = k2 f Xeq Yeq RA2T '
Jr3,s -- k3f Yeq RA3T
(13.30)
So that the system will be stable to small disturbances around the steady state if it satisfies the inequality p=l T ZJriAi
>- 0
(13.31)
i For the chemical system, the stability condition in Eq. (13.31) becomes T 2 p = klfSXeq ((~A1) 2 -nt- k2f XeqYeq ((~A2) 2 q- k3f Yeq ((~A3) 2 ~ 0
(13.32)
(ii) Assume that the system is far from equilibrium. If the reverse reactions in the system (13.27) are negligible, the kinetic relations become dX dt dY dt
-- klfS x - k2fX Y
(13.33)
- k2f X Y -
(13.34)
k3f Y
These kinetic relations yield the following steady-state solutions
Xs - k3f k2f'
Ys = klf' S k2f
(13.35)
This model resembles a Lotka-Volterra model, which may be used in studying the evolution of systems in time, such as biological clocks or the time-dependent properties of neural networks. In the vicinity of the steady state, X(t) and Y(t) may be X ( t ) = X s + xe at,
Y(t) = X s + ye at
with small enough magnitudes of perturbations of x and y
I+01 L0L 500] Plot[Evaluate[{XL[t],XD It] }/.Solnl ],{t,0,120}, Frame->True, FrameLabel->{"t", "XL, XD"}, PlotStyle->{{GrayLevel[0], Dashing[{0.02,0.025}]},Thickness [0.008], GrayLevel[0.3]}, DefaultFont->{"TimesRoman", ]2}] 13.7.3
Lotka-Volterra Model
In biological dissipative structures, self-organization may be related to the attractors in the phase space, which correspond to ordered motions of the involved biological elements (De la Fuenta, 1999). When the system is far from equilibrium, ordering in time or spontaneous rhythmic behavior may occur. The L o t k a - V o l t e r r a m o d e l of the predator-prey interactions is a simple example of the rhythmic behavior. The interactions are described by the following kinetics dX J1 -
- klX-k2XY
(13.63)
- - k 3Y - k 4 Y X
(13.64)
dt dY J~ -
dt
where the terms X and Y represent the number of individuals of species, and k 1 and k4 are the biological potentials, which are the difference between the birth and death rates, respectively. The terms k2 and k3 are the interactions between both populations. The flows shown by Jl and J2 have two stationary solutions; the first of these is X = Y = 0, and the second is at the stationary values o f % and Y~, which are given by
k4
k:
(13.65)
To see whether this state is stable or not, we add small perturbations of 6 X and 6 Y to Xs and Ys, so that X-A'~+aX
fix
and
Y-Y~+aY
(13.66)
We can introduce these expressions into Eqs. (13.63) and (13.64) and disregard the terms containing products of and 6 Y, and we obtain d(aX) dt
-- k l r ~ X - k 2
(Xs6Y + Y~aX)
d(ar) dt
-- k 3 ( X s ~ Y -4 Y s ~ X ) - k 4 ~ Y
(13.67)
(13.68)
654
13.
Organized structures
With Eq. (13.65), we can rewrite Eqs. (13.67) and (13.68) as follows d(aX)
(13.69)
- -kzXsaY
dt
d(aY) -
dt
(13.70)
k3YsaX
We now differentiate Eq. (13.69) with respect to time, and after combining with Eq. (13.70), we have d2(6X) dt 2
=
-k2k3Xs
aX -
-k &ax
(13.71)
This equation has the form of motion of a harmonic oscillator and the solution yields a harmonic oscillation 6X = 6X(O) cos (2'n'fi)
(13.72)
where f is the frequency given by
{1
( 1)
(13.73)
Therefore, the stationary states given by Eq. (13.65) are not stable. This means that after small perturbations, the system does not return to the original state. Instead, it oscillates. This oscillatory behavior can be explained with the following example: As X increases, species 2 has more food and tends to increase its population. As Y increases, the amount of species 1 consumed by species 2 also increases. Therefore, X begins to decrease, and hence the amount of food available for species 2 decreases; this leads to a decrease of Y. As the number of predators Y decreases, the population of prey X recovers. This causes the start of a new cycle. The amplitude and period of oscillation depend on the initial state.
Example 13.8 Prey-predator system" Lotka-Volterra model The Lotka-Volterra predator and prey model provides one of the earliest analyses of population dynamics. In the model's original form, neither equilibrium point is stable; the populations of predator and prey seem to cycle endlessly without settling down quickly. The Lotka-Volterra equations are dX dt dY dt
-¢¢Y-aYY
- -6 Y+ faXY
When the birth rate equals the death rate in the prey (host) population, so that/3 = 0, we have dX -
-olXY
dt dY dt
- Y(-6 + faX)
If we take the ratio of these equations, we have dY dX
--f+~
aX
13. 7
655
Biological structures
This equation shows that dY/dX may be negative, zero, or positive, according to the values and signs of (8laX). Because dX/dt is negative, Xdecreases with time. Even though the host population is dying out, the predator population can increase until X decreases to the critical value 6/fa; after that, the host population also begins to die out. If none of the predator eggs hatches, then f - O, and the Lotka-Volterra equations become dX dt dY -
-BY
dt
Here, the trajectories are determined by dY
dY w
dx
x(~-~Y)
The trajectories show that the prey population is minimum at Y-/3/ce.
13.7.4
Stability Properties of Lotka-Volterra Equations
To investigate the stability properties of the Lotka-Volterra equation in the vicinity of the equilibrium point (X, Y) = (0, 0), we linearize the equations ofXand Y appearing on the right side of Eqs. (13.63) and (13.64). These functions are already in the form of Taylor series in the vicinity of the origin. Therefore, the linearization requires only that we neglect the quadratic terms in XY, and the Lotka-Volterra equations become dX dt
dY
- fiX,
dt
- -8 Y
(13.74)
The first equation shows that X increases exponentially, and hence the equilibrium point (0, 0) is unstable. A relationship between the X and Y can be obtained by the method of separation of variables, which yields dX m
-
=
x ( ~ - ,~Y)
dY
(13.75)
Y (-~ + . f ~ x )
Furthermore this equation in separable form becomes dt-
(-8 + faX) dX
(fi - c~Y) d Y
X
Y
(13.76)
Integration of Eq. (13.76) yields f o~X-81nX - ~lnY-o~Y +C
(13.77)
X '~e./~x - C'Y~e -~r
(13.78)
In exponential form, Eq. (13.77) is
Here C' is the constant of integration, and is obtained from the initial condition. Equation (13.78) describes a family of closed trajectories in the X, Y space. Each trajectory is determined by a particular initial condition. Figure 13.10 displays such trajectories obtained with an m-file with the following MATLAB code. function lotka(time) tspan- [0:0,01 :time]; zo--[2;1]; kl=l; k2-1; k3=l; k4-1; [t, z] =ode4 5 (@fz,tsp an, zo): zo=[1,8;1];
656
13.
Organized structures
[t,y] =ode4 5 (@fy,t sp an, zo); zo:[1.6;1]; [t,x] : o d e 4 5 (@fx,tspan, zo); figure; plot(z(:, 1),z(:,2)),xlabel ('X'),ylabel('Y') figure; plot(y(:, 1),y(:,2)),xlabel ('X'),ylabel('Y') figure; plot(z(:, 1),z(:, 2),y(:, 1),y(:, 2),x(:, 1),x(:, 2)) xlabel ('X'),ylabel('Y') function dz = fz(t,z) dz- [k 1*z(1 )-k2*z ( 1)*z (2);-k3* z (2)+k4* z ( 1) *z (2)]; function dy = fy(t,y) dy= [k 1*y( 1)-k2 *y( 1) *y(2 );-k3 *y( 2) + k4*y( 1) *y( 2) ]; function dx = fx(t,x) dx= [k 1*x(1 )-k2*x (1)*x (2);-k3*x (2)+ k4*x (1)*x (2) ];
Example 13.9 Sustained oscillations of the Lotka-Volterra type An open system far from equilibrium could exhibit spontaneous self-organization by dissipating energy to the surroundings to compensate for the entropy decrease in the system. Prigogine called such systems dissipative structures. Sustained oscillations, therefore, require an open system capable of exchanging energy and matter with its surroundings. Since it cannot exchange matter with its environment, a closed system can exhibit transitory oscillations only, as it must approach equilibrium. These studies are helpful in understanding the spontaneous self-organization, independent from genes and natural selection, in biological systems. The Lotka-Volterra type of equations provides a model for sustained oscillations in chemical systems with an overall affinity approaching infinity. Perturbations at finite distances from the steady state are also periodic in time. Within the phase space (Xvs. Y), the system produces an infinite number of continuous closed orbits surrounding the steady state
dX dt
-
- X ( Y - 1)
and
dY = Y ( X - 1) dt
After dividing one of these equations by the other, we have the following equation for the trajectories in the X, Yspace
dY
Y ( X - 1)
dx
x ( r - 1)
Integration of this equation yields X+Y-lnX-lnY=
C
(13.79)
where C is an arbitrary constant, which is determined by the initial conditions. The critical point of marginal stability is reached in the limiting situation of infinite overall affinity. There is no mechanism for the decay of fluctuations. Equation (13.79) represents a family of cycles each corresponding to a given value of constant C around a steady state (Figure 13.10). Each cycle appears as a state of marginal stability where even a small perturbation can change the motion of the system to a new cycle, corresponding to a different frequency (Glansdorff and Prigogine, 1971). There is no average orbit in the vicinity where the system is maintained. The Lotka-Volterra model has properties similar to those of unstable systems at the marginal state. Only the orbits infinitesimally close to the steady state may be considered stable, according to Liyapunov's theory of stability. However, at a finite distance from the steady state, two neighboring points belonging to two distinct cycles tend to be far apart from each other because of differences in the period. Such motions are called stable in the extended sense of orbital stability. The average concentrations of X and Y over an arbitrary cycle are equal to their steady-state values (Xs = 1 and Ys = A = 1). Under these conditions, the average entropy production over one period remains equal to the steady-state entropy production.
13. 7
657
Biological structures
2.4 2.2
1.8 1.6 >- 1.4 1.2
1 0.8 0.6 0.4
0
0.5
1
1.5
2
2.5
X
Figure 13.10. State-space plots with various initial values of Xo for the Lotka-Volterra model.
Diffusive instability can appear in simple predator-prey models. Bartumeus et al. (2001) used the linear stability analysis and demonstrated that a simple reaction-diffusion predator-prey model with a ratio-dependent functional response for the predator can lead to Turing structures due to diffusion-driven instabilities.
Example 13.10 Lotka-Volterra model Solve the following equations and prepare a state-space plot where Xis plotted against Y using the solution. dX dt dY dt
- 0 . 2 X - 1.2XY - -0.5Y + 0.25XY
Initial conditions" at t = 0, X = 1.0, and Y = 0.01. Figure 13.11 display the trajectories of X and Y, and the state-space plot produced by the following MATHEMATICA code (*Lotka-Volterra Model*) k1-0.2;k2-1.2 ;k3-0.5;k4-0.25; Solnl =NDSolve [{X' It]-=kl *X[t]-k2*X [t]*Y[t], Y' It]- --k3*Y[t] + k4* X [t] *Y It], X[0]-= 1.0,Y[0]--0.01 },{X,Y},{t,0,100}, MaxSteps->4000] Plot[Evaluate[{X[t],Y[t]}/.Soln 1],{t,0, IL00}, Frame->True, FrarneLabel->{"t", "X,Y"}, PlotStyle->{{GrayLevel[0], Dashing[{0.02,0.025}]}, GrayLevel[0.3]}, DefaultFont->{"TimesRoman", 14}]; PararnetricPlot[Evaluate[{X[t],Y[t]}/.Soln 1],{t,0,100}, Frame->True, FrameLabel-> {"X", "Y"}, RotateLabel-> True,DefaultFont-> {"TimesRoman", 14}]
Example 13.11 Enzymatic reactions" Oscillations in the glycolytic cycle Biochemical chains are highly likely to exhibit limit cycles and dissipative structures. Oscillations appear in living systems in a variety of ways with
658
13.
6
'
'
;I
. . . . . . . . . . . . . . . Ii
I
5 >-
X-3
I
I
II
III
III
II
I
I
I Ii
I
II
f ~
,,,,"
I
i"
.....
20
ii
11
I
I
" ....
40
>" 0.4 0.2
|
,ll 7 .....
60
0.8 0.6
I
iI
'
I
I
l
0
Figure 13.11.
I
I
I
0
(a)
I
'
I I ii
I
I I
II
2 1
i'
iI
I
I
4
Organized structures
~i
80
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
100
t
2
3
4
.-
5
6
x
(b)
Plots from prey-predator in Example 13.10: (a) Dashed line displays X, (b) state space plot with kl = 0.2, k2 = 1.2, k3 = 0.5, k4 = 0.25, initial values: X(0) = 1.0, Y(0) = 0.01.
very different properties. Glycolysis is one the best examples of temporal structures, which show the richness and variety of self-organization far from thermodynamic equilibrium in a metabolic pathway. The temporal structures result when an instability-producing multienzymatic mechanism experiences variations in the domains of the initial functions (De la Fuenta, 1999). For example, oscillations of the concentrations of some metabolites of enzymatic reactions are possible at the molecular level or at the supracellular level (circadian ryttmas) (Goldbeter, 1996). Some characteristic examples of sustained chemical oscillations are (i) substrate- and product-inhibited enzymatic reactions and (ii) the product activated enzymatic reaction of phosphofructokinase in the glycolytic cycle. In a model by De la Fuenta (1999), the activity of three enzymes, namely hexokinase, phosphofructokinase, and pyruvatekinase, is considered, and the chaotic dynamics of a dissipative glycolytic subsystem is suggested. For a single range of the control parameter, this model predicts the coexistence of chaos with different periodic regimes and limit cycles. The enzyme phosphofructokinase is allosteric, that is, it is made up of equivalent units that possess specific reaction sites for the fixation of the substrate and product. Each unit exists in two conformational states: one active with more affinity for the substrate, and one inactive. The reaction products of phosphoffuctokinase (FDP and ADP) displace the conformational equilibrium in favor of the active form of the enzyme. This may create a destabilizing effect on the excess entropy production. In the glycolytic cycle, the allosteric properties of the phosphofructokinase may lead to oscillations. Consider the following simple model A--~C 1 C 1 Jr-D 1--* D 2 D 2 --~ C 2 + D 1
(13.80)
C 2 + D 3 ~ D1
C2-*F Where A is the initial product glucose, F the final product glyceraldehyde 3P, D1 and D 3 the active and inactive forms of the enzyme, respectively, D 2 the enzymatic complex, C1 and C2 the fructose 1P and fructose 2P, respectively. In the conformational equilibrium step (C2 + D3 ~ D1) FDP activates the enzyme. Experimental results indicate that under physiological conditions, the enzyme is controlled mainly by the ATP and ADP. Jrl
A 3 + D~ D2
,YA2 + D3 < A2
>A3
kfl >D2
(a) (b)
kbl
kf2 > A 2 + D 1 kf
(c)
3 )D1 kb3
(d)
>
(e)
k4
(13.81)
The reactant A3 is the ATP and enters the system at constant rate OfJr~ at step (a). The A2 is the ADP, which is the product of reaction and also the activator in step (d). The factor y (3' > 1) shows that the fixation of A2 on the enzyme activates more than one reacting site because of the concerted conformational equilibrium. The kinetics of the model above is
13. 7
659
Biological structures
OA3ot
02A3 - Yrl - kflA3D1 + kblD2 + DA 3 Or 2
OA2
at -
kfaD2 -
Nf3A~D3+ kb3D1 -
Old___.= .!'-" -kflA3D 1+(kbl Ot
+ kfz)D 2 +
02A2
k4A2 + DA2 Or"~
kf3A~D3 -
kb3D1
(13.82)
0 0 3 = - k f 3 A ~ D 3 + kb3D1 Ot
OD2 _ Ot
kflA3O 1 --(kbl
n t-
kfa)D 2
This model neglects the diffusion of the enzyme since it is much slower than for ATP and ADP. In agreement with experiments, the following conditions are assumed to hold (Glansdorff and Prigogoine, 1971)
kfl, kbl, kf2 , kf------~3kb3 >>1 A3 A~'
(13.83)
A3 , A2 >>1,Do = D 1 + D 2 + D 3 Do Do where D o is a constant. OA30t - Jrl - °llDoC~
°A2at-
A3A~ + DA3
02A3 Or 2
o24 a, Doc~~ A 3A~ - k 4A 2 + DA2 Or2
(13.84)
(13.85)
with kflkf2 , oe~ = kf3 oe5 = alD0C~~ kbl --}-kf2 kb3'
t9ll - - - ~
At steady state, we have
A2s- Jrl ,A3~ = j(1-rlY)~(kblq-kf2)k~kb3 k4
kflkf2kf3O 0
(13.86)
The linearization of the perturbation equations around the steady state yields the characteristic equation
A2 +[°es(A2s)'Y+ k4-T°esA3s(A2s)Y-I +DA2A+ DA3 2 ] (13.87) + ~(A~s) '
k4 + - - / -
+
< + - - r - - ~ A 3 s ( A ~ s ) '-~ = 0
The condition for homogeneous perturbations becomes
b7+l
_
J ~ < (Jrlc)7 - kbl -ff kf2 kb3 '~4
/~flkf2 kf3 Do
(13.88)
This condition corresponds to the vanishing of the coefficient A in Eq. (13.87). The frequency becomes
Ai - k4 ( T - 1)1/2 and a focus is followed by a limit cycle for the phosphofructokinase reaction.
(13.89)
660
13.
Organized structures
The condition for inhomogeneous perturbations becomes bY+l J~ < JYrlc -- kbl "+-kf2 kb3 ~4 DA3 (~-~_ 1)2
k lkf2 kf3 Do DA2
(13.90)
The critical wavelength (~Oc)is 2 % =
DA2
(13.91)
Comparing the conditions in Eqs. (13.88) and (13.89) indicates that both instabilities occur when DA3 ~.~ x ~ +1 D A2
(13.92)
xf-Y - 1
An approximate value of k4 is obtained from Eq. (13.86) as k4 = 4 x 10 -2 1/s (Glansdorff and Prigogine, 1971). The values of Jr1 and the steady-state value of ADP concentration are Jrl -- 6 X 10 -6 mole / s, A2s = 1.5 × 10 -4 mole, and k4 -- 4 × 10 -2 S- 1(approximate)
Setting y = 2 in Eq. (13.89), the frequency of oscillations is A = 2.4 min- 1 and the period is 2.6 min. This value is in agreement with the experimental value of 3-5 min. The spatial differentiation is related to k4 and the diffusion coefficient DA. The critical wavelength changes in the range 10 -4 cm < ~oc< 10 -2 cm.
Example 13.12 Long-wavelength instability in bacterial growth Growth conditions may lead to different morphologies. A nonlinear diffusion coefficient may cause the generation of patterns and a long-wavelength instability. Consider a two-dimensional reaction-diffusion system for the bacteria density B(r,t) with a nonlinear diffusion term, and nutrient density N(r,t) with a linear diffusion term OB - f B ( B , N ) + VDB(B)VB Ot ON Ot
- - f y (B,N) + DN V2N
(13.93)
(13.94)
where DN is the diffusion coefficient of the nutrient, and DB the bacteria-dependent diffusion coefficient of the bacteria, defined by
DB(B ) = Do Bk
(13.95)
Here, k describes the nonlinearity. For simplicity, it is assumed that
f B ( B , N ) = f N ( B , N ) = BN
(13.96)
In a simplified model, this representation reflects a bilinear autocatalytic reaction: B + N ~ 2B, and simply means that the bacteria needs nutrient (assume no shortage of nutrient) to double themselves (Muller and Saarloos, 2002). Any instability observed for k > 0 is due to the nonlinearity in the diffusion coefficient. With the relations above, the reaction-di~sion system becomes
OB _ BN + D ~2Bk+l Ot k+l
(13.97)
13.7
661
Biological structures
ON --BN+V2N Ot
(13.98)
where D is the rescaled diffusion coefficient: D Do/O N. The model in Eqs. (13.97) and (13.98) has two homogeneous states: a stable solution in which only bacteria are present, and an unstable solution with only nutrients. The propagation of stable state to unstable state can be studied. For k > 0 and beyond a critical diffusion (Dc) D < Dc(k), the planar front is unstable and has a long wavelength instability (Muller and Saarloos, 2002). In a more realistic model, one needs to be able to incorporate the bacteria growth properties into the effective diffusion coefficient. =
Example 13.13 Instability in a simple metabolic pathway Instability can occur in a system composed of interacting subsystems. The subsystems may be stable in isolation. Consider the following simple pathway consisting of three enzymes (e 1, e2, and e3). ÷ I
I
Xo< 1 >XI< 2 ~'X2 < 3 ~'X3
(13.99)
and the metabolites of Xo and X3, which are maintained at constant values. The other metabolites X and X2 have varying concentrations. The metabolites Xo and)(3 constitute the boundary conditions that keep the system under nonequilibrium conditions. The rate equation of enzyme i is denoted by Jri. The kinetics of the system is
dX1 ]
I
dX2
(13.100)
Jr2 -- Jr3
k--g-J The system dynamics is obtained by integrating these equations with some initial conditions. The system in Eq. (13.99) may be decomposed into two subsystems + I
Xo< 1 )~)(1( 2 )x2
(1)
X l ( 2 )x2 ( 3 )x3
(2)
(13.101)
The behavior of the subsystems can be described in isolation. In the first subsystem, X2 is kept constant, while X1 is constant in the second subsystem. If the whole system is at steady state then dX1/dt = 0 and dXz/dt = 0. If the system is stable, then any small perturbations in)(1 and X2 are corrected, and the system returns to its original state. We assume that the two subsystems are stable in isolation. The enzymes in the subsystems are sensitive to the metabolites and hence interact with the relevant subsystem. The stability of the subsystems assumes the following relationships
OJrl OX1
OJr2 < 0 OX 1
and
OJr2 OX2
OJr3 < 0 OX2
(13.102)
The first inequality indicates that the sensitivity of enzyme 1 to X1 is less than the sensitivity of enzyme 2 to X1. The second inequality indicates that the sensitivity of enzyme 2 to X2 is less than the sensitivity of enzyme 3 to X2. These partial differentials known as unscaled elasticities, are the component properties and quantify the sensitivity of a rate to a metabolite. The whole system is stable if the following conditions (tr < 0 and Det > 0) are satisfied
OJrl OX 1 I OJrl OX~
OJr2 OX 1
OJr2 OX2
-
OJr 3 < 0 (1) OX2
(13.103)
> 0
(2)
(13.104)
662
13.
Organized structures
Since we assumed that the subsystems are stable the first condition is satisfied. The condition is violated if we have
'r2) OY 1
>o
o ]
&
'
J~
'r3/ OX 2
>0
(13.105)
OX 2
&
The left side term indicates the interactions between the two component subsystems, while the fight term shows the interactions within the subsystem of each component. Thus, even with stable subsystems (in isolation), the system can be unstable if the interactions among subsystems are more significant than the interactions within subsystems. So, the enzymatic parameters and the boundary conditions can be controlled in such a way that systemic instability occurs. This particular phenomenon is known as a saddle-node bifurcation. Under slightly different conditions, oscillations appear when the first condition is violated. However, this would be unexpected because the subsystems are stable in isolation and contain only one variable metabolite. These conditions show that: (i) when one subsystem contains two variable metabolites arid the other subsystem has one variable metabolite, the system with dynamic subsystems in isolation oscillates, and (ii) instability is impossible with two dynamically stable subsystems if the influence is one-directional; they must influence each other mutually to become unstable. This simple example also displays the importance of nonlinear dynamics. The unscaled elasticities are constant only for a linear system and hence result in the same behavior: if a linear system is stable, it will remain stable when boundary conditions change, and if it is unstable, it will remain unstable. Under nonlinear equations, however, the values of elasticity depend on the system state, which varies with the boundary conditions. This may lead to a transition from a stable to unstable system, known as bifurcation or symmetry breaking. This means that nonlinearity causes a variety of new behaviors in a system.
Example 13.14 A model for an enzyme reaction inhibited by the substrate and product Consider the enzymatic reaction inhibited by the substrate and product Jri ) S 1 + E < S1E
kfl >$1E
kbl
kf2 )E_[_S2
Jrf )
kr3 >$1S1E Sl+SlE< k~
(a) (b)
(C)
(13.106) kf4 )
S2 + E
X
k2 > Y + E k3 >3X k4 )F
k2 k2 With the parameters kl = 1.3, k2 = 1.0, k3 = 1.0, k4 = 1.0, and the concentrations A = 1.1, B - 3.0, Prepare the trajectories of X and Y and state-space plot. 13.7
Consider the following chemical reaction system: A
B+X 2X+Y X
kl )X,
kl = 1.0
k2 ) Y + E , k3 >3X, k4 >F,
k2 = 1 . 0 k3 = 1 . 1
k4=l.1
The initial values of A and B are maintained at A - 0 . 6 M and B = 1.6 M, while the products E and F are removed. Prepare the trajectories and state-space plot of X and Y. 13.8
When the concentrations of A and B are controlled, the one-dimensional approach to complex reactiondiffusion systems with the spatial coordinate r under isothermal conditions yields the kinetic equations for Xand Y
OX -
Ot
k A- k BX + k3x r - k4X + Dx
02X Or
665
Problems
OY 02y c)t - k 2 B X - k 3 x Z Y + Dy Or2
where Dx and Dy are the respective diffusion coefficients. The boundary conditions are
Or
Or r--+L
r=-L
and
-0
For a specified value for A and some variation of B if one of the eigenvalues becomes positive, then Det < 0, and the system becomes unstable, and the propagating wave or the Turing structure occur
B-> 1l (k4 + +k°'2Dx 2 )(
k3(klA/k4)2)0 -2Dy
Solve the above equations and assess the structuring conditions. 13.9
The Belousov-Zhabotinsky reaction scheme is A+Y
k, > X + P
A+Y
k2 >2X+2Z
X+Y
k3 >2P
2X
k4 > A + P
B+Z Representative kinetic equations of the Belousov-Zhabotinsky reaction scheme based on Eqs. (13.21)-(13.25) are dX
- k l A Y + k 2 A X - k 3 X Y - 2k4 X2
dt dY
- -k~AY-
dt dZ dt
k3XY + ( f ) ksBZ
- 2k2A X _ ksBZ
By using the following data and maintaining the concentrations of A and B constant, find the oscillatory solutions of concentrations of AT, Y, and Z: k l - 1 . 2 8 , k 2 - 8 . 0 , k 3 - 8 . 0 × l0 s, k 4 = 2 × 103, ks - 1.0 L/(mol s); A - 0.06 M, B - 0.02 M , f = 1.6, and f = 1.4. 13.10
The kinetic equations of the two-variable Lengyel-Epstein model become (Horsthemke and Moore, 2004)
o
d[I-] dt
-0-
~d[ClO _ ] dt
du
- a-lJ-~
dt dv dt
_
b u-
4uv 1+ u 2
uv 1+ u 2
666
13.
Organized structures
where u and v are the dimensionless concentrations of [I-] and [C102, respectively. The constants a is proportional to [MA]/[C102] while the constant b is proportional to [I2]/[C102]. Display the oscillations for the Lengyel-Epstein model with different sets of parameters: o-= 2.0, a = 30.0, b = 6.0, the time interval is 0 0 and beyond a critical diffusion (Dc) D < Dc(k), the planar front is unstable and has a long wavelength instability, Miiller and Saarlos, 2002). 13.16
(a) Instability can occur in a system composed of interacting subsystems. Consider the following simple pathway consisting of three enzymes (e 1, e2, and e3). 4-
X°(
1 )kl(
I 2 >X2
(14.53)
where r is a specific rate of energy supply or energy lost and q the transport of internal energy due to conduction. Introducing the Helmholtz energy, A = u - Ts, and the following energy balance equation: du p ~ = - V.q - P : Vv + pr
(14.54)
dt
Equation ( 14.53) becomes
-P
d(A+sT)_p.v_ dt
1
T q VT -> 0
(14.55)
Here, P is the pressure tensor, and the velocity gradient splits into a symmetric (V) and an antisymmetric (W) part Vv = V + W
(14.56)
where V = 1 [Vv + (Vv)T ]. 2 m
w
1 = = [Vv - (Vv)T ]
2
Elements of the symmetric part of the velocity gradient are V/j = (l/2)[0vj/0xi + Ovi/Oxj]. Equation (14.55) is known as the Clausius-Duhem or the fundamental inequality for a single-component system. The selection of the constitutive independent variables depends on the type of system considered. For example, the density, velocity, and temperature fields in hydrodynamics are customarily chosen. A process is then described by solving the balance equations with a consideration of constitutive relations and the Clausius-Duhem inequality. For simplicity, a set of constitutive equations for a Stokesian fluid without memory is (14.57)
4~ = 4~(v,v,T, Vv, VT)
The dependence of + is expressed by ordinary functions instead of functionals. With Eq. (14.57) the Clausius-Duhem inequality becomes
/
/
-P ~Oa +s T - P ovOCt" v - P
oa
1
o(vT).(VT)-~q.VT-
U+P
) "V->0
(14.58)
where the mass conservation is expressed as
pv = V'v = V :U
(14.59)
When the constitutive equations are expressed in terms of functionals representing the whole history of the variables, a vast amount of information may be necessary.
14.5
EXTENDED NONEQUILIBRIUMTHERMODYNAMICS
Chemical process rate equations involve the quantity related to concentration fluctuations as a kinetic parameter called chemical relaxation. The stochastic theory of chemical kinetics investigates concentration fluctuations (Malyshev, 2005). For diffusion of polymers, flows through porous media, and the description liquid helium, Fick's and Fourier's laws are generally not applicable, since these laws are based on linear flow-force relations. A general formalism with the aim to go beyond the linear flow-force relations is the extended nonequilibrium thermodynamics. Polymer solutions are highly relevant systems for analyses beyond the local equilibrium theory.
681
14.5 Extended nonequilibrium thermodynamics
Extended nonequilibrium thermodynamics is concerned with the nonlinear region and deriving the evolution equations with the dissipative flows as independent variables, besides the usual conserved variables. Typical nonequilibrium variables such as flows and gradients of intensive properties may contribute to the rate of entropy generation. When the relaxation time of these variables differs from the observation time they act as constant parameters. The phenomenon becomes complex when the observation time and the relaxation time are of the same order, and the description of system requires additional variables. To coordinate components, the generalized flows and the thermodynamic forces can be used to define the trajectories of the evolution of nonequilibriun systems in time. A trajectory specifies the curve represented by the flow and force components as a function of time in the flow-force space. A useful trajectory can be found and analyzed by a variation principle. In thermodynamics, the variation principles lead to the least energy dissipation and minimum entropy generation at steady states. According to the most general evolutionary criterion, open chemical reaction systems are dissipative, and evolve toward an asymptotic state in time. Extended nonequilibrium thermodynamics is not based on the local equilibrium hypothesis, and uses the conserved variables and nonconserved dissipative fluxes as the independent variables to establish evolution equations for the dissipative fluxes satisfying the second law of thermodynamics. For conservation laws in hydrodynamic systems, the independent variables are the mass density, 9, velocity, v, and specific internal energy, u, while the nonconserved variables are the heat flux, shear and bulk viscous pressure, diffusion flux, and electrical flux. For the generalized entropy with the properties of additivity and convex function considered, extended nonequilibrium thermodynamics formulations provide a more complete formulation of transport and rate processes beyond local equilibrium. The formulations can relate microscopic phenomena to a macroscopic thermodynamic interpretation by deriving the generalized transport laws expressed in terms of the generalized frequency and wave-vector-dependent transport coefficients.
14.5.1
Extended Nonequilibrium Thermodynamics of Polymer Solutions
Extended nonequilibrium thermodynamics theory is often applied to flowing polymer solutions. This theory includes relevant fluxes and additional independent variables in describing the flowing polymer solutions. Other contemporary thermodynamic approaches for this problem are GENERIC formalism, matrix method, and internal variables (Jou and Casas-Vfizquez, 2001), which are summarized in the following sections. Polymer solutions may have the memory effects observed in viscoelastic phenomena. This requires additional relaxation terms in the constitutive equations for the viscous pressure tensor, which may be affected by the changes in the velocity gradient. Besides this, the orientation and stretching of the macromolecules may have an influence on the flow. The following summary is from Jou and Casas-Vfizquez (2001). In the extended nonequilibrium thermodynamics for a binary liquid mixture, the viscous pressure tensor pv and the diffusion flux J are considered as additional independent variables. The viscous pressure tensor, pv, by the simplest Maxwell model, is defined by the following constitutive equation:
d P " _ . .1p,, . . 2r/v dt
r
(14.60)
"r
where rl is the shear viscosity and r is the viscoelastic relaxation time for the viscous pressure tensor. In extended nonequilibrium thermodynamics of polymer solutions, the generalized extended Gibbs equation for a fluid characterized by internal energy U and viscous pressure pv is d, -:
1 T
1
1
T
T
d. + :- Pd,,-
= a.
-
dJ -
dpv
(14.61)
where T, P, u, and V are the temperature, pressure, internal energy, and volume, respectively, wl the mass fraction of the solute, A~ = tx~- ~2 the difference between the specific chemical potentials of the solute and the solvent, c~1 and OL2 the coefficients, and J the diffusion flux. In Eq. (14.61), T, P, and Atx depend on the fluxes J and Pv. Various formalisms on rheology indicate the connection between thermodynamics and dynamics. To integrate Eq. (14.61) T, P, and Atx depend also on the fluxes pv and J, while the corresponding definitions of values based on the local equilibrium quantities are defined as functions of only u, V, and wl. Assume the following entropy flux:
1 1 J, - -~-q- = A/~ J +/3P v .J 1
1
(14.62)
682
14. Nonequilibriumthermodynamics approaches
where [3 is a coupling coefficient for coupling between viscous pressure tensor and diffusion, and q the energy flux. The energy and mass balance equations are du p--=-V.q-p(V.v)-pv: dt
p
(Vv)
(14.63)
dw 1 = - V . J
(14.64)
dt
Combining Eqs. (14.62)-(14.64) yields the time derivative of the entropy 1
ds--1--V.qdt
T
(A/~)~7. J - - -
T
1 pv
"V
dJ
_
o/1J'---
T
o/2 P v "
dt
dP v
(14.65)
dt
Substituting Eqs. (14.62) and (14.65) in the general form of the balance equations of entropy ds p - - + V . J ~ =(P dt
we obtain the entropy production -o/1-d-7+V'(/3Pv) +
( P = q .V 1 + j . - V
• ---V-o/2
+/3VJ
(14.66)
For an isothermal process the simplest evolution equations for J and pv that are compatible with the positive chracter of the entropy production are --V
-- O/1 ~
1
- - -T - V - o/2
-+" V ' ( ~ e v ) -- [~1J
dP v - - ~ + flVJ = flzP v
(14.67)
(14.68)
where [31 and ~2 are the positive phenomenological coefficients which may be identified by comparing Eqs. (14.67) and (14.68) in the steady state and uncoupled forms with the Newton-Stokes' and Fick's equations PV =-2,qV,
J = -/)V(A/z)
(14.69)
where/) is related to the diffusion coefficient D by
Then, the parameters [3j and ~2 are 1
/31 - / ) T '
1
/32
2vtT
(14.70)
The respective relaxation times of diffusion flux and viscous pressure tensor are ~'1 -
o/1
~
- o/1 ( D T ) ,
T2 --
°/2
-- O/2 (2z/T)
(14.71)
Therefore, the evolution equations (14.67) and (14.68) for diffusion flux and viscous pressure tensor become dJ
'7"1 - - ~ --
- [ J +/)V(A/.,)] +/3/)TV. pv
(14.72)
683
14.6 Generic formulations
dp v 7" 2 ~
_ _(pv + 2~V) + 2/3T~VJ
(14.73)
_
dt
These equations describe the coupled phenomena between diffusion and viscous stresses, existing, for example, in diffusion of small molecules in polymer matrix. Other possible couplings occur in shear-induced diffusion and shear-induced separation. If pv is the sum of various viscous pressure contributions P(, each of them with its own relaxation time "r2i, Eq. (14.61) becomes
1 1 ds = l du +-- Pdv - -- A/x T T T
dw
vT.1 "~ V7.2i v. ~ J'dJPi dP~v DT ~i 2"qiT
1 -
(14.74)
By neglecting diffusion effects, integration of the above equation yields
s(u,v, p V ) _ Seq(u v ) _ VJ pv. pv 4T
(14.75)
'
where subscript "eq" refers to local equilibrium value and J (=~27q) is the steady-state flux, n2 being an averaged relaxation time. This equation is only a second-order expansion of the entropy in the viscous pressure. Often extended nonequilibrium thermodynamics with maximum-entropy formalism leads to more general expression for the entropy not limited to second order in fluxes.
14.6
GENERIC FORMULATIONS
GENERIC was formulated by Grmela and Ottinger (1997). The time evolution of the physical systems may be written in terms of two generators E and S and two matrices L and M to represent the essential features of the dynamics of the system
dx 6E 6S -L. +M.~ dt 6x 6x
(14.76)
where x represents a set of independent variables to describe the nonequilibrium system completely (namely hydrodynamic fields and additional structural variables), E and S are the total energy and entropy, respectively, and L and M the linear functional operators. The g/gx indicates functional derivatives, and the dot shows the multiplication of a vector by a matrix. The first term on the right of Eq. (14.76) describes the reversible part of the time evolution equations of x generated by the energy E and entropy S, while the second term represents the irreversible contributions. Equation (14.76) requires the following general and essential degeneracy conditions:
6E L.--= 6x
0,
6S M.~-0 6x
(14.77)
The first condition is for the reversible contribution of L to the time evolution of the system and requires that the functional form of the entropy is unaffected by the operator L responsible for the reversible dynamics. The second term is the conservation of the total energy by the contribution of the dynamics. In the GENERIC derivation, the following three brackets are defined:
{A, B I -
6A
6B)
6A
6B>
[A,B]- 7x,n. x
(14.78)
where {,) denotes the scalar product, the bracket {,} is the extension of the usual Poisson brackets of classical mechanics, and [,] describes the dissipative behavior. Using the chain rule and the brackets defined above, the evolution equation of an arbitrary function A becomes
dA dt
-{A,E}+[A,S]
(14.79)
684
14. Nonequilibriumthermodynamicsapproaches
The conditions of L and M become clear with the following properties of the brackets: {A,B} = - { B , A }
(antisymetric) (Jacobi identity)
{A,{B,C} } + {B,{C,A} } + {C,{A,B} } = 0
(14.80)
These properties indicate that L is antisymmetric, and restrict the possible forms of the connection mechanisms for the structural variables. The antisymmetric requirements of L guarantee the consistency of L with the structure of the equivalent Poisson bracket. The following properties:
[A,B]=[B,A],
[A,A]_>0
(14.81)
require that M is symmetric (assuming that all the variables x have the same time-reversal parity) and definite positive which leads to ds/dt >- O. The symmetry o f M is directly related to Onsager's reciprocal rules. Using two generators, E and S, provides more flexibility in the choice of variables. The behavior of the variables x under space transformation determines the matrix L. The information related to the dynamics of material describes the friction matrix M which is related to the transport coefficients. The GENERIC formulation describes the consistency between the generalized entropy and the corresponding evolution equation of the system. It suggests how to generalize entropy of extended nonequilibrium thermodynamics containing the terms beyond the second order in the viscous pressure tensor. The GENERIC formulations may be applied to polymer solutions, emulsions and blends, and polymer melts (Ottinger and Grmela, 1997; Jou and Casas-Vfizquez, 2001). The rational nonequilibrium thermodynamics and GENERIC are compared by Muschik et al. (2000).
14.7
MATRIX MODEL
In the matrix model (Jongschaap, 1990), the global thermodynamic system is composed of two separate physical parts, which are called the environment and the intemal variables. For the polymer solutions, for example, the pressure tensor pv may be the internal variables, and the classical variables density, velocity, and internal energy are the environment variables. In the matrix model, the power supplied to the system is characterized by a set of controllable extemal forces Fex and rate variables F and is given by dF pe = Fex @ ~ dt
(14.82)
where Q is the full-contracted product. The internal subsystem is characterized by a set of variables x i and the fundamental equation of the rate of change of energy is
(14.83)
Pw = I I @ dx~ dt
where II is the thermodynamic force conjugated to the respective variables xi. The total dissipation rate is the difference between the rate of power supplied to the system and the internal rate of storage of energy dF dx i A - Pwe -Pw - Fex @ - - ~ - - I I @ dt
(14.84)
When there are no internal variables, the whole supplied power would dissipate. The matrix model is derived from Eq. (14.84)
kdt)
/
(14.85)
where A = - 1/z. Here, it is assumed that the variables x i and the dissipation A remain unchanged under a reversal of sign of the rate variables dF/dt. This is called the principle of macroscopic time reversal. The upper part in Eq. (14.85) represents the flux-force relations and the bottom part is the time evolution equations for the internal variables.
685
14.8 Internal variables
For polymer solutions, and taking the p, v, and u as environment variables and pv as internal variable, the dissipation becomes
(')
dp v
6 - - ( V v ) . pv + Tq.V ~ - H . ~
->0
(14.86)
dt
The first two terms on the right are the dissipation (entropy production times the absolute temperature), same as in the linear nonequilibrium thermodynamics. The last term is the contribution of the internal variables. The pv acts as an internal variable and as a rate variable. The evolution equation for pv is dp v
1
,lq~)
"r
~"
_ _ _- ev _ - . , (Vv)
dt
(14.87)
Jongschaap et al. (1994) provided detailed examples for rheological problems with the matrix method by using the configuration function as variables. 14.8
INTERNAL VARIABLES
The theories with internal variables provide detailed description of microstructure by introducing additional variables relevant to the microstructure of the system, and enlarge the domain of application of thermodynamics. The theories of internal variables are applied in rheology, dielectric, and magnetic relaxation where the structure of the macromolecules plays a relevant role. In the theories of internal variables, it is usual to propose purely relaxational equations for the internal variables and associate the additional variables with some structure of underlying molecules. If we assume the configuration tensor W = (RR) as an independent variable, the Gibbs equation is ds -
-~ du + P
d v - ol
W'W
(14.88)
Therefore, the time derivative of the entropy becomes
It)
(14.89)
Combining mass and energy balances dv
--=V.v
P dt
du
'
p--=-V.q-P dt
v'v-pU'V
(14.90)
with Eq. (14.89), we have p-~-=V. Tq
-q.V
(14.91)
T -
Here, q / T i s the entropy flux and the first term on the right is the entropy production. The simplest constitutive equations satisfying the requirement of the positive entropy production are 1 1
dt
-~o
1 )paW
(Vv) - Lll
(14.92)
p~W
Here, the first equation is the usual Fourier law, the second relates the viscous pressure tensor to the internal variable W, and the last is the evolution of the internal variable. The matrix of the transport coefficients Lij is positive definite
686
14. Nonequilibriumthermodynamicsapproaches
with L10 = -L01 due to Onsager-Casimir reciprocal rules. Jou and Casas-Vfizquez (2001) compared the various formulations and the extended nonequilibrium thermodynamics and underlined the connection between thermodynamics and dynamics. Rubi and Perez-Madrid (2001) derived some kinetic equations of the Fokker-Planck type for polymer solutions. These equations are based on the fact that processes leading to variations in the conformation of the macromolecules can be described by nonequilibrium thermodynamics. The extension of this approach to the mesoscopic level is called the mesoscopic nonequilibrium thermodynamics, and applied to transport and relaxation phenomena and polymer solutions (Santamaria-Holek and Rubi, 2003). REFERENCES A.N. Beris and S.J. Edwards, Thermodynamics of Flowing Fluids with Internal Microstructure, Oxford University Press, New York (1994). K.R. Diller (Ed.), Biotransport, Heat and Mass Transfer in Living Systems, 858. Annals of the New York Academy of Sciences, New York Academy of Sciences, New York (1998). P. Glansdorff and I. Prigogine, Thermodynamics Theory of Structure, Stability and Fluctuations, Wiley, New York (1971). M. Grmela, D. Jou and J. Casas-Vfizquez, J Chem. Phys., 108 (1998) 7937. M. Grmela and H.C. Ottinger, Phys. Rev. E, 56 (1997) 6620. R.J.J. Jongschaap, Rep. Prog. Phys., 53 (1990) 1. R.J.J. Jongschaap, K.H. de Haas and C.A.J. Damen, J. Rheol., 38 (1994) 769. D. Jou, Extended Irreversible Thermodynamics, Springer-Verlag, New York (1996). D. Jou and J. Casas-Vfizquez, J. Non-Newton. Fluid Mech., 96 (2001) 77. D. Jou, J. Casas-Vfizquez and M. Criado-Sancho, Physica A, 262 (1999) 69. V.A. Malyshev, J. Stat. Phys., 119 (2005) 997. D.C. Mikulecky, Comp. Chem., 19 (1994) 999. D.C. Mikulecky, Comp. Chem., 25 (2001) 369. W. Muschik, S. Gumbel, M. Kroger, H.C. Ottinger, Physica A, 285 (2000) 448. H.C. Ottinger and M. Grmela, Phys. Rev. E, 56 (1997) 6633. R. Paterson, Network Thermodynamics, in: E.E. Bitter, Ed., Membrane Structure and Function, Vol. 2, Wiley, New York (1980). L. Peusner, Studies in Network Thermodynamics, Elsevier, Amsterdam (1986). H.V. Westerhoff and K.V. Dam, Thermodynamics and Control of Biological Free-Energy Transduction, Elsevier, Amsterdam (1987).
REFERENCES FOR FURTHER READING N. Despireux and G. Lebon, J. Non-Newton. Fluid Mech., 96 (2001) 105. B.J. Edwards, J. Non-Equilb. Thermodyn., 27 (2002) 5. E. Hernandez-Lemus and L.S. Garcia-Colin, J. Non-Equilib. Therm., 31 (2006) 397. G. Lebon, T. Desaive and P. Dauby, J. Appl. Trans.ASME, 73 (2006) 16. W. Muschik, C. Papenfuss and H. Ehrentraut, J. Non-Newton. Fluid Mech., 96 (2001) 255. H.C. Ottinger, Beyond Equilibrium Thermodynamics, Wiley, New York (2003). J.M. Rubi and A. Perez-Madrid, Physica A, 298 (2001) 177. I. Santamaria-Holek and J.M. Rubi, Physica A, 326 (2003) 284. ST Serdyukov, Phys. Lett. A, 324 (2004) 262.
APPENDIX APPENDIX A Tensors Scalars are specified by a single numerical value. Vectors come with directions as well as numerical values. In a threedimensional space, a tensor of rank n is determined by 3" elements. A scalar is a tensor with the rank zero, hence 30 = 1. A vector is tensor with the rank 1, hence 31 = 3. For a tensor with n = 2, we have 32 _ 9 elements. Differentiation of a tensor with respect to a scalar does not change its rank. The spatial differentiation of a tensor raises its rank by unity, and identical to multiplication by the vector V, called del or Hamiltonian operator or the nabla
V-
Ox
The gradient of a scalar field a is a vector Oa
grad a = m = Va Ox The derivative of a scalar a with respect to a vector is a vector. The gradient of a vector field v is a tensor of rank two 0V div v = m = Vv 0x
W h e n contraction is performed once (summation over repeated indices), the divergence is obtained instead of the gradient. The divergence of a vector field v is a scalar O divv---.vOx
V.v
The divergence of a tensor field T is a vector divT---.0 T=V.T Ox The Laplace operator or Laplacian is a scalar
V 2 - V. V - div grad -
0 ~
°
0 ~
Ox Ox Nonequilibrium thermodynamics often uses the Gauss-Ostrogradsky theorem, which states that the flux of a vector through a surface a is equal to the volume integral of the divergence of the vector v for the space of volume Vbounded by that surface
f v. da - f div vdV - I V. vdV a
l~
f"
Appendix
688 APPENDIX B
Table B1 Lennard-Jones (6-12) potential parameters and critical properties Species
MW
MW O)
Lennard-Jones parameters or (A)
e/K (K)
H2
2.016
-0.216
2.915
38.0
He
4.003
-0.390
2.576
10.2
Ne
20.183
2.789
35.7
Critical constants
Tc (K)
Pc atm
Vc (cm 3 mol)
33.3
12.80
65.0
34.7
2.26
57.8
25.4
5.26 44.5
26.9
/xc X 106 (g/(cm s))
41.7
156
kc X 106 (cal/(cm s K)) 79.2
Ar
39.948
0.000
3.432
1224.4
150.7
48.0
75.2
264
71.0
Kr
83.80
0.000
3.675
170.0
209.4
54.3
92.2
396
49.4
Xe
131.30
0.000
4.009
234.7
289.8
58.0
118.8
490
40.2
Air
28.9,7
0.035
3.617
97.0
132
36.4
86.6
193
90.8
N2
28.01
0.038
3.667
99.8
126.2
33.5
90.1
180
02
32.00
0.022
3.433
154.4
49.7
74.4
250.0
113
CO
28.01
0.048
3.590
110
132.9
34.5
93.1
190
CO2
44.01
0.224
3.996
190
304.2
72.8
94.1
343
86.8 105.3 86.5 122
NO
30.1
0.583
3.470
119
180
64
57
258
118.2
N20
44.01
0.141
3.879
220
309.7
71.7
96.3
332
131
SO2
64.06
0.245
4.026
363
430.7
77.8
F2
38.00
3.653
112
-
-
C12
70.91
4.115
357
417
Br2
159.82
4.268
520
584
102
I2
253.81
4.982
550
800
-
CH 4
16.04
C2H2
26.04
C2H 4
28.05
C2H 6
30.07
0.069
0.012
76.1
122
411
98.6
124
420
97
144
-
-
3.780
154
191.I
45.8
98.7
159
158
4.114
212
308.7
61.6
112.9
237
-
0.087
4.228
216
282.4
50.0
124
215
-
0.100
4.388
232
305.4
48.2
148
210
203
C3H 4
40.06
4.742
261
394.8
-
C3H 6
42.08
0.140
4.766
275
365.0
45.5
181
233
-
C3H8
44.10
0.152
4.934
273
369.8
41.9
200
228
-
n-C4Hlo
58.12
0.200
5.604
304
425.2
37.5
255
239
-
i-C4Hlo
58.12
0.181
5.393
295
408.1
36.0
263
239
-
n-CsH~2
72.15
0.252
5.850
326
469.5
33.2
311
238
-
i-CsH12 C(CH3) 4
72.15 72.15
5.812 5.759
327 312
460.4 433.8
33.7 31.6
306 304 370 432
248 254
-
492
n-C6HI4
86.18
0.301
6.264
342
507.3
29.7
n-CvH16 n-CsH18
100.20 114.23
0.350 0.400
6.663 7.035
352 361
540.1 568.7
27.0 24.5
259
-
n-C9H20 Cyclohexane
128.26 84.16
0.444 0.2t0
7.463 6.143
351 313
594.6 553
22.6 40.0
548
265
-
308
284
-
Benzene
78.11
0.210
5.443
387
562.6
48.6
260
312
-
CH3C1
50.49
0.153
4.151
355
416.3
65.9
143
338
-
CH2C12
84.93
0.199
4.748
398
510
60
CHC13
119.38
0.222
5.389
340
536.6
54
240
410
-
CC14
153.82
0.193
5.947
323
556.4
45.0
276
413
-
C2N2
52.04
4.361
349
400
59
COS
60.07
CS2
76.14
CClzF 2
0.111
120.92
4.130
336
378
61
4.483
467
552
78
170
404
-
5.116
280
384.7
39.6
218
-
-
Table B2 Collision integrals for predicting transport properties of gases at low densities KT/e or KT/gAB
~
= g~k for viscosity and
~'~D, AB = for diffusivity
thermal conductivity
KT/e or KT/SAB
f~. = fl~ for viscosity and
~-~D,AB = f o r diffusivity
thermal conductivity
0.30
2.840
2.469
2.7
1.0691
0.9782
0.35
2.676
2.468
2.8
1.0583
0.9682
0.40
2.531
2.314
2.9
1.0482
0.9588
(Contd:)
Appendix
689
Table B2 ( C o n t i n u e d )
KT/s or
{)u = ~k for viscosity and
KT/SAB
thermal conductivity
D.D.AB = for diffusivity
KT/e, or KT/eAB
flu = fla for viscosity and
~-'~D.AB =
for diffusivity
thermal conductivity
0.45
2.401
2.182
3.0
1.0388
0.9500
0.50
2.284
2.066
3.1
1.0300
0.9418
0.55
2.178
1.965
3.2
1.0217
0.9340
0.60
1.084
1.877
3.3
1.0139
0.9267
0.65
1.999
1.799
3.4
1.0066
0.9197
0.70 0.75
1.922 1.853
1.729
3.5 3.6
0.9996 0.9931
0.9131 0.9068
0.80 0.85
1.790 1.734
0.90
1.667 1.612
3.7
0.9868
0.9008
1.562
3.8
0.9809
0.8952
1.682
1.517
3.9
0.9753
0.8897
0.95
1.636
1.477
4.0
0.9699
0.8845
1.00
1.593
1.440
4.1
0.9647
0.8796
1.05
1.554
1.406
4.2
0.9598
0.8748
1.10
1.518
1.375
4.3
0.9551
0.8703
1.15
1.485
1.347
4.4
0.9506
0.8659
1.20
1.455
1.320
4.6
0.9462
0.8617
1.25
1.427
1.296
4.6
0.9420
0.8576
1.30
1.401
1.274
4.7
0.9380
0.8537
1.35
1.377
1.253
4.8
0.9341
0.8499
1.40
1.234
4.9
0.9304
0.8463
1.45
1.355 1.334
1.216
5.0
0.9268
0.8428
1.50 1.55
1.315 1.297
1.199
6.0
0.8962
0.8129
7.0
0.8727
0.7898
1.60
1.280
1 183 1 168
8.0
0.8538
0.7711
1.65
1.264
1 154
9.0
0.8380
0.7555
1.70
1.249
1141
10.0
0.8244
0.7422
1.75
1.235
1 128
12.0
0.8018
0.7202
1.80
1.222
1117
14.0
0.7836
0.7025
1.85
1.209
1 105
16.0
0.7683
0.6878
1.90 1.95
1 198
1.095
18.0
0.7552
0.6751
1 186
20.0
0.7436
0.6640
2.00
1 176
1.085 1.075
25.0
0.7198
0.6414
2.10 2.20 2.30 2.40 2.50
1 156
1.058 1.042 1.027 1.013 1.0006 0.9890
30.0
0.7010
0.6235
35.0 40.0 50.0 75.0
0.6854 0.6723 0.6510 0.6140
0.6088 0.5964 0.5763 0.5415
100.0
0.5887
0.5180
2.60
1 138 1 122 1.107 1.0933 1.0807
Table B3 Heat capacities of gases in the ideal-gas state a C'//R = a + b E + c r 2 + dT -2 T(K) from 298 to Tmax Chemical species
Tm,,~
C'/,.2~)sK/R
a
10 3 b
106 c
1O- 5 d
Paraffins Methane
CH 4
1500
4.217
1.702
9.081
Ethane
C2H~
1500
6.369
1.131
19.225
-2.164 - 5.561
Propane
C3Hs
1500
9.011
1.213
28.785
- 8.824 - 11.402
n-Butane
C4H~o
1500
11.928
1.935
36.915
Iso-butane
C4H~o
1500
11.901
1.677
37.853
- 11.945
n-Pentane
CsH~ 2
1500
14.731
2.464
45.351
- 14.111
n-Hexane
C~,H j4
1500
17.350
3.025
53.722
- 16.791
n-Heptane
CTH~(~
1500
20.361
3.570
62.127
- 19.486
CsH~ ~
1500
23.174
4.108
70.567
-22.208
C2H4
1500
5.325
1.424
14.394
- 4.392
n-Octane 1-Alkenes Ethylene
(Contd.)
Appendix
690 Table B3 ( C o n t i n u e d ) Tmax
Cp,298K/R
a
103 b
C3H6 C4H 8 CsH10 C6H12 C 7H14 C8H16
1500 1500 1500 1500 1500 1500
7.792 10.520 13.437 16.240 19.053 21.868
1.637 1.967 2.691 3.320 3.768 4.324
22.706 31.630 39.753 48.189 56.588 64.960
-6.915 -9.873 -12.447 -15.157 -17.847 -20.521
C2H40 C2H2 C 6H6 C4H6 C6H 12 C2H60 CsH 10 C2H40 CH20 CH40 CsHs CyH8
1000 1500 1500 1500 1500 1500 1500 1000 1500 1500 1500 1500
6.506 5.253 10.259 10.720 13.121 8.948 15.993 5.784 4.191 5.547 15.534 12.922
1.693 6.132 - 0.206 2.734 - 3.876 3.518 1.124 -0.385 2.264 2.211 2.050 0.290
17.978 1.952 39.064 26.786 63.249 20.001 55.380 23.463 7.022 12.216 50.192 47.052
-6.158
NH3 Br2 N20 NO NO 2 N204 O2 SO2 SO3 H20 CO CO2 CS2 C1z H2 HzS HC1
2000 1800 3000 2000 2000 2000 2000 2000 2000 2000 2000 2500 2000 1800 3000 3000 2300 2000
3.509 4.269 4.337 4.646 3.590 4.447 9.198 3.535 4.796 6.094 4.038 3.507 4.467 5.532 4.082 3.468 4.114 3.512
3.355 3.578 4.493 5.328 3.387 4.982 11.660 3.639 5.699 8.060 3.470 3.376 5.457 6.311 4.442 3.249 3.931 3.156
0.575 3.020 0.056 1.214 0.629 1.195 2.257 0.506 0.801 1.056 1.450 0.557 1.045 0.805 0.089 0.422 1.490 0.623
Chemical species Propylene 1-Butene 1-Pentene 1-Hexene 1 -Heptene 1 -Octene Organics Acetaldehyde Acetylene Benzene 1,3-Butadiene Cyclohexane Ethanol Ethylbenzene Ethylene oxide Formaldehyde Methanol Styrene Toluene Inorganics Air Ammonia Bromine Nitrous oxide Nitric oxide Nitrogen dioxide Dinitrogen tetroxide Oxygen Sulfur dioxide Sulfur trioxide Water Carbon monoxide Carbon dioxide Carbon disulfide Chlorine Hydrogen Hydrogen sulfide Hydrogen chloride
106 c
10-5d
- 1.299 -13.301 -8.882 -20.928 -6.002 -18.476 -9.296 -1.877 -3.450 -16.662 -15.716 -0.016 -0.186 -0.154 -0.928 0.014 -0.792 -2.787 • -0.227 -0.1015 -2.028 0.121 -0.031 -1.157 -0.906 -0.344 0.083 -0.232 0.151
aH.M. Spencer, Ind. Eng. Chem., 40 (1948) 2152; K.K. Kelly, US. Bur. Mines Buff., (1960) 584; L.B. Pankratz, U.S. Bur. Mines Buff., (1982) 672.
Table B4 Heat capacities of solids a C°p/R = a + bT 4- d T -2 T(K) from 298 to/max Chemical species
Tmax
C°p,298K/R
a
103 b
10-5 d
CaO CaCO3 Ca(OH)2 CaC2 CaCI2 C (graphite) Cu CuO Fe(O) FezO3 Fe304
2000 1200 700 720 1055 2000 1357 1400 1043 960 850
5.058 9.848 11.217 7.508 8.762 1.026 2.959 5.087 3.005 12.480 18.138
6.104 12.572 9.397 8.254 8.646 1.771 2.677 5.780 -0.111 11.812 9.594
0.443 2.637 5.435 1.429 1.530 0.771 0.815 0.973 6.111 9.697 27.112
- 1.047 - 3.120 - 1.042 -0.302 -0.867 0.035 -0.874 1.150 - 1.976 0.409
(Contd.)
Appendix Table B4
691
(Continued)
Chemical species
Tm~x
C'/,.29sx/R
a
103 b
10- 5 d
FeS
411
6.573
2.612
13.286
I2 LiC1
386.8 800
6.929 5.778
6.481 5.257
1.502 2.476
NHzC1
458
10.741
5.939
16.105 4.688
Na
-0.193
371
3.386
1.988
NaC1
1073
6.111
5.526
1.963
NaOH
566
7.177
0.121
16.316
NaHCO3
400
10.539
5.128
18.148
S (rhombic)
368.3
3.748
4.114
- 1.728
-0.783
SiO2 (quartz)
847
5.345
4.871
5.365
- 1.001
aK.K. Kelly,
1.948
US. But: Mines Bull., (1960) 584; L.B. Pankratz, US. Bul: Mines Bull., (1982) 672.
Table B5 Heat capacities of liquids a C~/R= a + bT + cT 2 T from 273.15 to 373.15 K Chemical species
C~_9~x/R
a
Ammonia
103 b
106 c
9.718
22.626
- 100.75
192.71
23.070
15.819
29.03
- 15.80
Benzene
16.157
-0.747
67.96
-37.78
1,3-Butadiene
14.779
22.711
- 87.96
205.79
Carbon tetrachloride
15.751
21.155
- 48.28
101.14
Chlorobenzene
18.240
11.278
32.86
-31.90
Aniline
Chloroform
13.806
19.215
-42.89
83.01
Cyclohexane
18.737
-9.048
141.38
- 161.62
Ethanol
13.444
33.866
- 172.60
349.17
Ethylene oxide
10.590
21.039
- 86.41
172.28
9.798
13.431
- 51.28
131.13
n-Propanol Sulfur trioxide
16.921 30.408
41.653 -2.930
-210.32 137.08
427.20 - 84.73
Toluene
18.611
15.133
6.79
16.35
9.069
8.712
1.25
- 0.18
Methanol
Water aJ.W. Miller, Jr., G.R. Schorr and C.L. Yaws,
Chem. Eng., 83 (1976) 129.
Table B6 Properties of some c o m m o n liquids a Substance
Boiling at 1 atm Tb (°C)
Ammonia
- 33.3
~ff-/, (kJ/kg) 1357
Freezing
Liquid properties
T~-(°C)
Z~r-/v (kJ/kg)
T (°C)
- 77.7
322.4
- 33.3
682
4.43
-20
665
4.52
639
4.60
602 1394
4.80 1.14
0 Argon Benzene Brine (20% NaC1 by mass) n-Butane Carbon dioxide
-185.9
161.6
28
5.5
126
20
879
1.72
-
20
1150
3.11
80.3
-0.5
601
2.31
0
298
0.59 2.46
80.2
394 -
-17.4
385.2
- 138.5
-0.5
230.5 (at 0°C)
C/, (kJ/(kg °C))
-189.3
103.9 -78.4
25 -185.6
p (kg/m 3)
-56.6
Ethanol
78.2
838.3
- 114.2
109
25
783
Ethyl alcohol
78.6
855
- 156
108
20
789
2.84
- 10.8
181.1
20
1109
2.84
18.9
200.6
20
1261
Ethylene glycol Glycerine
198.1
800.1
179.9
974
Helium
-268.9
22.8
Hydrogen
- 252.8
445.7
- 259.2
2.32
-
-268.9
146.2
22.8
59.5
- 252.8
70.7
10.0
(Contd.)
Appendix
692 Table B6
(Continued)
Substance
Boiling at 1 atm Tb (°C)
M-/v (kJ/kg)
Freezing
Liquid properties
Tf (°C)
M-/v (kJ/kg)
T (°C)
p (kg/m 3)
367.1 251 294.7 510.4
- 160 - 24.9 -38.9 - 182.2
105.7 11.4 58.4
- 11.7 20 25 - 161.5 - 100 25 - 195.8 -160 20 25 - 183 20 -42.1 0 50 -50 -26.1 0 25 0 25 50 75 100
593.8 820 13560 423 301 787 809 596 703 910 1141 640 581 529 449 1443 1374 1294 1206 1000 997 988 975 958
Iso-butane Kerosene Mercury Methane
- 11.7 204-293 356.7 - 161.5
Methanol Nitrogen
64.5 - 195.8
1100 198.6
-97.7 -210
99.2 25.3
124.8
306.3
-57.5
180.7
-218.8
13.7
- 187.7
80.0
Octane Oil (light) Oxygen Petroleum Propane
Refrigerant-134a
Water
ay.A. Cengel and M.A. Boles,
- 183 -42.1
-26.1
100
212.7 230-384 427.8
216.8
2257
-96.6
-
0.0
333.7
Cp (kJ/(kg °C)) 2.28 2.00 0.139 3.49 5.79 2.55 2.06 2.97 2.10 1.80 1.71 2.0 2.25 2.53 3.13 1.23 1.27 1.34 1.42 4.23 4.18 4.18 4.19 4.22
Thermodynamics.An EngineeringApproach, 4th ed., McGraw-Hill, New York (2002).
Table B7 Standard enthalpies and Gibbs energies of formation at 298.15 K Chemical species
State
M-/F (J/mol)
AGF (J/mol)
-46,110
- 16,450 -26,500 -64,900 - 1,128,790 - 748,100 - 8,101,900
Ammonia Ammonia Calcium carbide Calcium carbonate Calcium chloride Calcium chloride Calcium chloride Calcium hydroxide Calcium hydroxide Calcium oxide Carbon dioxide Carbon monoxide Hydrochloric acid Hydrogen cyanide
NH3 NH3 CaC 2 CaCO3 CaClz CaCI2 CaClz.6H20 Ca(OH)2 Ca(OH)2 CaO CO2 CO HC1 HCN
(g) (aq) (s) (s) (s) (aq) (s) (s) (aq) (s) (g) (g) (g) (g)
Hydrogen sulfide Iron oxide Iron oxide (hematite) Iron oxide (magnetite) Iron sulfide (write) Lithium chloride Lithium chloride Lithium chloride Lithium chloride
HzS FeO Fe203 Fe304 FeS2 LiC1 LiC1.H20 LiC1.2HzO LiC1.3H20
(g) (s) (s) (s) (s) (s) (s) (s) (s)
- 59,800 - 1,206,920 - 795,800 -2,607,900 -986,090 -635,090 - 393,509 - 110,525 -92,307 135,100 -20,630 -272,000 -824,200 - 1,118,400 - 178,200 -408,610 - 712,580 - 1,012,650 - 1,311.300
- 898,490 - 868,070 -604,030 - 394,359 137,169 -95,299 124,700 - 33,560 -742,200 - 1,015,400 - 166,900
(Contd.)
Appendix Table B7
693
(Continued)
Chemical species
State
Nitric acid
HNO~
Nitric acid Nitrogen oxides
HNO3 NO NO2 N20 N204 Na2CO3 Na2CO3.10H20 NaC1 NaC1 NaOH NaOH SO2 SO3 SOs H2SO 4 H2SO 4 H20 H20
(aq) (g) (g) (g) (g) (s) (s) (s) (aq) (s) (aq) (g) (g) (1) (1) (aq) (g) (1)
CH 4 C2H(, C3H~ C4Hi0 CsHI2 C6H14 CTHI~, CsHIs
Sodium carbonate Sodium carbonate Sodium chloride Sodium chloride Sodium hydroxide Sodium hydroxide Sulfur dioxide Sulfur trioxide Sulfur trioxide Sulfuric acid Sulfuric acid Water Water
(1)
~¢-/~ (J/mol) - 174,100 90,250 33,180 82,050 9,160 - 1,130,680 -4,081,320 -411,153 - 425,609 - 296,830 - 395,720 -441,040 - 813,989
AG~: (J/mol) - 80,710 - 111,250 86,550 51,310 104,200 97,540 - 1,044,440 -384,138 - 393,133 - 379,494 -419,150 - 300,194 - 371,060
- 241,818 - 285,830
-690,003 - 744,530 - 228,572 - 237.129
(g) (g) (g) (g) (g) (g) (g) (g)
- 74,520 - 83,820 - 104,680 - 125,790 - 146,760 - 166,920 - 187,780 - 208,750
- 50,460 - 31,855 -24,290 - 16,570 - 8,650 150 8,260 16,260
C2H4 C3H~ C4H ~ CsH~0 C6H12 C7H14
(g) (g) (g) (g) (g) (g)
52,510 19,710 - 540 -21,280 -41,950 -62,760
68,460 62,205 70,340 78,410 86,830
1,2-Ethanediol
C2H40 C2H402 C2H2 C6H~ C6H~, C4H(~ C6HI~ C6HI: C 2H(~O2
(g) (1) (g) (g) (1) (g) (g) (1) (1)
- 166,190 -484,500 227,480 82,930 49,080 109,240 - 123,140 - 156,230 - 454,800
- 128,860 - 389,900 209,970 129,665 124,520 149,795 31,920 26,850 - 323,080
Ethanol Ethanol Ethylbenzene Ethylene oxide Formaldehyde
C2H(~O C2H~,O CsHi0 C2H40 CH20
(g) (1) (g) (g) (g)
- 235,100 -277,690 29,920 -52,630 - 108,570
- 168,490 - 174,780 130,890 - 13,010 - 102,530
Methanol Methanol Methylcyclohexane Methylcyclohexane Styrene Toluene Toluene
CH40 CH40 C7H14 CTHt4 CsH~ CTHs CTH~
(g) (1) (g) (1) (g) (g) (1)
-
- 161,960 - 166,270 27,480 20,560 213,900 122,050 113,630
Paraffins Methane Ethane Propane n-Butane n-Pentane n-Hexane n-Heptane n-Octane
1-Alkenes Ethylene Propylene 1-Butene 1-Pentene 1-Hexene 1-Heptene Miscellaneous organics Acetaldehyde Acetic acid Acetylene Benzene Benzene 1,3-Butadiene Cyclohexane Cyclohexane
200,660 238,660 154,770 190,160 147,360 50,170 12,180
694
Appendix
Table B8 Selected state properties a
Compound (state) Glycerol (aq) Triglycerol-palmitate (s) Hydrogen H2 (g) Oxygen 02 (g) Oxygen 02 (aq) Water (1) Water (g) Hydrogen ion (aq) Hydroxyl ion (aq) Oxonium ion H 2 0 + (aq) Phosphoric acid (aq) Dihydrogen phosphate ion (aq) Hydrogen phosphate ion (aq) Phosphate ion (aq) Carbamide (urea) (aq) Carbon dioxide (g) Carbon dioxide (aq) Acetic acid (1) Acetic acid (aq) Acetate anion (aq) Ethanol (1) Ethanol (aq) Carbonic acid (aq) Bicarbonate ion (aq) Lactic acid (1) Lactic acid (aq) Lactate anion (aq) Pyruvic acid (aq) Pyruvate anion (aq) a-D-glucose (s) a-D-glucose (aq) Sucrose (s) Sucrose (aq) Palmitic acid (s) Glycerol (1)
M-/F (kJ/mol)
2~GF (kJ/mol)
S (J/(mol K))
Cp (J/(mol
K))
-488 - 2457 0 0 - 10 -286 -242 0 -230 -286
- 1277 - 319 -393 -413 -487 - 385 -485 -278 -266 -699 -690 -675 -687
- 1275 - 1264 - 1221 -882 -669
0 0 17 -237 -229 0 - 157 -237 - 1142 - 1130 - 1089 - 1019 - 204 -394 -386 -392 - 396 -369 - 175 - 182 -623
131 205 115 70 189 0 - 11 70
29 29 75 34 0 - 149 75
-222 174 214 121 160
37 123
87 161 222 191
- 518 -540 - 518 -489 -475 -910 -917 - 1544 - 1552 -305 -477
114
192
211 151 360
219 425
452
aGarby and Larsen (1995).
Table B9 Approximate standard reaction enthalpy and standard reaction Gibbs energy for some selected reactions at standard state T = 25°C, P = 1 atm a Reaction ATP and glucose reactions ATP ~ ADP + P (310K, pH 7.0, l m m o l Mg ++ Glucose to ethanol 298 K Glucose to ethanol 308 K Glucose to lactic acid (298 K, pH 7.0) Oxidation reactions (kJ/mol substrate) Ethanol 293 K Acetic acid 293 K Glucose 298 K, all components (aq) Glucose 310 K, all components (aq) Lactose Palmitic acid, all components (aq) Glycerol Triglycerides, average Protein urea, per gram, average
AH~ (kJ/mol)
AG~ (kJ/mol)
-20 -67 -66
-31 -235
-
100
-
198
-1371
-876 -2870 -2867 -5646 -9982 -1659 -34300
-2930 -9791
-17
(Contd.)
Appendix Table B9
695
(Continued)
Reaction
M-/I (kJ/mol)
Protein ---, a m m o n i a , per g r a m , average
AG,~ (kJ/mol)
- 19
Lactic acid
- 1363
Alanine
- 1621
A l a n i n e --, urea
- 1303
Urea
-634
G l u t a m i c acid --, urea
- 1930
Citric acid
- 1986
P r o t o n dissociation reactions N e u t r a l i z a t i o n by intracellular buffers
- 25
Acetic acid
-0.4
27
G l u t a m i c acid
25
Water, 298 K
56
D i s s o c i a t i o n o f strong electrolytes S o d i u m chloride
-407
"Garby and L a r s e n (1995).
APPENDIX C Table C1 Parameters for the thermal conductivity of alkanes in chloroform k(W/(mK))= Solute ( c o m p o n e n t 1)
Ei4oaixli
ao
a,
a2
a3
n-Hexane
0.11076
- 0 . 0 3 777
0.04917
0.01362
-0.02051
0.22
n-Heptane
0.11079
- 0.04462
0.11050
- 0.07782
0.02156
0.05
a4
%E ~
n-Octane
0.11081
-0.04457
0.14569
-0.14027
0.05211
0.16
3-Methylpentane
0.11079
-0.06830
0.10737
-0.04171
-0.00121
0.11
2,3-Dimethylpentane
0.11080
-0.04081
0.04925
-0.00950
-0.00453
0.17
2,2,4-Trimethylpentane
0.11080
-0.04350
0.01113
0.05298
-0.03542
0.08
"E = ~-Z,=, 1 ,,v
( f ....,~ - .[i.~xpt#;.~,,.
Table C2 Parameters for the mutual diffusion coefficients of alkanes in chloroform Solute ( c o m p o n e n t 1)
ao
al
D(10-9 m 2/s)= ~4i=o aixli
a2
a3
%E
n-Hexane
2.4394
- 1.0243
3.9930
-0.1670
-0.7710
n-Heptane
2.2368
- 0.7534
2.3321
2.7348
- 3.0324
0.08
n-Octane
2.0367
- 1.0923
3.2118
0.6468
- 1.7149
0.06 0.08
0.14
3-Methylpentane
2.2851
-0.2589
2.7260
- 1.3769
1.2063
2,3-Dimethylpentane
2.0936
- 1.1928
8.4556
- 11.7748
6.0184
0.08
2,2,4-Trimethylpentane
1.9529
-0.4573
1.2118
1.4582
- 1.0782
0.00
3 i Table C3 Parameters for the heats of transport of alkanes in chloroform - Q ( ' * ( k J / k g ) = ~_~i=oaiX~
Solute ( c o m p o n e n t 1)
a0
al
a2
n-Hexane
69.7765
- 89.6873
127.5539
- 78.3904
0.21
n-Heptane
63.8333
- 71.3086
68.3994
- 31.5108
0.21
n-Octane
61.3523
- 79.0694
110.1027
- 73.4022
0.10
3-Methylpentane
66.8969
- 75.9630
90.5502
-47.8497
0.08
2,3-Dimethylpentane
62.2974
- 71.8306
80.2944
- 47.6851
0.14
2,2,4-Trimethylpentane
59.7674
- 76.4213
99.8976
- 66.2341
0.45
%E
696
Appendix
4 Table C4 Parameters for the thermal conductivity of alkanes in carbon tetrachloride k(W/(mK)) = ~/--,;--0 aix]i
Solute (component 1)
a0
al
a2
a3
a4
%E
n-Heptane n-Octane 3-Methylpentane 2,3-Dimethylpentane 2,2,4-Trimethylpentane
0.0.9977 0.09979 0.09980 0.09969 0.09981
- 0 . 0 i 821 -0.04871 -0.06310 -0.03883 -0.07541
-0.02666 0.16253 0.15139 0.07447 0.18953
0.15875 -0.14561 -0.12861 -0.02811 -0.21637
-0.09150 0.05612 0.04580 0.0 0.10198
0.05 0.01 0.01 0.11 0.02
Table C5 Parameters for the mutual diffusion coefficients of alkanes in carbon tetrachloride D(10 -9 m2/s)-- Z4i = 0 Solute (component 1) n-Hexane n-Heptane n-Octane 3-Methylpentane 2,3-Dimethylpentane 2,2,4-Trimethylpentane
ao
al
1.5886 t.5027 1.4127 1.4940 1.3964 1.2280
1.9757 1.4798 0.3.903 1.7198 0.9422 0.4974
a2
a3
- 2.7016 -0.5978 2.3090 - 1.3199 1.1393 2.3803
6.6258 1.9214 - 1.0357 4.6769 -0.7047 - 1.9014
aixli
a4
%E
- 3.6101 - 1.1210 -0.5235 -2.5842 0.4628 0.4919
3
0.08 0.08 0.15 0.06 0.02 0.01
i
Table 06 Parameters for the heats of transport of alkanes in carbon tetrachloride -Q('* (10-9 (kJ/kg)= ~,;--oaiX~ Solute (component 1)
a0
al
a2
a3
%E
n-Hexane n-Heptane n-Octane 3-Methylpentane
68.5474 65.8806 63.0431 70.7849
-90.6118 - 85.9901 - 80.5948 - 122.0631
118.5839 111.5736 83.2332 221.6t95
-63.0289 -69.6799 - 50.3668 - 138.4118
0.01 0.01 0.13 0.07
2,3-Dimethylpentane 2,2,4-Trimethylpentane
61.3236 58.4977
- 75.3408 - 106.5203
88.1979 t 79.0495
- 55.9357 - 122.3974
0.40 0.01
APPENDIX D
Table D1 Saturated water-temperature table Temperature, T (°C)
Sat. press., /:'sat
Specific volume (m3/kg) Sat. liquid, VI
Sat. vapor, Vg
Internal energy (kJ/kg)
Enthalpy (kJ/kg)
Sat. liquid, U~
Sat. vapor, Ug
Sat. liquid, HI
Sat. vapor, Hg
Entropy (kJ/(kg K)) Sat. liquid, S~
Sat. vapor, Sg
(kPa) 0.01 5
0.6113 0.8721
0.001000 0.001000
206.14 147.12
0.0 20.97
2375.3 2382.3
0.01 20.98
2501.4 2510.6
0.000 0.0761
9.1562 9.0257
10 15
1.2276 1.7051
0.001000 0.001001
106.38 77.93
42.00 62.99
2389.2 2396.1
42.01 62.99
2519.8 2528.9
0.1510 0.2245
8.9008 8.7814
20 25 30 35
2.339 3.169 4.246 5.628
0.001002 0.001003 0.001004 0.001006
57.79 43.36 32.89 25.22
83.95 104.88 125.78 146.67
2402.9 2409.8 2416.6 2423.4
83.96 I04.89 125.79 146.68
2538.1 2547.2 2556.3 2565.3
0.2966 0.3674 0.4369 0.5053
8.6672 8.5580 8.4533 8.3531
40
7.384
0.001008
19.52
167.56
2430.1
167.57
2574.3
0.5725
8.2570
45 50 55 60
9.593 12.349 15.758 19.940
0.001010 0.001012 0.001015 0.001017
15.26 12.03 9.568 7.671
188.44 209.32 230.21 251.I1
2436.8 2443.5 2450.1 2456.6
188.45 209.33 230.23 251.13
2583.2 2592.1 2600.9 2609.6
0.6.387 0.7038 0.7679 0.8312
8.1648 8.0763 7.9913 7.9096
65
25.03
0.001020
6.197
272.02
2463.1
272.06
2618.3
0.8935
7.8310
70 75 80
31.19 38.58 47.39
0.001023 0.001026 0.001029
5.042 4.131 3.407
292.95 313.90 334.86
2469.6 2475.9 2482.2
292.98 313.93 334.91
2626.8 2635.3 2643.7
0.9549 1.0155 1.0753
7.7553 7.6824 7.6122
(Contd.)
Appendix Table D1
(Continued)
Temperature, T (°C)
Sat. press., P~,
Specific volume (m3/kg) Sat. liquid, V~
85 90 95
697
57.83 70.14 84.55
0.001033 0.001036 0.001040
Sat. vapor,
Internal energy (kJ/kg)
V,:
2.828 2.361 1.982
Enthalpy (kJ/kg)
Entropy (kJ/(kg K))
Sat. liquid, U~
Sat. vapor, ~
Sat. liquid, HI
Sat. vapor, ~
Sat. liquid, S~
Sat. vapor, Sg
355.84 376.85 397.88
2488.4 2494.5 2500.6
355.90 376.92 397.96
2651.9 2660.1 2668.1
1.1343 1.1925 1.2500
7.5445 7.4791 7.4159
418.94 440.02 461.14 482.30 503.50 524.74 546.02 567.35 588.74 610.18 631.68 653.24 674.87 696.56 718.33 740.17 762.09 784.10 806.19 828.37 850.65 873.04 895.53 918.14 940.87 963.73 986.74 1009.89 1033.21 1056.71 1080.39 1104.28 1128.39 1152.74 1177.36 1202.25 1227.46 1253.00 1278.92 1305.2 1332.0 1359.3 1387.1 1415.5 1444.6 1505.3 1570.3 164.19 1725.2 1844.0 2029.6
2506.5 2512.4 2518.1 2523.7 2529.3 2534.6 2539.9 2545.0 2550.0 2554.9 2559.5 2564.1 2568.4 2572.5 2576.5 2580.2 2583.7 2587.0 2590.0 2592.8 2595.3 2597.5 2599.5 2601.1 2602.4 2603.3 2603.9 2604.1 2604.0 2603.4 2602.4 2600.9 2599.0 2596.6 2593.7 2590.2 2586.1 2581.4 2576.0 2569.9 2563.0 2555.2 2546.4 2536.6 2525.5 2498.9 2464.6 2418.4 2351.5 2228.5 2029.6
419.04 440.15 461.30 482.48 503.71 524.99 546.31 567.69 589.13 610.63 632.20 653.84 675.55 697.34 719.21 741.17 763.22 785.37 807.62 829.98 852.45 875.04 897.76 920.62 943.62 966.78 990.12 1013.62 1037.32 1061.23 1085.36 1109.73 1134.37 1159.28 1184.51 1210.7 1235.99 1262.31 1289.07 1316.3 1344.0 1372.4 1401.3 1431.0 1461.5 1525.3 1594.2 1670.6 1760.5 1890.5 2099.3
2676.1 2683.8 2691.5 2699.0 2706.3 2713.5 2720.5 2727.3 2733.9 2740.3 2746.5 2752.4 2758.1 2763.5 2768.7 2773.6 2778.2 2782.4 2786.4 2790.0 2793.2 2796.0 2798.5 2800.5 2802.1 2803.3 2804.0 2804.2 2803.8 2803.0 2801.5 2799.5 2796.9 2793.6 2789.7 2785.0 2779.6 2773.3 2766.2 2758.1 2749.0 2738.7 2727.3 2714.5 2700.1 2665.9 2622.0 2563.9 2481.0 2332.1 2099.3
1.3069 1.3630 1.4185 1.4734 1.5276 1.5813 1.6344 1.6870 1.7391 1.7907 1.8418 1.8925 1.9427 1.9925 2.0419 2.0909 2.1396 2.1879 2.2359 2.2835 2.3309 2.3780 2.4248 2.4714 2.5178 2.5639 2.6099 2.6558 2.7015 2.7472 2.7927 2.8383 2.8838 2.9294 2.9751 3.0208 3.0668 3.1130 3.1594 3.2062 3.2534 3.3010 3.3493 3.3982 3.4480 3.5507 3.6594 3.7777 3.9147 4.1106 4.4298
7.3549 7.2958 7.2387 7.1833 7.1296 7.0775 7.0269 6.9777 6.9299 6.8833 6.8379 6.7935 6.7502 6.7078 6.6663 6.6256 6.5857 6.5465 6.5079 6.4698 6.4323 6.3952 6.3585 6.3221 6.2861 6.2503 6.2146 6.1791 6.1437 6.1083 6.0730 6.0375 6.0019 5.9662 5.9301 5.8938 5.8571 5.8199 5.7821 5.7437 5.7045 5.6643 5.6230 5.5804 5.5362 5.4417 5.3357 5.2112 5.0526 4.7971 4.4298
(MPa) 100 105 110 115 120 125 130 135 140 145 150 155 160 165 170 175 180 185 190 195 200 205 210 215 220 225 230 235 240 245 250 255 260 265 270 275 280 285 290 295 300 305 310 315 320 330 340 350 360 370 374.14
0.10135 0.12082 0.14327 0.16906 0.19853 0.2321 0.2701 0.3130 0.3613 0.4154 0.4758 0.5431 0.6178 0.7005 0.7917 0.8920 1.0021 1.1227 1.2544 1.3978 1.5538 1.7230 1.9062 2.104 2.318 2.548 2.795 3.060 3.344 3.648 3.973 4.319 4.688 5.081 5.499 5.942 6.412 6.909 7.436 7.993 8.581 9.202 9.856 10.547 11.274 12.845 14.586 16.513 18.651 21.03 22.09
0.001044 0.001048 0.001052 0.001056 0.001060 0.001065 0.001070 0.001075 0.001080 0.001085 0.001091 0.001096 0.001102 0.001108 0.001114 0.001121 0.001127 0.001134 0.001141 0.001149 0.001157 0.001164 0.001173 0.001181 0.001190 0.001199 0.001209 0.001219 0.001229 0.001240 0.001251 0.001263 0.001276 0.001289 0.001302 0.001317 0.001332 0.001348 0.001366 0.001384 0.001404 0.001425 0.001447 0.001472 0.001499 0.001561 0.001638 0.001740 0.001893 0.002213 0.003155
1.6729 1.4194 1.2102 1.0366 0.8919 0.7706 0.6685 0.5822 0.5089 0.4463 0.3928 0.3468 0.3071 0.2727 0.2428 0.2168 0.19405 0.17409 0.15654 0.14105 0.12736 0.11521 0.10441 0.09479 0.08619 0.07849 0.07158 0.06537 0.05976 0.05471 0.05013 0.04598 0.04221 0.03877 0.03564 0.03279 0.03017 0.02777 0.02557 0.02354 0.02167 0.019948 0.018350 0.016867 0.015488 0.012996 0.010797 0.008813 0.006945 0.004925 0.003155
698
Appendix
Table D2 Superheated steam T (°C)
V (m3/kg)
S~. 50 100 150 200 250 300 400 500 600 700 800 900 1000 1100 1200 1300
14.674 14.869 17.196 19.512 21.825 24.136 26.445 31.063 35.679 40.295 44.911 49.526 54.141 58.757 63.372 67.987 72.602
U (kJ/kg)
H (kJ/kg)
S (kJ/(kg K))
p = O.O01MPa(45.81°C) 2437.9 2584.7 2443.9 2592.6 2515.5 2687.5 2587.9 2783.0 2661.3 2879.5 2736.0 2977.3 2812.1 3076.5 2968.9 3279.6 3132.3 3489.1 3302.5 3705.4 3479.6 3928.7 3663.8 4159.0 3855.0 4396.4 4053.0 4640.6 4257.5 4891.2 4467.9 5147.8 4683.7 5409.7
8.1502 8.1749 8.4479 8.6882 8.9038 9.1002 9.2813 9.6077 9.8978 10.1608 10.4028 10.6281 10.8396 11.0393 11.2287 11.4091 11.5811
V (m3/kg)
3.240 3.418 3.889 4.356 4.820 5.284 6.209 7.134 8.057 8.981 9.904 10.828 11.751 12.674 13.597 14.521
U (kJ/kg)
H (kJ/kg)
P = O.05MPa(81.33°C) 2483.9 2645.9 2511.6 2585.6 2659.9 2735.0 2811.3 2968.5 3132.0 3302.2 3479.4 3663.6 3854.9 4052.9 4257.4 4467.8 4683.6
2682.5 2780.1 2877.7 2976.0 3075.5 3278.9 2488.7 3705.1 3928.5 4158.9 4396.3 4640.5 4891.1 5147.7 5409.6
S (kJ/(kg K))
7.5939 7.6947 7.9401 8.1580 8.3556 8.5373 8.8642 9.1546 9.4178 9.6599 9.8852 10.0967 10.2964 10.4859 10.6662 10.8382
Sat. 150 200 250 300 400 500 600 700 800 900 1000 1100 1200 1300
0.8857 0.9596 1.0803 1.1988 1.3162 1.5493 1.7814 2.013 2.244 2.475 2.705 2.937 3.168 3.399 3.630
P = 0.20 MPa 2529.5 2576.9 2654.4 2731.2 2808.6 2966.7 3130.8 3301.4 3478.8 3663.1 3854.5 4052.5 4257.0 4467.5 4683.2
(120.23°C) 2706.7 2768.8 2870.5 2971.0 3071.8 3276.6 3487.1 3704.0 3927.6 4158.2 4395.8 4640.0 4890.7 5147.5 5409.3
7.1272 7.2795 7.5066 7.7086 7.8926 8.2218 8.5133 8.7770 9.0194 9.2449 9.4566 9.6563 9.8458 10.0262 10.1982
0.6058 0.6339 0.7163 0.7964 0.8753 1.0315 1.1867 1.3414 1.4957 1.6499 1.8041 1.9581 2.1121 2.2661 2.4201
P = 0.30 MPa (133.55°C) 2543.6 2725.3 2570.8 2761.0 2650.7 2865.6 2728.7 2967.6 2806.7 3069.3 2965.6 3275.0 3130.0 3486.0 3300.8 3703.2 3478.4 3927.1 3662.9 4157.8 3854.2 4395.4 4052.3 4639.7 4256.8 4890.4 4467.2 5147.1 4683.0 5409.0
6.9919 7.0778 7.3115 7.5166 7.7022 8.0330 8.3251 8.5892 8.8319 9.0576 9.2692 9.4690 9.6585 9.8389 10.0110
S~. 200 250 300 350 400 500 600 700 800 900 1000 1100 1200 1300
0.3749 0.4249 0.4744 0.5226 0.5701 0.6173 0.7109 0.8041 0.8969 0.9896 1.0822 1.1747 1.2672 1.3596 1.4521
p = 0.50MPa(151.86°C) 2561.2 2748.7 2642.9 2855.4 2723.5 2960.7 2802.9 3064.2 2882.6 3167.6 2963.2 3271.9 3128.4 3483.9 3299.6 3701.7 3477.5 3925.9 3662.1 4156.9 3853.6 4394.7 4051.8 4639.1 4256.3 4889.9 4466.8 5146.6 4682.5 5408.6
6.8213 7.0592 7.2709 7.4599 7.6329 7.7938 8.0873 7.3522 8.5952 8.8211 9.0329 9.2328 9.4224 9.6029 9.7749
0.3157 0.3520 0.3938 0.4344 0.4742 0.5137 0.5920 0.6697 0.7472 0.8245 0.9017 0.9788 1.0559 1.1330 1.2101
P = 0.60MPa (158.85°C) 2567.4 2756.8 2638.9 2850.1 2720.9 2957.2 2801.0 3061.6 2881.2 3165.7 2962.1 3270.3 3127.6 3482.8 3299.1 3700.9 3477.0 3925.3 3661.8 4156.5 3853.4 4394.4 4051.5 4638.8 4256.1 4889.6 4466.5 5146.3 4682.3 5408.3
6.7600 6.9665 7.1816 7.3724 7.5464 7.7079 8.0021 8.2674 8.5107 8.7367 8.9486 9.1485 9.3381 9.5185 9.6906
S~. 100 150 200 250 300
1.6940 1.6958 1.9364 2.172 2.406 2.639
p = O.lOMPa(99.63°C) 2506.1 2675.5 2506.7 2676.2 2582.8 2776.4 2658.1 2875.3 2733.7 2974.3 2810.4 3074.3
7.3594 7.3614 7.6134 7.8343 8.0333 8.2158
(Contd.)
Appendix Table D2
(Continued)
T (°C)
V (m3/kg)
U (kJ/kg)
H (kJ/kg)
S (kJ/(kg K))
400
3.103
2967.9
3278.2
8.5435
500
3.565
600 700 800 900 1000 1100 1200
4.028 4.490 4.952
3131.6 3301.9 3479.2 3663.5 3854.8 4052.8
3488.1 3704.4 3928.2 4158.6 4396.1 4640.3
4257.3 4467.7 4683.5
4891.0 5147.6 5409.5
8.8342 9.0976 9.3398 9.5652 9.7767 9.9764 0.1659
1300
5.414 5.875 6.337 6.799 7.260
699
V (m3/kg)
U (kJ/kg)
H (kJ/kg)
S (kJ/(kg K))
6.5865 6.6940 6.9247
0.16333 0.16930 0.18234
P = 1.20 MPa (187.99°C) 2588.8 2784.8 2612.8 2815.9 2704.2 2935.0
7.1229
0.2138
2789.2
3045.8
7.0317
0.2345 0.2548
2872.2 2954.9
3153.6 3260.7
7.2121 7.3774
0.3463 0.5183
p = 0.40MPa(143.63°C) Sat. 150 200 250 30O 40O 500 600 7O0 800 900 1000 1100 1200 1300 Sat. 200 250 300 350 400 500 600 700 800 900 1000 1100 1200 1300 Sat. 200 250 300 350 400
0.4625 0.4708 0.5342 0.5951 0.6548 0.7726 0.8893 1.0055 1.1215 1.2372 1.3529 1.4685 1.5840 1.6996 1.8151 0.2404 0.2608 0.2931 0.3241 0.3544 0.3843 0.4433 0.5018 0.5601 0.6181 0.6761 0.7340 0.7919 0.8497 0.9076 0.1944 0.2060 0.2327
2553.6 2564.5
2738.6 2752.8
2646.8 2726.1 2804.8 2964.4 3129.2 3300.2 3477.9 3662.4 3853.9 4052.0
2860.5 2964.2 3066.8 3273.4 3484.9 3702.4 3926.5 4157.3 4395.1 4639.4
6.9299 7.1706 7.3789 7.5662 7.8985 8.1913 8.4558 8.6987 8.9244 9.1362 9.3360
4256.5 4467.0 4682.8
4890.2 5146.8 5408.8
9.5256 9.7060 9.8780
= 170.43°C) 2769.1 2839.3 2950.0 3056.5 3161.7 31267.1 3480.6 3699.4
6.6628 6.8158 7.0384 7.2328 7.4089 7.5716 7.8673 8.1333
3924.2 4155.6 4393.7 4638.2 4889.1 5145.9 5407.9
8.3770 8.6033 8.8153 9.0153 9.2050 9.3855 9.5575
P = 0.80MPa(T 2576.8 2630.6 2715.5 2797.2 2878.2 2959.7 3126.0 3297.9 3476.2 3661.1 3852.8 4051.0 4255.6 4466.1 4681.8
p = 1.00MPa(179.91°C) 2583.6 2778.1 2621.9 2827.9 2709.9
2942.6
0.2579 0.2825
2793.2 2875.2
3051.2 3157.7
0.3066
2957.3
6.8959
6.5233 6.5898 6.8294
3263.9
7.3011 7.4651
500 600
0.3541
3124.4
3478.5
7.7622
0.2946
3122.8
3476.3
7.6759
0.4011
3296.8
3697.9
8.0290
0.3339
3295.6
3696.3
7.9435
700
0.4478
0.3729
3474.4
3922.0
8.1881
0.4943
3923.1 4154.7
8.2731
800
3475.3 3660.4
8.4996
0.4118
3659.7
4153.8
8.4148
900 1000 1100 1200
0.5407 0.5871 0.6335 0.6798
3852.2 4050.5 4255.1 4465.6
4392.9 4637.6 4888.6 5145.4
1300
0.7261
4681.3
5407.4
8.7118 8.9119 9.1017 9.2822 9.4543
0.4505 0.4892 0.5278 0.5665 0.6051
3851.6 4050.0 4254.6 4465.1 4680.9
4392.2 4637.0 4888.0 5144.9 5407.0
8.6272 8.8274 9.0172 9.1977 9.3698
(Contd.)
700
Appendix
Table D2
(Continued)
T (°C)
V (m3/kg)
Sat. 225 250 300 350 400 500 600 700 800 900 1000 1100 1200 1300
0.12380 0.13287 0.14184 0.15862 0.17456 0.19005 0.2203 0.2500 0.2794 0.3086 0.3377 0.3668 0.3958 0.4248 0.4538
S~. 225 250 300 350 400 450 500 600 700 800 900 1000 1100 1200 1300
0.07998 0.08027 0.08700 0.09890 0.10976 0.12010 0.13014 0.13993 0.15930 0.17832 0.19716 0.21590 0.2346 0.2532 0.2718 0.2905
S~. 200 250 300 350 400 500 600 700 800 900 1000 1100 1200 1300
0.14084 0.14302 0.16350 0.18228 0.2003 0.2178 0.2521 0.2860 0.3195 0.3528 0.3861 0.4192 0.4524 0.4855 0.5186
Sat. 225 250 300 350 400 500 600
0.09963 0.10377 0.11144 0.12547 0.13857 0.15120 0.17568 0.19960
U (kJ/kg)
H (kJ/kg)
p = 1.60MPa(201.41°C) 2596.0 2794.0 2644.7 2857.3 2692.3 2919.2 2781.1 3034.8 2866.1 3145.4 2950.1 3254.2 3119.5 3472.0 3293.3 3693.2 3472.7 3919.7 3658.3 4152.1 3850.5 4390.8 4049.0 4635.8 4253.7 4887.0 4464.2 5143.9 4679.9 5406.0 p = 2.50MPa(223.99°C) 2603.1 2803.1 2605.6 2806.3 2662.6 2880.1 2761.6 3008.8 2851.9 3126.3 2939.1 3239.3 3025.5 3350.8 3112.1 3462.1 3288.0 3686.3 3468.7 3914.5 3655.3 4148.2 3847.9 4387.6 4046.7 4633.1 4251.5 4884.6 4462.1 5141.7 4677.8 5404.0 p = 1.40MPa(195.07)°C 2592.8 2790.0 2603.1 2803.3 2698.3 2927.2 2785.2 3040.4 2869.2 3149.5 2952.5 3257.5 3121.1 3474.1 3294.4 3694.8 3473.6 3920.8 3659.0 4153.0 3851.1 4391.5 4049.5 4636.4 4254.1 4887.5 4464.7 5144.4 4680.4 5406.5 P = 2.00 MPa (212.42) °C 2600.3 2628.3 2679.6 2772.6 2859.8 2945.2 3116.2 3290.9
2799.5 2835.8 2902.5 3023.5 3137.0 3247.6 3467.6 3690.1
S (kJ/(kg K))
V (m3/kg)
6.4218 6.5518 6.6732 6.8844 7.0694 7.2374 7.5390 7.8080 8.0535 8.2808 8.4935 8.6938 8.8837 9.0643 9.2364
0.11042 0.11673 0.12497 0.14021 0.15457 0.16847 0.19550 0.2220 0.2482 0.2742 0.3001 0.3260 0.3518 0.3776 0.4034
6.2575 6.2639 6.4085 6.6438 6.8403 7.0148 7.1746 7.3234 7.5960 7.8435 8.0720 8.2853 8.4861 8.6762 8.8569 9.0291
U (kJ/kg)
H (kJ/kg)
S (kJ/(kg K))
0.06668
P = 1.80MPa(207.15°C) 2598.4 2797.1 2636.6 2846.7 2686.0 2911.0 2776.9 3029.2 2863.0 3141.2 2947.7 3250.9 3117.9 3469.8 3292.1 3691.7 3471.8 3918.5 3657.6 4151.2 3849.9 4390.1 4048.5 4635.2 4253.2 4886.4 4463.7 5143.4 4679.5 5405.6 P = 3.00MPa(233.90°C) 2604.1 2804.2
6.1869
0.07058 0.08114 0.09053 0.09936 0.10787 0.11619 0.13243 0.14838 0.16414 0.17980 0.19541 0.21098 0.22652 0.24206
2644.0 2750.1 2843.7 2932.8 3020.4 3108.0 3285.0 3466.5 3653.5 3846.5 4045.4 4250.3 4460.9 4676.6
6.2872 6.5390 6.7428 6.9212 7.0834 7.2338 7.5085 7.7571 7.9862 8.1999 8.4009 8.5912 8.7720 8.9442
2855.8 2993.5 3115.3 3230.9 3344.0 3456.5 3682.3 3911.7 4145.9 4385.9 4631.6 4883.3 5140.5 5402.8
6.3794 6.4808 6.6066 6.8226 7.0100 7.1794 7.4825 7.7523 7.9983 8.2258 8.4386 8.6391 8.8290 9.0096 9.1818
6.4693 6.4975 6.7467 6.9534 7.1360 7.3026 7.6027 7.8710 8.1160 8.3431 8.5556 8.7559 8.9457 9.1262 9.2984 6.3409 6.4147 6.5453 6.7664 6.9563 7.1271 7.4317 7.7024
(Contd.)
Appendix
701
Table D2
(Continued)
T (°C)
V (m3/kg)
U (kJ/kg)
H (kJ/kg)
S (kJ/(kg K))
700 800 900 1000 1100
0.2232 0.2467 0.2700 0.2933 0.3166
3470.9 3657.0 3849.3 4048.0 4252.7
3917.4 4150.3 4389.4 4634.6 4885.9
7.9487 8.1765 8.3895 8.5901 8.7800
1200 1300
0.3398 0.3631
4463.3 5142.9 4679.0 5405.1 p = 3.50MPa(242.60°C)
8.9607 9.1329
Sat. 250 300 350 400 450 500 600 700 800 900 1000 1100 1200 1300
0.05707 0.05872 0.06842 0.07678 0.08453 0.09196 0.09918 0.11324 0.12699 0.14056 0.15402 0.16743 0.18080 0.19415 0.20749
2603.7 2803.4 2623.7 2829.2 2738.0 2977.5 2835.3 3104.0 2926.4 3222.3 3015.3 3337.2 3103.0 3450.9 3282.1 3678.4 3464.3 3908.8 3651.8 4143.7 3845.0 4384.1 4044.1 4630.1 4249.2 4881.9 4459.8 5139.3 4675.5 5401.7 P = 4.0 MPa (250.40°C)
6.1253 6.1749 6.4461 6.6579 6.8405 7.0052 7.1572 7.4339 7.6837 7.9134 8.1276 8.3288 8.5192 8.7000 8.8723
Sat. 275 300 350 400 450 500 600 700 800 900 1000 1100 1200 1300
0.04978 0.05457 0.05884 0.06645 0.07341 0.08002 0.08643 0.09885 O. 11095 0.12287 0.13469 0.14645 O. 15817 0.16987 0.18156
6.0701 6.2285 6.3615 6.5821 6.7690 6.9363 7.0901 7.3688 7.6198 7.8502 8.0647 8.2662 8.4567 8.6376 8.8100
0.04406 0.04730 0.05135 0.05840 0.06475 0.07074 0.07651 0.08765 0.09847 0.10911 0.11965 0.13013 O. 14056 0.15098 0.16139
P = 4.5 MPa (257.49°C) 2600.1 2798.3 2650.3 2863.2 2712.0 2943.1 2817.8 3080.6 2913.3 3204.7 3005.0 3323.3 3095.3 3439.6 3276.0 3670.5 3459.9 3903.0 3648.3 4139.3 3842.2 4380.6 4041.6 4627.2 4246.8 4879.3 4457.5 5136.9 4673.1 5399.4
6.0198 6.1401 6.2828 6.5131 6.7047 6.8746 7.0301 7.3110 7.5631 7.7942 8.0091 8.2108 8.4015 8.5825 8.7549
Sat. 300 350 400 450 500
0.03244 0.03616 0.04223 0.04739 0.05214 0.05665
2602.3 2801.4 2667.9 2886.2 2725.3 2960.7 2826.7 3092.5 2919.9 3213.6 3010.2 3330.3 3099.5 3445.3 3279.1 3674.4 3462.1 3905.9 3650.0 4141.5 3843.6 4382.3 4042.9 4628.7 4248.0 4880.6 4458.6 5138.1 4674.3 5400.5 P = 6.0 MPa (275.64°C) 2589.7 2784.3 2667.2 2884.2 2789.6 3043.0 2892.9 3177.2 2988.9 3301.8 3082.2 3422.2
5.8892 6.0674 6.3335 6.5408 6.7193 6.8803
0.02737 0.02947 0.03524 0.03993 0.04416 0.04814
P = 7.0 MPa (285.88°C) 2580.5 2772.1 2632.2 2838.4 2769.4 3016.0 2878.6 3158.1 2978.0 3287.1 3073.4 3410.3
5.8133 5.9305 6.2283 6.4478 6.6327 6.7965
V (m3/kg)
U (kJ/kg)
H (kJ/kg)
S (kJ/(kg K))
550
0.06101
3174.6
3540.6
7.0288
0.05195
3167.2
3530.9
6.9486
600
0.06525
3266.9
3658.4
7.1677
0.05565
3260.7
3650.3
7.0894
700
0.07352
3453.1
3894.2
7.4234
0.06283
3448.5
3888.3
7.3476
800 900 1000
0.08160 0.08958 0.09749
3643.1 3837.8 4037.8
4132.7 4375.3 4622.7
7.6566 7.8727 8.0751
0.06981 O.07669 0.08350
3639.5 3835.0 4035.3
4128.2 4371.8 4619.8
7.5822 7.7991 8.0020
1100 1200
0.10536 0.11321
4243.3 4454.0
4875.4 5133.3
8.2661 8.4474
0.09027 0.09703
4240.9 4451.7
4872.8 5130.9
8.1933 8.3747
1300
0.12106
4669.6 51396.0 P = 9.0 MPa (303.40°C) 2557.8 2742.1 2646.6 2856.0
8.6199
0.10377
4667.3
5393.7
8.5475
0.018026
P = IO.OMPa (318.06°C) 2544.4 2724.7 2610.4 2809.1
Sat. 325
0.02048 0.02327
5.6772 5.8712
0.019861
5.6141 5.7568
(Contd.)
Appendix
702 Table D2
(Continued)
T (°C)
V (m3/kg)
350 400 450 500 550 600 650 700 800 900 1000 1100 1200 1300
0.02580 0.02993 0.03350 0.03677 0.03987 0.04285 0.04574 0.04857 0.05409 0.05950 0.06485 0.07016 0.07544 0.08072
S~. 275 300 350 400 450 500 600 700 800 900 1000 1100 1200 1300
0.03944 0.04141 0.04532 0.05194 0.05781 0.06330 0.06857 0.07869 0.08849 0.09811 0.10762 0.11707 0.12648 0.13587 0.14526
S~. 300 350 400 450 500 550 600 700 800 900 1000 1100 1200 1300
0.02352 0.02426 0.02995 0.03432 0.03817 0.04175 0.04516 0.04845 0.05481 0.06097 0.06702 0.07301 0.07896 0.08489 0.09080
Sat. 350 400 450 500 550 600 650 700 800 900
0.013495 0.016126 0.02000 0.02299 0.02560 0.02801 0.03029 0.03248 0.03460 0.03869 0.04267
U (kJ/kg)
H (kJ/kg)
2724.4 2956.6 2848.4 3117.8 2955.2 3256.6 3055.2 3386.1 3152.2 3511.0 3248.1 3633.7 3343.6 3755.3 3439.3 3876.5 3632.5 4119.3 3829.2 4364.8 4030.3 4614.0 4236.3 4867.7 4447.2 5126.2 4662.7 5389.2 p = 5.0MPa(263.99°C) 2597.1 2794.3 2631.3 2838.3 2698.0 2924.5 2808.7 3068.4 2906.6 3195.7 2999.7 3316.2 3091.0 3433.8 3273.0 3666.5 3457.6 3900.1 3646.6 4137.1 3840.7 4378.8 4040.4 4625.7 4245.6 4878.0 4456.3 5135.7 4672.0 5398.2 P = 8.0 MPa (295.06°C) 2569.8 2758.0 2590.9 2785.0 2747.7 2987.3 2863.8 3138.3 2966.7 3272.0 3064.3 3398.3 3159.8 3521.0 3254.4 3642.0 3443.9 3882.4 3636.0 4123.8 3832.1 4368.3 4032.8 4616.9 4238.6 4870.3 4449.5 5128.5 4665.0 5391.5 P = 12.5 MPa (327.89°C) 2505.1 2673.8 2624.6 2826.2 2789.3 3039.3 2912.5 3199.8 3021.7 3341.8 3125.0 3475.2 3225.4 3604.0 3324.4 3730.4 3422.9 3855.3 3620.0 4103.6 3819.1 4352.5
S (kJ/(kg K))
V (m3/kg)
U (kJ/kg)
H (kJ/kg)
S (kJ/(kg K))
6.0361 6.2854 6.4844 6.6576 6.8142 6.9589 7.0943 7.2221 7.4596 7.6783 7.8821 8.0740 8.2556 8.4284
0.02242 0.02641 0.02975 0.03279 0.03564 0.03837 0.04101 0.04358 0.04859 0.05349 0.05832 0.06312 0.06789 0.07265
2699.2 2832.4 2943.4 3045.8 3144.6 3241.7 3338.2 3434.7 3628.9 3826.3 4027.8 4234.0 4444.9 4460.5
2923.4 3096.5 3240.9 3373.7 3500.9 3625.3 3748.2 3870.5 4114.8 4361.2 4611.0 4865.1 5123.8 5387.0
5.9443 6.2120 6.4190 6.5966 6.7561 6.9029 7.0398 7.1687 7.4077 7.6272 7.8315 8.0237 8.2055 8.3783
5.9734 6.0544 6.2084 6.4493 6.6459 6.8186 6.9759 7.2589 7.5122 7.7440 7.9593 8.1612 8.3520 8.5331 8.7055 5.7432 5.7906 6.1301 6.3634 6.5551 6.7240 6.8778 7.0206 7.2812 7.5173 7.7351 7.9384 8.1300 8.3115 8.4842 5.4624 5.7118 6.0417 6.2719 6.4618 6.6290 6.7810 6.9218 7.0536 7.2965 7.5182
(Contd.)
Appendix Table D2
(Continued)
T (°C)
V (m3/kg)
U (kJ/kg)
H (kJ/kg)
S (kJ/(kg K))
1000
0.04658
4021.6
4603.8
7.7237
1100
O.O5O45
4228.2
4858.8
7.9165
1200 1300
0.05430 0.05813
4439.3
.5118.0
8.0937 8.2717
Sat. 350 400 450 500 550
0.010337 0.011470 0.015649 0.018445 0.02080 0.02293
4654.8 5381.4 p = 15.0MPa(342.24°C) 2455.5 2610.5 2520.4 2692.4 2740.7 12975.5 2879.5 13156.2 2996.6 13308.6 3104.7 13448.6
600 650 700 800 900
0.02491 0.02680 0.02861 0.03210 0.03546
3208.6 3310.3 3410.9 3610.9 3811.9
1000 1100 1200 1300
0.03875 0.04200 0.04523 0.04845
375 400 425 450 500 550 600 650 700 800 900 1000 1100 1200 1300
0.0019731 0.006004 0.007881 0.009162 0.011123 0.012724 0.014137 0.015433 0.016646 0.018912 0.021045 0.02310 0.02512 0.02711 0.02910
375 400 425 450 500 550 600
0.0016407 0.0019077 0.002532 0.003693 0.005622 0.006984 0.008094
650
0.009063
700 800 900 1000
0.009941 0.011523 0.012962 0.014324
1100 1200 1300
703
V (m3/kg)
5.3098 5.4421
0.007920
5.8811 6.1404 6.3443 6.5199
0.012447 0.015174 0.017358 0.019288
13582.3 13712.3 13840.1 4092.4 4343.8
6.6776 6.8224 6.9572
0.02106 0.02274 0.02434
7.2040 7.4279
4015.4 4596.6 4222.6 4852.6 4433.8 5112.3 4649.1 5376.0 P = 25.0MPa 1798.7 11848.0 2430.1 2580.2 2609.2 2806.3 2720.7 2949.7 2884.3 3162.4 3017.5 3335.6 3137.9 3491.4 3251.6 3637.4 3361.3 3777.5 3574.3 4047.1 3783.0 4309.1 3990.9 4568.5 4200.2 4828.2 4412.0 5089.9 4626.9 5354.4 P = 40.0 MPa 1677.1 1742.8 1854.6 1930.9 2096.9 2198.1 2365.1 2512.8 2678.4 2903.3 2869.7 3149.1 3022.6 3346.4 3158.0 3520.6
U (kJ/kg)
H (kJ/kg)
P = 17.5 MPa (354.75°C) 2390.2 2528.8
0.02738 0.03031
2685.0 2844.2 2970.3 3083.9 3191.5 3296.0 3398.7 3601.8 3804.7
2902.9 3109.7 3274.1 3421.4 3560.1 3693.9 3824.6 4081.1 4335.1
7.6348 7.8283 8.0108 8.1840
0.03316 0.03597 0.03876 0.04154
4009.3 4216.9 4428.3 4643.5
4589.5 4846.4 5106.6 5370.5
4.0320 5.1418 5.4723 5.6744 5.9592 6.1765 6.3602 6.5229 6.6707 6.9345 7.1680 7.3802 7.5765 7.7605 7.9342
0.0017892 0.002790 0.005303 0.006735 0.008678 0.010168 0.011446 0.012596 0.013661 0.015623 0.017448 0.019196 0.020903 0.022589 0.024266
3.8290 4.1135 4.5029 9.9459 5.4700 5.7785 6.0144 6.2054
0.0015594 0.0017309 0.002007 0.002486 0.003892 0.005118 0.006112 0.006966
6.3750 6.6662 6.9150 7.1356
0.007727 0.009076 0.010283 0.011411
3283.6 3517.8 3739.4
3681.2 3978.7 4257.9
0.015642 0.016940
3954.6 4167.4 4380.1
4527.6 4793.1 5057.7
7.3364 7.5224
0.012496 0.013561
0.018229
4594.3
5323.5
7.6969
0.014616
Sat.
0.005834
p = 20.OMPa(365.81°C) 2293.0 2409.7
400
0.009942
2619.3
2818.1
4.9269 5.5540
450
0.012695
500 550
0.014768 0.016555
2806.2 2942.9
3060.1 3238.2
5.9017 6.1401
3062.4
3393.5
6.3348
P = 30.0 MPa 1737.8 1791.5 2067.4 2151.1 2455.1 2614.2 2619.3 2821.4 2820.7 3081.1 2970.3 3275.4 3100.5 3443.9 3221.0 3598.9 3335.8 3745.6 3555.5 4024.2 3768.5 4291.9 3978.8 4554.7 4189.2 4816.3 4401.3 5079.0 4616.0 5344.0 P = 50.0 MPa 1638.6 1716.6 1788.1 1874.6 1959.7 2060.0 2159.6 2284.0 2525.5 2720.1 2763.6 3019.5 2942.0 3247.6 3093.5 3441.8 3230.5 3479.8 3710.3 3930.5 4145.7 4359.1 4572.8
S (kJ/(kg K))
5.1419 5.7213 6.0184 6.2383 6.4230 6.5866 6.7357 6.8736 7.1244 7.3507 7.5589 7.7531 7.9360 8.1093 3.9305 4.4728 5.1504 5.4424 5.7905 6.0342 6.2331 6.4058 6.5606 6.8332 7.0718 7.2867 7.4845 7.6692 7.8432 3.7639 4.0031 4.2734 4.5884 5.1726 5.5485 5.8178 6.0342
3616.8 3933.6 4224.4 4501.1
6.2189 6.5290 6.7882 7.0146
4770.5 5037.2
7.2184 7.4058
5303.6
7.5808
(Contd.)
Appendix
704 Table D2 (Continued) T (°C)
V (m3/kg)
U (kJ/kg)
600 650 700 800 900 1000 1100 1200 1300
0.018178 0.019693 0.02113 0.02385 0.02645 0.02897 0.003145 0.03391 0.03636
375 400 425 450 500 550 600 650 700 800 900 1000 1100 1200 1300
0.0017003 0.002100 0.003428 0.004961 0.006927 0.008345 0.009527 0.010575 0.011533 0.013278 0.014883 0.016410 0.017895 0.019360 0.020815
375 400 425 450 500 550 600 650 700 800 900 1000 1100 1200 1300
0.0015028 0.0016335 0.0018165 0.002085 0.002956 0.003956 0.004834 0.005595 0.006272 0.007459 0.008508 0.009480 0.010409 0.011317 0.012215
H (kJ/kg)
S (kJ/(kg K))
3174.0 3537.6 3281.4 3675.3 3386.4 3809.0 3592.7 4069.7 3797.5 4326.4 4003.1 4582.5 4211.3 4840.2 4422.8 5101.0 4638.0 5365.1 P = 35.0MPa 1702.9 1762.4 1914.1 1987.6 2253.4 2373.4 2498.7 2672.4 2751.9 2994.4 2921.0 3213.0 3062.0 3395.5 3189.8 3559.9 3309.8 3713.5 3536.7 4001.5 3754.0 4274.9 3966.7 4541.1 4178.3 4804.6 4390.7 5068.3 4605.1 5333.6 P = 60.OMPa 1609.4 1699.5 1745.4 1843.4 1892.7 2001.7 2053.9 2179.0 2390.6 2567.9 2658.8 2896.2 2861.1 3151.2 3028.8 3364.5 3177.2 3553.5 3441.5 3889.1 3681.0 4191.5 3906.4 4475.2 4124.1 4748.6 4338.2 5017.2 4551.4 5284.3
V (m3/kg)
U (kJ/kg)
H (kJ/kg)
S (kJ/(kg K))
6.5048 6.6582 6.7993 7.0544 7.2830 7.4925 7.6874 7.8707 8.0442 3.8722 4.2126 4.7747 5.1962 5.6282 5.9026 6.1179 6.3010 6.4631 6.7450 6.9386 7.2064 7.4037 7.5910 7.7653 3.7141 3.9318 4.1626 4.4121 4.9321 5.3441 5.6452 5.8829 6.0824 6.4109 6.6805 6.9127 7.1195 7.3083 7.4837
APPENDIX E Table E1 Saturated refrigerant-134a properties-Temperature
T (°C) Psat(MPa)
-40 -36 -32 -28 -26 -24 -22
0.05164 0.06332 0.07704 0.09305 0.10199 0.11160 0.12192
Entropy (kJ/(kg K))
Enthalpy (kJ/kg)
Specific volume (m3/kg)
Internal energy (kJ/kg)
Sat. liquid
Sat. vapor
Sat. liquid
Sat. vapor
Sat. liquid
Evap.
Sat. vapor
Sat. liquid
Sat. vapor
0.0007055 0.0007113 0.0007172 0.0007233 0.0007265 0.0007296 0.0007328
0.3569 0.2947 0.2451 0.2052 0.1882 0.1728 0.1590
-0.04 4.68 9.47 14.31 16.75 19.21 21.68
204.45 206.73 209.01 211.29 212.43 213.57 214.70
0.00 4.73 9.52 14.37 16.82 19.29 21.77
222.88 220.67 218.37 216.01 214.80 213.57 212.32
222.88 225.40 227.90 230.38 231.62 232.85 234.08
0.0000 0.0201 0.0401 0.0600 0.0699 0.0798 0.0897
0.9560 0.9506 0.9456 0.9411 0.9390 0.9370 0.9351
(Contd.)
Appendix
705
Table E1 (Continued) T (°C)
-20 -18 -16 -12 -8 -4 0 4 8 12 16 20 24 26 28 30 32 34 36 38 40 42 44 48 52 56 60 70 80 90 100
Psat(MPa)
0.13299 0.14483 0.15748 0.18540 0.21704 0.25274 0.29282 0.33765 0.38756 0.44294 0.50416 0.57160 0.64566 0.68530 0.72675 0.77006 0.81528 0.86247 0.91168 0.96298 1.0164 1.0720 1.1299 1.2526 1.3851 1.5278 1.6813 2.1162 2.6324 3.2435 3.9742
Specific volume (m3/kg) Sat. liquid
Sat. vapor
Sat. liquid
Sat. vapor
0.0007361 0.0007395 0.0007428 0.0007498 0.0007569 0.0007644 0.0007721 0.0007801 0.007884 0.0007971 0.0008062 0.0008157 0.0008257 0.0008309 0.0008362 0.0008417 0.0008473 0.0008530 0.0008590 0.0008651 0.0008714 0.0008780 0.0008847 0.0008989 0.0009142 0.0009308 0.0009488 0.0010027 0.0010766 0.0011949 0.0015443
0.1464 0.1350 0.1247 0.1068 0.0919 0.0794 0.0689 0.0600 0.0525 0.0460 0.0405 0.0358 0.0317 0.0298 0.0281 0.0265 0.0250 0.0236 0.0223 0.0210 0.0199 0.0188 0.0177 0.0159 0.0142 0.0127 0.0114 0.0086 0.0064 0.0046 0.0027
24.17 26.67 29.18 34.25 39.38 44.56 49.79 55.08 60.43 65.83 71.29 76.80 82.37 85.18 88.00 90.84 93.70 96.58 99.47 102.38 105.30 108.25 111.22 117.22 123.31 129.51 135.82 152.22 169.88 189.82 218.6(i)
215.84 216.97 218.10 220.36 222.60 224.84 227.06 229.27 231.46 233.63 235.78 237.91 240.01 241.05 242.08 243.10 244.12 245.12 246.11 247.09 248.06 249.02 249.96 251.79 253.55 255.23 256.81 260.15 262.14 261.34 248.49
Entropy (kJ/(kg K))
Enthalpy (kJ/kg)
Internal energy (kJ/kg)
Sat. liquid 24.26 26.77 29.30 34.39 39.54 44.75 50.02 55.35 60.73 66.18 71.69 77.26 82.90 85.75 88.61 91.49 94.39 07.31 100.25 103.21 106.19 109.19 112.22 118.35 124.58 130.93 137.42 154.34 172.71 193.69 224.74
Evap.
Sat. vapor
Sat. liquid
Sat. vapor
211.05 209.76 208.45 205.77 203.00 200.15 197.21 194.19 191.07 187.85 184.52 181.09 177.55 175.73 173.89 172.00 170.09 168.14 166.15 164.12 162.05 159.94 157.79 153.33 148.66 143.75 138.57 124.08 106.41 82.63 34.40
235.31 236.53 237.74 240.15 242.54 244.90 247.23 249.53 251.80 254.03 256.22 258.35 260.45 261.48 262.50 263.50 264.48 265.45 266.40 267.33 268.24 269.14 270.01 271.68 273.24 274.68 275.99 278.43 279.12 276.32 259.13
0.0996 0.1094 0.1192 0.1388 0.1583 0.1777 0.1970 0.2162 0.2354 0.2545 0.2735 0.2924 0.3113 0.3208 0.3302 0.3396 0.3490 0.3584 0.3678 0.3772 0.3866 0.3960 0.4054 0.4243 0.4432 0.4622 0.4814 0.5302 0.5814 0.6380 0.7196
0.9332 0.9315 0.9298 0.9267 0.9239 0.9213 0.9190 0.9169 0.9150 0.9132 0.9116 0.9102 0.9089 0.9082 0.9076 0.9070 0.9064 0.9058 0.9053 0.9047 0.9041 0.9035 0.9030 0.9017 0.9004 0.8990 0.8973 0.8918 0.8827 0.8655 0.8117
Table E2 Saturated refrigerant-134a properties-Pressure P
T~.t
(MPa)
(°C)
0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.24 0.28 0.32 0.36 0.4 0.5 0.6 0.7 0.8 0.9
-37.07 -31.21 -26.43 -22.36 -18.80 -15.62 -12.73 -10.09 -5.37 -1.23 2.48 5.84 8.93 15.74 21.58 26.72 31.33 35.53
Specific volume (m3/kg) Sat. liquid 0.0007097 0.0007184 0.0007258 0.0007323 0.0007381 0.0007435 0.0007485 0.0007532 0.0007618 0.0007697 0.0007770 0.0007839 0.0007904 0.0008056 0.0008196 0.0008328 0.0008454 0.0008576
Internal energy (kJ/kg)
Enthalpy (kJ/kg)
Sat. vapor
Sat. liquid
Sat. vapor
Sat. liquid
Evap.
0.3100 0.2366 0.1917 0.1614 0.1395 0.1229 0.1098 0.0993 0.0834 0.0719 0.0632 0.0564 0.0509 0.0409 0.0341 0.0292 0.0255 0.0226
3.41 10.41 16.22 21.23 25.66 29.66 33.31 36.69 42.77 48.18 53.06 57.54 61.69
206.12 209.46 212.18 214.50 216.52 218.32 219.94 221.43 224.07 226.38 228.43 230.28 231.97
70.93 78.99 86.19 92.75 98.79
235.64 238.74 241.42 243.78 245.88
3.46 10.47 16.29 21.32 25.77 29.78 33.45 36.84 42.95 48.39 53.31 57.82 62.00 71.33 79.48 86.78 93.42 99.56
221.27 217.92 215.06 212.54 210.27 208.18 206.26 204.46 201.14 198.13 195.35 192.76 190.32 184.74 179.71 175.07 170.73 166.62
Entropy (kJ/(kg K)) Sat. vapor 224.72 228.39 231.35 233.86 236.04 237.97 239.71 241.30 244.09 246.52 248.66 250.58 252.32 256.07 259.19 261.85 264.15 266.18
Sat. liquid
Sat. vapor
0.0147 0.0440 0.0678 0.0879 0.1055 0.1211 0.1352 0.1481 0.1710 0.1911 0.2089
0.9520 0.9447 0.9395 0.9354 0.9322 0.9295 0.9273 0.9253 0.9222 0.9197 0.9177
0.2251 0.2399 0.2723 0.2999 0.3242 0.3459 0.3656
0.9160 0.9145 0.9117 0.9097 0.9080 0.9066 0.9054
(Contd.)
Appendix
706 Table E2 (Continued) P
Tsa t
Specific volume (m3/kg)
(MPa)
(°C)
Sat. liquid
1.0 1.2 1.4 1.6 1.8 2.0 2.5 3.0
39.39 46.32 52.43 57.92 62.91 67.49 77.59 86.22
Sat. vapor
0.0008695 0.0008928 0.0009159 0.0009392 0.0009631 0.0009878 0.0010562 0.0011416
0.0202 0.0166 0.0140 0.0121 0.0105 0.0093 0.0069 0.0053
Internal energy (kJ/kg)
Enthalpy (kJ/kg)
Entropy (kJ/(kg K))
Sat. liquid
Sat. vapor
Sat. liquid
Evap.
Sat. vapor
104.42 114.69 123.98 132.52 140.49 148.02 165.48 181.88
247.77 251.03 253.74 256.00 257.88 259.41 261.84 262.16
105.29 115.76 125.26 134.02 142.22 149.99 168.12 185.30
162.68 155.23 148.14 141.31 134.60 127.95 111.06 92.71
267.97 270.99 273.40 275.33 276.83 277.94 279.17 278.01
Sat. liquid
Sat. vapor
0.3838 0.4164 0.4453 0.4714 0.4954 0.5178 0.5687 0.6156
0.9043 0.9023 0.9003 0.8982 0.8959 0.8934 0.8854 0.8735
Table E3 Superheated refrigerant-134a T (°C)
S~. -20 -10 0 10 20 30 40 50 60 70 80 90
V (m3/kg)
0.31003 0.33536 0.34992 0.36433 0.37861 0.39279 0.40688 0.42091 0.43487 0.44879 0.46266 0.47650 0.49031
u (kJ/kg)
h (kJ/kg)
P = 0.06MPa (Tsat 206.12 217.86 224.97 232.24 239.69 247.32 255.12 263.10 271.25 279.58 288.08 296.75 305.58
--
-37.07°C) 224.72 237.98 245.96 254.10 262.41 270.89 279.53 288.35 297.34 306.51 315.84 325.34 335.00
s (kJ/(kg K))
V (m3/kg)
u (kJ/kg)
0.9520 1.0062 1.0371 1.0675 1.0973 1.1267 1.1557 1.1844 1.2126 1.2405 1.2681 1.2954 1.3224
0.19170 0.19770 0.20686 0.21587 0.22473 0.23349 0.24216 0.25076 0.25930 0.26779 0.27623 0.28464 0.29302
P = 0.10 MPa 212.18 216.77 224.01 231.41 238.96 246.67 254.54 262.58 270.79 279.16 287.70 296.40 305.27
0.9273 0.9362 0.9684 0.9998 1.0304 1.0604 1.0898 1.1187 1.1472 1.1753 1.2030 1.2303 1.2573
0.09933 0.09938 0.10438 0.10922 0.11394 0.11856 0.12311 0.12758 0.13201 0.13639 0.14073 0.14504 0.14932
0.9197
0.06322
0.9238 0.9566 0.9883 1.0192 1.0494 1.0789 1.1079 1.1364 1.1644 1.1920 1.2193
0.06576 0.06901 0.07214 0.07518 0.07815 0.08106 0.08392 0.08674 0.08953 0.09229
h (kJ/kg) (Tsa t -"
-26.43°C) 231.35 236.54 244.70 252.99 261.43 270.02 278.76 287.66 296.72 305.94 315.32 324.87 334.57
s (kJ/(kg K))
0.9395 0.9602 0.9918 1.0227 1.0531 1.0829 1.1122 1.1411 1.1696 1.1977 1.2254 1.2528 1.2799
100 SN. -10
0 10 20 30 40 50 60 70 80 90 100
0.10983 0.11135 0.11678 0.12207 0.12723 0.13230 0.13730 0.14222 0.14710 0.15193 0.15672 0.16148 0.16622
S~.
0.07193
0 10 20 30 40 50 60 70 80 90 100
0.07240 0.07613 0.07972 0.08320 0.08660 0.08992 0.09319 0.09641 0.09960 0.10275 0.10587
P = 0.18MPa(Tsat = -12.73°C) 219.94 239.71 222.02 242.06 229.67 250.69 237.44 259.41 245.33 268.23 253.36 277.17 261.53 286.24 269.85 295.45 278.31 304.79 286.93 314.28 295.71 323.92 304.63 333.70 313.72 343.63 P = 0.28 MPa (~at = - 1"23°C) 226.38 246.52 227.37 235.44 243.59 251.83 260.17 268.64 277.23 285.96 294.82 303.83 312.98
247.64 256.76 265.91 275.12 284.42 293.81 303.32 312.95 322.71 332.60 342.62
P = 0.20MPa(Tsat = -10.09°C) 221.43 241.30 221.50 241.38 229.23 250.10 237.05 258.89 244.99 267.78 253.06 276.77 261.26 285.88 269.61 295.12 278.10 304.50 286.74 314.02 295.53 323.68 304.47 333.48 313.57 343.43 P = 0.32 MPa (Tsat = 2.48°C) 228.43 248.66 234.61 242.87 251.19 259.61 268.14 276.79 285.56 294.46 303.50 312.68
255.65 264.95 274.28 283.67 293.15 302.72 312.41 322.22 332.15 342.21
0.9253 0.9256 0.9582 0.9898 1.0206 1.0508 1.0804 1.1094 1.1380 1.1661 1.1939 1.2212 1.2483 0.9177 0.9427 0.9749 1.0062 1.0367 1.0665 1.0957 1.1243 1.1525 1.1802 1.1076
(Contd.)
Appendix Table E3 T (°C)
707
(Continued) V (m3/kg)
u (kJ/kg)
h (kJ/kg)
352.78 363.08
s (kJ/(kg K))
1.2461 1.2727
V (m3/kg)
0.09503 0.09774
u (kJ/kg)
h (kJ/kg)
322.00 331.45
352.40 362.73
s (kJ/(kg K))
1.2345 1.2611
110
0.10897
322.27
120 130 140
0.11205
331.71
Sat.
0.13945
-20 -10 0 10
0.14549 0.15219 0.15875
223.03 230.55 238.21
243.40 251.86 260.43
0.9606 0.9922 1.0230
20 30
0.16520 0.17155
246.01 253.96
269.13 277.97
1.0532 1.0828
40 50 60 70 80 90 100
0.17783 0.18404 0.19020 0.19633 0.20241 0.20846 0.21449
262.06 270.32 278.74 287.32 296.06 304.95 314.01
286.96 296.09 305.37 314.80 324.39 334.14 344.04
1.1120 1.1407 1.1690 1.1969 1.2244 1.2516 1.2785
Sat. -10 0 10 20 30 40 50 60 70 80 90 100
0.08343
224.07
244.09
(t.9222
0.08574 0.08993 0.09339 0.09794 0.10181 0.10562 0.10937 0.11307 0.11674 0.12037 0.12398
228.31 236.26 244.30 252.45 260.72 269.12 277.67 286.35 295.18 304.15 313.27
248.89 257.84 266.85 275.95 285.16 294.47 303.91 313.49 323.19 333.04 343.03
0.9399 0.9721 1.0034 1.0339 1.0637 1.0930 1.1218 1.1501 1.1780 1.2055 1.2326
Sat.
0.05089
231.97
252.32
0.9145
0 10 20 30 40 50 60 70 80 90
0.05119 0.05397 0.05662 0.05917 0.06164 0.06405 0.06641 0.06873 0.07102
232.87 241.37 249.89 258.47 267.13 275.89 284.75 293.73 302.84
253.35 262.96 272.54 282.14 291.79 301.51 311.32 321.23 331.25
0.9182 0.9515 0.9837 1.01448 1.0452 1.0748 1.1038 1.1322 1.1602
100 110
0.07327 0.07550
312.07 321.44
341.38 351.64
1.1878 1.2149
120
0.07771
330.94
362.03
1.2417
130
0.07991
340.58
372.54
1.2681
140
0.08208
350.35
383.18
1.2941
Sat. 20
0.04086 0.04188
253.64 239.40
256.07 260.34
0.9117 0.9264
0.03408
238.74
259.19
0.9097
30 40 50
0.04416 0.04633 0.04842
248.20 256.99 265.83
270.28 280.16 290.04
0.9597 0.9918 1.0229
60
0.05043
274.73
299.95
1.0531
0.03581 0.03774 0.03958 0.04134
246.41 255.45 264.48 273.54
267.89 278.09 288.23 298.35
0.9388 0.9719 1.0037 1.0346
P = O.14MPa(C~. = -18.80°C) 216.52 236.04
0.9322
P = 0.24 MPa (T~ = -5.37°C)
P = 0.40MPa(~=
8.93°C)
P = 0 . 5 0 M P a ( ~ t = 15.74°C)
P = 0.60 MPa (Tsat = 21.58°C)
(Contd.)
Appendix
708 Table E3
(Continued) V(m3/kg)
(kJ/kg)
h (kJ/kg)
70 80 90 100 110 120 130 140 150 160
0.05240 0.05432 0.05620 0.05805 0.05988 0.06168 0.06347 0.06524
283.72 292.80 302.00 311.31 320.74 330.30 339.98 349.79
309.92 319.96 330.10 340.33 350.68 361.14 371.72 382.42
1.0825 1.1114 1.1397 1.1675 1.1949 1.2218 1.2484 1.2746
0.04304 0.04469 0.04631 0.04790 0.04946 0.05099 0.05251 0.05402 0.05550 0.05698
Sat. 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180
0.02547 0.02691 0.02846 0.02992 0.03131 0.03264 0.03393 0.03519 0.03642 0.03762 0.03881 0.03997 0.04113 0.04227 0.04340 0.04452
p = 0.80MPa(Tsa t 243.78 252.13 261.62 271.04 280.45 289.89 299.37 308.93 318.57 328.31 338.14 348.09 358.15 368.32 378.61 389.02
= 31.33°C) 264.15 273.66 284.39 294.98 305.50 316.00 326.52 337.08 347.71 358.40 369.19 380.07 391.05 402.14 413.33 424.63
0.9066 0.9374 0.9711 1.0034 1.0345 1.0647 1.0940 1.1227 1.1508 1.1784 1.2055 1.2321 1.2584 1.2843 1.3098 1.3351
0.02255 0.02325 0.02472 0.02609 0.02738 0.02861 0.02980 0.03095 0.03207 0.03316 0.03423 0.03529 0.03633 0.03736 0.03838 0.03939
Sat. 50 60 70 80 90 100 110 120 130 140 150 16O 170 180 190 200
0.01663 0.01712 0.01835 0.01947 0.02051 0.02150 0.02244 0.02335 0.02423 0.02508 0.02592 0.02674 0.02754 0.02834
P = 1.20MPa(Tsat 251.03 254.98 265.42 275.59 285.62 295.59 305.54 315.50 325.51 335.58 345.73 355.95 366.27 376.69
= 46.32°C) 270.99 275.52 287.44 298.96 310.24 321.39 332.47 343.52 354.58 365.68 376.83 388.04 399.33 410.70
0.9023 0.9164 0.9527 0.9868 1.0192 1.0503 1.0804 1.1096 1.1381 1.1660 1.1933 1.2201 1.2465 1.2724
0.01405
422.16
1.2980
0.02472 0.02541 0.02608
Sat.
0.02918
20 30 40 50 60 70 80 90 100 110 120
0.02979 0.03157 0.03324 0.03482 0.03634 0.03781 0.03924 0.04064 0.04201 0.04335
T (°C)
0.02912
387.21
P = 0.70MPa(~at = 26.72°C) 241.42 261.85 244.51 253.83 263.08 272.31 281.57 290.88 300.27 309.74 319.31 328.98
265.37 275.93 286.35 296.69 307.01 317.35 327.74 338.19 348.71 359.33
s (kJ/(kg K))
V (m3/kg)
0.01495 0.01603 0.01701 0.01792 0.01878 0.01960 0.02039 0.02115 0.02189 0.02262 0.02333 0.02403
u (kJ/kg)
h (kJ/kg)
282.66 291.86 301.14 310.53 320.03 329.64 339.38 349.23 359.21 369.32 P = 0.90 MPa ( T s a t 245.88 250.32 260.09 269.72 279.30 288.87 298.46 308.11 317.82 327.62 337.52 347.51 357.61 367.82 378.14 388.57 P = 1.40 MPa ( T s a t 253.74
308.48 318.67 328.93 339.27 349.70 360.24 370.88 381.64 392.52 403.51 = 35.53°C) 266.18 271.25 282.34 293.21 303.94 314.62 325.28 335.96 346.68 357.47 368.33 379.27 390.31 401.44 412.68 424.02 - " 52.43°C) 273.40
s (kJ/(kg K)) 1.0645 1.0938 1.1225 1.1505 1.1781 1.2053 1.2320 1.2584 1.2844 1.3100 0.9054 0.9217 0.9566 0.9897 1.0214 1.0521 1.0819 1.1109 1.1392 1.1670 1.1943 1.2211 t.2475 1.2735 1.2992 1.3245 0.9003
262.17 272.87
283.10 295.31
0.9297 0.9658
283.29 293.55 303.73 313.88 324.05 334.25 344.50 354.82 365.22 375.71 386.29 396.96 407.73
307.10 318.63 330.02 341.32 352.59 363.86 375.15 386.49 397.89 409.36 420.90 432.53 444.24
0.9997 1.0319 1.0628 1.0927 1.1218 1.1501 1.1777 1.2048 1.2315 1.2576 1.2834 1.3088 1.3338
0.9080 0.9197 0.9539 0.9867 1.0182 1.0487 1.0784 1.1074 1.1358 1.1637 1.1910
(Contd.)
Appendix Table E3 T (°C)
709
(Continued) V (m3/kg)
130
0.04468
140 150 160
0.04599 0.04729 0.04857
Sat. 40 50 60 70 80 90 100 110 120 I30 140 150 160 170 180
0.02020 0.02029 0.02171 0.02301 0.02423 0.02538 0.02649 0.02755 0.02858 0.02959 0.03058 0.03154 0.03250 0.03344 0.03436 0.03528
S~. 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200
0.01208
u (kJ/kg)
h (kJ/kg)
338.76
370.04
1.2179
380.86 391.79 402.82 - 39.39oc)
1.2444 1.2706 1.2963
267.97 268.68 280.19 291.36 302.34 313.20 324.01 334.82 345.65 356.52 367.46 378.46 389.56 400.74 412.02 423.40 P = 1.60MPa(~at = 57.92°C) 256.00 275.33
0.9043 0.9066 0.9428 0.9768 1.0093 1.0405 1.0707 1.1000 1.1286 1.1567 1.1841 1.2111 1.2376 1.2638 1.2895 1.3149
348.66 358.68 368.82 P = 1.OOMPa(~at 247.77 248.39 258.48 268.35 278.11 287.82 297.53 307.27 317.06 326.93 336.88 346.92 357.06 367.31 377.66 388.12
0.01233 0.01340 0.01435 0.01521 0.01601 0.01677 0.01750 0.01820 0.01887 0.01953 0.02017 0.02080 0.02142 0.02203 0.02263
258.48 269.89 280.78 291.39 301.84 312.20 322.53 332.87 343.24 353.66 364.15 374.71 385.35 396.08 406.90
278.20 291.33 303.74 315.72 327.46 339.04 350.53 361.99 373.44 384.91 396.43 407.99 419.62 431.33 443.11
s (kJ/(kg K))
V(m3/kg)
u
(kJ/kg)
h (kJ/kg)
s (kJ/(kg K))
0.8982 0.9069 0.9457 0.9813 1 0148 1 0467 1 0773 1 1069 1 1357 1 1638 1 1912 1.2181 1.2445 1.2704 1.2960 1.3212
Table E4 Ideal-gas properties of air T (R) 360 380 400 420 440 460 480 500 520 537 540 560 580
H (Bm/lbm) 85.97 90.75 95.33 100.32 105.11 109.90 114.69 119.48 124.27 128.10 129.06 133.86 138.66
P,. 0.3363 0.4061 0.4858 0.5760 0.6776 0.7913 0.9182 1.0590 1.2147 1.3593 1.3860 1.5742 1.7800
U (Btu/lbm) 61.29 64.70 68.11 71.52 74.93 78.36 81.77 85.20 88.62 91.53 92.04 95.47 98.90
S (Btu/(lbm R)) 396.6 346.6 305.0 270.1 240.6 215.33 193.65 174.90 158.58 146.34 144.32 131.78 120.70
0.50369 0.51663 0.52890 0.54058 0.55172 0.56235 0.57255 0.58233 0.59173 0.59945 0.60078 0.60950 0.61793
(Contd.)
710 Table E4 T (R) 600 620 640 660 680 700 720 740 760 780 800 820 840 860 880 900 920 940 960 980 1000 1040 1080 1120 1160 1200 1240 1280 1320 1360 1400 1440 1480 1520 1560 1600 1650 1700 1750 1800 1850 1900 1950 2000 2050 2100 2150 2200 2250 2300 2350 2400 2450 2500 2550 2600 2650 2700
Appendix (Continued) H (Btu/lbm) 143.47 148.28 153.09 157.92 162.73 167.56 172.39 177.23 182.08 186.94 191.81 196.69 201.56 206.46 211.35 216.26 221.18 226.11 231.06 236.02 240.98 250.95 260.97 271.03 281.14 291.30 301.52 311.79 322.11 332.48 342.90 353.37 363.89 374.47 385.08 395.74 409.13 422.59 436.12 449.71 463.37 477.09 490.88 504.71 518.71 532.55 546.54 560.59 574.69 588.82 603.00 617.22 631.48 645.78 660.12 674.49 688.90 703.35
P~ 2.005 2.249 2.514 2.801 3.111 3.446 3.806 4.193 4.607 5.051 5.526 6.033 6.573 7.149 7.761 8.411 9.102 9.834 10.61 11.43 12.30 14.18 16.28 18.60 21.18 24.01 27.13 30.55 34.31 38.41 42.88 47.75 53.04 58.78 65.00 71.13 80.89 90.95 101.98 114.0 127.2 141.5 157.1 174.0 192.3 212.1 223.5 256.6 281.4 308.1 336.8 367.6 400.5 435.7 473.3 513.5 556.3 601.9
U (Btu/lbm) 102.34 105.78 109.21 112.67 116.12 119.58 123.04 126.51 129.99 133.47 136.97 140.47 143.98 147.50 151.02 154.57 158.12 161.68 165.26 168.83 172.43 179.66 186.93 194.25 201.63 209.05 216.53 224.05 231.63 239.25 246.93 254.66 262.44 270.26 278.13 286.06 296.03 306.06 316.16 326.32 336.55 346.85 357.20 367.61 378.08 388.60 399.17 409.78 420.46 431.16 441.91 452.70 463.54 474.40 485.31 496.26 507.25 518.26
Vr
S (Btu/(lbm R))
110.88 102.12 94.30 87.27 80.96 75.25 70.07 65.38 61.10 57.20 53.63 50.35 47.34 44.57 42.01 39.64 37.44 35.41 33.52 31.76 30.12 27.17 24.58 22.30 20.29 18.51 16.93 15.52 14.25 13.12 12.10 11.17 10.34 9.578 8.890 8.263 7.556 6.9294 6.357 5.847 5.388 4.974 4.598 4.258 3.949 3.667 3.410 3.176 2.961 2.765 2.585 2.419 2.266 2.125 1.996 1.876 1.765 1.662
0.62607 0.63395 0.64159 0.64902 0.65621 0.66321 0.67002 0.67665 0.68312 0.68942 0.69558 0.70160 0.70747 0.71323 0.71886 0.72438 0.72979 0.73509 0.74030 0.74540 0.75042 0.76019 0.76964 0.77880 0.78767 0.79628 0.80466 0.81280 0.82075 0.82848 0.83604 0.84341 0.85062 0.85767 0.86456 0.87130 0.87954 0.88758 0.89542 0.90308 0.91056 0.91788 0.92504 0.93205 0.93891 0.94564 0.95222 0.95919 0.96501 0.97123 0.97732 0.98331 0.98919 0.99497 1.00064 1.00623 1.01172 1.01712
(Contd.)
Appendix
711
Table E4 (Continued)
T(R)
P,.
H (Btu/lbm)
2750
717.83
2800 2850 2900 2950 3000 3050 3100 3150 3200 3250 3300 3350 3400 3450 3500 3550 3600 3650 3700 3750 3800 3850 3900 3950 4000 4050 4100 4150 4200 4300 4400 4500 4600 4700 4800 4900 5000 5100 5200 5300
732.33 746.88 761.45 776.05 790.68 8O5.34 820.03 834.75 849.48 864.24 879.02 893.83 908.66 923.52 938.40 953.30 968.21 983 15 998 11 10131 10281 1043 1 10581 1073 2 1088.3 1103.4 1118.5 1133.6 1148.7 1179.0 1209.4 1239.9 1270.4 1300.9 1331.5 1362.2 1392.9 1423.6 1454.4 1485.3
U (Btu/lbm)
650.4 702.0 756.7 814.8 876.4 941.4 1011 1083 1161 1242 1328 1418 1513 1613 1719 1829 1946 2068 2196 2330 2471 2618 2773 2934 3103 3280 3464 3656 3858 4067 4513 4997 5521 6089 6701 7362 8073 8837 9658 10.539 11.481
S (Btu/(lbm R))
V,.
529.31
1.566
1.02244
540.40 551.52 562.66 573.84 585.04 596.28 607.53 618.82 630.12 641.46 652.81 664.20 675.60 687.04 698.48 709.95 721.44 732.95 744.48 756.04 767.60 779.19 790.80 802.43 814.06 825.72 837.40 849.09 860.81 884.28 907.8l 931.39 955.04 978.73 1002.5 1026.3 1050.1 1074.0 1098.0 1122.0
1.478 1.395 1.318 1.247 1.180 1.118 1.060 1.006 0.955 0.907 0.8621 0.8202 0.7807 0.7436 0.7087 0.6579 0.6449 0.6157 0.5882 0.5621 0.5376 0.5143 0.4923 0.4715 0.4518 0.4331 0.4154 0.3985 0.3826 0.3529 0.3262 0.3019 0.2799 0.2598 0.2415 0.2248 0.2096 0.1956 0.1828 0.1710
1.02767 1.03282 1.03788 1.04288 1.04779 1.05264 1.05741 1.06212 1.06676 1.07134 1.07585 1.08031 1.08470 1.08904 1.09332 1.09755 1.10172 1 10584 1 10991 111393 1 11791 1 12183 1 12571 1.12955 1.13334 1.13709 1 14079 1 14446 114809 1 15522 1 16221 1 16905 1 17575 1.18232 1.18876 1.19508 1.20129 1.20738 1.21336 1.21923
Table E5 Ideal-gas properties of carbon dioxide, C02 T (K) 0 220 230 240 250 260 270 280 290 298 300
H (kJ/kmol)
U (kJ/kmol)
S (kJ/(kmol K))
T (K)
H (kJ/kmol)
U (kJ/kmol)
S (kJ/(kmol K))
0 6.601 6.938 7.280 7.627 7.979 8.335 8.697 9.063 9.364 9.431
0 4.772 5.026 5.285 5.548 5.817 6.091 6.369 6.651 6.885 6.939
0 202.966 204.464 205.920 207.337 208.717 210.062 211.376 212.660 213.685 213.915
600 610 620 630 640 650 660 670 680 690 700
22.280 22.754 23.231 23.709 24.190 24.674 25.160 25.648 26.138 26.631 27.125
17.291 17.683 18.076 18.471 18.869 19.270 19.672 20.078 20.484 20.894 21.305
243.199 243.983 244.758 245.524 246.282 247.032 247.773 248.507 249.233 249.952 250.663
(Contd.)
Appendix
712 Table E5
(Continued)
T (K)
H (kJ/kmol)
U (kJ/kmol)
S (kJ/(kmol K))
r (K)
H (kJ/kmol)
U (kJ/kmol)
S (kJ/(kmol K))
310 320 330 340 350 360 370 380 390 400 410 420 430 440 450 460 470 480 490 500 510 520 530 540 550 560 570 580 590 1000 1020 1040 1060 1080 1100 1120 1140 1160 1180 1200 1220 1240 1260 1280 1300 1320 1340 1360 1380 1400 1420 1440 1460 1480 1500 1520 1540 1560
9.807 10.186 10.570 10.959 11.351 11.748 12.148 12.552 12.960 13.372 13.787 14.206 14.628 15.054 15.483 15.916 16.351 16.791 17.232 17.678 18.126 18.576 19.029 19.485 19.945 20.407 20.870 21.337 21.807 42.769 43,859 44.953 46.051 47.153 48.258 49.369 50.484 51.602 52.724 53.848 54.977 56.108 57.244 58.381 59.522 60.666 61.813 62.963 64.116 65.271 66.427 67.586 68.748 66.911 71.078 72.246 73.417 74.590
7.230 7.526 7.826 8.131 8.439 8.752 9.068 9.392 9.718 10.046 10.378 10.714 11.053 11.393 11.742 12.091 12.444 12.800 13.158 13.521 13.885 14.253 14.622 14.996 15.372 15.751 16.131 16.515 16.902 34.455 35.378 36.306 37.238 38.174 39.112 40.057 41.006 41.957 42.913 43.871 44.834 45.799 46.768 47.739 48.713 49.691 50.672 51.656 52.643 53.631 54.621 55.614 56.609 57.606 58.606 59.609 60.613 61.620
215.146 216.351 217.534 218.694 219.831 220.948 222.044 223.122 224.182 225.225 226.250 227.258 228.252 229.230 230.194 231.144 232.080 233.004 233.916 234.814 235.700 236.575 237.439 238.292 239.135 239.962 240.789 241.602 242.405 269.215 270.293 271.354 272.400 273.430 274.445 275.444 276.430 277.403 278.361 297.307 280.238 281.158 282.066 282.962 283.847 284.722 285.586 286.439 287.283 288.106 288.934 289.743 290.542 291.333 292.114 292.888 292.654 294.411
710 720 730 740 750 760 770 780 790 800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990 1760 1780 1800 1820 1840 1860 1880 1900 1920 1940 1960 1980 2000 2050 2100 2150 2200 2250 2300 2350 2400 2450 2500 2550 2600 2650 2700 2750 2800
27.622 28.121 28.622 29.124 29.629 20.135 30.644 31.154 31.665 32.179 32.694 33.212 33.730 34.251 34.773 35.296 35.821 36.347 36.876 37.405 37.935 38.467 39.000 39.535 40.070 40.607 41.145 41.685 42.226 86.420 87.612 88.806 90.000 91.196 92.394 93.593 94.793 95.995 97.197 98.401 99.606 100.804 103.835 106.864 109.898 112.939 115.984 119.035 122.091 125.152 128.219 131.290 134.368 137.449 140.533 143.620 146.713 149.808
21.719 22.134 22.522 22.972 23.393 23.817 24.242 24.669 25.097 25.527 25.959 26.394 26.829 27.267 27.706 28.125 28.588 29.031 29.476 29.922 30.369 30.818 31.268 31.719 32.171 32.625 33.081 33.537 33.995 71.787 72.812 73.840 74.868 75.897 76.929 77.962 78.996 80.031 81.067 82.105 83.144 84.185 86.791 89.404 92.023 94.648 97.277 99.912 102.552 105.197 107.849 110.504 113.166 115.832 118.500 121.172 123.849 126.528
251.368 252.065 252.755 253.439 254.117 254.787 255.452 256.110 256.762 257.408 258.048 258.682 259.311 259.934 260.551 261.164 261.770 262.371 262.968 263.559 264.146 264.728 265.304 265.877 266.444 267.007 267.566 268.119 268.670 301.543 302.217 302.884 303.544 304.198 304.845 305.487 306.122 306.751 307.374 307.992 308.604 309.210 310.701 312.160 313.589 314.988 316.356 317.695 319.011 320.302 321.566 322.808 324.026 325.222 326.396 327.549 328.684 329.800
(Contd.)
Appendix Table E5 T (K) 1580 1600 1620 1640 1660 1680 1700 1720 1740
713
(Continued) H (kJ/kmol) 76.767 76.944 78.123 79.303 80.486 81.670 82.856 84.043 85.231
U (kJ/kmol)
S (kJ/(kmol K))
62.630 63.741 64.653 65.668 66.592 67.702 68.721 69.742 70.764
295.161 295.901 296.632 297.356 298.072 298.781 299.482 300.177 300.863
T (K) 2850 2900 2950 3000 3050 3100 3150 3200 3250
H (kJ/kmol)
U (kJ/kmol)
152.908 156.009 159.117 162.226 165.341 168.456 171.576 174.695 177.822
129.212 131.898 134.589 137.283 139.982 142.681 145.385 148.089 150.801
S (kJ/(kmol K)) 330.896 331.975 333.037 334.084 335.114 336.126 337.124 338.109 339.069
APPENDIX F: THE LEE/KESLER GENERALIZED CORRELATION TABLES Table F1 Values of Z o Tr
Pr 0.0100
0.0500
0.1000
0.2000
0.4000
0.6000
0.8000
1.0000
0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.93 0.95 0.97 0.98
0.0029 0.0026 0.0024 0.0022 0.0021 0.9804 0.9849 0.9881 0.9904 0.9922 0.9935 0.9946 0.9954 0.9959 0.9961 0.9963 0.9965
0.0145 0.0130 0.0119 0.0110 0.0103 0.0098 0.0093 0.9377 0.9504 0.9598 0.9669 0.9725 0.9768 0.9790 0.9803 0.9815 0.9821
0.0290 0.0261 0.0239 0.0221 0.0207 0.0195 0.0186 0.0178 0.8958 0.9165 0.9319 0.9436 0.9528 0.9573 0.9600 0.9625 0.9637
0.0579 0.0522 0.0477 0.0442 0.0413 0.0390 0.0371 0.0356 0.0344 0.0336 0.8539 0.8810 0.9015 0.9115 0.9174 0.9227 0.9523
0.1158 0.1043 0.0953 0.0882 0.0825 0.0778 0.0741 0.0710 0.0687 0.0670 0.0661 0.0661 0.7800 0.8059 0.8206 0.8338 0.8398
0.1737 0.1564 0.1429 0.1322 0.1236 0.1166 0.1109 0.1063 0.1027 0.1001 0.0985 0.0983 0.1006 0.6635 0.6967 0.7240 0.7360
0.2315 0.2084 0.1904 0.1762 0.1647 0.1553 0.1476 0.1415 0.1366 0.1330 0.1307 0.1301 0.1321 0.1359 0.1410 0.5580 0.5887
0.2892 0.2604 0.2379 0.2200 0.2056 0.1939 0.1842 0.1765 0.1703 0.1656 0.1626 0.1614 0.1630 0.1664 0.1705 0.1779 0.1844
0.99 1.00 1.01 1.02 1.05 1.10 1.15 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.20 2.40 2.60 2.80 3.00 3.50 4.00
0.9966 0.9967 0.9968 0.9969 0.9971 0.9975 0.9978 0.9981 0.9985 0.9988 0.9991 0.9993 0.9994 0.9995 0.9996 0.9997 0.9998 0.9999 1.0000 1.0000 1.0000 1.0001 1.0001
0.9826 0.9832 0.9837 0.9842 0.9855 0.9874 0.9891 0.9904 0.9926 0.9942 0.9954 0.9964 0.9971 0.9977 0.9982 0.9986 0.9992 0.9996 0.9998 1.0000 1.0002 1.0004 1.0005
0.9648 0.9659 0.9669 0.9679 0.9707 0.9747 0.9780 0.9808 0.9852 0.9884 0.9909 0.9928 0.9943 0.9955 0.9964 0.9972 0.9983 0.9991 0.9997 1.0001 1.0004 1.0008 1.0010
0.9277 0.9300 0.9322 0.9343 0.9401 0.9485 0.9554 0.9611 0.9702 0.9768 0.9818 0.9856 0.9886 0.9910 0.9929 0.9944 0.9967 0.9983 0.9994 1.0002 1.0008 1.0017 1.0021
0.8455 0.8509 0.8561 0.8610 0.8743 0.8930 0.9081 0.9205 0.9396 0.9534 0.9636 0.9714 0.9775 0.9823 0.9861 0.9892 0.9937 0.9969 0.9991 1.0007 1.0018 1.0035 1.0043
0. 7471 0.7574 0.7671 0.7761 0.8002 0.8323 0.8576 0.8779 0.9083 0.9298 0.9456 0.9575 0.9667 0.9739 0.9796 0.9842 0.9910 0.9957 0.9990 1.0013 1.0030 1.0055 1.0066
0.6138 0.6355 0.6542 0.6710 0.7130 0.7649 0.8032 0.8330 0. 8764 0.9062 0.9278 0.9439 0.9563 0.9659 0.9735 0.9796 0.9886 0.9948 0.9990 1.0021 1.0043 1.0075 1.0090
0.1959 0.2901 0.4648 0.5146 0.6026 0.6880 0.7443 0.7858 0. 8438 0.8827 0.9103 0.9308 0.9463 0.9583 0.9678 0.9754 0.9865 0.9941 0.9993 1.0031 1.0057 1.0097 1.0115
Appendix
714 Table F2 Values of Z 1
rr
0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.93 0.95 0.97 0.98 0.99 1.00 1.01 1.02 1.05 1.10 1.15 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.20 2.40 2.60 2.80 3.00 3.50 4.00
er 0.0100
0.0500
O.1000
0.2000
0.4000
0.6000
0.8000
1.0000
-0.0008 -0.0009 -0.0010 -0.0009 -0.0009 -0.0314 -0.0205 -0.0137 -0.0093 -0.0064 -0.0044 -0.0029 -0.0019 -0.0015 -0.0012 -0.0010 -0.0009 -0.0008 -0.0007 -0.0006 -0.0005 -0.0003 0.0000 0.0002 0.0004 0.0006 0.0007 0.0008 0.0008 0.0008 0.0008 0.0008 0.0008 0.0007 0.0007 0.0007 0.0006 0.0006 0.0005 0.0005
-0.0040 -0.0046 -0.0048 -0.0047 -0.0045 -0.0043 -0.0041 -0.0772 -0.0507 -0.0339 -0.0228 -0.0152 -0.0099 -0.0075 -0.0062 -0.0050 -0.0044 -0.0039 -0.0034 -0.0030 -0.0026 -0.0015 0.0000 0.0011 0.0019 0.0030 O.0036 0.0039 0.0040 0.0040 0.0040 0.0040 0.0039 0.0037 0.0035 0.0033 0.0031 0.0029 0.0026 0.0023
-0.0081 -0.0093 -0.0095 -0.0094 -0.0090 -0.0086 -0.0082 -0.0078 -0.1161 -0.0744 -0.0487 -0.0319 -0.0205 -0.0154 -0.0126 -0.0101 -0.0090 -0.0079 -0.0069 -0.0060 -0.0051 -0.0029 0.0001 0.0023 0.0039 0.0061 0.0072 0.0078 0.0080 0.0081 0.0081 0.0079 0.0078 0.0074 0.0070 0.0066 0.0062 0.0059 0.0052 0.0046
-0.0161 -0.0185 -0.0190 -0.0187 -0.0181 -0.0172 -0.0164 -0.0156 -0.0148 -0.0143 -0.1160 -0.0715 -0.0442 -0.0326 -0.0262 -0.0208 -0.0184 -0.0161 -0.0140 -0.0120 -0.0102 -0.0054 0.0007 0.0052 0.0084 0.0125 0.0147 0.0158 0.0162 0.0163 0.0162 0.0159 0.0155 0.0147 0.0139 0.0131 0.0124 0.0117 0.0103 0.0091
-0.0323 -0.0370 -0.0380 -0.0374 -0.0360 -0.0343 -0.0326 -0.0309 -0.0294 -0.0282 -0.0272 -0.0268 -0.1118 -0.0763 -0.0589 -0.0450 -0.0390 -0.0335 -0.0285 -0.0240 -0.0198 -0.0092 0.0038 0.0127 0.0190 0.0267 0.0306 0.0323 0.0330 0.0329 0.0325 0.0318 0.0310 0.0293 0.0276 0.0260 0.0245 0.0232 0.0204 0.0182
-0.0484 -0.0554 -0.0570 -0.0560 -0.0539 -0.0513 -0.0487 -0.0461 -0.0438 -0.0417 -0.0401 -0.0391 -0.0396 -0.1662 -0.1110 -0.0770 -0.0641 -0.0531 -0.0435 -0.0351 -0.0277 -0.0097 0.0106 0.0237 0.0326 0.0429 0.0477 0.0497 0.0501 0.0497 0.0488 0.0477 0.0464 0.0437 0.0411 0.0387 0.0365 0.0345 0.0303 0.0270
-0.0645 -0.0738 -0.0758 -0.0745 -0.0716 -0.0682 -0.0646 -0.0611 -0.0579 -0.0550 -0.0526 -0.0509 -0.0503 -0.0514 -0.0540 -0.1647 -0.1100 -0.0796 -0.0588 -0.0429 -0.0303 -0.0032 0.0236 0.0396 0.0499 0.0612 0.0661 0.0677 0.0677 0.0667 0.0652 0.0635 0.0617 0.0579 0.0544 0.0512 0.0483 0.0456 0.0401 0.0357
-0.0806 -0.0921 -0.0946 -0.0929 -0.0893 -0.0849 -0.0803 -0.0759 -0.0718 -0.0681 -0.0648 -0.0622 -0.0604 -0.0602 -0.0607 -0.0623 -0.0641 -0.0680 -0.0879 -0.0223 -0.0062 -0.0220 0.0476 0.0625 0.0719 0.0819 0.0857 0.0864 0.0855 0.0838 0.0814 0.0792 0.0767 0.0719 0.0675 0.0634 0.0598 0.0565 0.0497 0.0443
1.0000
1.2000
1.5000
2.0000
3.0000
5.0000
7.0000
10.000
0.2892 0.2604 0.2379 0.2200 0.2056 0.1939 0.1842 O. 1765
0.3479 0.3123 0.2853 0.2638 0.2465 0.2323 0.2207 0.2113
0.4335 0.3901 0.3563 0.3294 0.3077 0.2899 0.2753 0.2634
0.5775 0.5195 0.4744 0.4384 0.4092 0.3853 0.3657 0.3495
0.8648 0.7775 0.7095 0.6551 0.6110 0.5747 0.5446 0.5197
1.4366 1.2902 1.1758 1.0841 1.0094 0.9475 0.8959 0.8526
2.0048 1.7987 1.6373 1.5077 1.4017 1.3137 1.2398 1.1773
Table F3 VaLues of Z °
0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65
2.8507 2.5539 2.3211 2.1338 1.9801 1.8520 1.7440 1.6519
(Contd.)
Appendix Table F3
715
(Continued)
Tr
/Or 1.0000
1.2000
1.5000
2.0000
3.0000
5.0000
7.0000
0.70
0.1703
0.2038
0.2538
0.3364
0.75
0.1656
0.1981
0.2464
0.3260
0.80
0.1626
0.1942
0.2411
0.85
0.1614
0.1924
0.90
0.1630
0.93
10.000
0.4991
0.8161
1.1341
1.5729
0.4823
0.7854
1.0787
1.5047
0.3182
0.4690
0.7598
1.0400
1.4456
0.2382
0.3132
0.4591
0.7388
1.0071
1.3943
0.1935
0.2383
0.3114
0.4527
0.7220
0.9793
1.3496
0.1664
0.1963
0.2405
0.3122
0.4507
0.7138
0.9648
1.3257
0.95
0.1705
0.1998
0.2432
0.3138
0.4501
0.7092
0.9561
1.3108
0.97
0.1779
0.2055
0.2474
0.3164
0.4504
0.7052
0.9480
1.2968
0.98
0.1844
0.2097
0.2503
0.3182
0.4508
0.7035
0.9442
1.2901
0.99
0.1959
0.2154
0.2538
0.3204
0.4514
0.7018
0.9406
1.2835
1.00
0.2901
0.2237
0.2583
0.3229
0.4522
0.7004
0.9372
1.2772
1.01
0.4648
0.2370
0.2640
0.3260
0.4533
0.6991
0.9339
1.2710
1.02
0.5146
0.2629
0.2715
0.3297
0.4547
0.6980
0.9307
1.2650
1.05
0.6026
0.4437
0.3131
0.3452
0.4604
0.6956
0.9222
1.2481
1.10
0.6880
0.5984
0.4580
0.3953
0.4770
0.6950
0.9110
1.15
0.7443
0.6803
0.5798
0.4760
0.5042
0.6987
0.9033
1.2232 1.2021
1.20
0.7858
0.7363
0.6605
0.5605
0.5425
0.7069
0.8990
1 1844
1.30
0.8438
0.8111
0.7624
0.6908
0.6344
0.7358
0.8998
1 1580
1.40
0.8827
0.8595
0.8256
0.7753
0.7202
0.7661
0.9112
1 1419
1.50
0.9103
0.8933
0.8689
0.8328
0.7887
0.8200
0.9297
1 1339 1 1320
1.60
0.9308
0.9180
0.9000
0.8738
0.8410
0.8617
0.9518
1.70
0.9463
0.9367
0.9234
0.9043
0.8809
0.8984
0.9745
1 1343
1.80
0.9583
0.9511
0.9413
0.9275
0.9118
0.9297
0.9961
1 1391
1.90
0.9678
0.9624
0.9552
0.9456
0.9359
0.9557
1.0157
1 1452
2.00 2.20
0.9754 0.9856
0.9715 0.9847
0.9664 0.9826
0.9599 0.9806
0.9550 0.9827
0.9772 1.0094
1.0328 1.0600
1.1516 1 1635
2.40
0.9941
0.9936
0.9935
0.9945
1.0011
1.0313
1.0793
1 1728
2.60
0.9993
0.9998
1.0010
1.0040
1.0137
1.0463
1.0926
1 1792
2.80
1.0031
1.0042
1.0063
1.0106
1.0223
1.0565
1.1016
1 1830
3.00
1.0057
1.0074
1.0101
1.0153
1.0284
1.0635
1.1075
1 1848
3.50
1.0097
1.0120
1.0156
1.0221
1.0368
1.0723
1.1138
1 1834
4.00
1.0115
1.0140
1.0179
1.0249
1.0401
1.0747
1.1136
1 1773
Table F4 Values of Z 1
~
Pr 1.0000
1.2000
1.5000
2.0000
3.0000
5.0000
7.0000
10.000
0.30
-0.0806
- 0.0966
- O. 1207
- O. 1608
- 0.2407
-0.3996
-0.5572
- 0.7915
0.35 0.40
-0.0921 -0.0946
-0.1105 -0.1134
-0.1379 -0.1414
-0.1834 -0.1879
-0.2738 -0.2799
-0.4523 -0.4603
-0.6279 -0.6365
-0.8663 -0.8936
0.45
-0.0929
-0.1113
-0.1387
-0.1840
-0.2734
-0.4475
-0.6162
-0.8608
0.50
-0.0893
-0.1069
-0.1330
-0.1762
-0.2611
-0.4253
-0.5831
-0.8099 -0.7521
0.55
-0.0849
-0.1015
-0.1263
-0.1669
-0.2465
-0.3991
-0.5446
0.60
-0.0803
-0.0960
-0.1192
-0.1572
-0.2312
-0.3718
-0.5047
-0.6928
0.65
-0.0759
-0.0906
-0.1122
-0.1476
-0.2160
-0.3447
-0.4653
-0.6346
0.70
-0.0718
-0.0855
-0.1057
-0.1385
-0.2013
-0.3184
-0.4270
-0.5785
0.75
-0.0681
-0.0808
-0.0996
-0.1298
-0.1872
-0.2929
-0.3901
-0.5250 -0.4740
0.80
-0.0648
-0.0767
-0.0940
-0.1217
-0.1736
-0.2682
-0.3545
0.85
-0.0622
-0.0731
-0.0888
-0.1138
-0.1602
-0.2439
-0.3201
-0.4254
0.90
-0.0604
-0.0701
-0.0840
-0.1059
-0.1463
-0.2195
-0.2862
-0.3788 -0.3516
0.93
-0.0602
-0.0687
-0.0810
-0.1007
-0.1374
-0.2045
-0.2661
0.95
-0.0607
-0.0678
-0.0788
-0.0967
-0.1310
-0.1943
-0.2526
-0.3339
0.97
-0.623
-0.0669
-0.0759
-0.0921
-0.1240
-0.1837
-0.2391
-0.3163
(Contd.)
716 Table F4
Appendix
(Continued)
rr 1.0000
1.2000
0.98
-0.0641
- 0.0661
0.99
-0.0680
-0.0646
1.00
-0.0879
-0.0609
1.01
-0.0223
-0.0473
1.02
-0.0062
1.5000
2.0000
3.0000
5.0000
7.0000
-0.0740
-0.0893
-0.0715
-0.0861
-0.0678 -0.0621
-0.0227
10.000
-0.1202
-0.1783
-0.2322
-0.3075
-0.1162
-0.1728
-0.2254
-0.2989
-0.0824
-0.1118
-0.1672
-0.2185
-0.2902
-0.0778
-0.1072
-0.1615
-0.2116
-0.2816
-0.0524
-0.0722
-0.1021
-0.1556
-0.2047
-0.2731 -0.2476
1.05
0.0220
0.1059
0.0451
-0.0432
-0.0838
-0.1370
-0.1835
1.10
0.0476
0.0897
0.1630
0.0698
-0.0373
-0.1021
-0.1469
-0.2056
1.15
0.0625
0.0943
0.1548
0.1667
0.0332
-0.0611
-0.1084
-0.1642
1.20
0.0719
0.0991
0.1477
0.1990
0.1095
-0.0141
-0.0678
-0.1231
1.30
0.0819
0.1048
0.1420
0.1991
0.2079
0.0875
0.0176
-0.0423
1.40
0.0857
0.1063
0.1383
0.1894
0.2397
0.1737
0.1008
0.0350
1.50
0.0854
0.1055
0.1345
0.1806
0.2433
0.2309
0.1717
0.1058
1.60
0.0855
0.1035
0.1303
0.1729
0.2381
0.2631
0.2255
0.1673
1.70
0.0838
0.1008
0.1259
0.1658
0.2305
0.2788
0.2628
0.2179
1.80
0.0816
0.0978
0.1216
0.1593
0.2224
0.2846
0.2871
0.2576
1.90
0.0792
0.0947
0.1173
0.1532
0.2144
0.2848
0.3017
0.2876
2.00
0.0767
0.0916
0.1133
0.1476
0.2069
0.2819
0.3097
0.3096
2.20
0.0719
0.0857
0.1057
0.1374
0.1932
0.2720
0.3135
0.3355
2.40
0.675
0.0803
0.0989
0.1285
0.1812
0.2602
0.3089
0.3459
2.60
0.0634
0.0754
0.0929
0.1207
0.1706
0.2484
0.3009
0.3475
2.80
0.0598
0.0711
0.0876
0.1138
0.1613
0.2372
0.2915
0.3443
3.00
0.0535
0.0672
0.0828
0.1076
0.1529
0.2268
0.2817
0.3385
3.50
0.0497
0.0591
0.0728
0.0949
0.1356
0.2042
0.2584
0.3194
4.00
0.0443
0.0527
0.0651
0.0849
0.1219
0.1857
0.2378
0.2994
Table F5 Values of
(HR)°/RTc
Tr
Pr 0.0100
0.0500
0.1000
0.2000
0.4000
0.6000
0.8000
1.0000
0.30
-6.045
-6.043
-6.040
-6.034
-6.022
-6.011
- 5.999
- 5.987
0.35
-5.906
-5.904
-5.901
-5.895
-5.882
-5.870
-5.858
- 5.845
0.40
- 5.763
- 5.761
- 5.757
- 5.751
- 5.738
- 5.726
- 5.713
- 5.700
0.45
- 5.615
- 5.612
- 5.609
- 5.603
- 5.590
- 5.577
- 5.564
- 5.551
0.50
- 5.465
- 5.463
- 5.459
- 5.453
- 5.440
- 5.427
- 5.414
- 5.401
0.55
-0.032
-5.312
-5.309
-5.303
- 5.290
-5.278
-5.265
-5.252
0.60
-0.027
-5.162
- 5.159
-5.153
-5.141
-5.129
-5.116
- 5.104
0.65
-0.023
-0.118
-5.008
-5.002
-4.991
-4.980
-4.968
-4.956
0.70
-0.020
-0.101
-0.213
-4.848
-4.838
-4.828
-4.818
-4.808
0.75
-0.017
-0.088
-0.183
-4.687
-4.679
-4.672
-4.664
-4.655
0.80
-0.015
-0.078
-0.160
-0.345
-4.507
-4.504
-4.499
-4.494
0.85
-0.014
-0.069
-0.141
-0.300
-4.309
-4.313
-4.316
-4.316
0.90
-0.012
-0.062
-0.126
-0.264
-0.596
-4.074
-4.094
-4.108
0.93
-0.011
-0.058
-0.118
-0.246
-0.545
-0.960
-3.920
-3.953
0.95
-0.011
-0.056
-0.113
-0.235
-0.516
-0.885
-3.763
-3.825
0.97
-0.011
-0.054
-0.109
-0.225
-0.490
-0.824
- 1.356
-3.658
0.98
-0.010
-0.053
-0.107
-0.221
-0.478
-0.797
- 1.273
-3.544
0.99
-0.010
-0.052
-0.105
-0.216
-0.466
-0.773
- 1.206
-3.376
1.00
-0.010
-0.051
-0.103
-0.212
-0.455
-0.750
- 1.151
-2.584
1.01
-0.010
-0.050
-0.101
-0.208
-0.445
-0.721
- 1.102
- 1.796
1.02
-0.010
-0.049
-0.099
-0.203
-0.434
-0.708
- 1.060
- 1.627
1.05
-0.009
-0.046
-0.094
-0.192
-0.407
-0.654
-0.955
- 1.359
1.10
-0.008
-0.042
-0.086
-0.175
-0.367
-0.581
-0.827
- 1.120
1.15
-0.008
-0.039
-0.079
-0.160
-0.334
-0.523
-0.732
-0.968
(Contd.)
Appendix Table F5
717
(Continued)
Tr
Pr 0.0100
0.0500
0.1000
0.2000
0.4000
0.6000
0.8000
1.0000
1.20
-0.007
-0.036
-0.073
-0.148
-0.305
-0.474
-0.657
1.30
-0.006
-0.031
-0.063
-0.127
-0.259
-0.399
-0.545
-0.857 -0.698
1.40
-0.005
-0.027
-0.055
-0.110
-0.224
-0.341
-0.463
-0.588
1.50
-0.005
-0.024
-0.048
-0.097
-0.196
-0.297
-0.400
-0.505
1.60
- 0.004
- 0.021
- 0.043
- 0.086
- 0.173
- 0.261
- 0.350
- 0.440
1.70
-0.004
-0.019
-0.038
-0.076
-0.153
-0.231
-0.309
-0.387
1.80
-0.003
-0.017
-0.034
-0.068
-0.137
-0.206
-0.275
-0.344 -0.307
1.90
-0.003
-0.015
-0.031
-0.062
-0.123
-0.185
-0.246
2.00
-0.003
-0.014
-0.028
-0.056
-0.111
-0.167
-0.222
-0.276
2.20
- 0.002
- 0.012
- 0.023
- 0.046
- 0.092
- 0.137
- 0.182
- 0.226
2.40
-0.002
-0.010
-0.019
-0.038
-0.076
-0.114
-0.150
-0.187
2.60
-0.002
-0.008
-0.016
-0.032
-0.064
-0.095
-0.125
-0.155
2.80
-0.001
-0.007
-0.014
-0.027
-0.054
-0.080
-0.105
-0.130
3.00
-0.001
-0.006
-0.011
-0.023
-0.045
-0.067
-0.088
-0.109
3.50
-0.001
-0.004
-0.007
-0.015
-0.029
-0.043
-0.056
-0.069
4.00
- 0.000
-0.002
-0.005
-0.009
-0.017
-0.026
-0.033
-0.041
Table F6 Values of
(HR)I/RTc
Tr
Pr 0.0100
0.0500
0.1000
0.2000
0.4000
0.6000
0.8000
1.0000
0.30
- 11.098
- 11.096
- 11.095
- 11.091
- 11.083
- l 1.076
- 11.069
- 11.062
0.35
- 10.656
- 10.655
- 10.654
- 10.653
- 10.650
- 10.646
- 10.643
- 10.640
0.40
- 10.121
- 10.121
- 10.121
- 10.120
- 10.121
- 10.121
- 10.121
- 10.121
0.45
-9.515
-9.515
-9.516
-9.517
-9.519
-9.521
-9.523
-9.525
0.50
- 8.868
- 8.869
- 8.870
- 8.872
- 8.876
- 8.880
- 8.884
- 8.888
0.55
-0.080
-8.211
-8.212
-8.215
-8.221
-8.226
-8.232
-8.238 - 7.596
0.60
-0.059
- 7.568
- 7.570
- 7.573
- 7.579
- 7.585
- 7.591
0.65
-0.045
-0.247
-6.949
-6.952
-6.959
-6.966
-6.973
-6.980
0.70
- 0.034
- 0.185
- 0.415
- 6.360
- 6.367
- 6.373
- 6.381
- 6.388
0.75
-0.027
-0.142
-0.306
-5.796
-5.802
-5.809
-5.816
-5.824
0.80
-0.021
-0.110
-0.234
-0.542
-5.266
-5.271
-5.278
-5.285
0.85
-0.017
-0.087
-0.182
-0.401
-4.753
-4.754
-4.758
-4.763
0.90
- 0.014
- 0.070
- 0.144
- 0.308
- 0.751
-4.254
-4.248
-4.249
0.93
-0.012
-0.061
-0.126
-0.265
-0.612
- 1.236
-3.942
-3.934
0.95
-0.011
-0.056
-0.115
-0.241
-0.542
-0.994
-3.737
-3.712
0.97
-0.010
-0.052
-0.105
-0.219
-0.483
-0.837
- 1.616
-3.470
0.98
-0.010
-0.050
-0.101
-0.209
-0.457
-0.776
- 1.324
-3.332
0.99
-0.009
- 0.048
-0.097
-0.200
-0.433
-0.722
- 1.154
-3.164 - 2.471
1.00
- 0.009
- 0.046
- 0.093
- 0.191
- 0.410
- 0.675
- 1.034
1.01
-0.009
-0.044
-0.089
-0.183
-0.389
-0.632
-0.940
- 1.375
1.02
-0.008
-0.042
-0.085
-0.175
-0.370
-0.594
-0.863
- 1.180
1.05
-0.007
-0.037
-0.075
-0.153
-0.318
-0.498
-0.691
-0.877
1.10
-0.006
-0.030
-0.061
-0.123
-0.251
-0.381
-0.507
-0.617
1.15
-0.005
-0.025
-0.050
-0.099
-0.199
-0.296
-0.385
-0.459
1.20
-0.004
-0.020
-0.040
-0.080
-0.158
-0.232
-0.297
-0.349
1.30
-0.003
-0.013
-0.026
-0.052
-0.100
-0.142
-0.177
-0.203
1.40
-0.002
-0.008
-0.016
-0.032
-0.060
-0.083
-0.100
-0.111
1.50
-0.001
-0.005
-0.009
-0.018
-0.032
-0.042
-0.048
-0.049
1.60
- 0.000
-0.002
-0.004
-0.007
-0.012
-0.013
-0.011
-0.005
-0.009
-0.017
-0.027
1.70
- 0.000
- 0.000
- 0.000
- 0.000
-0.003
1.80
0.000
0.001
0.003
0.006
0.015
0.025
0.037
0.051
1.90
0.001
0.003
0.005
0.011
0.023
0.037
0.053
0.070
(Contd.)
Appendix
718 Table F6
(Continued)
Tr
Pr 0.0100
0.0500
0.1000
0.2000
0.4000
0.6000
0.8000
1.0000
2.00
0.001
0.003
0.007
0.015
0.030
0.047
0.065
2.20
0.001
0.005
0.010
0.020
0.040
0.062
0.083
0.085 0.106
2.40
0.001
0.006
0.012
0.023
0.047
0.071
0.095
0.120
2.60
0.001
0.006
0.013
0.026
0.052
0.078
0.104
0.130
2.80
0.001
0.007
0.014
0.028
0.055
0.082
0.110
0.137
3.00
0.001
0.007
0.014
0.029
0.058
0.086
0.114
0.142
3.50
0.002
0.008
0.016
0.031
0.062
0.092
0.122
0.152
4.00
0.002
0.008
0.016
0.032
0.064
0.096
0.127
0.158
3.0000
5.0000
7.0000
Table F7 Values of
(HR)°/RTc
r~
Pr 1.0000
1.2000
1.5000
2.0000
10.000
0.30
- 5.987
- 5.975
- 5.957
- 5.927
- 5.868
- 5.748
- 5.628
- 5.446
0.35
-5.845
-5.833
-5.814
-5.783
- 5.721
-5.595
-5.469
-5.278
0.40
- 5.700
- 5.687
- 5.668
- 5.636
- 5.572
- 5.442
- 5.311
- 5.113
0.45
- 5.551
- 5.538
- 5.519
- 5.486
- 5.421
- 5.288
- 5.154
- 5.950
0.50
- 5.401
- 5.388
- 5.369
- 5.336
- 5.279
- 5.135
-4.999
-4.791
0.55
- 5.252
- 5.239
- 5.220
- 5.187
- 5.121
-4.986
-4.849
-4.638
0.60
- 5.104
- 5.091
- 5.073
- 5.041
-4.976
-4.842
-4.794
-4.492
0.65
-4.956
-4.949
-4.927
-4.896
-4.833
-4.702
-4.565
-4.353
0.70
-4.808
-4.797
-4.781
-4.752
-4.693
-4.566
-4.432
-4.221
0.75
-4.655
-4.646
-4.632
-4.607
-4.554
-4.434
-4.393
-4.095
0.80
-4.494
-4.488
-4.478
-4.459
-4.413
-4.303
-4.178
-3.974
0.85
-4.316
-4.316
-4.312
-4.302
-4.269
-4.173
-4.056
-3.857
0.90
-4.108
-4.118
-4.127
-4.132
-4.119
-4.043
-3.935
-3.744
0.93
-3.953
-3.976
-4.000
-4.020
-4.024
-3.963
-3.863
-3.678
0.95
- 3.825
- 3.865
- 3.904
- 3.940
- 3.958
- 3.910
- 3.815
- 3.634
0.97
-3.658
-3.732
-3.796
-3.853
-3.890
-3.856
-3.767
-3.591
0.98
- 3.544
- 3.652
- 3.736
- 3.806
- 3.854
- 3.829
- 3.743
- 3.569
0.99
- 3.376
- 3.558
- 3.670
- 3.758
- 3.818
- 3.801
- 3.719
- 3.548
1.00
- 2.584
- 3.441
- 3.598
- 3.706
- 3.782
- 3.774
- 3.695
- 3.526
1.01
- 1.796
-3.283
-3.516
-3.652
-3.744
-3.746
-3.671
-3.505
1.02
- 1.627
-3.039
-3.422
-3.595
-3.705
-3.718
-3.647
-3.484
1.05
- 1.359
-2.034
-3.030
-3.398
-3.583
-3.632
-3.575
-3.420 - 3.315
1.10
-
- 1.487
- 2.203
- 2.965
- 3.353
- 3.484
- 3.453
1.15
-0.968
1.120
- 1.239
- 1.719
- 2.479
- 3.091
- 3.329
- 3.329
- 3.211
1.20
-0.857
- 1.076
- 1.443
- 2.079
- 2.801
- 3.166
- 3.202
- 3.107
1.30
-0.698
-0.860
- 1.116
- 1.560
-2.274
-2.825
-2.942
-2.899
1.40
-0.588
-0.716
-0.915
- 1.253
- 1.857
-2.486
-2.679
-2.692
1.50
-0.505
-0.611
-0.774
- 1.046
- 1.549
-2.175
-2.421
-2.486
1.60
-0.440
-0.531
-0.667
-0.894
- 1.318
- 1.904
-2.177
-2.285
1.70
-0.387
-0.446
-0.583
-0.777
- 1.139
- 1.672
- 1.953
-2.091
1.80
-0.344
-0.413
-0.515
-0.683
-0.996
- 1.476
- 1.751
- 1.908
1.90
-0.307
-0.368
-0.458
-0.606
-0.880
- 1.309
- 1.571
- 1.736
2.00
-0.276
-0.330
-0.411
-0.541
-0.782
- 1.167
- 1.411
- 1.577
2.20
-0.226
-0.269
-0.334
-0.437
-0.629
-0.937
- 1.143
- 1.295
2.40
-0.187
-0.222
-0.275
-0.359
-0.513
-0.761
-0.929
- 1.058
2.60
-0.155
-0.185
-0.228
-0.297
-0.422
-0.621
-0.756
-0.858
2.80
-0.130
-0.154
-0.190
-0.246
-0.348
-0.508
-0.614
-0.689
3.00
-0.109
-0.129
-0.159
-0.205
-0.288
-0.415
-0.495
-0.545
3.50
-0.069
-0.081
-0.099
-0.127
-0.174
-0.239
-0.270
-0.264
4.00
-0.041
-0.048
-0.058
-0.072
-0.095
-0.116
-0.110
-0.061
Appendix
719
Table F8 Values of (HR)I/RTc
Tr
Pr 1.0000
1.2000
1.5000
2.0000
3.0000
5.0000
7.0000
10.000
0.30
- 11.062
- 11.055
- 11.044
- 11.027
- 10.992
- 10.935
- 10.872
- 10.781
0.35
- 10.640
- 10.637
- 10.632
- 10.624
- 10.609
- 10.581
- 10.554
- 10.529
0.40
- 10.121
- 10.121
- 10.121
- 10.122
- 10.123
- 10.128
- 10.135
- 10.150
0.45
-9.525
-9.527
-9.531
-9.537
-9.549
-9.576
-9.611
-9.663
0.50
-8.888
-8.892
-8.899
-8.909
-8.932
-8.978
-9.030
-9.111
0.55
- 8.238
- 8.243
- 8.252
- 8.267
- 8.298
- 8.360
- 8.425
- 8.531
0.60
- 7.596
- 7.603
- 7.614
- 7.632
- 7.669
- 7.745
- 7.824
- 7.950
0.65
-6.980
-6.987
-6.997
- 7.017
-7.059
- 7.147
- 7.239
- 7.381
0.70
-6.388
-6.395
-6.407
-6.429
-6.475
-6.574
-6.677
-6.837
0.75
- 5.824
- 5.832
- 5.845
- 5.868
-5.918
-6.027
-6.142
-6.318
0.80
-5.285
- 5.293
- 5.306
- 5.330
- 5.385
- 5.506
- 5.632
-5.824
0.85
-4.763
-4.771
-4.784
-4.810
-4.872
- 5.000
- 5.149
- 5.358
0.90
-4.249
-4.255
-4.268
-4.298
-4.371
-4.530
-4.688
-4.916
0.93
-3.934
- 3.937
-3.951
-3.987
-4.073
-4.251
-4.422
-4.662
0.95
-3.712
- 3.713
-3.730
- 3.773
-3.873
-4.068
-4.248
-4.497
0.97
- 3.470
- 3.467
- 3.492
- 3.551
- 3.670
- 3.885
-4.077
-4.336
0.98
- 3.332
- 3.327
- 3.363
- 3.434
- 3.568
- 3.795
- 3.992
-4.257
0.99
- 3.164
- 3.164
- 3.223
- 3.313
- 3.464
- 3.705
- 3.909
- 4.178
1.00
- 2.471
- 2.952
- 3.065
- 3.186
- 3.358
- 3.615
- 3. 8 2 5
- 4.100
1.01
- 1.375
- 2. 5 9 5
- 2. 8 8 0
- 3.051
- 3.251
- 3. 5 2 5
- 3. 7 4 2
- 4.023
1.02
- 1.180
- 1. 7 2 3
- 2.650
- 2.906
- 3.142
- 3.435
- 3.661
- 3.947
1.05
-0.877
-0.878
- 1.496
-2.381
-2.800
- 3.167
- 3.418
-3.722
1.10
- 0.617
- 0.673
- 0.617
- 1.261
-2.167
- 2. 7 2 0
- 3.023
- 3.362
1.15
-0.459
-0.503
-0.487
-0.604
- 1.497
- 2.275
- 2.641
- 3.019
1.20
-0.349
-0.381
-0.381
-0.361
-0.934
- 1.840
-2.273
-2.692
1.30
-0.203
-0.218
-0.218
- 0 . 178
-0.300
- 1.066
- 1.592
-2.086
1.40
-0.111
-0.115
-0.128
-0.070
-0.044
-0.504
- 1.012
- 1.547
1.50
-0.049
-0.046
-0.032
1.60
-0.005
0.004
0.023
0.008
0.078
0.065
0.151
-0.142 0.082
-0.556
- 1.080
-0.217
-0.689
1.70
0.027
0.040
0.063
0.109
0.202
0.223
0.028
- 0.369
1.80
0.051
0.067
0.094
0.143
0.241
0.317
0.203
-0.112
1.90
0.070
0.088
0.117
0.169
0.271
0.381
0.330
0.092
2.00
0.085
0.105
0.136
0.190
0.295
0.428
0.424
0.255
2.20
0.106
0.128
0.163
0.221
0.331
0.493
0.551
0.489
2.40
0.120
0.144
0.181
0.242
0.356
0.535
0.631
0.645
2.60
0.130
0.156
0.194
0.257
0.376
0.567
0.687
0.754
2.80
0.137
0.164
0.204
0.269
0.391
0.591
0.729
0.836
3.00
0.142
0.170
0.211
0.278
0.403
0.611
0.763
0.899
3.50
0.152
0.181
0.224
0.294
0.425
0.650
0.827
1.015
4.00
0.158
0.188
0.233
0.306
0.442
0.680
0.874
1.097
0.4000
0.6000
0.8000
1.0000
Table F9 Values of
(Sn)°/R
Tr
Pr 0.0100
0.0500
0.1000
0.2000
0.30
- 11.614
- 10.008
-9.319
-8.635
-7.961
-7.574
-7.304
-7.099
0.35
- 11.185
-9.579
-8.890
-8.205
-7.529
-7.140
-6.869
-6.663 -6.275
0.40
- 10.802
-9.196
- 8.506
- 7.821
- 7.144
-6.755
-6.483
0.45
- 10.453
- 8.847
- 8.157
- 7.472
-6.794
-6.404
-6.132
- 5.924
0.50
- 10.137
- 8.531
- 7.841
- 7.156
-6.479
-6.089
- 5.816
- 5.608 - 5.324
0.55
-0.038
- 8.245
- 7.555
-6.870
-6.193
- 5.803
- 5.531
0.60
- 0.029
- 7.983
- 7.294
- 6.610
- 5.933
- 5.544
- 5.273
- 5.066
0.65
- 0.023
- 0.122
- 7.052
- 6.368
- 5.694
- 5.306
- 5.036
-4. 830
(Contd.)
Appendix
720 Table F9
(Continued)
T~
/:'r 0.0100
0.0500
0.1000
0.2000
0.4000
0.6000
0.8000
1.0000
0.70
-0.018
-0.096
-0.206
-6.140
- 5.467
- 5.082
-4.814
-4.610
0.75
-0.015
-0.078
-0.164
- 5.917
- 5.248
-4.866
-4.600
-4.399
0.80
-0.013
-0.064
-0.134
-0.294
- 5.026
-4.694
-4.388
-4.191
0.85
-0.011
-0.054
-0.111
-0.239
-4.785
-4.418
-4.166
-3.976 - 3.738
0.90
-0.009
-0.046
-0.094
-0.199
-0.463
-4.145
-3.912
0.93
-0.008
-0.042
-0.085
-0.179
-0.408
-0.750
-3.723
-3.569
0.95
-0.008
-0.039
-0.080
-0.168
-0.377
-0.671
- 3.556
- 3.433
0.97
-0.007
-0.037
-0.075
-0.157
-0.350
-0.607
- 1.056
-3.259
0.98
-0.007
-0.036
-0.073
-0.153
-0.337
-0.580
-0.971
- 3.142
0.99
-0.007
-0.035
-0.071
-0.148
-0.326
-0.555
-0.903
- 2.972
1.00
-0.007
-0.034
-0.069
-0.144
-0.315
-0.532
-0.847
- 2.178
1.01
-0.007
-0.033
-0.067
-0.139
-0.304
-0.510
-0.799
- 1.391
1.02
-0.006
-0.032
-0.065
-0.135
-0.294
-0.491
-0.757
- 1.225
1.05
-0.006
-0.030
-0.060
-0.124
-0.267
-0.439
-0.656
-0.965
1.10
-0.005
-0.026
-0.053
-0.108
-0.230
-0.371
-0.537
-0.742
1.15
-0.005
-0.023
-0.047
-0.096
-0.201
-0.319
-0.452
-0.607
1.20
-0.004
-0.021
-0.042
-0.085
-0.177
-0.277
-0.389
-0.512
1.30
-0.003
-0.017
-0.033
-0.068
-0.140
-0.217
-0.298
-0.385
1.40
-0.003
-0.014
-0.027
-0.056
-0.114
-0.174
-0.237
-0.303
1.50
-0.002
-0.011
-0.023
-0.046
-0.094
-0.143
-0.194
-0.246
1.60
-0.002
-0.010
-0.019
-0.039
-0.079
-0.120
-0.162
-0.204
1.70
-0.002
-0.008
-0.017
-0.033
-0.067
-0.102
-0.137
-0.172
1.80
-0.001
-0.007
-0.014
-0.029
-0.058
-0.088
-0.117
-0.147
1.90
-0.001
-0.006
-0.013
-0.025
-0.051
-0.076
-0.102
-0.127
2.00
-0.001
-0.006
-0.011
-0.022
-0.044
-0.067
-0.089
-0.111
2.20
-0.001
-0.004
-0.009
-0.018
-0.035
-0.053
-0.070
-0.087
2.40
-0.001
-0.004
-0.007
-0.014
-0.028
-0.042
-0.056
-0.070
2.60
-0.001
-0.003
-0.006
-0.012
-0.023
-0.035
-0.046
-0.058
2.80
- 0.000
-0.002
-0.005
-0.010
-0.020
-0.029
-0.039
-0.048 -0.041
3.00
- 0.000
-0.002
-0.004
-0.008
-0.017
-0.025
-0.033
3.50
- 0.000
-0.001
-0.003
-0.006
-0.012
-0.017
-0.023
-0.029
4.00
-0.000
-0.001
-0.002
-0.004
-0.009
-0.013
-0.017
-0.021
Table FIO Values of
(SR)I/R
rr
er 0.0100
0.0500
0.1000
0.2000
0.4000
0.6000
0.8000
1.0000
0.30
- 16.782
- 16.774
- 16.764
- 16.744
- 16.705
- 16.665
- 16.626
- 16.586
0.35
- 15.413
- 15.408
- 15.401
- 15.387
- 15.359
- 15.333
- 15.305
- 15.278
0.40
- 13.990
- 13.986
- 13.981
- 13.972
- 13.953
- 13.934
- 13.915
- 13.896
0.45
- 12.564
- 12.561
- 12.558
- 12.551
- 12.537
- 12.523
- 12.509
- 12.496
0.50
- 11.202
- 11.200
- 11.197
- 11.092
- 11.082
- 11.172
- 11.162
- 11.153
0.55
-0.115
-9.948
-9.946
-9.942
-9.935
-9.928
-9.921
-9.914
0.60
-0.078
- 8.828
- 8.826
- 8.823
- 8.817
- 8.811
- 8.806
- 8.799
0.65
-0.055
-0.309
-7.832
-7.829
-7.824
-7.819
-7.815
-7.510
0.70
-0.040
-0.216
-0.491
-6.951
-6.945
-6.941
-6.937
-6.933
0.75
-0.029
-0.156
-0.340
-6.173
-6.167
-6.162
-6.158
-6.155
0.80
-0.022
-0.116
-0.246
-0.578
- 5.475
- 5.468
- 5.462
- 5.458
0.85
-0.017
-0.088
-0.183
-0.400
-4.853
-4.841
-4.832
-4.826
0.90
-0.013
-0.068
-0.140
-0.301
-0.744
-4.269
-4.249
-4.238
0.93
-0.011
-0.058
-0.120
-0.254
-0.593
- 1.219
-3.914
-3.894
0.95
-0.010
-0.053
-0.109
-0.228
-0.517
-0.961
-3.697
-3.658
0.97
-0.010
-0.048
-0.099
-0.206
-0.456
-0.797
- 1.570
-3.406
(Contd.)
Appendix Table FIO
721
(Continued)
Tr
Pr 0.0100
0.0500
0.1000
0.2000
0.4000
0.6000
0.8000
1.0000
0.98
- 0.009
- 0.046
- 0.094
- 0.196
- 0.429
- 0.734
- 1.270
- 3.264
0.99
- 0.009
- 0.044
- 0.090
- 0.186
- 0.405
- 0.680
- 1.098
- 3.093
1.00
-0.008
-0.042
-0.086
-0.177
-0.382
-0.632
-0.977
-2.399
1.01
-0.008
-0.040
-0.082
-0.169
-0.361
-0.590
-0.883
- 1.306
1.02
-0.008
-0.039
-0.078
-0.161
-0.342
-0.552
-0.807
- 1.113
1.05
-0.007
-0.034
-0.069
-0.140
-0.292
-0.460
-0.642
-0.820
1.10
-0.005
-0.028
-0.055
-0.112
-0.229
-0.350
-0.470
-0.577
1.15
-0.005
-0.023
-0.045
-0.091
-0.183
-0.275
-0.361
-0.437
1.20
-0.004
-0.019
-0.037
-0.075
-0.149
-0.220
-0.286
-0.343
1.30
-0.003
-0.013
-0.026
-0.052
-0.102
-0.148
-0.190
-0.226
1.40
-0.002
-0.010
-0.019
-0.037
-0.072
-0.104
-0.133
-0.158
1.50
-0.001
-0.007
-0.014
-0.027
-0.053
-0.076
-0.097
-0.115
1.60
-0.001
-0.005
-0.011
-0.021
-0.040
-0.057
-0.073
-0.086
1.70
-0.001
-0.004
-0.008
-0.016
-0.031
-0.044
-0.056
-0.067
1.80
-0.001
-0.003
-0.006
-0.013
-0.024
-0.035
-0.044
-0.053
1.90
-0.001
-0.003
-0.005
-0.010
-0.019
-0.028
-0.036
-0.043
2.00
- 0.000
-0.002
-0.004
-0.008
-0.016
-0.023
-0.029
-0.035
2.20
- 0.000
-0.001
-0.003
-0.006
-0.011
-0.016
-0.021
-0.025
2.40
- 0.000
-0.001
-0.002
-0.004
-0.008
-0.012
-0.015
-0.019
2.60
- 0.000
-0.001
-0.002
-0.003
-0.006
-0.009
-0.012
-0.015
2.80
- 0.000
-0.001
-0.001
-0.003
-0.005
-0.008
-0.010
-0.012
3.00
- 0.000
-0.001
-0.001
-0.002
-0.004
-0.006
- 0.008
-0.010
3.50
- 0.000
- 0.000
- 0.001
- 0.001
- 0.003
- 0.004
- 0.006
- 0.007
4.00
- 0.000
- 0.000
-0.001
-0.001
-0.002
-0.003
-0.005
-0.006
Table F l l
Values of
(SR)°/R
Tr
Pr 1.0000
1.2000
1.5000
2.0000
3.0000
5.0000
7.0000
10.000
0.30
- 7.099
-6.935
-6.740
-6.497
-6.180
- 5.847
- 5.683
- 5.578
0.35
-6.663
-6.497
-6.299
-6.052
- 5.728
- 5.376
-5.194
- 5.060
0.40
-6.275
-6.109
- 5.909
- 5.660
- 5.330
-4.967
-4.772
-4.619
0.45
- 5.924
- 5.757
- 5.557
- 5.306
-4.974
-4.603
-4.401
-4.234
0.50
- 5.608
- 5.441
- 5.240
-4.989
-4.656
-4.282
-4.074
- 3.899
0.55
- 5.324
- 5.157
-4.956
-4.706
-4.373
- 3.998
-3.788
-3.607
0.60
-5.066
-4.900
-4.700
-4.451
-4.120
-3.747
-3.537
-3.353
0.65
- 4.830
-4.665
-4.467
-4.220
- 3.892
- 3.523
- 3.315
- 3.131
0.70
-4.610
-4.446
-4.250
-4.007
-3.684
-3.322
-3.117
-2.935
0.75
- 4.399
- 4.238
- 4.045
- 3.807
- 3.491
- 3.138
- 2.939
- 2.761
0.80
-4.191
-4.034
-3.846
-3.615
-3.310
-2.970
-2.777
-2.605
0.85
- 3.976
- 3.825
-3.646
- 3.425
- 3.135
-2.812
-2.629
-2.463
0.90
-3.738
-3.599
- 3.434
- 3.231
-2.964
-2.663
-2.491
-2.334
0.93
- 3.569
- 3.444
- 3.295
- 3.108
-2.860
-2.577
-2.412
-2.262
0.95
- 3.433
- 3.326
- 3.193
- 3.023
-2.790
-2.520
-2.362
-2.215
0.97
-3.259
-3.188
- 3.081
-2.932
-2.719
-2.463
-2.312
-2.170
0.98
- 3.142
-3.106
- 3.019
-2.884
-2.682
-2.436
-2.287
-2.148
0.99
- 2.972
- 3.010
- 2.953
- 2.835
- 2.646
- 2.408
- 2.263
- 2.126
1.00
-2.178
-2.893
-2.879
-2.784
-2.609
-2.380
-2.239
-2.105
1.01
- 1.391
-2.736
-2.798
- 2.730
-2.571
-2.352
-2.215
-2.083
1.02
- 1.225
-2.495
- 2.706
-2.673
-2.533
-2.325
-2.191
-2.062
1.05
-0.965
- 1.523
-2.328
-2.483
-2.415
-2.242
-2.121
-2.001
1.10
-0.742
- 1.012
- 1.557
-2.081
-2.202
-2.104
-2.007
- 1.903
1.15
-0.607
-0.790
- 1.126
- 1.649
- 1.968
- 1.966
- 1.897
- 1.810
(Contd.)
Appendix
722 Table F l l
(Continued)
Tr
Pr
1.0000
1.2000
1.5000
2.0000
3.0000
5.0000
7.0000
10.000
1.20
-0.512
-0.651
-0.890
- 1.308
- 1.727
- 1.827
- 1.789
- 1.722
1.30
-0.385
-0.478
-0.628
-0.891
- 1.299
- 1.554
- 1.581
- 1.556
1.40
-0.303
-0.375
-0.478
-0.663
-0.990
- 1.303
- 1.386
- 1.402
1.50
-0.246
-0.299
-0.381
-0.520
-0.777
- 1.088
- 1.208
- 1.260
1.60
-0.204
-0.247
-0.312
-0.421
-0.628
-0.913
- 1.050
- 1.130
1.70
-0.172
-0.208
-0.261
-0.350
-0.519
-0.773
-0.915
- 1.013
1.80
-0.147
-0.177
-0.222
-0.296
-0.438
-0.661
-0.799
-0.908
1.90
-0.127
-0.153
-0.191
-0.255
-0.375
-0.570
-0.702
-0.815
2.00
-0.111
-0.134
-0.167
-0.221
-0.625
-0.497
-0.620
-0.733
2.20
-0.087
-0.105
-0.130
-0.172
-0.251
-0.388
-0.492
-0.599
2.40
-0.070
-0.084
-0.104
-0.138
-0.201
-0.311
-0.399
-0.496
2.60
-0.058
-0.069
-0.086
-0.1 I3
-0.164
-0.255
-0.329
-0.416
2.80
-0.048
-0.058
-0.072
-0.094
-0.137
-0.213
-0.277
-0.353
3.00
-0.041
-0.049
-0.061
-0.080
-0.116
-0.181
-0.236
-0.303
3.50
-0.029
-0.034
-0.042
-0.056
-0.081
-0.126
-0.166
-0.216
4.00
-0.021
-0.025
-0.031
-0.041
-0.059
-0.093
-0.123
-0.162
Table F12 Values of (S R)1/R
rr
er 1.0000
1.2000
1.5000
2.0000
3.0000
5.0000
7.0000
10.000
0.30
- 16.586
- 16.547
- 16.488
- 16.390
- 16.195
- 15.837
- 15.468
- 14.925
0.35
- 15.278
- 15.251
- 15.211
- 15.144
- 15.011
- 14.751
- 14.496
- 14.153 - 13.144
0.40
- 13.896
- 13.877
- 13.849
- 13.803
- 13.714
- 13.541
- 13.576
0.45
- 12.496
- 12.482
- 12.462
- 12.430
- 12.367
- 12.248
- 12.145
- 11.999
0.50
- 11.153
- 11.143
- 11.129
- 11.107
- 11.063
- I0.985
- 10.920
- 10.836
0.55
-9.914
-9.907
-9.897
-9.882
-9.853
-9.806
-9.769
-9.732
0.60
- 8.799
- 8.794
- 8.787
- 8.777
- 8.760
- 8.736
- 8.723
- 8.720
0.65
-7.810
-7.807
-7.801
-7.794
-7.784
-7.779
-7.785
-7.811
0.70
-6.933
-6.930
-6.926
-6.922
-6.919
-6.929
-6.952
-7.002
0.75
-6.155
-6.152
-6.149
-6.147
-6.149
-6.174
-6.213
-6.285
0.80
- 5.458
- 5.455
- 5.453
- 5.452
- 5.461
- 5.501
- 5.555
- 5. 6 4 8
0.85
-4.826
-4.822
-4.820
-4.822
-4.839
-4.898
-4.969
- 5.082
0.90
-4.238
-4.232
-4.230
-4.236
-4.267
-4.351
-4.442
-4.578 -4.300
0.93
- 3.894
- 3.885
- 3.884
- 3.896
- 3.941
-4.046
-4.151
0.95
- 3.658
- 3.647
- 3.648
- 3.669
- 3.728
- 3.851
- 3.966
-4.125
0.97
- 3.406
- 3.391
- 3.401
- 3.437
- 3.517
- 3.661
- 3.788
- 3.957
0.98
- 3.264
- 3.247
- 3.268
- 3.318
- 3.412
- 3.569
- 3.701
- 3.875
0.99
-3.093
-3.082
-3.126
-3.195
-3.306
-3.477
-3.616
-3.796
1.00
-2.399
-2.868
-2.967
-3.067
-3.200
-3.387
-3.532
-3.717
1.01
- 1.306
-2.513
-2.784
-2.933
-3.094
-3.297
-3.450
-3.640
1.02
- 1.113
- 1.655
-2.557
-2.790
-2.986
-3.209
-3.369
-3.565
1.05
-0.820
-0.831
- 1.443
- 2.283
- 2.655
- 2.949
- 3.134
- 3.348
1.10
-0.577
-0.640
-0.618
- 1.241
-2.067
-2.534
-2.767
- 3.013
1.15
-0.437
-0.489
-0.502
-0.654
- 1.471
-2.138
-2.428
-2.708
1.20
-0.343
-0.385
-0.412
-0.447
-0.991
- 1.767
-2.115
-2.430
1.30
-0.226
-0.254
-0.282
-0.300
-0.481
- 1.147
- 1.569
- 1.944
1.40
-0.158
-0.178
-0.200
-0.220
-0.290
-0.730
- 1.138
- 1.544
1.50
-0.115
-0.130
-0.147
-0.166
-0.206
-0.479
-0.823
- 1.222
1.60
-0.086
-0.098
-0.112
-0.129
-0.159
-0.334
-0.604
-0.969
1.70
-0.067
-0.076
-0.087
-0.102
-0.127
-0.248
-0.456
-0.775
1.80
-0.053
-0.060
-0.070
-0.083
-0.105
-0.195
-0.355
-0.628
1.90
-0.043
-0.049
-0.057
-0.069
-0.089
-0.160
-0.286
-0.518
(Contd.)
Appendix Table F12
(Continued)
T~
2.00 2.20 2.40 260 2.80 3.00 3.50 4.00
723
Pr 1.0000
1.2000
1.5000
2.0000
3.0000
5.0000
7.0000
10.000
-0.035 -0.025 -0.019 -0.015 -0.012 -0.010 -0.007 -0.006
-0.040 -0.029 -0.022 -0.018 -0.014 -0.012 -0.009 -0.007
-0.048 -0.035 -0.027 -0.021 -0.018 -0.015 -0.011 -0.009
-0.058 -0.043 -0.034 -0.028 -0.023 -0.020 -0.015 -0.012
-0.077 -0.060 -0.048 -0.041 -0.025 -0.031 -0.024 -0.020
-0.136 -0.105 -0.086 -0.074 -0.065 -0.058 -0.046 -0.038
-0.238 -0.178 -0.143 -0.120 -0.104 -0.093 -0.073 -0.060
-0.434 -0.322 -0.254 -0.210 -0.180 -0.158 -0.122 -0.100
Table F13 Values of ~h° Tr
Pr 0.0100
0.0500
0.1000
0.2000
0.4000
0.6000
0.8000
1.0000
0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.93 0.95 0.97 0.98 0.99 1.00 1.01 1.02 1.05 1.10 1.15 1.20 1.30 1.40 1.50 1.60
0.0002 0.0034 0.0272 O. 1321 0.4529 0.9817 0.9840 0.9886 0.9908 0.9931 0.9931 0.9954 0.9954 0.9954 0.9954 0.9954 0.9954 0.9977 0.9977 0.9977 0.9977 0.9977 0.9977 0.9977 0.9977 0.9977 0.9977 1.0000 1.0000
0.0000 0.0007 0.0055 0.0266 0.0912 0.2432 0.5383 0.9419 0.9528 0.9616 0.9683 0.9727 0.9772 0.9795 0.9817 0.9817 0.9817 0.9840 0.9840 0.9840 0.9840 0.9863 0.9886 0.9886 0.9908 0.9931 0.9931 0.9954 0.9954
0.0000 0.0003 0.0028 0.0135 0.0461 0.1227 0.2716 0.5212 0.9057 0.9226 0.9354 0.9462 0.9550 0.9594 0.9616 0.9638 0.9638 0.9661 0.9661 0.9683 0.9683 0.9705 0.9750 0.9795 0.9817 0.9863 0.9886 0.9908 0.9931
0.0000 0.0002 0.0014 0.0069 0.0235 0.0625 0.1384 0.2655 0.4560 0.7178 0.8730 0.8933 0.9099 0.9183 0.9226 0.9628 0.9290 0.9311 0.9333 0.9354 0.9376 0.9441 0.9506 0.9572 0.9616 0.9705 0.9772 0.9817 0.9863
0.0000 0.0001 0.0007 0.0036 0.0122 0.0325 0.0718 0.1374 0.2360 0.3715 0.5445 0.7534 0.8204 0.8375 0.8472 0.8570 0.8610 0.8650 0.8690 0.8730 0.8770 0.8872 0.9016 0.9141 0.9247 0.9419 0.9550 0.9638 0.9727
0.0000 0.0001 0.0005 0.0025 0.0085 0.0225 0.0497 0.0948 0.1626 0.2559 0.3750 0.5188 0.6823 0.7551 0.7709 0.7852 0.7925 0.7980 0.8035 0.8110 0.8166 0.8318 0.8531 0.8730 0.8892 0.9141 0.9333 0.9462 0.9572
0.0000 0.0001 0.0004 0.0020 0.0067 0.0176 0.0386 0.0738 0.1262 0.1982 0.2904 0.4018 0.5297 0.6109 0.6668 0.7112 0.7211 0.7295 0.7379 0.7464 0.7551 0.7762 0.8072 0.8318 0.8531 0.8872 0.9120 0.9290 0.9441
0.0000 0.0000 0.0003 0.0016 0.0055 0.0146 0.0321 0.0611 0.1045 0.1641 0.2404 0.3319 0.4375 0.5058 0.5521 0.5984 0.6223 0.6442 0.6668 0.6792 0.6902 0.7194 0.7586 0.7907 0.8166 0.8590 0.8892 0.9141 0.9311
1.70 1.80 1.90 2.00 2.20 2.40 2.60 2.80 3.00 3.50 4.00
1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
0.9977 0.9977 0.9977 0.9977 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
0.9954 0.9954 0.9954 0.9977 0.9977 1.0000 l. 0000 1.0000 1.0000 1.0000 1.0000
0.9886 0.9908 0.9931 0.9954 0.9977 0.9977 1.0000 1.0000 1.0000 1.0023 1.0023
0.9772 0.9817 0.9863 0.9886 0.9931 0.9977 1.0000 1.0000 1.0023 1.0023 1.0046
0.9661 0.9727 0.9795 0.9840 0.9908 0.9954 0. 9977 1.0000 1.0023 1.0046 1.0069
0.9550 0.9661 0.9727 0.9795 0.9886 0.9931 0. 9977 1.0023 1.0046 1.0069 1.0093
0.9462 0.9572 0.9661 0.9727 0.9840 0.9931 0.9977 1.0023 1.0046 1.0093 1.0116
Appendix
724 Table F14 Values of ~h1
rr
0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.93 0.95 0.97 0.98 0.99 1.00 1.01 1.02 1.05 1.10 1.15 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.20 2.40 2.60 2.80 3.00 3.50 4.00
er 0.0100
0.0500
0.1000
0.2000
0.4000
0.6000
0.8000
1.0000
0.0000 0.0000 0.0000 0.0002 0.0014 0.9705 0.9795 0.9863 0.9908 0.9931 0.9954 0.9977 0.9977 0.9977 0.9977 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
0.0000 0.0000 0.0000 0.0002 0.0014 0.0069 0.0227 0.9311 0.9528 0.9683 0.9772 0.9863 0.9908 0.9931 0.9931 0.9954 0.9954 0.9954 0.9977 0.9977 0.9977 0.9977 1.0000 1.0000 1.0023 1.0023 1.0046 1.0046 1.0046 1.0046 1.0046 1.0046 1.0046 1.0046 1.0046 1.0023 1.0023 1.0023 1.0023 1.0023
0.0000 0.0000 0.0000 0.0002 0.0014 0.0068 0.0226 0.0572 0.9036 0.9332 0.9550 0.9705 0.9795 0.9840 0.9885 0.9908 0.9908 0.9931 0.9931 0.9931 0.9954 0.9977 1.0000 1.0023 1.0046 1.0069 1.0069 1.0069 1.0069 1.0093 1.0069 1.0069 1.0069 1.0069 1.0069 1.0069 1.0069 1.0069 1.0046 1.0046
0.0000 0.0000 0.0000 0.0002 0.0014 0.0068 0.0223 0.0568 0.1182 0.2112 0.9057 0.9375 0.9594 0.9705 0.9750 0.9795 0.9817 0.9840 0.9863 0.9885 0.9908 0.9954 1.0000 1.0046 1.0069 1.0116 1.0139 1.0163 1.0163 1.0163 1.0163 1.0163 1.0163 1.0139 1.0139 1.0139 1.0116 1.0116 1.0023 1.0093
0.0000 0.0000 0.0000 0.0002 0.0014 0.0066 0.0220 0.0559 0.1163 0.2078 0.3302 0.4774 0.9141 0.9354 0.9484 0.9594 0.9638 0.9683 0.9727 0.9772 0.9795 0.9885 1.0023 1.0116 1.0163 1.0257 1.0304 1.0328 1.0328 1.0328 1.0328 1.0328 1.0304 1.0304 1.0280 1.0257 1.0257 1.0233 1.0209 1.0186
0.0000 0.0000 0.0000 0.0002 0.0014 0.0065 0.0216 0.0551 0.1147 0.2050 0.3257 0.4708 0.6323 0.8953 0.9183 0.9354 0.9440 0.9528 0.9594 0.9638 0.9705 0.9863 1.0046 1.0186 1.0280 1.0399 1.0471 1.0496 1.0496 1.0496 1.0496 1.0496 1.0471 1.0447 1.0423 1.0399 1.0375 1.0352 1.0304 1.0280
0.0000 0.0000 0.0000 0.0002 0.0013 0.0064 0.0213 0.0543 0.1131 0.2022 0.3212 0.4654 0.6250 0.7227 0.7888 0.9078 0.9225 0.9332 0.9440 0.9528 0.9616 0.9840 1.0093 1.0257 1.0399 1.0544 1.0642 1.0666 1.0691 1.0691 1.0666 1.0666 1.0642 1.0593 1.0568 1.0544 1.0496 1.0471 1.0423 1.0375
0.0000 0.0000 0.0000 0.0002 0.0013 0.0063 0.0210 0.0535 0.1116 0.1994 0.3168 0.4590 0.6165 0.7144 0.7797 0.8413 0.8729 0.9036 0.9311 0.9462 0.9572 0.9840 1.0163 1.0375 1.0544 1.0716 1.0815 1.0865 1.0865 1.0865 1.0840 1.0815 1.0815 1.0765 1.0716 1.0666 1.0642 1.0593 1.0520 1.0471
Table F15 Values of ~o
Tr
0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65
/'r 1.0000
1.2000
1.5000
2.0000
3.0000
5.0000
7.0000
0.0000 0.0000 0.0003 0.0016 0.0055 0.0146 0.0321 0.06 i 1
0.0000 0.0000 0.0003 0.0014 0.0048 0.0127 0.0277 0.0527
0.0000 0.0000 0.0003 0.0012 0.0041 0.0107 0.0234 0.0445
0.0000 0.0000 0.0002 0.0010 0.0034 0.0089 0.0193 0.0364
0.0000 0.0000 0.0002 0.0008 0.0028 0.0072 0.0154 0.0289
0.0000 0.0000 0.0002 0.0008 0.0025 0.0063 0.0132 0.0244
0.0000 0.0000 0.0002 0.0009 0.0027 0.0066 0.0135 0.,0245
10.000 0.0000 0.0000 0.0003 0.0012 0.0034 0.0080 0.0160 0.0282
(Contd.)
Appendix Table F15
(Continued)
Tr
0.70 0.75 0.80 0.85 0.90 0.93 0.95 0.97 0.98 0.99 1.00 1.01 1.02 1.05 1.10 1.15 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.20 2.40 2.60 2.80 3.00 3.50 4.00
725
Pr 1.0000
1.2000
1.5000
2.0000
3.0000
5.0000
7.0000
0.1045 0.1641 0.2404 0.3319 0.4375 0.5058 0.5521 0.5984 0.6223 0.6442 0.6668 0.6792 0.6902 0.7194 0.7586 0.7907 0.8166 0.8590 0.8892 0.9141 0.9311 0.9462 0.9572 0.9661 0.9727 0.9840 0.0031 0.9977 1.0023 1.0046 1.0093 1.0116
0.0902 0.1413 0.2065 0.2858 0.3767 0.4355 0.4764 0.5164 0.5370 0.5572 0.5781 0.5970 0.6166 0.6607 0.7112 0.7499 0. 7834 0.8318 0.8690 0.8974 0.9183 0.9354 0.9484 0.9594 0.9683 0.9817 0.9908 0.9977 1.0023 1.0069 1.0116 1.0139
0.0759 0.1188 0.1738 0.2399 0.3162 0.3656 0.3999 0.4345 0.4529 0.4699 0.4875 0.5047 0.5224 0.5728 0.6412 0.6918 0.7328 0.7943 0.8395 0.8730 0.8995 0.9204 0.9376 0.9506 0.9616 0.9795 0.9908 0.9977 1.0046 1.0069 1.0139 1.0162
0.0619 0.0966 0.1409 0.1945 0.2564 0.2972 0.3251 0.3532 0.3681 0.3828 0.3972 0.4121 0.4266 0.4710 0.5408 0.6026 0.6546 0.7345 0.7925 0.8375 0.8710 0.8995 0.9204 0.9376 0.9528 0.9727 0.9886 0.9977 1.0069 1.0116 1.0186 1.0233
0.0488 0.0757 0.1102 0.1517 0.1995 0.2307 0.2523 0.2748 0.2864 0.2978 0.3097 0.3214 0.3334 0.3690 0.4285 0.4875 0.5420 0.6383 0.7145 0.7745 0.8222 0.8610 0.8913 0.9162 0.9354 0.9661 0.9863 1.0023 1.0116 1.0209 1.0304 1.0375
0.0406 0.0625 0.0899 0.1227 0.1607 0.1854 0.2028 0.2203 0.2296 0.2388 0.2483 0.2576 0.2673 0.2958 0.3451 0.3954 0.4446 0.5383 0.6237 0.6966 0.7586 0.8091 0.8531 0.8872 0.9183 0.9616 0.9931 1.0162 1.0328 1.0423 1.0593 1.0666
0.0402 0.0610 0.0867 0.1175 0.1524 0.1754 0.1910 0.2075 0.2158 0.2244 0.2328 0.2415 0.2506 0.2773 0.3228 0.3690 0.4150 0.5058 0.5902 0.6668 0.7328 0.7907 0.8414 0.8831 0.9183 0.9727 1.0116 1.0399 1.0593 1.0740 1.0914 1.0990
10.000 0.0453 0.0673 0.0942 0.1256 0.1611 0.1841 0.2000 0.2163 0.2244 0.2328 0.2415 0.2500 0.2582 0.2844 0.3296 0.3750 0.4198 0.5093 0.5943 0.6714 0.7430 0.8054 0.8590 0.9057 0.9462 1.0093 1.0568 1.0889 1.1117 1.1298 1.1508 1.1588
Table F16 Values of ~bl Tr
0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.93 0.95 0.97
Pr 1.0000
1.2000
1.5000
2.0000
3.0000
5.0000
7.0000
0.0000 0.0000 0.0000 0.0002 0.0013 0.0063 0.0210 0.0536 0.1117 0.1995 0.3170 0.4592 0.6166 0.7145 0.7798 0.8414
0.0000 0.0000 0.0000 0.0002 0.0013 0.0062 0.0207 0.0527 0.1102 0.1972 0.3133 0.4539 0.6095 0.7063 0.7691 0.8318
0.0000 0.0000 0.0000 0.0002 0.0013 0.0061 0.0202 0.0516 0.1079 0.1932 0.3076 0.4457 0.5998 0.6950 0.7568 0.8185
0.0000 0.0000 0.0000 0.0002 0.0012 0.0058 0.0194 0.0497 0.1040 0.1871 0.2978 0.4325 0.5834 0.6761 0.7379 0.7998
0.0000 0.0000 0.0000 0.0001 0.0011 0.0053 0.0179 0.0461 0.0970 0.1754 0.2812 0.4093 0.5546 0.6457 0.7063 0.7656
0.0000 0.0000 0.0000 0.0001 0.0009 0.0045 0.0154 0.0401 0.0851 0.1552 0.2512 0.3698 0.5058 0.5916 0.6501 0.7096
0.0000 0.0000 0.0000 0.0001 0.0008 0.0039 0.0133 0.0350 0.0752 0.1387 0.2265 0.3365 0.4645 0.5470 0.6026 0.6607
10.000
0.0000 0.0000 0.0000 0.0001 0.0006 0.0031 0.0108 0.0289 0.0629 0.1178 0.1954 0.2951 0.4130 0.4898 0.5432 0.5984
(Contd.)
Appendix
726 Table F16
(Continued)
/'r
0.98 0.99 1.00 1.01 1.02 1.05 1.10 1.15 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.20 2.40 2.60 2.80 3.00 3.50 4.00
er 1.0000
1.2000
1.5000
2.0000
3.0000
5.0000
7.0000
0.8730 0.9036 0.9311 0.9462 0.9572 0.9840 1.0162 1.0375 1.0544 1.0715 1.0814 1.0864 1.0864 1.0864 1.0839 1.0814 1.0814 1.0765 1.0715 1.0666 1.0641 1.0593 1.0520 1.0471
0.8630 0.8913 0.9204 0.9462 0.9661 0.9954 1.0280 1.0520 1.0691 1.0914 1.0990 1.1041 1.1041 1.1041 1.1015 1.0990 1.0965 1.0914 1.0864 1.0814 1.0765 1.0715 1.0617 1.0544
0.8492 0.8790 0.9078 0.9333 0.9594 1.0186 1.0593 1.0814 1.0990 1.1194 1.1298 1.1350 1.1350 1.1324 1.1298 1.1272 1.1220 1.1143 1.1066 1.1015 1.0940 1.0889 1.0789 1.0691
0.8298 0.8590 0.8872 0.9162 0.9419 1.0162 1.0990 1.1376 1.1588 1.1776 1.1858 1.1858 1.1858 1.1803 1.1749 1.1695 1.1641 1.1535 1.1429 1.1350 1.1272 1.1194 1.1041 1.0914
0.7962 0.8241 0.8531 0.8831 0.9099 0.9886 1.1015 1.1858 1.2388 1.2853 1.2942 1.2942 1.2883 1.2794 1.2706 1.2618 1.2503 1.2331 1.2190 1.2023 1.1912 1.1803 1.1561 1.1403
0.7379 0.7674 0.7962 0.8241 0.8531 0.9354 1.0617 1.1722 1.2647 1.3868 1.4488 1.4689 1.4689 1.4622 1.4488 1.4355 1.4191 1.3900 1.3614 1.3397 1.3183 1.3002 1.2618 1.2303
0.6887 0.7178 0.7464 0.7745 0.8035 0.8872 1.0186 1.1403 1.2474 1.4125 1.5171 1.5740 1.5996 1.6033 1.5959 1.5849 1.5704 1.5346 1.4997 1.4689 1.4388 1.4158 1.3614 1.3213
10.000 0.6266 0.6546 0.6823 0.7096 0.7379 0.8222 0.9572 1.0864 1.2050 1.4061 1.5524 1.6520 1.7140 1.7458 1.7620 1.7620 1.7539 1.7219 1.6866 1.6482 1.6144 1.5813 1.5101 1.4555
SUBJECT INDEX
A Absolute activity, 6 Active transport, 495-497, 525-527, 531,533-534, 592 Activity coefficient, 32 Actual cogeneration plant, 218-219 Actual regenerative Rankine cycle, 204-207 Actual reheat Rankine cycle, 198-201 Actual reheat regenerative cycle, 211 Adiabatic mixer, 108-109 Affinity, 28, 90 Alveoli, 542 Anisotropic diffusion, 357 Annular packed bed, 166-167 Antiport, 551-552 Antoine equation, 34 Approach temperature, 254 Arrhenius group, 458 Arrow of time, 7 Atomic volumes, 82 ATP synthase, 551 Availability, 286-287 Average velocity, 115 Azeotropes, 38 B Bacterial growth, 660 Balance equations, 14-15, 112-113,456, 470-471 Bejan number, 165 Belousov-Zhabotinsky reaction scheme, 625,642, 643 B6nard cells, 634 Bifurcation, 632-633 Bioenergetics, 548-549 Biomembranes, 526-529 Biological, fuels, 544 gels, 359 structures, 650-651 Boltzmann's H-theorem, 56 Bronsted work principle, 20 Brownian motion, 84 Brusselator system, 616, 638,640 Butterfly effect, 635 C Caloric equation, 21 Caratheodory's theory, 13 Cash flow diagram, 314-315 Cell electric potentials, 542 Chapman-Enskog formula, 70 Chemical, affinity of real gases, 28 equilibrium, 8
exergy, 243-244 potential, 6, 46 stability, 601 Chiral symmetry breaking, 651-652 Clapeyron equation, 28-29 Claude process, 227 Cogeneration, 215 Collision integral, 60 Column grand composite curves, 260 Combined energy flow, 63 Combined momentum flow, 59 Compartmental structure, 529 Composite, curves, 246-248 membranes, 520 wall, 64 Compressibility factor, 45 Conductance coefficients, 132 Conductance matching, 571-573 Configurational heat capacity, 603-604 Conjugate flows and forces, 123 Conservation law, 114-115 Conservation of mass, 3 Conservation of energy, 3, 118 Corresponding state correlation, 60 Couette flow, 161-166 Coupled processes, 90-92, 564 Coupled reactions, 447-448, 677 Cristae, 549-551 Cross relations, 47-48 Cumulative degree of thermodynamic perfection, 280-281 Cumulative exergy consumption, 280-282 Curie-Prigogine principle, 143 Cyclic process, 11 D Damk6hler number, 473 Dead state, 184-185 Decrease of exergy principle, 187 Degree of coupling, 371,475-476, 534, 553,568-570 Degrees of freedom, 506 Depletion number, 244-245 Diabatic distillation configuration, 298 Diffusion, barrier, 83 coefficients, 75, 78 in mixtures, 329 potential, 91 velocity, 73 Discontinues systems, 131 Dissipation function, 124-125,379, 529-532 Dissipation with conductance matching, 571
728 Dissipative processes, 7 Dissipative structures, 55,634 Distillation column targets, 260 Distribution coefficient, 349 Distribution of entropy production, 171 Dufour effect, 91,365 Dynamic equilibrium, 9 E
Ecological cost, 285-286 Effective diffusivity, 453-455 of cellular systems, 566 Effectiveness factor, 458-462 Effective thermal conductivity, 456 Efficiency of coupling, 376 Efficiency of energy conversion, 377, 475-476, 541,554, 570 Elasticity coefficient, 558 Electrical conductance, 340-343 Electric field intensity, 88 Electrochemical affinity, 105 Electrochemical potential, 337 Electrokinetic effect, 90 Electron transport in mitochondria, 549 Electrosmotic pressure, 91 Energy, balance, 15, 117 dissipation, 106-110 expenditure, 544-547 of transport, 370 Enthalpy, 45-48 deficit, 261-263 of blood, 543 Entropy, 3 balance 109, 110, 156 distribution, 14 of melting, 100 of vaporization, 100-102 of transport, 370 production, 16-19, 104-113, 121,426-443 production in distillation, 237 source strength, 16-17 Equation of state, 22 Equilibrium, 6-14 constant of reaction, 563 Equipartition principle, 176-177, 236-237, 289 Essergy, 287 Eucken formula, 70 Euler equations, 21 Evolutionary criterion, 595,681 Evolution of couple systems, 481 Excess, enthalpy, 32, 37 entropy, 32, 33 Gibbs energy, 32 volume, 32 Exergy, 184-195 analysis, 184-185 analysis of distillation, 234 analysis of power plant, 194 analysis of refinery, 242 analysis of refrigeration cycle, 225 analysis procedure, 192 balance, 186-187 cost, 275-277 destruction number, 285 efficiency, 235-236, 582, 590
Subject index factor, 187-188 loss, 184-192, 581 loss profiles, 234-242 use in bioenergetics, 581 Exhaustion of nonrenewable resources, 284 Expansivity, 4-5 Expression of Kohlrausch, 343 Extended exergy, 275-276 Extended nonequilibrium thermodynamics, 680-681 Extensive properties, 2 Extent of reaction, 90, 415 Extremum principles, 48 F
Facilitated transport, 485-487,489-494, 496, 525-526 Faraday constant, 84 Fick's first law, 73 First law of thermodynamics, 11-12 Flow control coefficients, 558-559 Flow exergy, 189 Fokker-Planck equation, 395 Fourier's law, 63 Fugacity, 30-31 Fugacity coefficient, 31, 46 Fundamental equations, 10 G
Gas permeation, 508-509 General stability condition, 602 Generic cubic equation of state, 25 Generic equation of state, 25-26 GENERIC formulations, 683-684 Generalized chemical kinetics, 501 Generalized correlations, 45 Generalized matrix method, 328 Geothermal power plants, 222 Gibbs, -Duhem relation, 20 equation, 20-21 free energy, 46-48 stability theory, 599-601 Global warming potential, 286 Glycolysis pathway, 552 Gradient of chemical potential, 126 Grand composite curves, 248 H
Heat capacities for real gases, 22 Heat exchanger network synthesis, 248 Heat of transport, 369-371,383 Helmholtz free energy, 46-48 Henry's law, 34-35 Heterogeneous azeotrope, 43 Hittorf transference number, 341 Hydrostatic equilibrium, 8 I
Ideal regenerative Rankine cycle, 201-202 Ideal reheat regenerative cycle, 208 Increase of entropy principle, 156 Infinitesimal processes, 2 Information capacity, 245-246 Intensive properties, 2 Internal, energy, 5, 47 variables, 681,684-685 Irreversible process, 3, 6-8
Subject index Isentropic efficiency, 156 Isomerization reaction, 547 Isothermal compressibility, 3-5 d Joule Thomson coefficient, 26
machines, 593 Momentum, balance, 116-119 flow, 57-59 Mosaic nonequilibrium thermodynamics, 678-679 Multidimensional inflection points, 564 Mutual diffusion coefficients, 85,324
K
Kinematic viscosity, 57 Kinetic energy, 2 Kirkwood's procedure, 515 Knudsen coefficient, 347 Kxonecker delta, 58 L Leaks, 537, 575-578 Lengyel-Epstein model, 647 Level flow, 371 Lewis number, 87 Limit cycle, 633-635,637-640 Linear nonequilibrium thermodynamics, 54-55 Linear pathway, 562 Linear stability analysis, 614-615 Lineweaver-Burk plot, 444, 445 Liquid-like root, 25 Liquid-liquid equilibrium, 33 Local entropy production, 123-124 Local equilibrium, 54-55, 98 Local gross exergy loss, 283 Lorenz equations, 635-636 Lotka-Volterra model, 657 Lyapunov function, 599-600 M
Macroscopic behavior, 613 Macroscopic state, 2, 14 Mass action law, 89 Mass average velocity, 73, 115 Mass balance, 15, 116 Matrix model, 684 Maxwell-Cattaneo equation, 89 Maxwell relations, 47 Maxwell-Stefan equations, 86, 319 Mechanical, filtration coefficient, 511 stability, 601 Membrane equilibrium, 505-507 Metabolic control analysis, 558 Metabolic pathway, 661 Michaelis-Menten equation, 444, 445, 581 Microbial growth, 445-446 Microscopic domain, 56 Microscopic reversibility, 7, 97, 421-423 Minimum, entropy production, 146-149, 442 separation work, 182 temperature level, 246 utilities, 250 Mitochondria, 549-552 Mixing functions, 37 Mixing rules, 42 Modified Graetz problem, 390 Modified Raoult's Law, 34 Molar average velocity, 73 Molecular, evolution, 593
N
Natural processes, 2 Nernst-Haskell equation, 84 Nernst-Planck equation, 349 Network thermodynamics, 671 Newtonian fluids, 56-57 Newton's law of viscosity, 56 Nonequilibrium systems, 53-55,629 Nonlinear transport and rate processes, 394 Non-Newtonian fluids, 56-57 Normal stress, 58 Nozzle flow, 102 NRTL model, 36-37 Nusselt number, 463-464 O
Ohm's law, 87-88 Onsager's reciprocal relations, 132 Open circuit, 532 Optimization problem, 279, 287 Oscillating systems, 616-617 Osmotic, diffusion, 83 equilibrium, 43-44 pressure, 44-45,404 temperature, 404 Oxidation of glucose, 546 Oxidative phosphorylation, 550-551 P
Packed duct flow, 168 Partial excess properties, 37 Peltier effect, 91,407 Peltier heat, 409-410 Phase, diagrams, Tyx, Pyx, 39, 41 rule, 33 stability, 604 Phenomenological, coefficients, 127-128, 132-135, 379-383,476 equations, 127-129, 397, 474-475, 532-535 stoichiometric coefficient, 371 Photosynthesis, 551,554 Pinch analysis, 246-249 Polytropic compression, 104 Potassium channels, 578-579 Potential energy, 2 Power plant analysis, 229 Prandtl number, 64 Prefractionation arrangements, 183 Principle of detailed balance, 419 Production functions, 573 Proper pathways, 557-559 Pseudocritical properties, 61 R
Raoult's law, 34 Rational thermodynamics, 679-680
729
730 Rayleigh number, 634-635 Reaction-diffusion systems, 453,622, 650 Reaction velocity, 425,475-479, 547 Reduced, phenomenological coefficients, 343 pressure, 60 temperature, 60 viscosity, 60 Reflection coefficient, 511-514, 523-524 Regulation in bioenergetics, 574-575 Reheat Rankine cycle, 196 Relaxation length, 528 Relaxation times, 89 Residual enthalpy, 45,529 Residual entropy, 45 Resistance coefficients, 132 Retrofit of distillation, 299 Reverse diffusion, 83 Reversible process, 6-8 Rheopectic fluids, 57 S Saxen relations, 521 Scalar flows, 129 Schmidt number, 87 Second law analysis, 155 Seebeck effect, 406 Self-assembly, 632 Self-diffusivity, 76 Self-organization, 631-632 Sensitivity of enzymatic reaction, 563 Separation work, 182 Second law efficiency, 181-182 Second law of thermodynamics, 13-14, 98-99 Shear stresses, 58 Short Circuit, 532 Slippage, 575-576 Solute permeability coefficient, 511 Sonic velocity, 72 Soret effect, 91,363-364 Spurious drift term, 355 Stability of chemical reactions, 606 Stability of stationary states, 612 State function, 2-3 Static head, 371 Stationary states, 111,430-431,609-613 Stokes-Einstein equation, 83 Streaming current, 91 Stress tensors, 58 Substantial differential derivative, 113 Sylvester expansion, 334 Symport, 552 Synergetics, 632
T Temperature interval method, 257-258 Thermal, conductivity, 63-64, 66-72, 379-380 diffusion coefficient, 364 diffusion factor, 365 diffusion ratio, 365-367, 381 diffusion regime, 398 diffusivity, 64 effusion, 403 equation, 21 expansion coefficient, 26, 51
Subject index field-flow fractionation, 387 stability, 601 Thermodynamic, branch, 54-56 cost, 275 coupling, 19 efficiency, 191-192 factor, 82 fluctuations, 607 forces, 131 optimum, 287 potentials, 46 Thermoeconomics, 275-276 of distillation, 298 of extraction, 291 of latent heat storage, 307 Thermoelectric effect, 91 Thermomechanical coupling, 161, 164 Thermomolecular pressure, 404 Thermoosmosis, 404 Thiele modulus, 460 Thixotropic fluids, 57 Thomson heat, 407 Throttling, 157-159 Time variation of affinity, 440-441 Total differentials, 47 Total energy flow, 389 Transference numbers, 344, 519 Transport coefficients, 87 Transport in biological cells, 581 Transport number, 523 Tricarboxylic acid cycle, 550 Turing structures, 650-651 U Ultrafiltration coefficient, 511 Uncoupling, 574-575 V Van der Pol's equation, 637 van der Waals, equation of state, 49 isotherms, 23, 49 van Kampen's hopping model, 355 van't Hoff equation, 418 Vapor-like root, 25 Vapor-liquid equilibrium, 33 Variation of coupling, 572 Vectorial flows, 129 Virial equation, 27, 40, 42, 49, 628 Viscosity of gases, 59-60 Viscous dissipation, 123 Volume average velocity, 115 Volumetric rate of entropy production, 124 W Wall friction factors, 348 Weisz modulus, 463 Wetted wall column, 330, 332-334 Wilke-Chang equation, 82 Wilson equation, 36 Work, 1-2 Z Zeroth law of thermodynamics, 11
