Advanced Fluid Mechanics

Advanced Fluid Mechanics

Chapter 1 Introduction 1.1 Classification of a Fluid (A fluid can only substain tangential force when it moves) 1.) 2.) 3.) 4.)

By viscous effect: inviscid & Viscous Fluid. By compressible: incompressible & Compressible Fluid. By Mack No: Subsonic, transonic, Supersonic, and hypersonic flow. By eddy effect: Laminar, Transition and Turbulent Flow.

The objective of this course is to examine the effect of tangential (shearing) stresses on a fluid. Remark: For a ideal (or inviscid) flow, there is only normal force but tangential force between two contacting layers. 1.2 Simple Notation of Viscosity

U

h

F

(tangential force required to move upper plate at velocity of U )

Fluid (e.g. water)

y

x u(y) = y/h U From observation, the tangential force per unit area required is proportional to U/h, or du/dy. Therefore

τ ≡ shear stress = tangential force per unit area (F/A) ∝

U h

or

τ =µ

U ∂u = µ ∂y h

〝Newton’s Law of function〞

(1.1)

µ : Constant of proportionality The first coefficient of viscosity Remark: E.g. (1.1) provides the definition of the viscosity and is a method for measuring the viscosity of the fluid. Chapter1- 1

Advanced Fluid Mechanics

In generally, if ε XY represent the strain rate, then

τ xy = f (ε xy )

(1.2)

plastic

τ

yielding fluids Dilatent fluid

Non- Newtonian fluid

Pesudoplastic fluid

Newtonian fluid

Yield stress

ε Newtonian fluid: linear relation between τ and ε Pesudoplastic fluid: the slope of the curve decrease as ε increase (shear-thinning) of the shear-thinning effect is very strong. The fluid is called plastic fluid. Dilatent fluid: the slope of the curve increases as ε increases (shear-thicking). Yielding fluid: A material, part solid and part fluid can substain certain stresses before it starts to deform. Note 1 Pa (Pascal) ≡

Newton m2

(Pascal, a French philosopher and Mathematist)

(a unit of pressure ) [µ] = [pa · sec]

(=

kg ⋅ m m

2

s2 ⋅ s =

kg g = 10 ) m⋅s cm ⋅ s

The metric unit of viscosity is called the poise (p) in honor of J.L.M. Poiseuille (1840), who conducted pioneering experiment on viscous flow in tubes. 1P≡

1g (cm)(s ) = 0.1 pa ⋅ sec

Chapter1- 2

Advanced Fluid Mechanics

The unit of viscosity:   F   2 τ [µ ] = αu  =  L L   ∂y   T L 

   F   =  L2 T      

← (Old -English Unit: F-L-T)

or

[µ ]

Denote:

 ML 2  M  =  2T ⋅T  =    L   LT   

← (international system SI unit: M-L-T)

N ≡ Pa, then M2

µ water , 20°c = 1.01 × 10 3 Pa ⋅ sec (liquid): T

µ water ,100° c = 283 Pa ⋅ sec

→ µ

µ air , 20°c = 17.9 Pa ⋅ sec (gas): T

µ air ,100° c = 22.9 Pa ⋅ sec For dilute gas:

T  µ  ≈  µ° T  °  T µ ≈  µ °  T°

  

n

3

(Power- law) 2

T° + S T + S

(Sutherland’s law)

Where µ 0 , T0 and S depends on the nature of the gases. Kinematics Viscosity υ ≡

µ ρ

Chapter1- 3

→ µ

Advanced Fluid Mechanics

Exp: (Effect of Viscosity on fluid) Flow past a cylinder Foe a ideal flow:

u

U∞

θ

R

w

  R2 u (r , θ ) = U ∞ cos θ  2 − 1  r 1

  R2 v(r , θ ) = U ∞ sin θ  2 + 1  r

At r = R, u=0, v = 2U ∞ sin θ

-3

The Bernoulli e.g. along the surface is:

1 1 ρU ∞2 + P∞ = ρv 2 + p 2 2

(Incompressible flow)

P − P∞ v2 1 Cp = = 1 − 2 = 1 − sin 2 θ 1 4 U∞ ρU ∞2 2 D’Albert paradox: No Drag. For a real flow: (viscous effect in) ρVD Re =

µ

(前後幾乎對稱) Re=0.16 (fig 6)

(前後不對稱) Re=1.54 (fig 24)

Chapter1- 4

Advanced Fluid Mechanics

Separation occur (pair of recirculating eddies) Re=9.6 (fig 40) (6 < Re 90 °

Subcritical θ sep < 90 °

Chapter1- 5

Advanced Fluid Mechanics

The pressure distribution then becomes:

θ

Cp

1

θ

-1

-2 -3

Supercritical (separation)

Subcritical (separation) Theoretical (invuscid)

(White. P.9. Fig. 1-5)

Chapter1- 6

Advanced Fluid Mechanics

Remark:

Newtonian Fluid Non-Newton Fluid For a Newtonian fluid: ↔

↔

↔

τ = −µ ε

τ : stress tension ↔

ε : rate of strain tension

µ = a constant for a given temp, pressure and composition Lf µ is not a constant for a given temp, pressure and composition, then the fluid is called Non-Newtonian fluid. The Non-Newtonian fluid can be classified into several kinds depending on how we model the viscosity. For example: （I）Generalized Newtonian fluid ↔

↔

↔

τ = −η ε

η = a function of the scalar invariants of ε

(i) The Carreau-Yusuda Model ( n −1) η − η∞ a = [1 + (λε ) ] a η0 − η∞

↔

ε: magnitude of the

ε

(ii) power-Law model

η = mε

n −1

n1: dilatant (shear thickening) （II）Linear Viscoelastic Fluid → polymeric fluids （III）Non-linear Viscoelastic Fluid → The fluid has both 〝viscous〞 and 〝elastic〞 properties.

By 〝elasticity〞one usually means the ability of a material to return to some unique, original shape on the other hand, by a 〝fluid〞, one means a material that will take the shape of any container in which it is left, and thus does not possess a unique, original shape. Therefore the viscoelastic fluid is often returned as 〝memory fluid〞. Chapter1- 7

Advanced Fluid Mechanics

FIGURE 2.2- 1 Tube flow and “shear thinning.” In each part, the Newtonian behavior is shown on the left ○ N ; the behavior of a polymer on the right

P . ○

(a) A tiny

sphere falls at the same through each; (b) the polymer out faster than Newtonian fluid.

Chapter1- 8

Advanced Fluid Mechanics

○

FIGURE2.3-1. fixed cylinder with rotating rod N . The Newtonian liquid, glycerin, P the polymer solution, polyacrylamide in glycerin, climbs the rod. shows a vortex; ○ The rod is rotated much faster in the glycerin than in the polyacrylamide solution. At comparable low rates of rotation of the shaft, the polymer will climb whereas the free surface of the Newtonian liquid will remain flat. [Photographs courtesy of Dr. F. Nazem, Rheology Research Center, University of Wisconsin- Madison.]

Chapter1- 9

Advanced Fluid Mechanics

P*= (P+ τyy )1 - (P+ τyy)2 > 0

P1=P2

FIGURE 2.3-4 A fluid is flowing from left to right between two parallel plates across a deep transverse slot. “Pressure”are measured by flush-mounted transducer “1.”and recessed transducer “2.”○ N For the Newtonian fluid P1=P2. ○ P For

polymer fluids (P+ τyy )1 > (P+ τyy)2. The arrows tangent to the streamline indicate how the extra tension along a streamline tends to “lift ”the fluid out of the holes, so that the recessed transducer gives a reading that is lower than that of the flush mounted transducer.

Chapter1- 10

Advanced Fluid Mechanics

FIGURE 2.4-2 Secondary flows in the disk-cylinder system. ○ N The Newtonian fluid moves up at the center, whereas ○ P the viscoelastic fluid , polyacrlamid (Separan 30)-glycerol-water, moves down at the center. [Reproduced from C. T. Hill, Trans. Soc. Rheol , 213-245 (1972).]

Chapter1- 11

Advanced Fluid Mechanics

N A stream of Newtonian FIGURE 2.5-1 Behavior of fluids issuing from orifices. ○ fluid (silicone fluid) shows no diameter increase upon emergence from the capillary − tube ;○ P a solution of 2.44g of polymethylmethacrylate ( M = 10 6 g

mol

) in 100

cm3 of dimethylphthalate shows an increase by a factor of 3 in diameter as it flows downward out of the tube. [Reproduced from A. S. Lodge, Elastic Liquid, Academic Press, New York (1964), p. 242.]

Chapter1- 12

Advanced Fluid Mechanics

N When the siphon tube is lifted out of the FIGURE 2.5-2 the tubeless siphon. ○ P the macromolecular fluid continues to fluid, the Newtonian liquid stops flowing; ○ be siphoned.

FIGURE 2.5-8 AN aluminum soap solution, made of aluminum dilaurate in decalin and m-cresol, is (a) poured from a beaker and (b) cut in midstream. In (c), note that the liquid above the cut springs back to the beaker and only the fluid below the cut falls to the container.[Reproduced from A. S. Lodge, Elastic liquids, Academic Press, New York (1964), p. 238. For a further discussion of aluminum soap solutions see N. Weber and W. H. Bauer, J. Phys. Chem., 60, 270-273 (1956).] Chapter1- 13

Advanced Fluid Mechanics

320 熱傳學

ReD < 5 無分離流動

5 到 15 ≤ ReD< 40 渦脊中具 Foppl 渦旋

4 ≤ ReD< 90 和 90 ≤ ReD< 150 渦旋串（Vortex street）為層流

150 ≤ ReD< 300 300 ≤ ReD< 3×105

3×105 < ReD < 3.5×106 層流邊界層變成紊流

6

3.5×10 ≤ ReD< ∞(?)

完全紊流邊界層

圖 7-6 正交流過圓柱之流動情形

Chapter1- 14

Advanced Fluid Mechanics

圖 7-7 長圓柱及球體之阻力係數 C p 與 Re 數關係 以下討論不同雷諾數下的物理現象： （1） 雷諾數的數量級為 1 或更小時，流場沒有分離現象，黏滯力是阻力的唯一 因素，此時流場可由勢流理論(Potential flow theory)來導證，在圖 7-7 中阻力係數隨著雷諾數的提高而直線變化下降。 （2） 雷諾數的數量級為 10 時，流場漸漸發生渦流，在圓柱後面有小渦旋 (Vertex) 出現，此時阻力的因素除了邊界層阻力外尚有渦流的因素，阻 力係數依雷諾數的提高而下降。 （3） 雷諾數介於 40 到 150 之間時，圓柱後形成渦旋串(Vertex street) ，產 生渦旋的頻率 fv 與流場雷諾數大小有關，定義 Strouhal 數 Sh：

Sh =

fv ⋅ D u∞

(7-32)

Sh 與雷諾數 Re D 的關係如下圖 7-8；此時阻力主要由係由渦流造成。 （4） 雷諾數介於 150〜300 時，渦流串由層性漸漸轉變成紊性，雷諾數 300 到

Chapter1- 15

Advanced Fluid Mechanics 5

3×10 之間，渦旋串是完全紊性的，流場非二次元性質，必須三次元才能 完全描述流場，分離點的位置變化不大，由 θ = 80 ° 到 θ = 85° ，且自圓柱 前端到分離點的流場維持屬性，所以此時阻力係數也幾乎維持在固定值。 雷諾數不影響阻力係數也就是說黏滯力對阻力的影響很小。

圖 7-8 Sh 數與 Re 數關係

5

（5） 雷諾數大於 3×10 ，阻力係數 C D 急遽下降，也就是阻力下降，這是因為 分離點往圓柱後面移動的緣故，見圖 7-9[7]；分離點會再往前移。此時 阻力係數會回升。渦脊變窄無次序，不再出現渦流串。阻力的形成可分為 兩個因數，邊界層存在時沿邊界層的地方有黏滯阻力存在，分離點之後面 產生渦流，這是低壓地區，以致有反流的現象，造成圓柱前後壓力不平衡， 是阻力產生的最大原因；當 Re= 10 6 之後流場本身穩定性不足以維持流場 的穩定，分離點再度向前移，使阻力係數再回升；這些流場現象不僅影響 到阻力，也影響到對流熱傳係數。

Chapter1- 16

Advanced Fluid Mechanics

圖 7-9 管之分離點位置與 Re 數關係[7] (二)熱傳係數 （1） 圓柱四周流場變化多端，要求得各點熱傳係數的解析值是很困難的， 圖 7-10 是 W. H. Giedt [8] 所做的實驗結果，當雷諾數大於 1.4×10

5

之後，熱傳係數 Nu (θ ) 出現兩個最低點，第一個最低點是邊界層由層 性過渡到紊性時發生，第二個最低點則是由於分離現像。

圖 7-10 圓柱之 Nu 與自停滯點 計起角度 θ 關係

Chapter1- 17

Advanced Fluid Mechanics

1.3 Properties of Fluids

Property is a point function, not a point function.

There are four types of properties: 1. Kinematic properties

(Linear velocity, angular velocity, vorticity, acceleration, stain, etc.) —strictly speaking, these are properties of the flow field itself rather than of the

fluid. 2. Transport properties (Viscosity, thermal conductivity, mass diffusivity) Transport phenomena: Molecular Transport

Macroscopic cause

Macroscopic Reset

Non uniform flow velocity

Momentum

Viscosity

Non uniform flow temp

Energy

Heat conduction

Non uniform flow composition

Mass

Diffusion

e.g.: τ = µ

du1 dx dT , g = −K , ΓA = − D AB A du 2 dx 2 dx 2

3. Thermodynamic properties (pressure, density, temp, enthalpy, entropy, specific heat, prandtl number, bulk modulus, etc) —Classical thermodynamic, strictly speaking, does not apply to this subject, since

a viscous fluid in motion is technically not in equilibrium. However, deviations from local thermodynamic equilibrium are usually not significient except when flow residence time are short and the number of molecular particles, e.g., hypersonic flow of a rarefied gas. 4. Other miscellaneous properties (surface tension, vapor pressure, eddy-diffusion coeff, surface-accommodation coefficients, etc.) Chapter1- 18

Advanced Fluid Mechanics

1.4 Boundary Conditions (1) Fluid In permeable solid interface →

(i) No slip: V

→

fluid

= V solid

(ii) No temperature jump:

when the thermal contact between solid-fluid is good, i.e. Bi =

T fluid = Tsol

hL >> 1 ) k

or equality of heat flux (when the solid heat flux is known)

(K

∂T ) fluid = q (from solid to fluid) ∂n

Remark: If fluid is a gas with large mean free path (Normally in high Mach number & low Reynolds No.), there will is velocity jump and temperature jump in the interface. (2) Fluid-permeable Wall interface (Vt ) fluid = (Vt ) wall

(no slip)

(Vn ) fluid ≠ (Vn ) wall

(flow through the wall)

T fluid = Twall

k

(Suction)

dT ∣wall ≈ ρ fluid Vn C p (Twall − Tqluid ) dn

(injection)

Remark: (1) ρ fluidVn is the mass flow of coolant per unit area through the wall. The

actual numerical value of Vn depends largely the pressure drop across the k

wall. For example: Darcy’s Law given

V =−

 k xx u  1    k yx or  v  = − µ   k zx  w  

 ∂p ∂x  ∂p ∂y     ∂p ∂z 

k xy k yy k zy

k xz   k yz  k zz 

Chapter1- 19

µ

⋅ ▽p

Advanced Fluid Mechanics

(3) Free liquid Surface

z η = η(x, y, t)

Liquid,

R

P

y

x

(i) At the surface, particles upward velocity (w) is equal to the motion of the free surface w(x, y, z) =

Dη ∂η ∂η ∂η = +u +v Dt ∂y ∂t ∂x

(ii) Pressure difference between fluid & atmosphere is balanced by the surface Pa

tension of the surface.

P

1 1 ) + P(x, y,η ) = Pa − σ ( Rx Ry

( P>Pa)

Pa

Remark:

P

( P<Pa)

In large scale problem, such as open-channel or river flow, the free surface deforms only slightly and surface-tension effect are negligible, therefore W≈

∂η , ∂t

P ≈ Pa

(4) Liquid-Vapor or Liquid-Liquid Interface

V

interface

(1)

(2)

Liquid

V1 = V2

( Vn1 = Vn 2 , Vt1 = Vt 2 )

P1 = P2

(if surface tension is neglected)

τ1 = τ 2 or µ1

—(*)

∂Vt1 ∂V ∂V = µ 2 t 2 , this is the slope t ∂n ∂n ∂n

need not be equal

Chapter1- 20

Advanced Fluid Mechanics T

T1 = T2 q1 = q 2 (Since interface has vanishing mass,

interface

(1)

it can’t store momentum or energy.)

(2)

or

− k1

∂T1 ∂T = −k 2 2 ∂n ∂n

—(**)

Remark: (1) If region (1) is vapor, its µ & k are usually much smaller than for a liquid, therefore, we may approximate E.g. (*) & (**) as

(

∂V t ∂T ) liq ≈ 0 ) liq ≈ 0 , ( ∂n ∂n

(2) If there is evaporation, condensation, or diffusion at the interface, the mass ⋅

⋅

flow must be balance, m1 = m2 .

D1

∂C 2 ∂C1 = D2 ∂n ∂n

(5) Inlet and Exit Boundary Conditions For the majority of viscous-flow analysis, we need to know V , P, and T at every point on inlet & exit section of the flow. However, through some approximation or simplification, we can reduce the boundary condition s needed at exit.

Chapter1- 21

Advanced Fluid Mechanics

Supplementary Remarks (1) Transports of momentum, energy, and mass are often similar and sometimes genuinely analogous. The analogy fails in multidimensional problems become heat and mass flux are vectors while momentum flux is a tension. (2) Viscosity represents the ability of a fluid to flow freely. SAE30 means that 60 ml of this oil at a specific temperature takes 30s to run out of a 1.76 cm hole in the bottom of a cup. (3) The flow of a viscous liquid out of the bottom of a cup is a difficult problem for which no analytic solution exits at present. (4) For some non-Newtonian flow, the shear stress may vary w.r.t time as the strain rate is held constant, and vice versa.

τ

Rheopectic

ε

=const

Thixotropic

t

Chapter1- 22

Advanced Fluid Mechanics

Chapter1- 23

Advanced Fluid Mechanics

Chapter1- 24

Advanced Fluid Mechanics

Chapter1- 25

Advanced Fluid Mechanics

Chapter1- 26

Advanced Fluid Mechanics

Chapter 2 Derivation of the Equations of motion 2.1 Description of fluid motion Consider a specific particle At t=0, x = X, y = Y, z = Z At t>0,

y

"! R

dx ) dt x=X+ ∫ ( 0 dt t

∫

y=Y+

or

t 0

t

z=Z+

∫

r = R +

∫

0 t

0

(

dy ) dt dt

(

dz ) dt dt

(

dr ) dt dt

( X ,Y , Z ) t =0

! r

(x, y, z) t>0 (same particle)

x z (2.1)

r = r(R,t) material position vector (become it represents the coordinate, used to “tag” on identify a given particle) spatial position vector (become it locate a particle in space) velocity of a particle = time rate of change of the spatial position vector for this particle.

V = ( Where

dr Dr )R ≡ dt Dt

(2.2)

D denote the time derivation is evaluated with the material coordinate held Dt

constant, it is called a material derivative. In this approach, we describe the fluid particle as if we are siding on this fluid particle. The fluid motion is described by material coordinate and time and is often referred to as the Lagrangian description. In general, if Q is a property of the fluid, we have

Q = Q (R ,t) That is, we measure the properties Q while moving with a particle. The time rate of change of Q is

Chapter2- 1

Advanced Fluid Mechanics

(

dQ Q (t +#t ) − Q (t ) "! DQ ) k = lim[ ]R = # t →∞ dt #t Dt

Note that Q(t +#t ) and Q(t) us the properties of Q for the same fluid particle. However, Q may be measured at a point fixed in space by a instrument. That is ! Q = Q ( x, y , z , t ) = Q ( r , t ) (2.3)

This is called a 〝Euler Description〞. ! If the spatial coordinate r are held constant while we take the limit dQ ∂Q dy Q (t +#t ) − Q (t ) ! y ( ) r! = = lim [ ]r dt ∂t dx # t →∞ #t The relation between (

(x, y, z)

dQ ! dQ ) r and ( ) "R! is as follows: dt dt

! ! "! Q = Q(r , t ) = Q(r ( R, t ), t )

x z

"! "! "! = Q[ x( R, t ), y ( R, t ), z ( R, t ), t ]

(

dQ "! DQ ∂Q dx ∂Q dy ∂Q dz ∂Q = ( )( ) "R! + ( )( ) "R! + ( )( ) "R! + )R = dt Dt ∂x dt ∂y dt ∂z dt ∂t u

∴

(

v

d Q "! d Q ! "! )R = ( ) + V i∇ Q dt dt r

w

=(

dQ ! ) dt r (2.4a)

or

DQ ∂Q "! = + V i∇Q ∂t Dt

(2.4b)

(Convective derivative) (Unsteady derivative or local derivative) (Material Derivative substantial Euler derivative) If moves with the same does stay in a stationary location, nor moves with same "! ""! velocity as the fluid particle (V ) , but moves with velocity Vb , then

! Q = Q(r , t )

dQ ∂Q ∂Q dx ∂Q dy ∂Q dz = + + + dt ∂t ∂X dt ∂y dt ∂z dt Chapter2- 2

Advanced Fluid Mechanics

"! ! V (r , t )

! r (t ) "! Vb

•

•

"! ! ! V (r +#r , t +#t )

! r (t +#t )

Inertia coordinate system

and

DQ ∂Q "! = + V i∇Q ∂t dt

(2.5)

(

dQ "! ∂Q ) = dt r ∂t

(

dQ ! ) observer = The time rate of change of fluid property Q(r , t ) measured by dt the observer. ∂Q "! ""! = + (V − Vb )i∇Q ∂t

Similarly: ! a = acceleration of a fluid particle = time rate of change of the fluid particle "! "! "! dV "! DV ∂V "! "! = + V i∇V = ( )R = (2.6) dt Dt ∂t Note: (1) Observer riding with the fluid particle would describe his acceleration in "! ! "! terms of a single vector a ; the fixed observer would note the V , ▽ V , "! ∂V , and from these quantities be would deduce the acceleration. ∂t "! ∂V = 0) , the acceleration is not necessarily zero. (2) If the flow is steady ( ∂t Since, from (2.6)

! "! "! a = V ⋅ ∇V

Chapter2- 3

Advanced Fluid Mechanics

2.2 Transport Theorem Consider a volume e.g. a sphere V(t) moving through space so that the velocity of "! "! each point of the volume is given by V . The velocity V may be a function of the spatial coordinate. (if the volume is deforming) and time (if the volume is accelerating "! or decelerating). V %

I (t ) = ∫∫∫

V (t )

! Q ( r , t )dτ

n

Vol. of "" cylinder = !

V(t)

(V s # t ) ⋅ n% d s

dτ

dI =? dt

V(t+# t)

dI I (t +#t ) − I (t ) = lim dt #t →0 #t

""! Vs#t

""! (VS #t ) ⋅ n%

ds

! ! 1 [ ∫∫∫ Q(r , t +#t )dτ − ∫∫∫ Q (r , t )dτ ] V ( t + #t ) V (t ) # t → 0 #t "! ( V : fluid velocity as seen by a fixed observer) = lim

Leibnitz’s Rule in Calculus: B ∂f ( x, t ) d B dB dA f ( x, t ) dt = ∫ dt + f ( x, B) − f ( x, A) ∫ A dt A ∂x dx dx

where A=A(x), B=B(x) and A' ( x), B ' ( x) are continuous in (a, b), with a ≤ x ≤ b and A ≤ t ≤ B ∵ ∫∫∫

V ( t + #t )

! Q(r , t +#t )dτ = ∫∫∫

V (t )

= ∫∫∫

! Q(r , t +#t )dτ + part changing be cause of volume.

V (t )

∴

! Q ( r , t +#t ) dτ +#t ∫∫

S (t )

"! % QV i nds

! ! "! dI 1 % } = lim {∫∫∫ [Q ( r , t +#t ) − Q (r , t )]dτ +#t ∫∫ QV inds V ( t ) S ( t ) # t → 0 #t dt

By Taylor’s expansion

! ! ∂Q Q ( r , t +#t ) = Q ( r , t ) + #t + h.o.T ∂t ! "! d 1 ∂Q 1 # # Q ( r , t ) d lim [ t ] d lim t QV τ = τ + ∴ ∫∫∫V (t ) ∫∫S ⋅ nds # t → 0 #t ∫∫∫V ( t ) ∂t # t → 0 #t dt 1 [ ](# t ) 2 # t → 0 # t ∫∫

+ lim

Chapter2- 4

Advanced Fluid Mechanics

so

! "! & d ∂Q Q r t d d QV ( , ) τ = τ + ∫∫∫V (t ) ∂t ' ∫∫ s (t ) ⋅ nds dt ∫∫∫V (t )

(2.7)

〞General Transport Theorem, 3-D Leibnitz’s Rule〞 Special Cause:

"! (1) If the volume is fixed in space. ( V =0 on the S(t), V(t)= fixed ≡V)

d ∂Q τ = Qd ∫∫∫V ∂t dτ dt ∫∫∫V (2) If the mass is fixed. (closed system,

(2.8) d D = ) dt Dt

"! & D ∂Q τ = τ + Qd d QV ∫∫∫V (t ) ∂t ' ∫∫ S (t ) ⋅nds Dt ∫∫∫V (t )

(2.9)

〞Reynolds Transport Theorem 〞 By Divergence Theorem "! "! & ▽ ⋅ Ad τ = ∫∫∫ ' ∫∫ A ⋅ nds V

We obtain

"! D ∂Q τ = [ + ▽ ⋅ ( VQ )]dτ Qd ∫∫∫V (t ) ∂t Dt ∫∫∫V ( t ) As V(t)→0

"! ∂Q D + ▽ ⋅ (VQ )]#τ [Q#τ ] = [ ∂t Dt As Q=1

"! 1 D (#τ ) = ∇ ⋅V #τ Dt take limit

"! 1 D (#τ ) = ∇ ⋅V #τ → 0 #τ Dt lim

Rate of the volume change = dilatation Therefore:

"! ∇ ⋅V = 0

↔ volume strain is zero ↔ incompressible This is the basic definition of 〞incompressible〞.

if

Chapter2- 5

(2.10)

Advanced Fluid Mechanics

Supplementary material "! V : Fluid velocity seen by a fixed observer. "! V C .V : c.v velocity seen by a fixed observer. "! V C .S : c.s velocity seen by a fixed observer. "! Vr : Fluid velocity seen by a fixed observer moving with the c.s. → (see note 5-1 back )

"! V (Fluid particle velocity) Control volume and system at time t

t

:control volume at t+#t

t+#t "!

Control volume velocity V C .V

:System at t+#t

"! If the absolution fluid velocity is V , then the fluid velocity relative to moving control "! "! surface Vr is "! "! "! V Vr = V − V C .V (4.7) "! "! "! "! Vr = V − V C .V V C .V "! That is, Vr is the velocity of the flow as seen by an observer moving with velocity "! V C .V . For this observer, the control volume is fixed, this E.g. (4.5)or(4.6) can be "! "! applied if V is replaced by Vr , that is "! & D ∂ ρ bdV = ∫ ρ bdV + ∫ ρ bV r ⋅ ndA (4.8) ∫ cs Dt sys ∂t cv "! Where Vr is given in E.g. (4.7) "! (3) If the control volume is moving with V C .V and the volume is deforming. Then the "! "! volume of the control surface V C .S will not be the same as V C .V , we then have Reynold Transport Theorem as (4.8) except that "! "! "! Vr = V − V C .S (4.9)

Chapter2- 6

Advanced Fluid Mechanics

2.3 Conservation of Mass ⋅

(1) For a closed system: ( m = 0 , Lagrangian Description )

D ρ dτ = 0 Dt ∫∫∫V (t )

"! & ∂ρ dτ + ' ρ V ∫∫ s (t ) ⋅ nds V ( t ) ∂t ""! ∂ρ

= ∫∫∫

(2.9)

=

∫∫∫V (t ) [ ∂t + ▽ ⋅ (ρ V )]dτ

if V(t) is arbitrary and the integrand is continuous, then "! ∂ρ + ∇ ⋅ ( ρV ) = 0 〞Continuity equation〞 ∂t

(2.11)

(Since it is continuous in the 1st order) or "! ∂ρ ""! + V ⋅∇ ρ + ρ∇ ⋅ V = 0 ∂t

d ρ "! D ρ ) = dt R Dt "! Dρ + ρ∇ ⋅V = 0 Dt =(

⇒

(2.12)

Special Cases: (a) For a steady flow: ( ⇒

∂ = 0) ∂t

"!

▽ ⋅ ( ρV ) = 0

"! (b) For a incompressible flow: ( ∇ ⋅ V = 0 ) Dρ =0 E.g. (2.12) ⇒ (This implies that ρ is constant along a streamline. ρ is Dt ! not a constant everywhere, but ρ = ρ ( x, t ) in general.)

(2) For a fixed region: Time rate of increase of mass within the C.V

=

Not influx of mass across the control surface

"! & d ρ dτ = − ' ∫∫ S . fixed ρV ⋅ nds dt ∫∫∫V . fluid

Chapter2- 7

"! V

n

Advanced Fluid Mechanics

Since V, S is fixed, from E.g. (2.8) with Q= ρ , we have

d ∂ρ dτ ρ dτ = ∫∫∫ ∫∫∫ V ∂t dt V Therefore

∫∫∫

V

"! &

ρ dτ = − ' ∫∫ ρV ⋅ nds S

fixed

〞conservation of mass〞 (2.13)

fixed

Chapter2- 8

Advanced Fluid Mechanics

Supplementary material "! V : Velocity of fluid particle seen by a fixed observer. "! V C .V : Velocity of control volume seen by a fixed observer. "! V C .S : Velocity of control surface seen by a fixed observer. "! Vr : Velocity of fluid particle seen by a observer moving with the control volume. 1 For non-deforming, no-moving control volume ○ "! "! "! V C .V =0, Vr = V "! V C .S =0 2 For non-deforming, moving control volume ○

"! "! V C .V = V C .S "! "! "! "! "! "! "! "! V = Vr + V C .S or Vr = V - V C .V = V - V C .S "! "! = Vr + V C .V 3 For deforming, moving control volume ○

"! ! V C .V ≠ V C . S "! "! "! Vr = V − V C .S

"! "! "! but Vr ≠ V − V C .V

"! If the C.V is non-deformed and moving with a velocity of V C .V , then we have derive in chapter 4 that

"! "! "! V = Vr + V C .V

(5.5) "! Where V is the absolute velocity of the fluid seen by a stationary observer in a fixed "! coordinate system, and Vr is the fluid velocity seen by an observer moving with the control volume. The control volume expression of the continuity equation is "! & ∂ ρ dV + ∫ ρV r ⋅ ndA = 0 (5.6) ∫ C .S ∂t C .V "! If the control volume is deforming and moving, then the velocity of the surface V C .S "! and the velocity of the control volume V C .V as seen by a fixed observer in a stationary "! coordinate. System will not be the same. The relation between V (absolution fluid "! velocity.) and Vr (relative velocity referenced to the control surface.) is "! "! "! V = V r + V C .S (5.7) and the control volume, expression of the continuity equation is remained the same as

equation. (5.6) Chapter2- 9

Advanced Fluid Mechanics

2.4 Equation of Change for momentum Newton’s second low "! "! ! dV F = ma = m dt applies only for a point particle of fixed mass m. For a closed system (Lagrangian description), it become "! "! D Vd ρ τ = (2.14) ∑F Dt ∫∫∫V (t ) The external not forces include forces acting on the body (volume) and on the surface, namely.

"! "! "! F = F + F surface body ∑

Neglecting magnetic & electrical effect, the only body force is due to the gravitational force, thus

"! F body = ∫∫∫

V (t )

"!

ρ f dτ

"! Where f represent the body force per unit mass. For any arbitrary position, the surface stresses (surface force/area) not only depend on the direction of the force, but also on the orientation of the surface. (! Therefore, the surface stress is a second order tension, and is denoted by σ .

"!

Before we involve on the derivation of F

su r fa c e

, we need to know more about

tension. 〞pressure〞means the normal force per unit area acted on the fluid particle> As the fluid is static, the pressure of the fluid is called hydrostatic pressure. Since the fluid is motionless, the fluid is in equilibrium, therefore the (Hydrostatic pressure = thermodynamic pressure) As the fluid is in motion, the 3 principal normal stresses are not necessary equal, and the fluid is not in equilibrium. Therefore, the hydrodynamic pressure is defined by

1 (Hydrostatic pressure) ≡ (σ xx + σ yy + σ zz ) 3 and which is not equal to the thermodynamic pressure either. Later we will prove that (Hydrostatic pressure) = thermodynamic pressure +

1 ' λ 3

= (Hydrostatic pressure)

Chapter2- 10

Advanced Fluid Mechanics

Supplementary material

5.2 Conservation of momentum Consider a particular moment when the control volume is coincide with the control volume, then "! "! (5.7) ∑ F system = ∑ F C.V Newton’s 2nd law for the control mass system is

"! "! D ρ = VdV ∑ F sys Dt ∫sys = time rate of change of the linear momentum of the system

(5.8)

= Sum of external forces acting on the system

From Reynolds Transport Theorem for a fixed spaced, non-deforming control volume.

"! "! "! "! & D ∂ ρ = ρ + VdV V dV V ∫C .S ρV ⋅ ndA ∂t ∫C .V Dt ∫sys

Apply (5.7) & (5.8) into above equation, we can get the momentum equation for a control volume.

"! "! "! & "! ∂ + ⋅ = V ρ dV V ρ V ndA ∑ F C .V ∫C .S ∂t ∫C .V = time rate of change of the linear momentum of the system

(5.9)

= not rate of flow of linear momentum through the C.S.

Remark:

"! If the control volume is non-deforming, but moves with a velocity of V C .V , then "! we may take b= V in equation (4.8), and get

"! "! "! "! & D ∂ V dV = V dV + V ρ ρ ∫C.S ρV r ⋅ ndA Dt ∫sys ∂t ∫C .V

Combined with (5.7) & (5.8), the above equation can be written as

"! "! "! & "! ∂ V ρ dV V ρ V + r ⋅ ndA = ∑ F C .V ∫C .S ∂t ∫C .V

Chapter2- 11

(5.10)

Advanced Fluid Mechanics

"!

"!

"!

Since V = Vr + V C .V Equation (5.10) ⇒

"! "! "! "! "! & "! ∂ ( V ( V r + V C .V ) ρ dV + r + V C .V ) ρ V r ⋅ ndA = ∑ F C .V ∫C.S ∂t ∫C .V

(5.11)

(Non-deforming + moving C.V) If the flow is steady, then "! "! ∂ (V r + V C .V )ρ dV = 0 ∫ ∂t C .V and from the continuity equation "! & ∂ + ρ dV ρ V ∫C.S r ⋅ ndA = 0 ∂t ∫C .V

(*)

The momentum equation of (5.11) reduces to

∫

C .S

"! "! & "! "! & "! V r ρV r ⋅ ndA + ∫ V C .V ρV r ⋅ ndA = ∑ F C .V C .S

"! = V C .V ∫

C .S

"! &

ρV r ⋅ ndA = 0 from equation(*)

or

∫

C .S

"! "! & "! V r ρV r ⋅ ndA = ∑ F C .V

(5.12)

(For a non-deforming C.V moving with a constant velocity in a steady state flow)

Chapter2- 12

Advanced Fluid Mechanics

"!"! Aside: A second order tension, called a dyad and denoted as AB , satisfies the following properties: "!"! "! "! "! "! ( AB) ⋅ C = A( B ⋅ C ) "! "!"! "! "! "! C ⋅ ( AB) = (C ⋅ A) B ("! A unit tension, II , is a tension with (! "! "! "! (! "! Ⅱ⋅ C = C , C ⋅Ⅱ = C In a Cartesian coordinate system, (! !! ! ! ! ! Ⅱ = ii + j j + k k

Now, back to the issue of surface forces, as the fluid is in static equilibrium, the only stress is the normal stresses, thus   1 0 0  (! (! (!    σ = − pⅡ Ⅱ= 0 1 0 = (hydrostatic pressure)

 

   0 0 1   

If the fluid is in motion, we assume:

(!

(! )

σ = − pⅡ+ τ

(2.15)

Viscous stress Thermodynamic pressure Question: Are 〞hydrostatic pressure〞, 〞hydrodynamic pressure〞and 〞thermodynamic pressure〞the same? We will answer this question later.

The surface forces thus become

"! F =' ∫∫

s (t )

=' ∫∫

s (t )

=

∫∫∫ *

& (! (n ⋅ σ )ds

& − pnds + ' ∫∫

R (t )

s (t )

) & (τ ⋅ n) ds

) [ −∇ p + ∇ ⋅ τ ]d τ

Equation (2.14) thus become

Chapter2- 13

Advanced Fluid Mechanics

D Dt

R (t )

"!

R (t )

ρ f dτ + ∫∫∫

R (t )

) [ −∇ p + ∇ ⋅ τ ]dτ

"! "! "! & ∂(ρV ) = ∫∫∫ dτ + ' ( ρ V ∫∫ S ( t ) )V ⋅ nds R (t ) ∂t "! "! "! ∂(ρV ) = ∫∫∫ [ + ∇ ⋅ ( ρ V V )]d τ R (t ) ∂t

(2.9)

⇒

∫∫∫

"!

ρ V dτ = ∫∫∫

"! "! "! "! ) ∂(ρV ) + ∇ ⋅ ( ρ V V ) = ρ f − ∇ p + ∇ ⋅τ ∂t

(2.16)

momentum flux tensor or D Dt

"! D ∫∫∫R ( t ) ρ V dτ = Dt

∫∫∫

R (t )

"! V dm =

∵

∫∫∫

R (t )

"! DV dm = Dt

"! DV ∫∫∫R ( t ) ρ Dt dτ

D means we follow a fluid particle, Dt

thus the mass dm is a constant, and not a function of time & location. Equation (2.16) thus has another form of "! "! ) DV ρ = ρ f − ∇ p + ∇ ⋅τ Dt Equation (2.17) may be derived from (2.16) either "! "! "! ∂( ρV ) L.H.S = + ∇ ⋅ (ρV V ) ∂t (2.16) "! "! "! "! ∂V ∂ρ "! "! =ρ + V + V∇ ⋅ ( ρ V ) + ( ρ V )(∇ ⋅ V ) ∂t ∂t "! "!  "! ∂ ρ "!  ∂V "! = ρ  + V ⋅ ∇V  + V  + ∇ ⋅ ( ρ V )   ∂t   ∂t 

(2.17)

"! (= 0 from continuity equation) DV =ρ Dt By the tenor operation, we show that the left-hand side of the equation (2.16) & (2.17) are identical.

Chapter2- 14

Advanced Fluid Mechanics

Chapter 3 Exact Solution of N-S Equation 1 Constant Density (Incompressible Flow) Assumptions: ○ 2 Constant µ, k, C , C ○ v p e = CvT 3 No body forces ○

3.1 Parallel Flow V = ui + v j + wk =0

=0

or v = w = 0 , but

u = u ( x, y , z , t ) , p = p ( x, y , z , t ) , T = T ( x , y , z , t ) 1)

Continuity equation: ∇ ⋅V = 0 ⇒

⇒

or 2)

0

0

∂u ∂v ∂w + + =0 ∂x ∂y ∂z

∂u = 0 ⇒ u does not depend on x ∂x u = u ( y, z, t )

Momentum equation:  ∂V  + V ⋅ ∇V  = −∇P + µ∇2 V  ∂t 

ρ or

 ∂ 2u ∂ 2u ∂ 2u   ∂u ∂u ∂u ∂u  ∂p +u +v +w  = − +µ 2 + 2 + 2  ∂x ∂y ∂z  ∂x ∂y ∂z   ∂t  ∂x

ρ

 ∂ 2v ∂ 2v ∂ 2v   ∂v ∂v ∂v ∂v  ∂p µ +u +v +w  = − +  2 + 2 + 2  ∂x ∂y ∂z  ∂y ∂y ∂z   ∂t  ∂x

ρ

 ∂2w ∂2w ∂2w   ∂w ∂w ∂w ∂w  ∂p +u +v +w =− +µ 2 + 2 + 2  ∂x ∂y ∂z  ∂z ∂y ∂z   ∂t  ∂x

ρ ⇒

∂p ∂p = = 0 ⇒ p = p ( x, t ) ∂y ∂z

3- 1

Advanced Fluid Mechanics

ρ

 ∂ 2u ∂ 2u  ∂u ∂p = − +µ 2 + 2  ∂t ∂x ∂z   ∂y

∂ ∂x

if we

(3.1)

 ∂u  ∂ 2u ∂ 2u   ∂p ρ µ = − +   2 + 2  ∂x ∂z    ∂y  ∂t

0

0 0 2 2  ∂ ∂u ∂ ∂p ∂ ∂u ∂ ∂u  (− ) + µ  2 ( ) + 2 ( )  ⇒ ρ ( )= ∂t ∂x ∂x ∂x ∂z ∂x   ∂y ∂x

3)

⇒

∂2 p ∂p = 0 or = function of t only. 2 ∂x ∂x

∴

u = u (t , y, z ) , p = p ( x, t ) ,

∂p = f n (t ) ∂x

Energy equation: (v=0) (w=0) Eq. (2.40) ⇒  ∂T  ∂u ∂T ∂T ∂T  ∂v ∂w ρ Cv  +u +v +w = 2µ ( )2 + ( )2 + ( )2  ∂x ∂y ∂z  ∂y ∂z  ∂t  ∂x 1  ∂u ∂v ∂u ∂w ∂v ∂w   ∂ 2T ∂ 2T ∂ 2T + ( + )2 + ( + )2 + ( + )2   + k ( 2 + 2 + 2 ) 2  ∂y ∂x ∂z ∂x ∂z ∂y   ∂x ∂y ∂z

⇒ ρ Cv (

 ∂ 2T ∂ 2T ∂ 2T   ∂u ∂T ∂T ∂u  + u ) = µ ( ) 2 + ( ) 2  + k  2 + 2 + 2  ∂t ∂x ∂z  ∂y ∂z   ∂y  ∂x

(3.2)

T = T (t , x, y, z )

∴

3.1.1 Steady, Parallel, 2-D Flow (

∂ ∂ = 0, = 0) ∂t ∂z

From the pressure discussion, we know u = u ( y ) , p = p ( x ) , T = T ( x, y ) ,

∂p dp = = constant ∂x dx

The Equation of motion become

µ

∂ 2u dp = = constant ∂y 2 dx

ρ Cv u

(3.3a)

 ∂ 2T ∂ 2T  ∂T ∂u 2 µ ( ) = +k 2 + 2  ∂x ∂y ∂y   ∂x

3- 2

(3.3b)

Advanced Fluid Mechanics

integrate Eq (3.3a), we have u( y) =

y 2 dp ( ) + C1 y + C2 2 µ dx

(3.4)

a) Poiseuille (pressure-deriver) duct flows: y

u(b)=0

u(b)=u(-b)=0

x

2b

u(-b)=0

Eq. (3.4) ⇒ 1 dp 2 u( y) = ( )(b − y 2 ) 2µ dx

parabolic profile

The shear stress is

τ ij = 2 µε ij = µ (

τ 11 = τ xx = µ (

∂V j ∂ xi

+

∂ Vi ) ∂x j

∂u ∂u ∂u ) = 0 (∵ = 0 from continuity equation) + ∂x ∂x ∂x

⇒ No normal shearing stresses ∂v ∂u du τ 12 = τ 21 = µ ( + ) = µ dy ∂x ∂y

∴ τ =µ

du dp = y dy dx

Thus the wall function is ( τ w

τw =

= τ y =± b )

dp b dx

From the energy equation:

ρ Cvu

∂T dp ∂ 2T ∂ 2T ) = µ ( )2 y 2 + k ( 2 + dx ∂x ∂x ∂y 2

If the channel is infinitely long, we may assume that the temperature distribution is fully-developed, i.e. ∂T = 0 or T = T ( y ) only ∂x

Energy equation become k

d 2T 1 dp = − ( )2 y 2 2 µ dx dy

3- 3

Advanced Fluid Mechanics

integrate twice T ( y) = −

1 dp 2 y 4 + C3 y + C4 ( ) µ k dx 12

If the B.C’S are: T ( b ) = T w+ , T ( − b ) = Tw− , then

Tw+ + Tw− Tw+ + Tw− y b4 dp 2 y4 T ( y) = + + ( ) (1− 4 ) b 2 2 b 12k µ dx When we calculate u(y), we are actually interested in the value of τ w . Similarly, as we solve the temperature distribution, we want to know the heat transfer on the walls. Aside: In the temperature section, we mentioned that

(k

∂T ) fluid = qsolid → fluid ∂n

For the current case

n

q = q (−n) = −qn

On upper surface

On lower surface

q

q = −qn

therefore q = − k∇T

− qe n = − k

∂T ∂n

en ⇒ q s → F = k ∂T

∂n

fluid

fluid

However, this is not a good way become e n always change its direction for a fixed coordinate frame. Therefore, we may take the positive valve of q as heat transfer in the direction of the positive-coordinate axis, then q = −k∇p

⇒

qx i + q y j + qz k = − k (

⇒

qx = − k

∂T ∂T ∂T i+ j+ k) ∂x ∂y ∂z

∂T ∂T ∂T , q y = −k , qz = − k ∂x ∂y ∂z 3- 4

Advanced Fluid Mechanics

At any point, if qx > 0 , it means the x-component of the heat transfer at this point is in the +x-axis direction. For this case: y

2

q

x 1

Therefore, we set q in the direction of +y, then q = −k

(i) If q a t

p o in t ①

dT dy

> 0 , it means q is transferred upward, therefore, it is from

the lower wall to the fluid. (ii) If q a t p o in t ① < 0 , the heat is transferred downward, therefore, it is from the fluid to the lower wall. (iii)If q a t p o in t ② > 0 ⇒ fluid to upper wall. (iv) If q a t p o in t ② < 0 ⇒ upper wall to the fluid # Take y q=qj

x

q>0, heat transfer upward q 0 dx

τ lower wall

(∵ (

parallel flow

dp ) < 0) dx

is acted on the negative direction

of i . Similarly:

τ upper wall = + j ⋅ (τ mn e m e n ) = +τ jn e n = +τ 21 i = + (τ w )upper i = + (

Therefore,

dp )bi < 0 dx

τ upper wall is still in the direction of –x axis. 3- 6

(j=2, n=1,2,3)

Advanced Fluid Mechanics +

−

(3) q (b) ≠ q (−b) because Tw ≠ Tw

However, if Tw+ = Tw− , we know the results that q+ = +

b3 dp 2 ( ) 3µ dx

q− = −

b3 dp 2 ( ) 3µ dx

Why these is a difference in sign? Does it mean that one wall is received heat while the other given away the heat? The answer is that q + > 0 ⇒ heat transfer upward ⇒ from fluid to the upper wall q − < 0 ⇒ heat transfer downward ⇒ from fluid to the lower wall To understand the flow in more detail, let’s see the temperature profile for the case of: Tw+ = Tw− = Tw T ( y ) = Tw +

b 4 dp 2 y4 ( ) (1 − 4 ) 12k µ dx b ≥0

Tw

τw

Tw

τw

The shearing stress is du dp τ =µ = y dy dx

3- 7

0

Advanced Fluid Mechanics

Question: Why the temperature is highest but the shearing stress is minimum (zero) along the centerline? Answer: The high viscous force along the walls will produce a large amount of dissipation energy. In turn, it will increase the internal energy of the fluid near the wall. Partial internal energy transport to the wall due to dissipation  b3 dp 2  gradient,  qw = ( )  , the rest of viscosity. Along the centerline, the 3 µ dx  

fluid received the diffused energy from upper & lower surface, thus it has the max temperature.

b) Poiseuille (pressure-driven) pipe flow: (Parallel flow: planar (2D) flow, or Axisymmetric flow.) In cylindrical coordinate: V = ue x + ve r + weφ

r

R

x

and

∂ = 0 (2-D) ∂φ

u = u (r , x) , v = w =0 (parallel), P = P(r , x) T = T (r , x)

(may be! Write down in this way temperature)

Continuity:

∂  α ∂  ∂r (h2 h3v) + ∂φ (h1h3 w) + α x (h1h2u )    1 ∂  =  (ru )  = 0 r  ∂x  u ≠ f n ( x) → ∴ u = u(r)

∇ ⋅V =

1 h1h2 h3

Momentum:

 ∂V

 + V ⋅ ∇V  = −∇P + µ∇2 V  ∂t 

ρ

∇= =

1 ∂ 1 ∂ 1 ∂ e1 + e2 + e3 h1 ∂x1 h2 ∂x2 h3 ∂x3 ∂ 1 ∂ ∂ er + eφ + eX r ∂xφ ∂xr ∂X

3- 8

h1=h3=1 h2= r x1= r x2= φ x3=X

Advanced Fluid Mechanics

∇V = ∇(ue x ) = (∇ u ) e x + u∇( e x ) = (

du du e r )e x = er e x dr dr

0 (∵ e x is fixed, however e r , eφ are not )

V ⋅ ∇V = (ue x ) ⋅ (

du du e r e x ) = (u )( e x ⋅ e r ) e x = 0 dr dr

∇p =

∂p ∂p er + e x ∂r ∂x

∇2 =

1 h1h2 h3

 ∂ h2 h3 ∂ ∂ h1h3 ∂ ∂ h1h2 ∂  ( )+ ( )+ ( )  x h x x h x x h x ∂ ∂ ∂ ∂ ∂ ∂ 1 1 2 2 2 3 3 3   1

1 ∂ ∂ ∂ 1 ∂ ∂ ∂  ( ) + (r )  =  (r ) + r  ∂r ∂r ∂φ r ∂φ ∂x ∂x  1  ∂ ∂V ∂ 1 ∂V ∂ ∂V  )+ ( ) + (r ) ∇2 V =  (r r  ∂r ∂r ∂φ r ∂φ ∂x ∂x 

, V = u ( r )e x

0 ( V = u ( r )e r ) 0

1  ∂ ∂u  =  (r )  e x r  ∂r ∂r  ∴ Momentum equation: x-dir: 0 = − r-dir:

∂p µ ∂ ∂u + (r ) ∂x r ∂r ∂r

(3.5)

∂p = 0 ⇒ p=p(x) ∂r

Eq. (3.5) ⇒ dp µ d du = (r ) = constant dx r dr dr f n ( x) f n (r )

∵ L.H.S = f n ( x) R.H.S = f n (r ) ∴the only solution is that it is a constant

integrate twice with the B.C.’s: (i) u(r) = 0 (ii) u (r ) = −

1 dp 2 2 (R − r ) 4µ dx

umax = u r =0 = −

du dr

= 0 , we obtain r =0

(parabolic profile)

1 dp 2 R 4µ dx

3- 9

Advanced Fluid Mechanics R

Volume flow rate, Q = ∫ u (r )(2π r )dr = − 0

The mean velocity, u =

Q R 2 dp umax = − = π R2 8µ dx 2

Shear stress at wall τ w = µ

τw

Cf ≡

1 ρu 2

=

2

π R 4 dp 8µ dx

du 1 dp 4µ u = R(− ) = dr 2 dx R

16 ρ uD , where ReD = Re D µ

which agrees wall with the experiment data for laminar flow

16 Re d

experiment

Cf turbulent

laminar

transition

Re D

2000 Energy equation:

 ∂T  + V ⋅ ∇T  = − p∇ ⋅ V + 2µε : ε + k∇2T  ∂t 

ρ Cv 

V = u ( r )e x ∇V =

ε =

du er e x , dr

(∇V ) t =

du e x er dr

1 1 du ∇V + (∇V ) t  =  2 dr ( e r e x + e x e r ) 2

ε :ε = (

1 du 2  ) er e x + e x er  : er e x + e x er     2 dr 

3- 10

(2.40)

Advanced Fluid Mechanics

=(

1 du 2  ) e r e x : e r e x + 2e x e r : e r e x + e x e r : e x e r   2 dr  = (e x ⋅ e x )(e r ⋅ e r ) = (e r ⋅ e x )(e r ⋅ e x ) =0 =1

1 du 2 ( ) 2 dr

=

∇ ⋅ V = 0 (Incompressible flow) ∂T ∂T V ⋅∇T = (u (r )e x ) ⋅ ( e x + ..) = u ∂x ∂x 1  ∂ ∂T ∂ 1 ∂T ∂ ∂T  ∇2T =  (r )+ ( ) + (r ) r  ∂r ∂r ∂φ r ∂φ ∂x ∂x  Assume T = T (r ) only then Eq. (2.40) becomes 0 = µ( Sub. (

du 2 k ∂ dT ) + (r ) dr r ∂r dr

(fully-developed in temperature)

du ) into the above equation, and integrate twice with the B.C.’s: dr

1 T (r ) = T ○ w

2 ○

T (r ) = Tw +

dT dr

= 0 , we have r =0

1 dp 2 4 4 ( ) (R − r ) 64k µ dx

Remark: 1 Can discuss q ○ wall ~ (

dp dp dp 2 ) , while τ w ~ ( ) and Q ~ ( ) , Q ~ R 4 dx dx dx

c) Couette (Wall Driven) Duct Flow: u = U , T = T1 (a constant)

y

x

2h

(p= constant)

continuity:

fixed u = 0, T = T0 (a constant)

∂u =0 ∂x

momentum: 0 = µ

d 2u dy 2

Since the plate is infinite long with constant wall temperature, the temperature 3- 11

Advanced Fluid Mechanics

can be assumed fully developed. Thus T=T(y) only. The energy equation reduces to 0 = µ(

du 2 d 2T ) +k 2 dy dy

From momentum equation & B.C.’s, we have velocity distribution u( y) =

U y (1 + ) 2 h

shear stress at any point. ∂u ∂v µU τ = µ( + ) = = const 2h ∂y ∂x

1.0

y h

C f ≡ function coefficient ≡

Knowing

0

-1

0

U

u(y)

τ µ 1 ρUh = = , where Re = h 1 ρU 2 ρUh Re h µ 2

∂u , we can get T(y) from energy equation & B.C.’s: ∂y 2 y2  T + T T + T y  µU (1 − 2 ) T ( y) =  1 0 + 1 0  + 2 h  8k k  2

(Due to conduction of fluid)

(Due to viscous dissipation)

Define: Brinkman Number, Br

µU 2

Br ≡

k (T1 − T0 )

=

dissipation effect conduction effect

µC p

U2 = Pr Ec k C p (T1 − T0 )

T1

1 Br=0

y 0 h -1

=

T0

8

(Br=0 means the flow is pure conduction

16

since dissipation effect is zero)

T(y)

3- 12

Advanced Fluid Mechanics

Question: From velocity profile, the u m ax occurs at y = h (upper plate) and τ is

constant at any point. It looks that the viscous dissipation should have equal magnitude every where or at least near the upper plate. But as we can see from temperature profile, the T m ax does not occur at hot upper plate, why? Explain this from physical phenomena? Answer: Energy dissipation is independent of y, as well as τ . But since the wall

temperature is different, therefore, q u p p e r is lower while qlower is higher as can be seen from qwall on next page. Thus, Tm ax occurs in upper half region.

As the given example in p.108 of white, except for giving oils, we commonly neglect dissipation effect in low speed flow temperature analysis. (∵Br is very small) Heat transfer at the walls: qw = k

∂T ∂y

= ±h

k µU 2 (T1 − T0 ) ± 2h 4h

(*)

the heat convection coefficient, hc , is defined as hc =

qw qw = T T1 − T0

(**)

Define Nusselt N 0 ≡ N u ≡

hc L k

Take characteristic length L=2h, we have Nu =

hc (2h) Br = 1± k 2

Since Br =0 means the dissipation effect is zero, the flow is pure conduction heat transfer. (Nu = 1+0 = 1 here) Thus, the numerical value of Nu represents the ratio of convection heat transfer to conduction for the same value of T .

3- 13

Advanced Fluid Mechanics

d) Couette (Wall Driven) Pipe Flow For the flow between two concentric cylinders rotating at angular velocity w1 and w2 , the fluid has velocity of V = ur e r + uθ eθ + u z e z

Assume: ur = u z = 0 uθ = u (r )

paralle

p = p(r ) T = T (r ) ρ = constant

2-D + fully-developed

z

r

z

r1

ω1

y

θ

r2

x

ω2

r

The continuity equation is identically satisfied. The momentum equation can be reduced as

ρu 2 r

=

dp dr

(in r-dir)

(3.6a)

(in θ -dir)

(3.6b)

and d 2u d u + ( )=0 dr 2 dr r

With the B.C.’s: (i) u ( r1 ) = w1 r1 (ii) u ( r2 ) = w 2 r2 Eq. (3.6b) becomes u (r ) =

1 2 r2 − r12

2 2   r1 r2 2 2 ( ) ( w2 − w1 )  − − r w r w r  2 2 1 1 r  

3- 14

(3.7)

Advanced Fluid Mechanics

Remarks: (1) If r2 → ∞ , w2 → 0 2

2

(i) r2 − r1 ≈ r2

2

(ii) w2 → 0 , r2 → ∞ 2

∴ w2 r2 → uncertain No: however

2

2

w2 r2 → ∞ , ∴ w2 r2 − w1 r1 ≈ w2 r2 u (r ) ≈

1 2 r2

2 2 2 2   r1 r2 r1 w1 r1 w1 2 rw r w rw + = + =  2 2 1 2 r r r  

Γ ≡ circulation ≡

∴ u (r ) =

2

∫ u (r )d = (

2

r1 w1 2 )(2π r ) = 2π w1r1 = const k

Γ0 2π r

u

potential vortex (free vortex)

∵∇ × V = 0 ,No vorticity

r

+

+ Orientation of + Will not change

(2) If r1 = w1 = 0 (No inner cylinder) 1 2 u ( r ) = 2 rw2 r2 = rw2 r2

{

}

u Rigid body rotation (froce vortex)

∵∇× V ≠ 0

r

+ Orientation of + Of the cross Will change

3- 15

Advanced Fluid Mechanics

∇ ×V =

er

r eφ

ez

1 ∂ r ∂r 0

∂ ∂φ rw2

∂ 1  ∂ ( rw2 )  w2 =  = =Ω ∂z r  ∂ r  r 0

∴ Ω r =0 → ∞

(3) A 〝Tornado〞is a combination of potential vortex & Rigid-body rotation.

∇× V

u (r ) r The viscous stress of the fluid can be stress as  du u  −  dr r  The moment on the outer cylinder of unit height is

τ rθ = τ θ r = µ 

M

2

=

∫∫

S2

(3.8)

r × ( n ⋅ τ ) ds

By Eq. (3.8), we can show that M 2 = 4πµ

2

2

r1 r2 ( w2 − w1 ) 2 2 r2 − r1

(3.9)

From the energy equation; we can derive the temperature distribution as T (r ) =

where

α =−

α r

2

+ n(

1 r2

 α r2 r ∂ (T1 − 2 ) n + (T2 − 2 ) n  r1 r1  r1 r2 )

r1

µ  r12 r2 2 ( w1 − w2 )   k

2

2

r2 − r1

(3.10)

2

 

Remark: The derivation of Eqs. (3.6)~(3.10) can be left as a homework problem for the students.

3- 16

Advanced Fluid Mechanics

e) Combined Couette and Poiseuilli Duct Flow U1

u ( h ) = U 1 , T ( h ) = T1

h

x

u (0) = 0, T (0) = T0

Then the solution of the momentum of (3.4) becomes u( y) =

U1 h 2 dp y y y− (1 − ) h 2 µ dx h h ≡ U1Ρ = pressure gratient parameter

or u y y y = + Ρ (1 − ) U1 h h h

The velocity profile is:

Ρ = −3

-2

h

-1

0

As Ρ < −1 1 2

0

3

;backflow occurs. This is called the separation of the flow

1.0

The function along the upper & lower

τ w± = µ (

du ) y = 0 ,h dy

(C f ) y =0 =

2 ρU1h (1 + Ρ) , where Re = Re µ

(C f ) y = h =

2 (1 − Ρ) Re

, Cf ≡

τw

1 ρU 2 2 1

We have

∴ C f = C f (R e, Ρ )

3- 17

Advanced Fluid Mechanics

In general: C f = C f (Re, Ρ ,

Tw+ µ C p , = Pr) Tw− k

(If consider conductivity) If the flow is compressible, connected with the energy equation For the energy equation, if we assume also T=T(y) only, then T − T0 y 1 µU1 y y = + (1 − ) T1 − T0 h 2 h(T1 − T0 ) h h 2

=

µC p k

2

×

U1 ≡ Pr⋅ Ec Cp ( T ) 0

Where ( T0 ≡ T1 − T0 ) Pr ≡

µC p k

≡ Prandtl No. =

viscous diffusion rate thermal diffusion rate

2

Ec ≡

U1 2( T)ad T ≡ Eckert No. = = (r − 1) M 2 ∞ , M→Mach No. C p ( T )0 ( T )0 ( T )0

≈ work of compression (or the absolution temperature of the free stream)/(temperature difference) (Ec is important when the velocity is comparable with sound speed.) also then

y h T − T0 1 = η + Pr⋅ Ec η (1 − η ) T1 − T0 2

η=

Pr⋅ Ec = 0 y h

1

4

8 T − T0 T1 − T 0

3- 18

Advanced Fluid Mechanics

3.2 Simple Unsteady Flow Assumptions: (1) Constant density, ρ = constant (2) µ , λ , k constants (3) planar parallel flow

∂ = 0, v = w = 0 ∂z

With the assumption above, we can only have u = u (t , x, y ) , p = p (t , x, y ) , T = T (t , x, y ) By continuity equation:

∂u = 0 ⇒ u = u (t , y ) ∂x

By y-momentum equation:

∂p = 0 ⇒ p = p (t , x) ∂y

By x-momentum & continuity:

∂2 p ∂p =0 ⇒ = fn(t ) only 2 ∂x ∂x ∂p = fn(t ) ∂x

∴

(3.11)

The x-momentum and energy equations become ∂u 1 dp ∂ 2u =− +ν 2 ∂t ∂y ρ dx Cv ρ (

(3.12)

∂T ∂T ∂u ∂ 2T ∂ 2T + u ) = µ ( )2 + k ( 2 + 2 ) ∂t ∂x ∂y ∂x ∂y

(3.13)

3.2.1 Stokes First problem (Rayleigh’s problem) Consider a semi-infinite space, y ≥ 0

y

x The air (or any medium) is still for y ≥ 0 , t < 0. At t = 0, become wall impulsively moves to speed U0 U (t , y = 0)

U0

t 3- 19

Advanced Fluid Mechanics

The question is: What is the subsequent motion for y>0, and t>0 ? First of all, we notice that there is 2 equations ((3.12) & (3.13)) but 3 unknown (p, u, T). One unknown should be removed first. From Eq. (3.11), we have get only, i.e. at a certain time (a fixed time) the value of

dp = fn(t ) dx

dp is not a function of position, dx

dp is the same at any point of the flow domain. For our problem, as dx y → ∞ , u should approach to zero velocity for t >0, therefore,

or the value of

∂u ∂t

y →∞

∂u = ∂y

y →∞

1 dp =− ρ dx

y →∞

∂ 2u = 2 ∂y

=0 y →∞

Eqn (3.12) ⇒

∂u ∂t ⇒

dp dx

∂ 2u +ν 2 ∂y y →∞

y →∞

=0 y →∞

From our arguments that

dp is not a function of position, hence dx

dp =0 dx

everywhere

(3.14)

And the momentum equation becomes du ∂ 2u =ν 2 dt ∂y

with B.C.’S u (t , y = 0) = U 0

(3.15)

u (t , y → ∞) = 0

Eg. (3.15) is a P.D.E, we try to change it into a O.D.E, which will be easier to be solved. Sinceν is a constant, Eq.(3.15) become

∂u ∂ 2u = ∂ (νt ) ∂y 2

3- 20

Advanced Fluid Mechanics

the independent variables areνt and y, therefore, a new variable should contain this two parameters. Furthermore, we want the new variable to be dimensionless. In this way, if we also set a new dependent variable to replace the old dependent variable u, the equation will be dimensionless. We therefore can solve the O.D.E easier and once for ever. Thus, define

η = yα (νt ) β

,

u = f (η ) u0

try the dimensional analysis    τ ∂u  F2 ⋅ L ML ⋅ LT 2  µ   ∂y  L L T FLT T = = = = = ν [ ]   M 3 M M [ρ] ρ  L 2

=L

T

[νt ] = L2

, [y] = L

So if want to dimensionlizeη , we should Take ∂ =1

Thus

η=

&

β =−

y 2 νt

Recall

1 2 The coefficient 2 in the denominator is taken to make the final O.D.E. more easier to be integrated

u = f (η ) u0

(3.16)

(3.17)

u = U 0 f (η ) ∂η 1 y −η ∂η 1 =− = , = ∂t ∂y 2 νt 2t 2 νt 2t ∂u df ∂η df η = U0 = −U 0 ∂t dη ∂t dη 2t ∂u df ∂η df 1 = U0 = U0 ∂y dη ∂y dη 2 νt ∂ 2u d2 f 1 = U0 ∂y 2 dη 2 4νt

3- 21

Advanced Fluid Mechanics

Eq. (3.15) ⇒ df η d2 f 1 −U 0 =νU0 dη 2t dη 2 4νt

d2 f df + 2η =0 2 dη dη

⇒

(3.18)

1 η = 0 , f (0) = 1 with B.C’S: ○ 2 η → ∞ , f (∞ ) = 0 ○

Eq. (3.18) ⇒ df ' + 2η f ' = 0 ' f

⇒ ⇒

or

df ' = −2η dη f'

n f ' = −η 2 + C1

f ' = Ae −η

2

integrate again

f = A∫ e−η dη + B 2

1 η = 0 , f (0) = 1 ○

0

1=A ∫ e-η dη + B 2

∴ B=1

0

2 η → ∞ , f (∞ ) = 0 ○

∞

0=A ∫ e-η dη + 1 2

∴A=

0

−1

∫

∞

0

∴

or

f (η ) = 1 −

f (η ) = 1 −

2

π 2

π

∫e

∫

−η 2

η

0

2

e-z dz

=

−1

π

2

dη

2

e − z dz

(3.19a)

erf (η )

3- 22

Advanced Fluid Mechanics

f (η ) = 1 − erf (η ) ≡ erfc (η )

∴

(3.19b)

Complementary error function

y t increase

η u U0

1.0

1.0

u U0

−U 0 −η 2 du df (η ) dη 2 −η 2 1 = U0 = U 0 (− )= e )( e dy dη dy π 2 νt πνt

τw = µ

du dy

=− y =0

µU 0 πνt

(3.20)

The displacement thickness δ (t ) is defined as ∞

U 0δ (t ) = ∫ u ( y, t )dy 0

y

At given t

=

δ

U0 ∞

or

U

∞

u dy δ (t ) = ∫ ( y, t )dy = ∫ erfc(η ) dη U d η 0 0 0 ∞

= 2 νt ∫ erfc(η )dη 0

integrate by parts ∞ ∞ (η erfc η 0 − ∫0 η d (erfc η ) dη ) dη

0 3- 23

Advanced Fluid Mechanics

erf (η )

erfc(η ) = 1 − erf (η )

1

η

2

η

d (erfc (η )) df (η ) 2 −η 2 = =− e dη dη π ∞

2

0

π

∴δ (t ) = −2 νt ∫ η (− =2

=2

νt

π

∫

∞

0

2

e−η )dη

2

2η e−η dη =2

νt

π

∫

∞

0

2

e −η d (η 2 )

ν t  −η 2  ∞ νt −e =2 0 π  π

(3.21)

for example: Kg/m s at t=10 sec Air,40℃

1.71 × 10-5

Water,℃

6.61 × 10-7

Lubricating Oil,℃

µ (Pa Sec)

δ (m) 0.0147 0.0029 0.0357

ν(m2/s)

-4

1 × 10

17.1 655 ----------------

Remark: At the first glance, it seems strength that the strength of the momentum transport (or the speed of the propagation of the external disturbance) in three different fluid is: Oil > Air > water While the µ of there is in the order of Oil > water > Air However, it is reasonable, since δ ~ ν1/2 ~ µ , not only depend on µ . ρ

3- 24

Advanced Fluid Mechanics

How about the temperature change if we imposed suddenly a temperature to the boundary? Similarly, we will obtain

δ Τ (t ) = 2

αt π

(3.22)

where

α =

∴

κ ρ Cp

δ µ (t ) ν = = δ T (t ) α

Pr

(Pr=

µ Cp k

)

(3.23)

Remarks: (1) as Pr>1 , the δ µ is larger than δ T (2) Typical values of Pr for different fluid are Fluid

Mercury

He

Air

F-12

Methyl alcohol(甲醇)

Water

Ethyl alcohol(乙醇)

Pr

0.025

0.7

0.72

3.7

6.8

7.0

16

Fluid

SAE 30 oil

Pr

3500

(The

δµ are in the order of Air < Water < Oil, now!) δT

3.2.2 Stokes Second Problem---Oscillating plate y

x Governing equation:

B.C.’s :

∂u ∂ 2u =ν 2 ∂t ∂y u (y = 0, t) = U0 cosωt

u (y→∞, t) = 0

3- 25

(3.24)

Advanced Fluid Mechanics

It is convenient (and make the procedure easier) to use a complex variable to solve the problem. Furthermore, if we are doing the problem of u (0, t ) = U 0 sin wt , we can take the imaginary part of the solution and it is no need to do the problem twice.

∵ e iω t = cosω t + i sin t we take the B.C. as

u ( 0 , t ) = U 0 e iω t

(3.25)

Use separation of variables, we assume

u ( y , t ) = U 0 e iω t f ( y )

(3.26)

 Note : The solution fo this problem will be the real part of Eg.(3.26).    Which is the solution under the B.C. of Eg.(3.25)  

∂u = iω U 0 e iω t f ∂t ∂u ∂ 2u iω t ′ = U 0 e f , 2 = U 0 e iω t f ′′ ∂y ∂y

sub into egn.(3.24) yields:

iω U 0 e iω f = ν U 0 e iω t f ′′ f ′′ −

iω f =0 ν

(3.27)

Use characteristic equation to solve, i.e. we assume

f = e λ y ⇒ f ′ = λ e λ y , f ′′ = λ 2 e λ y sub into Eg (3.27)

λ2 −

iω ω i =0⇒λ =± ν ν 1/ 2

 i π2  ∵ i = e    ∴λ = ±

i

π

= e 4 = cos

ω

(1 + i) 2ν

3- 26

π 4

+ i sin

π 4

=

2 (1 + i) 2

Advanced Fluid Mechanics

(3.26) ⇒

u ( y, t ) = U 0e

 iω t + = U 0  Ae 

iω t λ y

e

 = U 0  Ae 

ω 2ν

y i (ω t +

e

ω 2ν

y)

ω 2ν

(1+ i ) y

+ Be

−

+ Be

ω 2ν

iω t

y i (ω t −

e

e

−

ω 2ν

ω 2ν

(1+ i ) y

  

y) 

 

ω   y 2ν →∞   as y → ∞, e    but u(∞, t) = 0,∴ A = 0 

Also

u (0, t ) = U 0 e iω t = BU 0 e iω t ⇒ B = 1

Thus

u ( y, t ) = U 0e = U 0e

−

ω

−

2ν

ω 2ν

y

y

e

w y) 2ν

i (ω t −

  cos (ω t − 

ω 2ν

y ) + i sin (ω t −

 y) 2ν 

ω

Since we have only the real part,∴

u ( y , t ) = U 0e

−

ω 2ν

y

cos (ω t −

ω 2ν

y)

(3.28)

Decaying Amplitude The velocity distribution is

6 η = y

0

② :π

①

③

② −1.0

ωt 0 π ① : 2

w 2ν

1

④

1.0

0

3- 27

u U0

③ : 3π 2 ④ : 2π

Advanced Fluid Mechanics

Remarks: (1) This is similar to the temp. varies on the earth every day due to the sunrise and sunset.. (or, if we take u as the average temp. of a day, the distribution will similar to the temp. on the earth every year due to the seasons.) (2)

e

w y 2ν

−

One wave length wave length =

λ=

2π

ω 2ν

= 2π (

2ν

ω

2π frequency

)1 / 2 ≡ depth of penetration

3- 28

Advanced Fluid Mechanics

3.3 steady, 2-D stagnathion flow (Hiemenz Flow) V

?

y

 Z-direction is infinite , but the distributin of V in     x-span is finite, therefore it will have a stagnation   pt on the plate, where we take as the origin of the     coord. system. Our objective is to understand the   flowfield near the stagnation pt!   

? x

For 2-D, steady, incompressible. Flow with constant µ , the G..E’S are:

 ∂u ∂u =0  + x y ∂ ∂   ∂u  ∂p ∂ 2u ∂ 2 u   ∂u  ρ u + v  = − + µ ( 2 + 2 ) ∂y  ∂x ∂x ∂y   ∂x  2 2  ρ u ∂v + v ∂v  = − ∂p + µ ( ∂ v + ∂ v )   ∂x ∂y  ∂y ∂x 2 ∂y 2 if we consider a particular solution, say u = ax  v = −ay

(3.29)

(3.30)

Continuity equation: a-a = 0 (√ ) y-momentum: ρ [ 0 + (− ay )(− a ) ] = − x-momentum: ρ [ (ax)(a) + 0] = − ∴P =

∂p −1 ⇒ P = a 2 x 2 + g (u ) 2 ∂x

−1 2 2 1 2 2 a y − a x + const 2 2

v2 or

∂p −1 ⇒ P = a 2 y 2 + f (x) ∂y 2

u2

1 P + (u 2 + v 2 ) ≡ P = const. 2

This is the Bernoulli equation, that is the given velocity distribution is for a inviscid flow. The streamline is given as:

V ×ds = 0

(parallel each other)

3- 29

Advanced Fluid Mechanics

i j k u v 0 = (udy − vdx)k = 0 dx dy 0 ⇒ ⇒

dx dy dx dy = = ⇒ ⇒ u v ax −ay

n x = − n y+C

n ( xy ) = C ⇒ xy = constant

family of hyperbalas

Therefore, the streamline looks like:

Remarks: (1) though the given velocity distribution satisfies the N-S equation, it can’t satisfy the no slip B.C’S. ( @ y = 0, v = 0 but u = ax ≠ 0, except for x = 0 (2)

we , therefore , want to modify the u, v, such that it can satisfies the no slips boundary condition

To modify v= -ay, let us assume a similar form of v = − f ( y)

(3.31a)

To satisfy the continuity equation, ∂u ∂v ∂u + =0⇒ − f ' ( y ) ⇒ u = xf ' ( y ) ∂x ∂y ∂x or u = xf ' ( y )

(3.31b)

3- 30

Advanced Fluid Mechanics

In order to satisfy the no-slip B.C’S:

u y =0 = 0 ⇒ f ' (0) = 0 v

y =0

= 0 ⇒ f (0) = 0

(3.32a, b)

As the y → ∞ , we want u back to the inviscid case, that is u = ax, thus f ' (∞ ) = a

(3.32c)

In the inviscid flow , the pressure is p = p0 −

1  2 2 ρ a x + a2 y2   2 

Now, we modify the pressure as 1 P = P0 − ρ  a 2 x 2 + a 2 F ( y )  2

(3.33)

Not that u, v, p are replaced by two unknown function f (y) and F(y). However, we still have two momentum equations. the problem is closure. Sub. u, v, and p into the x-momentum equation, we have

f

'2

− ff '' = a 2 +ν f

'''

(3.34)

Sub u, v ∆p into y-momentum equation: 1 2 ' a F -νf '' 2

⇒ ff ' = ' or F =

or F =

2 ν f '' + ff '  a2 

2 a2

 ' f2 ν f + 2  + const  

(3.35)

In summary, we have

ν f ''' + ff '' − f F=

2 a2

'2

+ a2 = 0

 ' f2 ν f + 2  + const  

' 1 2 f ' (0 ) = 0 ○ 3 f (∞ ) = a f (0) = 0 ○ with B.C’S. ○

3- 31

(3.34) (3.35)

Advanced Fluid Mechanics

with eq.(3.34) and B.C’S, we can solve the unknown function f. We want to use similarity method, introduce η =α y, f ( y ) = Aφ (η ) then

ν Aα 3φ ''' + ( Aφ )( Aα 2φ ''' ) − A 2α 2φ ' + a 2 = 0 2

A2α 2φφ ''

To let the equation non-dimensionalized, i.e., let the coefficients of the above equation become all identically equal to unity , we put

νAα 3 = a 2 ∴ A = νa

and

A2α 2 = a 2

and

α=

a

ν

Thus, the new independent variables are

η=

a

ν

y , F ( y ) = νaφ (η )

(3.36)

The G.E’s become

φ ''' + φφ '' − φ ' + 1 = 0 2

with B.C' s :

(3.37)

φ (0) = 0, φ ' (0) = 0, φ ' (∞) = 1 ⇐ Hiemenz Flow

also get F=

ν 2 (φ + 2φ ' ) a

(3.38)

Eqn (3.37) is solved by Hiemenz, and tabulate as Table 5.1 in the p.p98 of schlichting.

η=

a y ν

0 0.2 ： 2.4 ： 4.0 ： 4.6

dφ u = dη U

φ 0 0.0233 ： 1.7553 ： 3.3521 ： 3.9521

(φ ' = d φ = dη

η

0 0.2266 ： 0.9905 ： 1.000 ： 1.000

d( f d(

) aν = a y)

ν

φ ∼v φ' =

u U

1.0 1

1 ' 1 u df u = = f = a dy a a x U aν ⋅

ν

3- 32

)

Advanced Fluid Mechanics

Remarks: (1) As η = 2.4, u / U = 0.9905 . We consider the corresponding distance from a

the wall as the boundary layer δ , therefore (η =

δ = ηδ

ν

= 2.4

ν

ν )

y =η

y

a

ν

(3.39)

a a Note also that δ is independent of x.

(The boundary-layer thickness is constant because the thinning due to stream acceleration exactly balances the thicknessing due to viscous dissipation)

y

U = ax 2

U = ax1

δ x1 (2) As x → ∞ , v = −ay & u → ∞ for y ≠ 0 As y → ∞ (or η → ∞ ), u = ax and v → ∞ (∵ φ → ∞ ) That is the modified solution, though satisfies the no slip condition, still can’t satisfy the condition at infinite. We will see this problem in the “boundary layer theory”. (Localized solution)

Wrong, the sol satisfies the B.C at infinite. The sol. Match the inviscid flow solution when it is far away from walls.

Corresponding problem: Vj

2-D or axis-symmetric stagnation jet: If jet fluid is the same as the surrounding fluid How about the flow field? How does it look like? How we specified the boundary condition?

h

d y

x

(The potential flow solution u = ax v = − ay , which is a ideal, theoretical flow case, will not be the outer solution of the present problem. We need to solve this problem by Numerical method.) 3- 33

Advanced Fluid Mechanics

2-D or axis-symmetric Spraying.

Vj

Liquid fuel

How does the spray looked like?

Air

x

3- 34

Advanced Fluid Mechanics

3.4 Flow over a rotating disk (White, 3-8.2, p. 163)

ez , w

Z

ω

Infinite plane disk rotating with angular velocity ω

eˆr , u

∂ Symmetric with respect to θ ⇒ =0 ∂θ

θ Continuity:

1 ∂ ∂ (ru ) + ( w) = 0 r ∂r ∂z

r

x

r- Momentum: u

 ∂ 2u ∂ u ∂ 2u  ∂u ∂u v 2 1 ∂p +w − =− +ν  2 + ( ) + 2  ∂r ∂z r ∂r r ∂z  ρ ∂r  ∂r

θ -momentum: u

 ∂ 2v ∂ v ∂ 2v  ∂v ∂v v + w + u =ν 2 + ( ) + 2  ∂r ∂z ∂r r ∂z  r  ∂r

z- Momentum: u

 ∂ 2 w 1 ∂w ∂ 2 w  ∂w ∂w 1 ∂p +w =− +ν 2 + ( ) + 2  ∂r ∂z ∂z  ρ ∂z r ∂r  ∂r

4 equations, 4 unknowns

eθ , v

(√ )

How many B.C’s do we need? --- second order in u, v, w and 1st order in p; thus we need 7 boundary conditions. B.C.’s: (1) At z = 0, u = w = 0, v = rω p = 0 (a convenient constant) (2) At z =∞, u = v = 0

(3) (1) (2)

w = ? ( w ≠ 0 , because the fluid near the rotating disk will be

pumped out, so we expected there are fluid coming from the top of the rotating disk.) Need one more boundary condition. (3)

∂p = 0 (so that p is bounded, otherwise p → ±∞ as r → ∞) ∂r

(The flow would move in circular streamlines if the pressure increased radially to balance the inward centripetal acceleration.)

3- 35

y

Advanced Fluid Mechanics

Compare inertial & viscous term in the r-momentum: ∂u ∂ 2u u ∼ν 2 ∂r ∂z O [ (ωr )(ω) ] ∼ O νω ( r ) / δ 2  ⇒ δ ∼ (ν ) ω

1

2

Therefore, we may non-dimensional z by the use of δ . Introduce a new variable

ζ =

z

ω1 = z( ) 2 δ ν

(White: z*)

Also, try to use separation variables method by assuming

u = ω r F( ζ ) v = ω r G( ζ ) w = (νω )

1

2 H(

(so that the effect of r and z are separated; u = vr; v = vq; w = vz)

ζ )←

p = ρν ω P( ζ )

function of z only since r & z are assumed separated

← Since

∂p = 0 ∴ p is function of z only ) ∂r

The B.C.’s become:

ζ= 0, F(0) = H(0) = P(0) = 0, G(0) = 1

(z = 0)

ζ =∞, F(∞) = G(∞) = 0

(z→∞)

(

∂p = 0 cancel one term in r-momentum equation!) ∂r

The G.E.’s becomes Continuity: 2F + H' = 0 r: F2 － G 2 + HF ' = F '' θ: 2FG + HG '－G '' = 0 z: P ' + HH '－H '' = 0

(3.40) (3-185)

(3.41a~d ) (3-184)

Equation (3.41a-c) with B.C. (3.40) is sufficient to solve F, H, and G the results can be applied to equation (3.41d) to solve P. For small value of z, such that ζ is small seek a solution in powers of ζ

F = a0 +a1ζ + a2ζ 2 + a3ζ 3+ h.o.T

neglecting high order terms

G = b0 +b1ζ +b2ζ 2 + b3ζ 3+ h.o.T H = c0 +c1ζ +c2ζ 2 + c3ζ 3+ h.o.T

3- 36

(3.42)

Advanced Fluid Mechanics

a0 ,……c3 (12 unknowns)

Try to determine

From B.C’s on ζ = 0, F = 0 ⇒ a0 = 0 G = 1 ⇒ b0 = 1 H = 0 ⇒ c0 = 0 Apply the G.E. at ζ = 0, with F = H = 0 and G =1, we have Continuity: 0 + H '(0) = 0

⇒ H '(0) = 0

⇒ c1 = 0

r:

0 – 1 + 0 = F ''(0) ⇒ F ''(0) = -1

⇒ a2 = −

θ:

0 + 0 - G ''(0) = 0 ⇒ G ''(0) = 0

⇒ b2 = 0

1 2

Now differentiate original equations ω.r.t. ζ

2 F ' + H '' = 0 2 FF ' − 2GG ' + H ' F ' + HF '' − F ''' = 0

(3.43a-c)

2 F 'G + 2 FG ' + H 'G ' + HG '' − G ''' = 0 Sub. (3.42) into (3.43) again for, ζ = 0, and use the previous results (i.e. a0 = 0, b0 = 0, c0 = 0, c1 = 0, a2 = -1/2, b2 = 0), we get 2a1 + 2c2 = 0 -2b1 – 6a3 = 0 2a1 – 6b3 =0

(3.44 a-c)

Differentiate Eq. (3.43a) again and evaluate at ζ = 0:

2F '' + H ''' = 0 ''

(at ζ = 0, F = 2a2 = -1, H ⇒

-2 + 6c3 = 0

'''

= 6c3 + 24c4ς + ... ς =0 = 6c3 ) ⇒ c3 =

1 3

We have get 3+3+1=7 coefficients, therefore 5 unknowns left. However, we have 3 equations (Eq 3.44a-c), thus, we can express 3 unknown (c2, a3, b3) in terms of the other 2 unknowns (a1, b1). From (3.44) we have C2 = -a1, a3 = -b1/3, b3 = a1/3

3- 37

Advanced Fluid Mechanics

The solution thus become 1 2 b1 3 ζ − ζ + .... 2 3 a G = 1 + b1ζ + 1 ζ 3 + ...... 3 1 H = − a1ζ 2 + ζ 3 + ...... 3 F = a1ζ −

(3.45)

Two unknowns: a 1 & b 1 . Also note that Eq. (3.45) will not suitable for ζ→∞, because F, G, H will→∞ Now, let’s look at the equation. At ζ → ∞ , where F(ζ ) = G(ζ )=0 is the known B.C.’s Continuity: 2 F + H ' = 0

⇒ H '= 0 → H (ζ ) = -C (∵w < 0 at ζ→∞)

r:

F2 － G 2 + HF ' = F '' ⇒ HF ' = F ''

θ:

2FG + HG '－G '' = 0

⇒ HG '= G ''

⇒

F ' ∼ e H ∞ζ ⇒ F ' (ζ ) ∝ e − cζ ⇒ F (ζ ) ∝ e − cζ

G ' ∼ e H ∞ζ ⇒ G ' (ζ ) ∝ e − cζ ⇒ G (ζ ) ∝ e − cζ Thus, in the far away region, we seek solution of the form of

F = A1e-cζ + A2e-2cζ +…… G = B1e-cζ + B2e-2cζ +……

(3.46)

H = – C + C1e-cζ + C2e-2cζ +…… Sub (3.46) into the G.E.s Continuity:

{2A1e-Cζ + O[e-2Cζ] + …} + {– CC1e-Cζ + O[e-2Cζ] +……} = 0 -Cζ

⇒ e

: 2A1 – CC1 = 0

⇒ C1 = 2A1/C

r-momentum:

{A1e-Cζ + A2e-2Cζ}2 – { B1e-Cζ +…}2 + [– C + C1e-Cζ +…][–A1Ce-Cζ –2CA2e-2Cζ +…] = + A1C2 e-Cζ + 4C2A2C2 e-2Cζ +… -2cζ

⇒ e

: A12 – B12 - CC1A1 + 2C2A2 – 4C2A2= 0

2 2 1 ) ⇒ A2 = – (A1 +B 2

2C

3- 38

Advanced Fluid Mechanics

θ -momentum B2 = 0

and

2 2 1 ) C2 = – (A1 +B 3

2C

The solution near ζ→∞ is thus

F = A1e-Cζ + (A12 +B12) / 2C2e-2Cζ +…… G = B1e-Cζ + O[e-3Cζ] +…… H=–C+

(3.47)

2 2 2A1 -Cζ e – (A1 +B3 1 ) e-2Cζ +…… C 2C

Unknowns: A1, B1, C By matching the “inner” solution for small ζ to an “outer” solution for large ζ. That is, take small value of ζ in Eq. (3.47a). Numerically, we finally obtain a1 = 0.51, b1 = -0.616 C= 0.886, A1 = 0.934, B1 = 1.208 These values may not be unique, but they have been verified in laminar flow experiment. The velocity distribution is 1.0

−H =−w

G ~ν

F ~ u 1

ξ =Z

ω ν

4

Or , see Table 3.5 on page 166 of the book of White, “Viscous Fluid Flow”, 2nd ed. At ζ = 5.4, F ≈ G ≈ 0.01 . Therefore the boundary layer δ is 5.4 ≈ δ

ω ⇒ ν

δ = 5.4

ν ω

or

3- 39

(3.48) (3-187)

Advanced Fluid Mechanics

Z

u δ δ

v = ωr

-

see also Fig. 3-28 (White)

pumping outward near the disk by centrifugal action, replenished from above at constant (at Z→∞) downward velocity.

Supplementary data for rotating disk:

∵ H(∞) = -0.8838, thus vz (∞) = -0.8838 ων

(disk draw fluid toward it)

The circumferential wall shear stress on the disk is

τ zθ = µ

∂uθ ∂z

=

ρ rG ' (0) νω 3 = −0.6159 ρ r νω 3

z =0

Remarks: (1) If we apply the above result, to find the torque required to turn a disk of radius r0 . them r0

M = ∫ τ zθ r (2π r )dr = -0.967ρr0 0

4

3.87 ω r02 νω ; CM ≡ , Re = ≅ 1 ρω 2 r05 ν Re 2 3

3- 40

−2 M

Advanced Fluid Mechanics

The equation agrees well with experimental data for Re < 3 × 10

5

5

For Re < 3 × 10 , the flow becomes turbulent. (2) If we stir tea in a cup; the flow pattern will be reversed. Thus these exists an inversed radial flow. (3) Rogers & Lance (1960) used a Runge-Kutta method to solve eqns (3.41), by defining Y1 = H, Y3 = F, Y5 = G Y2 = F ' ,

Y4 = G ' , Y6 = Р

with I.C’s: Y1(0) = Y3(0) = Y6(0) = 0 , Y5(0) = 1

(3.40a)

The two unknown conditions of Y2 (0) & Y4 (0) must be chosen to satisfy the end B.C.’s (3.40b). Namely Y3 → 0, Y5 → 0, at ζ → ∞. The fortrain statements for (3.41) are simply six statements, as described in p.165 of White’s book. By numerical iteration, we can find the I.C.’s to be Y2 (0) = F ' (0) = 0.5102 Y4 (0) = G ' (0) = -0.6159 The numerical results agree well with those obtained by asymptotic expansion.

3- 41

Advanced Fluid Mechanics

3.5 Flow in a channel (3-8.3 Jeffery-Hamel Flow in a Wedge-Shaped Region)

φ

φ

2α

u0

divergent (source) flow

convergent (sink) flow 1 ○ 2 ○ 3 ○

4 ○

The velocity distributions are Re = u0r /ν 1 : Re = 5000 ○ 2 : Re = 1342 ○ convergent 3 : Re = 684 ○ 5 : Re = 684 ○ 6 : Re = 1342 ○ 7 : Re = 5000 ○

divergent

5 ○ 6 ○ 7 ○

5 6

5

4

3

2

1

7

φ

−5

3- 42

0

u

u0

1

Advanced Fluid Mechanics

3.6 Stream Function For a 2-D, constant density flow (incompressible flow) ∇ ⋅ V = 0 ⇒ ∂u + ∂v = 0 ⇒ u = ∂ψ ; v = − ∂ψ ∂y ∂x ∂x ∂y

(3.49)

∂ ∂ψ ∂ ∂ψ ∂ 2ψ ∂ 2ψ ( ) + (− )= − =0 ∂x ∂y ∂y ∂x ∂x∂y ∂x∂y

⇒

By introducing the stream function “ψ ”, the continuity equation is automatically satisfied. (By introducing theψ , the independent variable & the governing equation are reduced by one, however, the order of the P.D.E. increases by one.) In 3-D flow, the eqn of streamline is

V ×ds = 0 Where

V = u iˆ + vˆj + w kˆ ,

Thus

dx dy dz = = u v w

or

dy v ( x, y , z ) = , dx u ( x, y , z )

d s = dxi? + dyj + dzkˆ

dz w( x, y, z ) = dx u ( x, y , z )

The stream functions will be ψ 1(x,y,z) = C1 = constant,

ψ 2(x,y,z) = C2 =constant

z

ψ 2 = const ∇ψ 1

V (stream line )

∇ψ 2

ψ 1 = const y

x

Since ∇ψ 1 × ∇ψ 2 has the same direction as V , so we can say V = k ( ∇ψ 1 × ∇ψ 2 ) proportional constant 3- 43

Advanced Fluid Mechanics

Or ҜV = ∇ψ 1 × ∇ψ 2 Since

∇ ⋅ (∇ψ 1 × ∇ψ 2 ) = 0

(Mathematically)

∴ ∇ ⋅ (ҜV ) = 0 But for steady flow, we know div( ρV) = 0 , so we can pick up Ҝ = ρ, then ρV = ∇ψ 1 × ∇ψ 2

(3.50)

If the flow is constant density, we know div( ρV) = 0 , so we can pick up Ҝ = 1, then V = ∇ψ 1 × ∇ψ 2

(3.51)

Remarks: There are only a few exact solutions for N-S equation unless the physical problem and geometry is easy. The N-S equation may be simplified as Re >>1 or Re

→ ∞, where the exact solution may also be exist. In the next two chapters, we will consider the flow fluid when Re >>1 or Re → 0.

3- 44

Advanced Fluid Mechanics

Chapter4 Very Slow Motion 4.1 Equations of motion Consider a constant density flow, the equations of motion are:

v ∇ ⋅V = 0 v v v  ∂V v Momentum: ρ  + V ⋅ ∇V  = −∇p + ∇ 2V  ∂t  Continuity:

Introduce the

characteristic velocity : U∞

characteristic length : d characteristic pressure : p0 characteristic time : t0 then the non-dimensional properties become

v ~v V V = , U∞ and

v ~v r r = , d

p ~ , p= p0

U∞

d

t ~ t = t0

~ ∂ 1 ∂ ∇ ~ ▽ = v = ⇒ ∇ = d∇ ~v = ∂r d ∂r d (The magnitude of “〜” order 1)

~ ~v ∇ ⋅V = 0

Continuity:

(continuity equation is invarant for non-dimensionalization) Momentum:

~v ~v ~ ~v P0 ~ ~ 1 ∂V ~ 2 ~v + V ⋅ ∇V = − ∇ p + ∇ V ~ U ∞t0 ∂t ρU ∞ d / µ ρU ∞ 2 d

If we denote: Reynolds No. = And pick up:

P0

= ρU∞2

ρU ∞ d µ =

dynamic pressure

Equation becomes

~v ~v ~ ~v 1 ~ 2 ~v ∂V ~ + V ⋅ ∇V = − ∇ ~ p + ∇ V ~ U ∞t0 ∂t Re d

unsteady part

convective part

inertia forces

pressure forces Chapter 4-1

viscous forces

Advanced Fluid Mechanics

1)

~v ~v ~ ~v ∂V ~ If Re→∞ ⇒ + V ⋅ ∇V = − ∇ ~ p ~ U ∞t0 ∂t

2)

If Re→ 0 ⇒ ∇ 2V = 0

d

~ v

(4.1)

(4.2)

Note that there is no balance term. We want to have a balance term. Multiply (4.11) by Re

~v ~v ~ ~v P0 ∂V ~~ ~ 2 ~v Re + Re V ⋅ ∇ V = − Re ∇ p + V ∇ U ∞ t 0 ∂~ t ρU ∞ 2 d

(0, as Re→0) (i)

The unsteady term coefficient: For a oscillation body flow, w = frequency of oscillation we can choose: t0 =

1

ω

the first coefficient: Re

d U ∞ t0

= Re

dω → 0, as Re → 0 and ω is not very large U∞

1 if there is no body oscillation, we may pick t0 = U∞/d Remark: ○ ’ 2 For a highly oscillation body, the unsteady term can t neglected. ○

(ii) The pressure coefficient We want to pick up P0 such that

P0

ρU ∞ 2

Re → 1, and this term can be left to

balance the viscous term. Therefore P0 =

ρU ∞ 2 Re

=

ρU ∞ 2

ρU ∞ d / µ

=

µU ∞ d

And the momentum equation (4.1) become

~ ~ ~v 0 = −∇ ~ p + ∇ 2V Summary:

(4.3)

For a steady, constant density, slow flow (Re→0)

~ ~v ∇V = 0

~ ~ ~v 0 = −∇ ~ p + ∇ 2V Also name as: Slow flow, Creeping flow, or stokes′flow. Chapter 4-2

Advanced Fluid Mechanics

4.2 Slow flow past a sphere consider a steady, constant density flow with Re→0. ur

uθ

U∞

r

θ

a

Z

v V = 0 on r = a

v v V = u r evr + uθ evθ + uφ eφ 0 在 upstream 方向上無 φ 方向之分量，故在 sphere 附近 uφ 幾乎為零。但 uθ 和 u r 則由 U ∞ eˆ z 變化而來，故不可忽略

v

div V = 0

Mass:

⇒

∂ ∂ 2  ( sin ) ( sin ) + r θ u r θ u r θ   =0 ∂θ r 2 sin θ  ∂r 1

∂ψ ∂θ

−

∂ψ ∂r

The streamfunction are takes such that the continuity equation is satisfied automatically. Thus r2sinθur = or

ur =

∂ψ , ∂θ

rsinθuθ= −

∂ψ , 2 ∂ θ r sin θ 1

uθ= −

∂ψ ∂r

1 ∂ψ r sin θ ∂r

Momentum equation:

v

0 = − ∇p − µ∇ 2V Since

v v v ∇ 2V = grad (div V ) – cuel curl V 0

⇒

v

0 = − ∇p -μcurl curl V

Chapter 4-3

(4.4)

Advanced Fluid Mechanics

take curl on both side

v

0 = 0 -μcurl curl curl V

(4.5)

v ≡Ω eˆr v v ∂ 1 Ω = cuel V = 2 r sin θ ∂r ur

reˆθ ∂ ∂θ ruθ

r sin θ eˆφ ∂ ∂φ 0

=

  ∂ (ruθ ) ∂u r   ˆr (0) + eˆθ (0) + r sin θ eˆφ  e −   ∂θ    ∂r r 2 sin θ 

=

eˆφ  ∂uθ ∂u r  −  uθ + r r  ∂θ  ∂r

=

eˆφ  ∂ ∂ ∂ψ  1 ∂ψ 1 ∂ψ 1 − + r (− )− ( 2 )  ∂θ r sin θ ∂θ  ∂r r sin θ ∂r r  r sin θ ∂r

=

eˆφ  1 ∂ψ 1 ∂ 1 ∂ψ  r − 1 ∂ψ 1 ∂ 2ψ ( 2 ) ( ) − + − − r  r sin θ ∂r sin θ r ∂r r ∂r 2 r 2 ∂θ sin θ ∂θ 

=

− eˆφ  ∂ 2ψ sin θ ∂ 1 ∂ψ  ( )  2 + 2 r sin θ  ∂r r ∂θ sin θ ∂θ 

1

sub. Eqn (4.4)

= Ω eˆφ Where Ω ≡ −

1 Dψ r sin θ

D ≡ differential operator ≡

∂2 ∂r 2

+

sin θ ∂ 1 ∂ ( ) r 2 ∂θ sin θ ∂θ

Then

eˆr v ∂ 1 Curl Ω = r 2 sin θ ∂r 0 =

reˆθ ∂ ∂θ 0

r sin θ eˆφ ∂ ∂φ Ω

∂Ω   ∂Ω ˆr − reˆθ e  ∂r  r 2 sin θ  ∂θ 1

Chapter 4-4

Advanced Fluid Mechanics

Finally, we can obtain

eˆφ

v

curl curlΩ = Eq. (4.5) ⇒

r sin θ

D2ψ

D2ψ = 0 or

2  ∂ 2 sin θ ∂ 1 ∂  ( ) ψ =0  2+ 2 r ∂θ sin θ ∂θ   ∂r

B.C’S in terms of ψ : (Recall ur = (i)

(4.6)

1 ∂ψ ∂ψ , uθ= − ) r sin θ ∂r r 2 sin θ ∂θ 1

On r = a:

∂ψ =0 ∂θ ∂ψ uθ= 0 → =0 ∂r

ur = 0 →

(4.7a)

(ii) Infinity condition

v

∵ V∞ = U∞ eˆ z = U∞[( cosθ) eˆr + (-sinθ) eˆθ ]

∂ψ r sin θ ∂θ 1 ∂ψ uθ= − r sin θ ∂r

∴ ur =

1

2

→ U∞cosθ

as r→∞

→ －U∞sinθ as r→∞

(4.8a) (4.8b)

integrate (4.8a) and (4.8b), we obtain

ψ ~ U∞

r2 sin2θ 2

as r→∞

Assume: ψ (r,θ) = f(r)sin2θ, then the B.C’S become (4.7a) r = a → f ' (a ) = f (a) = 0 (4.7b) r→∞ → f (r) ~ U∞

r2 2

as r→∞

Chapter 4-5

(4.7b)

Advanced Fluid Mechanics

Sub. Into Eq. (4.6), we get

(

d2 dr

2

−

Aside: r 4

2

r

2

d2

2

) f =(

d4 f dr 4

+ ar 3

dr

2

d3 f dr 3

−

2

r

2

)(

+ br 2

d2 f dr

2

d2 f dr 2

−

2f

r2

+ cr

)=0

df + df = 0 dr n

we can assume solution to be the form of f = Ar , we will have 4 roots for n, n= 1, -1, 2, 4. f=

∴

A + Br + Cr2 +Dr4 r

(4.9)

B.C’S: (1) f (r) ~ U ∞

r2 2

as r→∞

compare with (4.9), we observe that we need to take C = to satisfy f (r) ~ U ∞

U∞ and D = 0 2

r2 for r →∞. ( The value of A, B are not important, 2

since they are not the highest order term, and r2 >> r as r→∞) Eq. (4.9) ⇒ f=

U A + Br + ∞ r2 r 2

(2) f (a) = 0 → '

(4.10)

U A + Ba + ∞ a2 = 0 a 2 A

f (a) = 0 → －

a2

+ B + U∞ a = 0

1 4

⇒

3 4

A = a3U∞ , B = − aU∞ ∴

1 a 3 r 1 r  ψ (r,θ) = a2 U∞ sin2θ  ( ) − ( ) + ( ) 2  + const 4 r 4 a 2 a  

3 a

1 a 



3 a

1 a 

(4.11a)

ur = U∞cosθ 1 − ( ) + ( ) 3  2 r   2 r uθ= -U∞sinθ 1 − ( ) − ( ) 3   4 r 4 r  Chapter 4-6

(4.11 b, c)

Advanced Fluid Mechanics

The streamlines are:

ψ = 10 ψ =5

θ =π

ψ =0 ψ = −5

θ =0

ψ = −10

Remark: (1) The streamlines possess perfect forward – and – backward symmetry: there is no wake. It is the role of the convective acceleration terms, here neglected, to provide the strong flow asymmetry typical of higher Reynolds number flows. (2) The local velocity is everywhere retarded from its freestream value: there is no faster region such as occurs in potential flow. (3) The effect of the sphere extent to enormous distance: at r = 10a, the velocity are still 10 percent below their freestream values. (4) The streamlines and velocity are entirely independent of the fluid viscosity.

The pressure distribution is

v

0 = -▽p -μcurl Ω or

1 1 ∂Ω 1 ∂p ∂p ∂Ω = －μ ; =－ 2 r ∂θ ∂r r sin θ ∂θ r sin θ ∂θ

integrate the eqns with the known value of Ω, we finally obtain P = P∞－

3 cos θ aμU∞ 2 2 r

(4.12)

The shear stress in the fluid is

τ rθ =μ(

1 ∂u r ∂uθ Uµ sin θ + )= − r ∂θ ∂r r

 3 a 5 a 3 1 − 4 ( r ) + 4 ( r ) 

Chapter 4-7

(4.13)

Advanced Fluid Mechanics

The drag force on the sphere is thus π

π

D = － ∫ τ rθ r = a sin θ dA － 0

∫0

p r = a cos θ dA

dA = 2πa2 sinθdθ τ rθ sin θ

τ rθ θ

p

a

θ p cos θ

θ

dA = 2π(a sinθ)(a dθ) = 2πa2 sinθdθ π π ∴ D = 3πμaU∞  sin 3 θdθ + cos 2 θ sin θdθ  ∫ ∫



0



0

4/3

2/3

= 4πμa U∞ + 2πμa U∞ due to friction or

due to pressure force ’

D = 6πμaU∞

Define:

Re =

Then

CD =

“ Stoke s Formula “

(4.14)

U ∞ 2a

ν

D 1 2

2

2

ρU ∞ (π a )

=

24 Re

Chapter 4-8

(4.15)

Advanced Fluid Mechanics

Remarks: (1) Stokes formula : D=6πμa U∞ provides a method to determine the viscosity of a fluid by observing the terminal velocity U ∞ of a small falling ball of radius a. (2) Stokes formula valid only for Re>1) can be neglected ○ lU Re νl 1 l 2

1 = where ○

4 ○

=

(

)

~ 0(1) from the result of x-momentum equation

Re δ Uδ 2 U l 2 － >> 0(1) , so that we can see that this term are larger than ○ V δ other terms, the y-momentum equation can be written contained only dominant term as ∂p

∂y

=0

Chapter 5-2

Advanced Fluid Mechanics

i.e. we can conclude p = p(x) only. So we conclude: For a flow with Re >>1, the flow very near the wall is governed by the equation (incompressible flow)

∂u ∂v + =0 ∂x ∂y

(5.1)

∂ 2u ∂u ∂u ∂u 1 dp +u +v = － +ν ∂t ∂x ∂y ρ dx ∂y 2 ∂p

and ∂y = 0. This equation is called”boundary layer equation”

Remark: Compare the Navier-stokes equation and the boundary layer equation, and explain why the latter is easier to be solved numerically? (Ans:)

For simplicity, let’s consider the incompressible flow as an example: Navier-stokes equation:

∂u ∂v + =0 ∂x ∂y ∂ 2u ∂ 2u ∂u ∂u ∂u 1 dp + +u +v = － +ν( ) ∂t ∂x ∂y ρ dx ∂x 2 ∂y 2 ∂v ∂ 2v ∂ 2v ∂v ∂v 1 ∂p + +u +v = － +ν( ) ∂t ∂x ∂y ρ ∂y ∂x 2 ∂y 2 Boundary layer equation:

∂u ∂v + =0 ∂x ∂y ∂ 2u ∂u ∂u ∂u 1 ∂p +u +v = － +ν ∂t ∂x ∂y ρ ∂x ∂y 2 ∂p ∂y

=0 Chapter 5-3

Advanced Fluid Mechanics

There are many things to be noticed: (1) Continuity equation is not affected by the consideration of Reynolds number. (2) p=pe(x) in Boundary-layer equation, and is determined by the Bernoulli equation outside the boundary layer..

dpe dU e = －ρUe dx dx

(5.2)

where x is the coordinate parallel to the wall. (3) The equation becomes parabolic in B-L theory, with x as the marching variable. In computer, parabolic equation is easier to solve than the elliptic equation, which the N-S equation belongs to. (4) Boundary conditions: In B-L equation (i)

∂ 2v ∂y 2

∂v ∂ 2 v ∂v , have been discarded, only left. Therefore, we need ∂x ∂x 2 ∂y

,

only one boundary condition of v on y-direction. The obvious condition to retain is no slip: v = 0 at y=o. (ii)

∂ 2u ∂x 2

has been discarded. Therefore, one condition of u in x-direction (to

satisfy

∂u ) is sufficient. The best choice is normally the inlet plane, and the ∂x

u in the exit plane will yield the correct value without our specifying them. (iii) Boundary condition of u on y has no change. There are two conditions to satisfy

∂ 2u ∂y 2

Namely

.

u=0

as y = 0

∂u =0 ∂y

as y =δ

Chapter 5-4

U∞

δ

Advanced Fluid Mechanics

Steady state

∂u ∂v + =0 ∂x ∂y ∂ 2u ∂u ∂u 1 ∂p u +v = － +ν ∂x ∂y ρ ∂x ∂y 2 0 =－

∂p ∂y

→ P=Pe(x)

B.C’s u(x, y = 0) = 0 v (x, y = 0) = 0 u(x, y =δ) = Ue

This condition must match the inner limit of the outer (inviscid) flow

←

v(x, y =∞) = Ue

(not the same as U∞) inviscid

U∞ ∞

Chapter 5-5

Advanced Fluid Mechanics

5.2 Flat plate case (infinite far) inviscid

U∞ ∞ Ue = U∞ = constant; pe = p∞ = constant Known the inviscid properties, we next go to the boundary layer problem.

∂u ∂u + =0 ∂x ∂y ∂ 2u ∂u ∂u u +v =ν 2 ∂x ∂y ∂y

u=0 v=0 y=∞ u = u∞ →(since here, we stand in the Boundary layer. We can see

: y= 0

only B.C, so the edge of the B.C seen ∞ for me.) Introduce stream function

u=

∂ψ ∂ψ ,v= − ∂y ∂x

the continuity equation can be satisfied automatically. Sub into momentum equation , we can get one depended variable → mess equation. (hard to be solved) Similarity solution: Introduce (try)

η =

y U∞ x α ( ) x ν

find α, so that a single variable differential equation is obtained in terms ofηonly. Can we determine a similarity variable

η=η(x, y), u = ũ (η) so that we can reduce a PDE → ODE. Assume Chapter 5-6

Advanced Fluid Mechanics

η=

y U∞ x α ( ) → α=1/2 x ν

Dimension of length: x, y,

ν U∞ y

Dimensionless of length: ỹ =

ν

U∞ y

=

ν

U∞

U = fn(η) = f ' (η) U∞

U x , ~ x = ∞

ν

anticipate f (η) as a dimensionless stream function.

∂ψ =U∞ f ' (η) ∂y ∂ψ ∂η =U∞ f ' (η) ∂η ∂y ∂ψ U ∞ 1/2 1 ( ) = U∞ f ' (η) 1 ∂µ ν x 2

ψ =0

∂ψ ν )1/2 x1/2 f ' (η) = U∞( ∂µ U∞

ψ = U∞( ν )1/2 x1/2 f (η) + const U∞

∴

η =0 so that ψ =0 represent the body shape(∵along body (y=0)→ η =0, but f(0)=0 from B.C.)

y U∞ x α ) x ν

η= (

ψ = U∞( ν )1/2 x1/2 f (η) U∞

u=

∂ψ =U∞ f ' (η) ∂y

v =−

∂ψ ν )1/2  1 f (η) + x1/2 df ∂η  = -U∞(  12 dη ∂x  ∂x U∞  2x

 1 y  ∂η η = ( ν )1/2 − =− 1  ∂x 2x U∞  2x x 2   f

v = -U∞( ν )1/2  U∞

 2 x

1

− 2

ηf ' 

1  2 x 2 

Chapter 5-7

Advanced Fluid Mechanics

1 v = -U∞( ν )1/2 1 [ f-η f ' ] U∞

2x

2

Sub. Into equation, we finally get

2 f ' ' ' +f f ' ' = 0 u = f ' (η) = 0 at y = 0 orη= 0) U∞ v ( =0 atη= 0) U∞

f ' (0) = 0

B.C.

(

f (0) = 0 f ' ( ∞) = 1

(y→∞,

u = U∞)

This is called”Blasius Problem”

Blasius Equation:

η=

y y U∞ x = x ν νx U∞

ψ = U∞( νx )1/2 f (η) U∞

u = f ' (η), U∞

ν U∞

=

1 ν (η f ' － f ) 2 U∞ x

∂ 2u ∂u ∂u Sub. into the momentum equation: u +v =ν 2 ∂x ∂y ∂y ∂( u

) − 12 y U∞ d u dη η '' = ( ) =f ( ) = - f '' ∂x dη U ∞ dx 2x x νx U∞ ) ∂( u U∞ 1 d u dη = ( ) = f '' ( ) dη U ∞ dy ∂y νx U∞ ) ∂2 ( u U∞ 1 = f ''' ( ) νx ∂y 2 U∞ Chapter 5-8

(5.3)

Advanced Fluid Mechanics

⇒

2 f '' ' + f f ' ' = 0 ”Blasius equation”

1 f(0)=0 B.C’S: ○

Solve Blasius equation by series express

' 2 f (0)=0 ○

f = A0 +A1η+A2

1 .2

2

+….

→ or using Runge-Kutta numerical method to

' 3 f (∞)=0 ○

f ' (η ) = u

η2

solve it.

U∞

1 .0

0 .8

0.6 0.4

0.2 0

0

1

2

3

4

5

6

7

8

η=

y

νx

( p.136)

U∞

Fig.7.7 or Table 7.4 on p.139 of H. Schlichting

(or Table 4.1 Fig. 4 - 6 on p.236 of white) Boundary Layer thickness: Engineering Argumemt: y =δwhen u

U∞

=0.99 from Blasius table, we find f ' (η)

=0.99 whenη=5.

y

∴ 5 =η =

νx U∞

∴

δ x

=

5 = U∞x

=

δ νx U∞

5 Re x

ν or δ=5

νx

(5.4)

U∞ Chapter 5-9

Advanced Fluid Mechanics

Surface friction:

τ=μ

∂u ∂y

τw =μ(

∂u )y=0 ∂y

∵ u = U∞ f ' (η)

1 ∂u ∂η = U∞ f ' ' (η) = U∞ f ' ' (η) ∂y ∂y νx ∂u ∂y

= y =0

∴ τw =

U ∞ 2 f ' ' ( 0) νx U∞

µU ∞ 2 f ' ' (0) νU ∞ x τw

Cf =

U∞

1 ρU ∞ 2 2

=

2 f ' ' (0) 2 f ' ' (0) = Re x U∞ x

ν

Since

f ' ' (0) = 0.332 ∴

Cf =

0.644 Re x

τw = 0.332μU∞(

(5.5a)

U ∞ 1/2 ) νx

(5.5b)

The Drag on the flat plate is D=

L ∫0 τw(x)dx

(for unit depth)

= 0.644 U∞ µρLU ∞

(5.6a)

And for a plate wetted on both side

D ' = 2D = 1.328 U∞ µρLU ∞ Chapter 5-10

(5.6b)

Advanced Fluid Mechanics

As Remarked on White book, the boundary-layer approximation is not realized until Re ≥ 1000. For ReL (= UL/ν) ≤ 1 , the Oseen theory is valid. In the Range of 1