Fundamentals of Reaction Engineering

RAFAEL KANDIYOTI FUNDAMENTALS OF REACTION ENGINEERING ‐ EXAMPLES DOWNLOAD FREE TEXTBOOKS AT BOOKBOON.COM Rafael Kan...

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RAFAEL KANDIYOTI

FUNDAMENTALS OF REACTION ENGINEERING ‐ EXAMPLES

DOWNLOAD FREE TEXTBOOKS AT

BOOKBOON.COM

Rafael Kandiyoti

Fundamentals of Reaction Engineering Worked Examples

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Fundamentals of Reaction Engineering – Worked Examples © 2009 Rafael Kandiyoti & Ventus Publishing ApS ISBN 978-87-7681-512-7


Fundamentals of Reaction Engineering – Worked Examples

Contents

Contents Chapter I:

Homogeneous reactions – Isothermal reactors

5

Chapter II:

Homogeneous reactions – Non-isothermal reactors

13

Chapter III:

Catalytic reactions – Isothermal reactors

22

Chapter IV:

Catalytic reactions – Non-isothermal reactors

31

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I. Homogenepus reactions - Isothermal reactors

Worked Examples - Chapter I Homogeneous reactions - Isothermal reactors Problem 1.1 Problem 1.1a: For a reaction AoB (rate expression: rA = kCA ) , taking place in an isothermal tubular reactor, starting with the mass balance equation and assuming plug flow, derive an expression for calculating the reactor volume in terms of the molar flow rate of reactant ‘A’. Solution to Problem 1.1a: Assuming plug flow, the mass balance over a differential volume element of a tubular reactor is written as:

(MA nA )V Taking the limit as 'VR 0

- (MA nA)V+'V

dn A rA dVR

MA rA 'VR

-

0 ; rA

dn A ; VR dVR

= 0

n Ae

dn A rA

³

n A0

Problem 1.1b: If the reaction is carried out in a tubular reactor, where the total volumetric flow rate 0.4 m3 s-1, calculate the reactor volume and the residence time required to achieve a fractional conversion of 0.95. The rate constant k=0.6 s-1. Solution to Problem 1.1b: The molar flow rate of reactant “A” is given as,

n A0 ( 1 x A ) . kn A . Substituting, kC A vT

nA Also nA

C AvT . The rate is then given as rA

v T k

VR

VR

n Ae

³

n A0

dn A nA

vT k

x Ae

³

0

n A0 dx A n A0 ( 1 x A )

vT k

x Ae

³

0

dx A ( 1 xA )

vT ln( 1 x Ae ) # 2 m3 k

In plug flow the residence time is defined as

W{

VR vT

ª m3 º « 3 1 » ¬« m s ¼»

> s@ ;

W

2 / 0.4

5 s

Problem 1.2 The gas phase reaction Ao2S is to be carried out in an isothermal tubular reactor, according to the rate expression rA = k CA. Pure A is fed to the reactor at a temperature of 500 K and a pressure of 2 bar, at a rate of 1000 moles s-1. The rate constant k= 10 s-1. Assuming effects due to pressure drop through the reactor to be negligible, calculate the volume of reactor required for a fractional conversion of 0.85.



Data: R, the gas constant P the total pressure nT0 total molar flow rate at the inlet k, reaction rate constant


: 8.314 J mol-1 K-1 = 0.08314 bar m3 kmol-1 K-1 : 2 bar : 1000 moles s-1= 1 kmol s-1 : 10 s-1

Solution to Problem 1.2

In Chapter I we derived the equation for calculating the volume of an isothermal tubular reactor, where the first order reaction, A o 2S, causes volume change upon reaction: ½° ª 1 º RT ° (Eq. 1.37) VR nT 0 ®( 1 y A0 )ln « » y A0 x Ae ¾ Pk °¯ °¿ ¬ 1 x Ae ¼ In this equation, for pure feed “A”, yA0 { nA0/nT0 = 1. ( 0.08314 )( 500 ) VR ( 1 )^2 ln( 6.67 ) 0.85` ( 2 )( 10 ) VR

6.1 m3

Let us see by how much the total volumetric flow rate changes between the inlet and exit of this reactor. First let us review how the total molar flow rate changes with conversion: = nI0 nI nA0 – nA0 xA nA = nS0 + 2 nA0 xA nS = ----------------------------------= nT0 + nA0xA nT

for the inert component for the reactant for the product nT is the total molar flow rate

nT 0 RT ( 1 )( 0.08314 )( 500 ) / 2 20.8 m3 s-1 P vT ,exit nT ,exit RT / P; but nT,exit nT0 n A0 x A,exit vT 0

and nT 0 vT ,exit

n A0 ; therefore

( 1 0.85 )( 0.08314 )( 500 ) / 2 38.5 m3 s 1

vT 0

20.8 m3 s 1 ; vT ,exit

38.5 m3 s 1

The difference is far from negligible! In dealing with gas phase reactions, rates are often expressed in terms of partial pressures: rA= kpA, as we will see in the next example.




Problem 1.3 The gas phase thermal decomposition of component A proceeds according to the chemical reaction k1 ZZZ X A YZZ Z BC k2

with the reaction rate rA defined as the rate of disappearance of A: rA k1C A k2 CB CC .The reaction will be carried out in an isothermal continuous stirred tank reactor (CSTR) at a total pressure of 1.5 bara and a temperature of 700 K. The required conversion is 70 %. Calculate the volume of the reactor necessary for a pure reactant feed rate of 4,000 kmol hr-1. Ideal gas behavior may be assumed. Data k1 108 e10,000 / T s-1 (T in K)

k2

108 e8,000 / T


Gas constant, R = 0.08314

m3 kmol-1 s-1 (T in K) bar m3 kmol-1 K-1




Solution to Problem 1.3 Using the reaction rate expression given above, the “design equation” (i.e. isothermal mass balance) for the CSTR may be written as:

VR

n A0 n A k1C A k2CB CC

n A0 x AvT , where k2 k1n A ( )nB nC vT

ni vT

Ci

(Eq. 1.1.A)

We next write the mole balance for the reaction mixture:

nA

nA0 nA0 x A

nB

nA 0 x A

nC

nA0 x A

nT

nA0 nA0 x A

nA0 (1 x A ) .

This result is used in the “ideal gas” equation of state:

vT

nT RT PT

n A0 RT ( 1 xA ) . PT

Substituting into Eq. 1.1.A derived above, we get:

VR

nA0 x A (nA0 RT )(1 x A ) . k2 PT nA2 0 x A2 ½ PT ®k1nA0 k1nA0 xA ¾ nA0 RT (1 x A ) ¿ ¯

Simplifying, § nA0 RT · x A (1 x A ) . ¨ ¸ 2 © PT ¹ k k x k2 PT xA ½ ® 1 1 A ¾ RT 1 x A ¿ ¯ kP To further simplify, we define a 2 T . RT VR

VR

§ nA0 RT · x A (1 x A ) 2 ¨ ¸ 2 2 © PT ¹ k1 k1 x A k1 x A ax A k1 x A

(Eq. 1.1.B)

Next we calculate the values of the reaction rate constants k1 62.49 s 1 ; xA,exit = 0.70 ; T = 700 K k2 = 1088 m3s-1kmol-1; PT = 1.5 bar

Substituting the data into Eq. 1.1.B 2 4000 0.08314 700 ª 0.7 1 0.7 VR « 2 2 3600 1.5 «¬ 62.49 k1 x A a 0.7 VR

º »;a »¼

1088 1.5 0.08314 700

28

43.11 0.11 # 4.74 m3




Problem 1.4 The liquid phase reaction for the formation of compound C k1 A B o C

with reaction rate rA1

k1C AC B , is accompanied by the undesirable side reaction k2 A B o D

with reaction rate rA2

k2 C ACB . The combined feed of 1000 kmol per hour is equimolar in A and B. The two

components are preheated separately to the reactor temperature, before being fed in. The pressure drop across the reactor may be neglected. The volumetric flow rate may be assumed remain constant at 22 m3 hr-1. The desired conversion of A is 85 %. However, no more than 20 % of the amount of “A” reacted may be lost through the undesirable side reaction. Find the volume of the smallest isothermally operated tubular reactor, which can satisfy these conditions. Data k1 = 2.09u1012 e-13,500/T k2 = 1017 e-18,000/T

m3 s-1 kmol-1 m3 s-1 kmol-1

; ;

(T in K) (T in K)

Solution to Problem 1.4 For equal reaction orders: (n1 / n2 ) (k1 / k2 ) , where n1 is moles reacted through Reaction 1, and n2 is moles reacted through Reaction 2. Not allowing more than 20 % loss through formation of by-product “D” implies: (n1 / n2 ) (k1 / k2 ) t 4 . Since 'E2 > 'E1, the rate of Reaction 2 increases faster with temperature than the rate of Reaction 1. So the maximum temperature for the condition ^ n1 / n2 t 4` will be at (n1 / n2 ) (k1 / k2 ) 4 . This criterion provides the equation for the maximum temperature:

2.09 u 10 12 4500 / T e 10 17

4 ;

4 u 1017 ½ . Solving, T ln ® 12 ¾ ¯ 2.09 u 10 ¿

4500 T

4500 12.16

370 K .

The temperature provides the last piece of information to write the integral for calculating the volume. 0.85

n dx

A ; ³0 k1 A0 k2 C ACB

VR

VR

0.85

³0

vT2 n A0 dx A kn 2A0 1 x A

2

define k k1 k2

vT2

0.85

dx

A ³ 0 kn A0 1 x A 2

° ½° 22 1 1¾ VR ® k 3600 (1000) 0.5 ¯° 1 0.85 ¿° 2.69 u 104 5.67 ; (k + k ) = 2.98u10-4 + 0.74u10-4 = 3.72u10-4 2

VR

k

1

2

VR = 4.1 m3




Problem 1.5 A first order, liquid phase, irreversible chemical reaction is carried out in an isothermal continuous stirred tank reactor A o products. The reaction takes place according to the rate expression rA = k CA. In this expression, rA is defined as the rate of reaction of A (kmol m-3 s-1), k as the reaction rate constant (with units of s-1) and CA as the concentration of A (kmol m-3). Initially, the reactor may be considered at steady state. At t = 0, a step increase takes place in the inlet molar flow rate, from nA0,ss to nA0. Assuming VR (the reactor volume) and vT (total volumetric flow rate) to remain constant with time, a. Show that the unsteady state mass balance equation for this CSTR may be expressed as:

n A0 dn A ª 1 º « k » nA , dt ¬W W ¼ Also write the appropriate initial condition for this problem. The definitions of the symbols are given below. b. Solve the differential equation above [Part a], to derive an expression showing how the outlet molar flow rate of the reactant changes with time. c. How long would it take for the exit flow rate of the reactant A to achieve 80 % of the change between the original and new steady state values (i.e. from nA,ss to nA) ? DEFINITIONS and DATA: nA0,ss steady state inlet molar flow rate of reactant A before the change : 0.1 kmol s-1 new value of the inlet molar flow rate of reactant A : 0.15 kmol s-1 nA0 reactor volume : 2 m3 VR k reaction rate constant : 0.2 s-1 total volumetric flow rate : 0.1 m3 s-1 vT nA,ss nA t

W

rA CA

steady state molar flow rate of reactant A exiting from the reactor, before the change, kmol s-1 the molar flow rate of reactant A exiting from the reactor at time t, kmol s-1 time, s average residence time, VR/vT , s rate of reaction of A , kmol m-3 s-1 concentration of A, kmol m-3




Solution to Problem 1.5 a. CSTR mass balance with accumulation term: material balance on component A, over the volume VR for period ' t:

^n

A0

`

(t ) n A (t ) r A (t )VR 't

N A t ' t N A t

Bars indicate averages over the time interval 't, and NA(t) is the total number of moles of component A in the reactor at time t. Also C A n A / vT N A / VR , and the average residence time, W VR / vT . Taking the limit as 't o dt in the above equation, we get:

dNA dt

nA0 nA rAVR Combining this with N A { VR C A

VR (nA / vT ) leads to the equation: VR dnA nA0 nA kC AVR vT dt

Rearranging,

nA0 nA

knAVR vT

dnA , and, dt

W

dnA ª 1 º « k » nA dt ¬W ¼

nA 0

W

with initial condition: at t=0,

ª1

nA=nA,ss

º

b. We define E { « k » and write the complementary and particular solutions for the first order ordinary ¬W ¼ differential equation derived above. nA,complementary = C1e -E t ; nA,particular = C2 (const) where the solution is the sum of the two solutions: nA (t) = nA,c (t)+ nA,p (t). We substitute the particular solution into differential equation to evaluate the constants:

ª1 º «¬W k »¼ C2

nA0

W

The most general solution can then be written as n A

; C2

C1e E

t

nA0

EW n A0

EW

. . To derive C1, we substitute the

initial condition: nA=nA,ss at t=0, into the general solution:

nA, ss

C1

nA0

EW

;

C1

nA, ss

nA0

EW

The solution of the differential equation can then be written as:

nA

c. lim n A t of

n A0

EW

nA0 ½ E t nA0 ®nA, ss ¾e EW ¿ EW ¯ where EW =1+kW.

n A0 The new steady state exit molar flow rate of A 1 kW Download free books at BookBooN.com 11



The relationship linking t, nA,ss and nA is derived from the solution of the differential equation § § nA 0 ·½ nA 0 · ½ nA, ss ¸ ° ° ¨ nA, ss ° ¨ ¸° EW ¹° W ° © W ° © EW ¹° t ®ln ¾ ®ln ¾ 1 kW ° § 1 kW ° § nA0 · ° nA 0 · ° nA ¸ n °¯ ¨© A E W ¸¹ °¿ °¯ ¨© E W ¹ °¿

W

2 0.1

W 1 kW

nA, final

20 s

;

nA, ss

20 1 (0.2)(20) nA 0

EW

nA0 1 kW

nA0, ss 1 kW

0.1 1 (20)(0.2)

0.02 kmol s 1

4s

0.15 1 (20)(0.2)

0.03 kmol s 1

For 80 % of the change to be accomplished: nA = nA,ss + 0.8 (nA,final - nA,ss) nA= 0.8 nA,final + 0.2 nA,ss = 0.8(0.03) + 0.2 (0.02) = 0.028 kmol s-1 º ªn º °½ W ° ª nA0 nA, ss » ln « A0 nA » ¾ t ®ln « 1 kW ¯° ¬ E W ¼ ¬EW ¼ ¿° t

4 ^ln > 0.03 0.02@ ln > 0.03 0.028@`

6.44 s

6.44 s


t

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II. Homogeneous reactions - Non-isothermal reactors


Worked Examples - Chapter II Homogeneous reactions – Non-isothermal reactors Problem 2.1 Problem 2.1a: Identify the terms in the energy balance equation for a continuous stirred tank reactor: T

¦ ni0 ³ C pi dT VR ¦ H fi ( T )ri Q 0 T0

1st term: heat transfer by flow 2nd term: heat generation/absorption by the reaction(s) 3rd term: sensible heat exchange Steady state: no accumulation

Solution to Problem 2.1a:

Problem 2.1b: What form does the equation take for the single reaction: A o B + C Solution to Problem 2.1b:

6 Hfi ri= HfA rA + HfB rB + HfC rC

and rA = - rB = -rC.

= rA {HfA – HfB – HfC} = - rA'Hr

Then the energy balance equation takes the form: T

¦ ni0 ³ C pi dT VR ' H r rA Q 0 T0

Problem 2.1c: For a simple reversible-exothermic reaction A B, qualitatively sketch the locus of maximum reaction rates on an xA vs. T diagram (together with the xA,eq line). Solution to Problem 2.1c:

xA xA,eq

Conversion

xA,OPT

Temperature

T Download free books at BookBooN.com

13

II. Homogeneous reactions - Non-isothermal reactors


Problem 2.1d: The chemical reaction A o B is carried out in a continuous stirred tank reactor. Calculate the adiabatic temperature rise for 100 % conversion. The mole fraction of A in the input stream yA0 = (nA0/nT0) = 1. The total heat capacity, C p = 80 J mol-1 K-1 The heat of reaction, ' H r 72 kJ mol-1 Solution to Problem 2.1d: From notes, the simplified energy balance for an adiabatic CSTR

xA Solving for T T0

° C p ½° ® ¾ T T0 . °¯ y A0 ' H r °¿

'H r

( T T0 )

72,000 80

Cp

900 K

Problem 2.2 Problem 2.2a: Sketch the operating line on an xA vs. T diagram (i) for an endothermic reaction carried out in an adiabatic CSTR; (ii) for an exothermic reaction carried out in an adiabatic CSTR. Solution to Problem 2.2a: xA 'Hr > 0

'Hr < 0

To

T

Problem 2.2b: Sketch the operating line on an xA vs. T diagram for an exothermic reaction carried out in a CSTR, with cooling (Q < 0 ): Solution to Problem 2.2b: xA

Q5. Now let us calculate ) by assuming that K # 1 / ) . From the main text (Chapter 6), we have: C ½ V p k [ C A,s ] n ° A,s ° n ® ³ DA,eff k [ C ] dC ¾ Sx 2 °¯ 0 °¿

)

)

V p kV [ C A,s ] n ° k [ C A,s ] n1 ½° D ® A,eff ¾ Sx n1 2 °¯ °¿

1 / 2

1 / 2

V p ° 2Deff 1 ½° ® n1 ¾ S x °¯ n 1 kV C A,s °¿

(Eq. 6.74) 1 / 2

For n = 2, we get 1/ 2

½° d p ° 3 k C ® V A,s ¾ 6 °¯ 2Deff °¿

)

/2 1.9 u 10 2 C 1A,s

(Eq. A)

Neglecting external transport resistances, at the reactor inlet:

CA

p A,partial

nA vT

RT

0.2( 1 ) ( 0.08314 ) ( 600 )

4 u 10 3 kmol m 3

At the reactor exit:

CA

( 0.05 )( 1 ) 1 u 10 3 kmol m 3 ( 0.08314 )( 600 )

Substituting these values into Eq. A above, we get

) | 12 at the reactor inlet and ) | 6 at the reactor exit. So we must conclude that diffusion resistances are not negligible within this reactor. Part (b): At steady state, the rate of reaction within a particle must be equal to the net rate of diffusion of the reactant.

net flux

From jD

5 u 10 3 ( 10 )

3333 1.5 u 10 5 ( 0.46 / H ) (Re)0.4 we can calculate jD 0.043 . Meanwhile jD G 1 km 2/ 3 Ug ° P ½° ® ¾ ¯° U g DA ¿°

Let us calculate km am .

Re

d pG

km am ( Cb Cs )

P

° P °½ ® ¾ °¯ U g DA °¿

2/ 3

° 1.5 u 10 5 °½ ® 4 ¾ °¯ 1 u 1 u 10 °¿

2/ 3

0.28 Download free books at BookBooN.com

26

III. Catalytic reactions - Isothermal reactors


( 0.043 )( 10 ) 1 1.54 and 1 0.28 4S rp2 3 6 am 3 4 3 S rp U p rp U p 5 u 10 u 2200 3 km am 0.84 m3 kg 1 s 1

km

0.545 m 2 kg 1

The catalyst bed density is calculated from U B ( 1 H ) U p ( 0.58 ) u ( 2200 ) 1276 kg m 3 Then we use the concept that, at steady state, net diffusion of reactant to the catalyst pellet equals the amount of reactant that has been converted. 2 U B km am ( C Ab C As ) K k C As

Define a constant D . At the inlet D

kmol /( m2 s ) U B km am ( 1276 ) u ( 0.84 ) 0.372 kmol m 3 Kk § 1 · 4 ¨ ¸ ( 3.46 u 10 ) © 12 ¹

The equation to solve now takes the form: 2 D ( C Ab C As ) C As

(Eq. B)

and at the inlet C Ab 4 u 10 3 kmol m 3 . Eq. B can be solved as a quadratic equation or by trial and error.

C As # 3.95 u 10 3 kmol m 3 Comparing C Ab and C As , we can see the difference is small. Therefore, in this case, the bulk-stream to pellet surface mass transport resistance appears negligible at the reactor inlet.

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Problem 3.3 Differential reactors are used to simulate an infinitesimally thin (“slice”) section of, say, a tubular reactor. In the laboratory, this is done to measure reaction rates and other parameters of catalytic reactions, in the context of a small amount of conversion.

The first order irreversible reaction A o B C will be carried out over 0.002 kg packed catalyst in a differential test reactor. The intrinsic rate expression for the reaction is

kmol ( kg cat u s )1

rA 7 u 1015 exp ^18,000 / T ` C A

where the temperature T is given in K and CA, the concentration of the reactant “A”, is expressed in kmol m-3. Assuming isothermal operation at 500 K and given the data listed below, (a) Estimate whether intraparticle diffusion resistances would be expected to affect the overall rate. (b) Estimate whether bulk to surface diffusion resistances would be expected to affect the overall rate. (c) Calculate the conversion through the reactor. Data: Catalyst pellet diameter, dp:

0.001

m

Catalyst pellet density Up : Effective diffusivity of A within the pellet, DA,eff :

1500

kg m-3

1 x 10-6

m2 s-1

Molecular diffusivity of A in the bulk stream, DA :

1 x 10-4

Mass flow rate, G: Viscosity of the gas mixture, μ: Density of the gas mixture, Ug: Void fraction of the bed, H: Total volumetric flow rate, vT :

0.1 1 x 10-6 1 kg m-3 0.42 0.005

m2 s-1 kg m-2 s-1 kg m-1 s-1

m3 s-1

The dimensionless mass flux jD is given by the correlation

jD

( 0.46 / H ) (Re)0.4 , where Re

d pG / P

where jD is related to km, the bulk stream to particle surface mass transfer coefficient, by

jD

° P ½° ( km U g / G ) ® ¾ ¯° U g DA ¿°

2/ 3

.

The Thiele modulus for spherical catalyst pellets is given by:

· ° k U p ¸® ¹ °¯ DA,eff

1/ 2

°½ )s ¾ ¿° where Rs is the pellet radius, k the reaction rate constant; Up and Deff have been defined above. § Rs ¨ © 3

(Eq. A)




Solution to Problem 3.2 Part (a)

Rs

0.0005 m .

At 500 K, the reaction rate constant k

0.0005 3

)s

( 1.62 )( 1500 ) 1 u 10 6

1.62. Substituting into Eq. A, we can calculate ) s .

8.225

) s is large. We conclude that intraparticle diffusion resistances are significant and would be expected to affect the overall rate. Part (b):

4S rp2

am

3 rp U p

4 3 S rp U p 3

3 5 u 10

4

u 1500

4.0 m 2 kg 1

To calculate km , we first need the Reynolds number. Re

jD

( 0.001 )( 0.1 ) 1 u 10 6

100 . Using this value we can get

0.174. Then

km

jD G ° P ½° ® ¾ U g ¯° U g DA ¿°

2 / 3

0.375

m s 1

The basic equation to use for comparing external and intraparticle diffusion effects is:

rp

1 1 1 K k km am

km am

C Ab . We now compare

1 1 and . Kk km am

1 0.66 km am 1 1.62 / 8.225 0.197; 5.1 Kk

( 4 )( 0.375 ) 1.5 m3 s 1 ;

Kk # k / )s

External diffusion resistance less significant! Part (c): For a “differential reactor” the mass balance equation dW

'W #

'n A rp

, where 'n A

n A0 ' x A . We also have rp

dn A can be written as: rp

1 C Ab and 1 1 K k km am




CA

n A / vT . Thus we can write C Ab

n A0 ( 1 x A ) and rp vT

In the case of the differential reactor, x A # ' x A . Thus

1 1 1 K k km am

u

n A0 ( 1 x A ) . vT

1 1 ½ ® ¾ ¯K k km am ¿ 0.005 ^5.1 0.667` 14.42 0.002

1 x A vT # 'W xA

'W

1 1 xA n A0 ' x A 1 ½ u® ¾ vT ; then xA n A0 ( 1 x A ) ¯K k km am ¿

xA

1 0.066 15.41

xA

0.066

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IV. Catalytic reactions - No isothermal reactors


Worked Examples - Chapter IV Catalytic reactions – Non-isothermal reactors Problem 4.1 The exothermic catalytic gas phase cracking reaction A o light products is carried out with a feed of 10 % “A” in an inert diluent, “B”. The intrinsic rate for the reaction is given by:

rA

kmol s 1 ( kg cat )1

125 exp ^5000 / T ` p A

where pA denotes the partial pressure (bar) of “A” and T denotes the local temperature in K. The catalyst pellets (dia. 0.004 m) may be considered isothermal and the value of the effectiveness factor is estimated as 0.08 throughout the reactor. We select a point within the reactor where, G, the superficial mass velocity of the gas stream is 0.8 kg m-2 s-1, the bulk temperature, Tb, is 500 K, and the partial pressure of “A” in the bulk, pA,b, is 0.05 bar. Calculate: (i) (ii) (iii)

the global rate of reaction, the external concentration gradient (expressed in terms of the partial pressures), and, the temperature gradient between the bulk gas stream and external catalyst pellet surfaces,

at the selected point in the reactor. Bulk gas properties may be taken as those of the diluent. The pressure drop along the reactor axis may be neglected. Data: Effectiveness factor, K : Total reactor pressure, pT :

0.08 1 bar

Void fraction of the packed bed, HB : Surface area of catalyst pellets, am: Heat of reaction, 'Hr : Molecular diffusivity of A in B, DAB :

(see statement of problem)

0.5 0.4 m2 kg-1 20 000 kJ (kmol A reacted)-1 2 u 10-4 m2s-1

Density of “B”, U B :

0.1 kg m-3

Viscosity of “B”, P B : Heat capacity of “B”, Cp,B :

1.0 u 10-5 kg m-1 s-1 1.0 kJ kg-1 K-1

Thermal conductivity of “B”, OB :

jD

jD

1.5 u 10-5 kJ m-1 s-1 K-1

( 0.7 ) jH

0.46

HB

km U B P B ½ ® ¾ G ¯ Ub DAB ¿

Re 0.4 ; Re

2/ 3

;

jH

d p G / PB

C p PB ½ ® ¾ C p G ¯ OB ¿ h

2/ 3

.




In these equations, km, the bulk to surface mass transfer coefficient is defined in terms of the equation: rA,p = (kmam/RTb) ( pb - ps ), and, h is the surface to bulk heat transfer coefficient, defined by h am (Ts - T b) = ( 'Hr) rA,p . Finally, R (gas constant) = 8.314 kJ kmol-1K-1= 0.08314 bar m3 K-1 kmol-1 Solution to Problem 4.1

( 0.004 ) u ( 0.8 )

Re

d p G / PB

jD

9.16 u 10 2

320 ; substituting this value into the correlations, we get

1 u 10 5 and jH 0.131 .

Next, we calculate the Prandtl and Schmidt numbers

Pr

C pB P B

1 u 1 u 10 5

OB

1.5 u 10 5

PB

0.67 and Sc

U B DAB

1 u 10 5

0.5 .

0.1 u 2 u 10 4

On the other hand

ham ( Ts Tb ) rA,p u ' H r , leads to

h

jH C p G (Pr)2 / 3

( Ts Tb )

rA,p ( ' H r )(Pr)2 / 3

( rA,p ) ( 20,000 ) ( 0.67 )2 / 3

jH C p G am

( 0.131 )( 1 )( 0.8 )( 0.4 )

recalling that Tb

3.65 u 10 5 rA,p

500 K , Ts

500 3.65 u 10 5 rA,p .

(Eq. A)

Similarly

km am ( pb ps )A RTb

rA,p and km

jD G

1

U B Sc 2 / 3

Then

( pb ps )A

leads to ( pb ps )A

rA,p U B ( Sc 2 / 3 ) R Tb

rA,p ( 0.1 ) ( 0.5 )2 / 3 ( 0.08314 )( 500 )

am jD G

( 0.4 )( 9.16 u 10 2 )( 0.8 )

89.1 u rA,p , where p A,b

0.05 leads to p A,s

The intrinsic reaction rate was given as rA

0.05 89.1 u rA,p

(Eq. B)

125 exp ^5000 / T ` p A




Since rA,p

K rA,s K rA ( at surface conditions ) , and the intrinsic reaction rate expression was

given in the problem statement as rA can be written as

rA,p K rA,s

( 125 ) exp ^5000 / T ` p A , the global reaction rate expression

( 0.08 ) u ( 125 ) exp ^5000 / Ts ` p A,s .

Substituting expressions for Ts from Eq. (A) and p A,s from Eq. (B), we get rA,p

K rA,s

^

`

( 0.08 ) u ( 125 ) u ª¬0.05 89.1 u rA,p º¼ u exp 5000 /( 500 3.65 u 10 5 rA,p )

p

p

p A,s

Ts

This equation can be solved by a simple trial and error procedure. rA,p 2.6 u 10 5 kmol ( kg cat u s )1 Then Ts

500 3.65 u 10 5 [ 2.6 u 10 5 ] # 509.5

'T ps

0.05 89.1 rA,p

9.5 K

Not negligible!

4.77 u 10 2

' p 0.05 0.0477 0.0023 bar o ~ 5 % of total pressure.


Probably negligible…just.




Problem 4.2 The exothermic gas phase reaction

A+BoC is carried out over a porous catalyst, in a non-isothermal/non-adiabatic fixed bed catalytic reactor. The feed stream is equimolar in A and B. (a) Using a one-dimensional, pseudo-homogeneous model framework, derive steady-state mass and energy balance equations and state boundary conditions to describe the behaviour of the reactor. Define your terms carefully. (b) Calculate the maximum temperature that the exit gas stream would attain if the reactor were to be operated adiabatically. Judging by this result, what change would you recommend in reactor operating conditions. (c) Discuss briefly for the present case, advantages and disadvantages of using a one-dimensional reactor model over a two-dimensional model.

Data: Heat of reaction, 'Hr : Average heat capacities

CpA # CpB : CpC :

300 000

kJ (kmol A reacted)-1

30

kJ kmol-1 K-1 kJ kmol-1 K-1

60

Solution to Problem 4.1 Part A: Mass balance: The mass balance over an infinitesimally thin (“slice”) element of the catalyst packed reactor (dW) gives: dn A rA dW , where n A is defined by the equation n A n A0 ( 1 x A ) and dn A n A0 dx A . In these equations W denotes the mass of catalyst and nA the molar flow rate of “A”. nT 0 C p is the total heat capacity of the inlet stream. It is assumed not to change with conversion; so we shall use it as the total heat capacity of the reaction mixture. From dn A

rA dW we can then write the full mass balance equation as:

Energy balance:

ª ¦ ni C pi º dT ¬ ¼

where Q is the external heat transfer term and

dx A dW

rA . n A0

(Eq. A)

dW ( Q rA ' Hr )

¦ ni C pi #

nT 0 C p as explained in Chapter 3 of the main

text. The energy balance equation may then be written as:

dT dW

Q rA ' H r nT 0 C p

(Eq. B)

The initial condition is given as: At W = 0, T = T0 = Twall .




There are many ways of formulating Q, the heat loss term. One common way would be: Q A h ( T Twall ) , where A denotes the contact surface area and h the heat transfer coefficient. Part B: Combining Eqs. A and B, we can write

n A0 rA

dT dx A For adiabatic operation, Q = 0 .

dT dx A

° Q r ' H ½° A r . ® ¾ ¯° nT 0 C p ¿°

'H r [ nT 0 C p ]

n A0

Integrating

'T

' H r n A0 C p nT 0

( 0.5 )( 3 u 10 5 ) 30

5000 K

Unacceptably high! Improve cooling and/or reduce reactant concentrations and/or dilute the catalyst. Part C: The main difficulty arises from ignoring the radial temperature distribution. In dealing with highly exothermic reactions, when the maximum temperature along the central axis turns out to be much larger than the radially averaged temperature, runaway may occur under conditions where the 1-dimensional model would predict “safe” operation. The disadvantage of 2-dimensional operation is the requirement for accurate additional data for radial heat (and mass) transport: Orad ,effective , Drad ,effective .

Problem 4.3 The exothermic reaction

A+BoC

was carried out using two different catalyst particle sizes: 8 mm diameter catalyst pellets and 50 μ (average) diameter catalyst particles. Both sets of experiments were carried out under conditions where external temperature and concentration gradients were negligible. Global rates of reaction, measured as a function of temperature, using approximately similar external surface concentrations of A and B for all experiments, are given in the table below. Assuming the 50 μ particles to be isothermal under all operating conditions, (a) Calculate from this data, the energy of activation most likely to approximate the true value for the chemical reaction step at the catalytic sites. (b) For the 8 mm catalyst pellets, calculate the effectiveness factor at each temperature. State your assumptions clearly. c. Briefly explain observed changes in the values of effectiveness factors as a function of the experimental temperature. What do values above unity imply?




Table A. Experimental global reaction rates as a function of catalyst particle size and temperature.

Catalyst Particles Rate, kmol s-1

Temp (K) 425.5 458.7 495.1 549.5 637.0

9.2 x 10-8 2.5 x 10-7 6.95 x 10-7 1.85 x 10-6 4.12 x 10-6

8 mm pellets rA,p (kg cat)-1 5.12 x 10-7 5.95 x 10-7 7.18 x 10-7 8.58 x 10-7 9.39 x 10-7

Solution to Problem 4.3 Part (a): “Series 2” Arrhenius plot for 50 Pm particles (Figure A) shows the line bending above 495 K. The slope of the straight component of the line gives the correct ('Ea/R) value for chemical control, i.e. for K=1. Using

values of the rate between 425 and 495 K, we can calculate the slope of the straight part of the line. ln r1 ln r2 1.42 1.62 ° °½ Slope ® 3 ¾ 1 1 °¯ ( 2.02 2.35 ) u 10 °¿ T2 T2

' Ea R

8.314 kJ /( kmol u K )

5.04 kJ kmol 1


' Ea

6060 K ; the gas cons tan t R




ln r vs. 1000 x (1/T) 0 0

0.5

1

1.5

2

2.5

ln r

-5 Series1

-10

Series2

-15 -20 1000 x (1/T) [T in K]

Figure A. Arrhenius plot based on data for 50 Pm particles in Table A

Part (b): The first three rates in rows 1-3 of column-2 in Table A and Table B are the same. The data corresponds to the part of the Series 2 curve in Figure A before it begins to bend. Since the line is straight up to and including 495 K, we can safely conclude we have chemical control and K = 1. Above that temperature, the line bends. If there had been no diffusive limitations the line would have continued as a straight line. That is why the extrapolated line still corresponds to K = 1. We can read off (in an enlarged diagram) that the extrapolated rate for 549.5 K is 2.26u10-6 and for 637 K, the rate corresponding to K = 1 is 1.07u10-5. We then use Eq. 6.76 (Chapter 6) from the main text.

(robs )1 (Eq. 6.76) Ș1 /Ș2 (robs )2 where (robs )1 is the second column in Table B and (robs )2 third column in Table B. With K1 taken as unity, we can calculate the effectiveness factors for the 8 mm pellets at different temperatures. Table B. Experimental global reaction rates as a function of catalyst particle size and temperature.

Rate, kmol s-1 Temp (K) 425.5 458.7 495.1 549.5 637.0

rA,p (kg cat)-1 8 mm pellets (data from Table A)

50 Pm catalyst particles (K1=1) 9.2 x 10-8 2.5 x 10-7 6.95 x 10-7 2.26 x 10-6 1.07 x 10-6

5.12 x 10-7 5.95 x 10-7 7.18 x 10-7 8.58 x 10-7 9.39 x 10-7

K Effectiveness factor for 8 mm pellets

5.56 2.38 1.03 0.38 0.228




Part (c): For exothermic reactions K=1 means that the temperature at the centre of the pellet is greater than the surface temperature Ts. That in turn implies that the rate at the centre of the pellet is greater than the rate at the

surface. In this example, the effect was more visible at relatively low surface temperatures. This explains the change in the nature of the value of effectiveness factor with increasing temperature.

Problem 4.4 Differential reactors are used to simulate an infinitesimally thin (“slice”) section of, say, a tubular reactor. In the laboratory, this is done to measure reaction rates and other parameters of catalytic reactions, in the context of a small amount of conversion.

The intrinsic rate for the endothermic catalytic gas phase cracking reaction “A o products”, is given by the first order reaction rate expression

rA

4,000 ^exp( 10,000 / T )` p A

kmol /( kg cat u s )1 ,

where pA is the partial pressure (bar) of A, and T denotes the local temperature in K. The reaction is carried out in a differential reactor, where reactant “A” is supplied, mixed 1:1 with an inert diluent. The superficial mass velocity of the gas stream flowing over the catalyst particles, G, is 1 kg m-2 s-1, and the temperature of the bulk stream is 750 K. The overall conversion in the reactor has been determined experimentally as 3.6 %. The catalyst pellets have a diameter of 0.004 m and may be considered isothermal. Calculate the global rate of reaction, the magnitude of the effectiveness factor, and the external concentration and external temperature gradients. The bulk gas properties can be taken as those of the diluent, B. The pressure drop through the reactor may be neglected. Data: Molecular diffusivity of A in B, DAB: Density of diluent B, UB: Viscosity of B, μB: Heat capacity of B, Cp,B: Thermal conductivity of B, OB: Reactor pressure, pT: Void fraction of the packed bed, H: Surface area of the catalyst pellets, am: Heat of reaction, 'HR: Pellet effective diffusivity, Deff: Pellet density, Uc:

jD

jD

( 0.7 ) jH

3 x 10-4 0.1 1.5 x 10-5 1.0 2.5 x 10-5 1 0.48 0.4 25,000 1 x 10-6 2,000

0.46

HB

km U B P B ½ ® ¾ G ¯ Ub DAB ¿

m2 s-1 kg m-3 kg m-1 s-1 kJ kg-1 K-1 kJ m-1 s-1 K-1 bar m2 kg-1 kJ (kmol A)-1 m2 s-1 kg m-3

Re 0.4 ; Re

2/ 3

;

jH

d p G / PB

C p PB ½ ® ¾ C p G ¯ OB ¿ h

2/ 3

.

In these equations, km, the bulk to surface mass transfer coefficient is defined in terms of the equation: rA,p = (kmam/RTb) ( pA,b – pA,s ), Download free books at BookBooN.com 38



and, h is the surface to bulk heat transfer coefficient, defined by h am (Ts - T b) = ( 'Hr) rA,p . Finally, R (gas constant) = 8.314 kJ kmol-1K-1= 0.08314 bar m3 K-1 kmol-1 Solution to Problem 4.4

h am ( Ts Tb ) rA,p ( ' H r ) , where h ( Ts Tb ) Pr

Re

C pB P B

OB

1 u 1.5 u 10 5

d pG

2.5 u 10 ( 0.004 )( 1 )

PB

1.5 u 10 5

5

0.6;

( Ts Tb )

(Pr)2 / 3

(Eq. A)

jH C pB G am 0.71

( rA,p ) u ( 25000 ) u ( 0.71 ) ( 0.147 )( 1 )( 1 )( 0.4 ) 750 3.02 u 10 5 rA,p

0.103 and jH

0.147 .

3.02 u 10 5 rA,p

(Eq. B)


Ts

(Pr)2 / 3

rA,p ( ' H r )(Pr)2 / 3

267 ; from equations in the data section jD

Substituting in Eq. A, we get

jH C p G




km am ( p A,b p A,s ) ; we can write RTb

Using the mass transfer equation: rA,p

(Eq. C)

2 / 3

km km

jD G P B ½ jD G ( Sc )2 / 3 ® ¾ U B ¯ Ub DAB ¿ UB º ( 0.103 )( 1 ) ª 1.5 u 10 5 1.59 « 4 » 0.1 ¬« ( 0.1 ) u ( 3 u 10 ) ¼»

Substituting this result in Eq. C, we get:

p A,b p A,s

rA,p RTb

( 0.08314 )(750 )rA,p

km am

( 1.59 )( 0.4 )

98.04 rA,p

Reactor pressure: 1 bar. Conversion is 3.6%. Exit stream composition assumed equal to reactor composition. This basically means that we treat the “differential reactor” as a mini-CSTR. The feed is 50 % “A”. Therefore:

pb

( 0.5 )( 1 0.036 ) 0.482 bar . Thus: ps 0.482 98.04 rA,p rA,p K rA,s and K

1 1 1 ½ ® ¾ ; rearranging ) s ¯ tanh 3) s 3) s ¿ 1/ 2

° k RTs Uc ½° )s ® ¾ °¯ Deff °¿ ) s in the above equation is form of the Thiele modulus appropriate for the given reaction rate Rs 3

expression.

)s

0.002 3

1/ 2

ª 10000 º ( 0.08314 )( 2000 ) °½ ° u u T ®4,000 exp « ¾ » s 1 u 10 6 ¬ Ts ¼ ¯° ¿°

)s

1/ 2

544 ^Ts exp( 10000 / Ts )`

(Eq. D)

At Ts 750 K , ) s # 14 ; therefore we can safely consider K # 1 / ) s . Going back to the global reaction rate expression: rA,p 4000 K u exp ^10,000 / Ts ` u p A,s (Eq. E)

rA,p

^

`

4000 K u exp 10,000 /(750 3.02 u 10 5 rA,p u ( 0.482 98.04 rA,p )

(Eq. F)

We need to substitute for K in Eq. E. As already mentioned, K # 1 / ) s ; ) s is given by Eq. C, and Ts by

Eq. B. Then: 1/ 2

ª º °½ 10,000 ° 5 »¾ ) s 544 ®(750 3.02 u 10 rA,p ) u exp « «¬ (750 3.02 u 10 5 rA,p ) »¼ ° °¯ ¿ The easier way to solve for rA,p is by trial-and-error between K # 1 / ) s , Eq. G and Eq. F.

(Eq. G)




Close enough answers:

rA,p # 1.15 u 10 4

K 0.075 Ts

'T p A,s

715.3 K 34.7 K 0.47 bar

p A,b p A,s

0.011 bar

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