Notes-cfd Viscous Flow

8/13/2019 Notes-cfd Viscous Flow

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COMPUTATIONAL FLUID DYNAMICS

VISCOUS FLUID FLOW

Introduction

Computation Fluid Dynamics CFD is the analysis of systems involving fluid flow, heattransfer and associated phenomena (such as chemical reactions) by means of computer

simulation. CFD is becoming an important component in the design of industrial products and

processes.

The finite volume method was originally developed as a special finite difference formulation.

This method is used to develop the main commercially available CFD codes: FLUENT,

PHOENICS, FLOW 3D, STAR-CD, CFX etc.

The finite volume algorithm consists of 3 main steps:

1. Divide the solution domain into finite number of control volumes with one grid pointwithin each control volume.

2. Integrate the governing equations over all the control volumes and apply the initial andboundary conditions.

3. Solve the resulting algebraic equations to find the dependent variable in all solutiondomains.

The Fluid Flow Equations

The momentum and continuity equations for a flowing fluid can be written in many co-ordinatesystems. In this chapter the Cartesian co-ordinate will be considered.

The Conservation of Mass

Consider the element shown in Figure 1 with the velocity components (u, v, w) at point (x, y, z)

in the x, y and z directions respectively. The density is at that point also.


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dx dy

dz

y

x

z

2

dx

x

xxxx

2

dx

x

xxxx

2

dz

z

xzxz

2

dz

z

xzxz

Figure 2

xx , yy and zz are similar to negative pressures; xy , yz and zx etc are shear stresses. In

addition, by writing an angular momentum balance on the element it is easy to show that

xy = yx ; yz = zy ; zx = xz

The total force on the element in the direction is made up of three terms.

dxdydzx

dydzx

x

dx

x

xxxxxx

xxxx

)]

2()

2[(

dxdydzy

dxdzy

y

dy

y

xyxy

xy

xy

xy

)]

2()

2[(

And

dxdydzz

dxdyz

z

dz

z

xzxzxz

xzxz

)]

2()

2[(

In addition a body force, such as gravity, with component xg may exist giving a force on the

element dxdydzgx .

This must balance the acceleration


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dxdydzDt

Du

Thus the force balance in the x direction becomes,

dxdydzDtDu

zyxg xz

xyxxx

Where

zw

yv

xu

tDt

Du

Similar balances can be written for the y and z equations to give

DtDug

zyx x

xzxyxx

Dt

Dvg

zyx y

yzyyyx

(2)

Dt

Dwg

zyx z

zzzyzx

These constitute the conservation of momentum equations. Before they can be used the equation

of rs linking with the velocity components of u, v and w.

The Rheological Equation

The normal method of obtaining this equation is to consider the stresses in a cube as in Elasticity

Theory (e.g. the Mohr Circle). A simpler method using simple tensor calculus is outlined below.

Using matrix notation, the stress tensor rs is given by,

zzzyzx

yzyyyx

xzxyxx

rs

We have stated above that rs = rs ; thus rs is a symmetric tensor. Furthermore, the trace of

rs (viz. zzyyxx ) is variant with respect to the orientation of the axis. Thus a scalar

quantity can be identified that usually called as the pressure P as


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)(3

1zzyyxxP

Now every tensor can be resolved into an isotropic component and a deviator. Thus,

P

P

P

P

P

P

zzzyzx

yzyyyx

xzxyxx

rs

00

00

00

The first matrix corresponds to the isotropic pressure, the second to the residual stresses causing

distortion or deviation of the fluid element. Defining the Kronecker delta rs , as

100010

001

rs

rs can be written as,

orsrs P (3)

Where o is the deviator component

Consider again the tensor made up of velocity gradients s

u r

to form,

z

w

y

w

x

wz

v

y

v

x

vz

u

y

u

x

u

s

ur

In general, this tensor is not symmetric, but can be made so by taking averages of diagonallyopposite terms to give the rate of strain tensor,


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z

w

y

w

x

wz

v

y

v

x

v

x

w

z

u

x

v

y

u

x

u

ers

)(2

1)(

2

1

The rheological equation which we are seeking is of the form,

rsrs ef

Since in various flow we are interested in relating only the deviator components of the stress and

strain components, the tensor rse is split up into two as above, by noting that

)(3

1

z

w

y

v

x

ueV

Is the rate of change of volumetric strain, a scalar quantity, so that

V

V

V

rsVrs

e

z

w

y

w

z

v

x

w

z

uy

w

z

ve

y

v

x

v

y

ux

w

z

u

x

v

y

ue

x

u

ee

)(

2

1)(

2

1

)(2

1)(

2

1

)(2

1)(

2

1

Or using similar nomenclature to above,

orsVrs eee (4)

Now the simplest rheological equation relating the deviator components of rs and rse is to say

that they are proportional to one another. Thus, let us assume that

oo e 2 (5)

Where is a constant and;


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z

w

y

v

x

u

xz

u

y

u

x

u

x

P

3

12

2

2

2

2

2

Similar substitutions can be made for y and z direction momentum balances to give,

z

w

y

v

x

u

xz

u

y

u

x

ug

x

P

z

uw

y

uv

x

uu

t

ux

3

12

2

2

2

2

2

z

w

y

v

x

u

yz

v

y

v

x

vg

y

P

z

vw

y

vv

x

vu

t

vy

3

12

2

2

2

2

2

(8)

z

w

y

v

x

u

zz

w

y

w

x

wg

z

P

z

ww

y

wv

x

wu

t

wz

3

12

2

2

2

2

2

Where v is the kinematic viscosity / .

These are the NavierStokes equations which form the basis of modern fluid mechanics. They

must be solved along with the continuity equation (1).

The Navier-Stokes equations also can be written in the polar coordinate system as stated below:

In r-direction:

v

rr

v

x

vv

rr

v

rr

vg

r

P

r

w

x

vu

v

r

w

r

vv

t

vr 222

2

2

2

22

22 2111

In -direction:

v

rr

w

x

ww

rr

w

rr

wg

P

rr

vwwu

w

r

w

r

wv

t

w222

2

2

2

22

2 2111

In x-direction:

2

2

2

2

22

2 111

x

uu

rr

u

rr

ug

x

Puu

u

r

w

r

uv

t

ux

Special Cases of the Fluid Flow Equations

Equations (1) and (8) have been solved for only a few special cases, mainly because equations(8) are nonlinear in the velocities. Fluid mechanics tends to split into two parts (1) viscous flow

and (2) inviscid flow. In the former solutions are available for some special cases.


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(1a) Exact solutions where the quadratic terms in (8) are identically zero e.g. steady

incompressible flow through a pipe with v = w = 0 and thus 0

x

ugiving,

2

21

x

u

x

P

(1b) Approximate solutions in which the quadratic terms may be neglected e.g. theory of

lubrication or stokes flow past a sphere.

(1c) problems in which the viscosity effects are assumed to occur close to a surface only e.g.

boundary layer theory in which the equations (8) are simplified.

In (1c) curvature of the surface is assumed to effect the flow only through the pressure gradient

termx

P

. Surface shape influences the flow outside the boundary layer where the flow is

assumed to be inviscid. In this second category of fluid flow problems further simplifying

assumptions are made.

(2a) The flow is inviscid and incompressible so that,

0

z

w

y

v

x

u

x

P

z

uw

y

uv

x

uu

t

u

1

(9)

y

P

z

vw

y

vv

x

vu

t

v

1

z

P

z

ww

y

wv

x

wu

t

w

1

These are known as Eulers equations.

(2b) In addition the flow is two dimensional so that equations (9) reduce to

0

yv

xu

x

P

y

uv

x

uu

t

u

1

y

P

y

vv

x

vu

t

v

1


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Governing differential equations

The differential forms of the governing equations in the fluid flow and heat transfer in twodimension are

1.

Continuity equations

0)()(

v

yu

xt

(10)

2. x-momentum equations.)()()()()(

y

u

yx

u

xx

Pvu

yuu

xu

t

(11)

3. y-momentum equations.

)()()()()(y

v

yx

v

xy

Pvv

yuv

xv

x

(12)

4. Energy equations

(13)

Equations (2), (3) and (4) can be written in general form of:

Syyxx

vy

uxt

)()()()()(

Or

Syyxxy

vx

ut

)()(

(14)

Where is the general dependent variable

= for momentum equations= k/Cpfor energy equations.

Convection and Diffusion

For steady 1-D convection and diffusion, the equation (14) can be reduced to:

Sy

T

C

k

yx

T

C

k

xvT

yuT

xT

t pp

)()()()()(


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Sxx

ux

)()(

Convection Diffusion Source

Or, in general (by neglecting the source term)

)()(xx

ux

(15)

The flow must satisfy continuity, so

0)(

u

x

In this stage, no reference will be made to the evaluation of velocities. It is assumed that they are

somehow known. The method of computing velocities will be discussed later.

Integration of the transport equation (15) over the control volume shown in Figure 3 gives:

)()()()( WPw

w

PE

e

e

wexx

uu

wx ex

ew

x

W P E

Figure 3 Staggered grids

It is convenient to define two variables F and D to represent the convection mass flux per unit

area and diffusion conductance at cell faces.

uF andx

D

Then equation (23) becomes:

)()( WPePEewwee DDFF

If we assume linear profile of (central difference) as

)(2

1PEe and )(

2

1PWw , we get


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WWEEPP AAA (16)

Where

2

eeE

FDA ,

2

wwW

FDA ,

)(22

weWEw

we

eP FFAAF

DF

DA

This assumption of linear profile of to represent eand whave some difficulties. For

example assume

De= Dw= 1 and Fe= Fw= 4 (continuity Fe- Fw= 0 satisfied)

Then AE= 1, AW= 3, AP= 2,

Now if E= 200 and W= 100, equation (16) gives P= 50 ?? and if E= 100 and W= 200, equation (16) gives P= 250 ??

Since Pmust be in the range 100 P 200. We must seek better formulation for convection-

diffusion problems.

In general the convection-diffusion formulation can be written as

WWEEPP AAA

With

)0,()( eeeP FMaxPADA

)0,()( wwwW FMaxPADA

)( weWEP FFAAA

Where the new function A(|P|) is defined as follows for different schemes

Table 1 The function A(|P|) for different schemes

Scheme A ( | P | )

Central difference

Upwind

HybridPower law

Exponential

1 - 0.5 |P|

1

Max (0, 1 - 0.5 |P|)Max {0, (1-0.1 |P|)

5}

|P| / {exp (|P| - 1}


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Where P= F/D is the Peclet number. The Peclet number at the control volume faces can be

calculated as

e

ee

e

e

e

xu

D

FP

)(

andw

ww

w

w

w

xu

D

FP

)(

For the 2-D problems, the Peclet number at the north and south faces of the control volume can

be calculated as

n

n

nD

FP and

s

ss

D

FP

Solution of the linear algorithm equations using Tri-diagonal matrix algorithm (TDMA)

Generally TDMA is an efficient tool to solve a set of linear equations in the form of equation(17) i.e.

iiiiiii DTCTBTA 11 , i= 2, 3, 4, , N-1 (17)

For the steady 1-D conduction problem, assume that the boundary conditions are given as T1and

TN; or

At i= 1; A1= 1, B1= C1= 0, D1= T1.

At i= N; AN= 1, BN= CN= 0, DN= TN.

The solution seek a relation

iiii QTPT 1 (18)

Or

111 iiii QTPT (19)

Substituting (19) into (17) to find

iiiiiiiii DQTPCTBTA )( 111

Rearrange the above equation in the form of (18) as

1

iii

i

iPCA

BP and

1

1

iii

iii

iPCA

QCDQ , i= 2, 3, 4, , N-1 (20)


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At i= 1; P1= B1/A1and Q1=D1/A1 (21)

At i= N; PN= 0 and from (18) TN= QN (22)

Summary of TDMA

1.

Calculate P1and Q1from (21)2. Calculate Piand Qifor i= 2, 3, 4, .., N from (20)3. Set TN= QN4. Use (18) to find TN-1, TN-2, T3, T2, T1by back substitution.

Unsteady 2-D convection and diffusion

The 2-D form of equation (14) can be written as

S

yyxx

v

y

u

xt

)()()()()(

It can be written as

Sy

J

x

J

t

yx

)(

(23)

Wherex

uJx

and

xvJy

are the total fluxes

(i.e. convection and diffusion).

N

S

EWew

s

nny

sy

y

x

wJ eJ

nJ

sJ

Figure 4 Control Volume


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Integration of equation (23) over the control volume shown above would give

tyxStxJJtyJJxy ysynxwxeo

p

o

ppp )()()(

Let xJJyJJ yssxee ;.....; (the integrated total fluxes) then the above equations becomes

yxSJJJJ

t

yxsnwe

o

p

o

ppp

(24)

Again here the superscript o refers to the dependent variable at time t; at time level t+t

are not superscripted.

The above equation (24) must be solved with the continuity equation

0)()(

vyuxt

Or

0)()(

yx F

yF

xt

Integrating the above equation over the control volume in Figure 4 gives

0)()()( txFFtyFFxy ysynxwxeo

pp

Let xFFyFF yssxee ;.....; (mass flow rate through the faces), then

0

snwe

o

ppFFFF

t

yx

(25)

Multiply equation (25) by pand substrate it from (24) to find

yxSFJFJFJFJ

t

yxpsspnnpwwpee

o

po

pp

)()()()()(

(26)

Where (J - F p) can be written as

)( epEpee AFJ (27)

)( PWWpww AFJ

)( NpNpnn AFJ


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)( PSSPss AFJ

Where

)0,()( eeeE FMaxPADA

(28)

)0,()( wwwW FMaxPADA

)0,()( nnnN FMaxPADA

)0,()( sssS FMaxPADA

Where A(|P|) is defined in Table 1.

Substituting (27) in (26) and rearranging to find

bAAAAA SSNNWWEEPP (29)

where AE, AW, AN, and ASare defined in (28) and

t

yxA

o

po

P

o

P

o

PTAyxSb o

PSNWEP AAAAAA

and the mass flow rate through the faces of the control volume and the corresponding diffusion

conductance are defined asyuF ee )( ;

e

ee

x

yD

yuF wn )( ;w

ww

x

yD

yuF nn )( ;n

nn

x

yD

yuF ss )( ;s

ss

x

yD

And the Peclet numbers are

e

ee

D

FP ;

w

ww

D

FP ;

n

n

nD

FP ;

s

ss

D

FP

The function A(|P|) can be selected from Table 1. The power law scheme is recommended by

Patankar, for which


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5

1.01(,0)( PMaxPA w

The resulting set of algebraic equations can be solved using line-by-line TDMA same as in the

case of unsteady 2-D conduction problem.

Flow-Field Calculation (The main difficulties)

1. Pressure gradient termConsider first the steady 1-D x-momentum equation (11)

)()(x

u

xx

Puu

x

Integrate this over the control volume as shown in Figure, we have

EW ew

Control volume of

pressure

Figure 5 Control volume

)()()()()( ewwewe PPy

u

x

uuuuu

This we assume a piecewise-Linear profile for pressure, we find

)(2

1)(

2

1)(

2

1eweppwew PPPPPPPP

This means that the momentum equation will contain the pressure difference between twoalternate grid points and not between adjacent points.

100 500 100 500 100 500 100

Figure 6 Pressure difference between two alternate grid points

If the pressure field is given as in above, for any point P, the corresponding PwPecan be seen

to be zero. Thus such a worry pressure field will be like a uniform pressure field (dp/dx = 0) bythe momentum equation???

2. Representation of the continuity equation


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A similar difficulty arises in the integration of the steady 1-D continuity equation (du/dx=

0) to find uwue= 0 or

0)(2

1)(

2

1 pwep uuuu

0 pe uu ??

A Remedy: The Staggered Grid

In the staggered grid, the velocity is calculated in the points that lie on the faces of thepressure control volume as shown in Figure 7 below.

Control volume of

U-velocity

Control volume of

pressure

1iu iu 1iu

1iPiP 1i

P

Figure 11.8 Control volumes of pressure and velocity

In this way, we have the main control volumes for the pressure (and any other scalar

variable such as density temperature etc.) and different control volumes for the velocities.

Advantages:

i. The mass flow rate across the main control volume faces (FeFw) can now becalculated without any interpolation for the velocity component.

ii. The continuity equation would contain the differences of adjacent velocities.iii. The pressure difference between two adjacent pressure grid points now becomes

the driving force for the velocity component located between these pressure gridpoints.

The price here in the difficulty in the programming since the velocity grid is staggered the newnotation based on grid line and cell numbering must be used.

Solution algorithm using staggered grid (Two-Dimensional)

The momentum equations


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x

P

yyxxvu

yuu

xu

t

)()()()()(

integration of this equation over the u-control volume shown below, with the center at i, j gives

yPPuAuAuAuAuA jijijiuSjiuNjiuwjiuEjiuP )( ,1,1,1,,1,1,

or it can be written as

yPPuAuA jijinbu

nbji

u

P )( ,1,, (30)

Similarly the integration of the y-momentum equation over the v-control volume with the center

at i, j gives

xPPvAvA jijinbvnbjivP )( ,1,, (31)

where Aus and A

vs are calculated in the same way as in (29)

N

S

EWew

s

n

J-2

j

j+1

J

J+1

I-1 i I i+1 I+1i-1I-2

j-1

J-1

P

v-cell

u-cell

Figure 8 Two-dimensional staggered grid


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Semi-Implicit Method for Pressure-Linked Equations (SIMPLE)

It is an efficient algorithm put by Patankar in 1972 and it is essentially a Guess-and-Correct

procedure for calculation of pressure on the staggered grid arrangement.

To start the SIMPLE algorithm, the pressure field P*is guessed. Then the momentum equations

(30) and (31) are solved using the guessed pressure field to find the velocities such an imperfectvelocity field based on P

*will be denoted by u

*and v

*

yPPuAuA jijinbu

nbji

u

ji )( *,* ,1**,, (32)

xPPvAvA jijinbv

nbji

v

ji )( *,* 1,**,, (33)

Now, let us propose that the correct pressure P is obtained from

PPP *

(34)

where Pwill be called pressure correction.

Similarly, we define the velocity corrections uand vto the guessed velocities u*and v

*

uuu *

vvv *

subtraction of Equation (32) from Equation (30) gives

yPPuAuA jijinbu

nbji

u

ji )( ',' ,1'',, (35)

Similarly for the y-momentum equation

xPPvAvA jijinbv

nbji

v

ji )( ',' 1,'',, (36)

At this point an approximation is introduced to ensure

xPPvAuA jijinbv

nbnb

u

nb )( ',' 1,''

Then equations (35) and (36) are simplified to

)( ','

,1,

'

, jiji

u

jiji PPdu

)( ','

1,,

'

, jiji

v

jiji PPdv

yPPPPuuAuuA jijijijinbnbunbjijiuji )()()()( *,,* ,1,1**,,,


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whereu

ji

u

jiA

yd

,

,

and

v

ji

v

jiA

xd

,

,

Then add these corrections to the guessed values in Equations (37) and (38) gives

)( ','

,1,*,, jiji

ujijiji PPduu (37)

)( ','

1,,

*

,, jiji

v

jijiji PPdvv (38)

Similarly expressions exit for ui+1,jand vi,j+1as

)( ' ,1'

,,1

*

,1,1 jiji

u

jijiji PPduu (39)

)( ' 1,'

1,1,

*

1,1, jijiv

jijiji PPdvv (40)

To find pressure correction P

, the continuity equation is used

0)()(

v

yu

xt

Continuity is satisfied is discretized form for the main (scalar) control volume shown in Figure

below.

N

S

EWew

s

nny

sy

y

x

eu

J-1

j

j+1

J

J+1

I-1 i I i+1 I+1 Figure 11.10 Main scalar control volume

0)()()()( ,1,,,1

xvvyuu

t

yxjijijiji

o

pp

(41)


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Substituting of the corrected velocities equation equations (37)-(39) into equation (40) gives

(after arranging)

p

jiji

p

jiji

p

jiji

p

jiji

p

jiji

p

ji bPAPAPAPAPA ,1,1,1,1,,1,1,1,1,, (42)

where

ydA jiup

ji ,1, )(

ydA jiup

ji ,,1 )(

xdA jivp

ji 1,1, )(

xdA jivp

ji ,1, )( p

ji

p

ji

p

ji

p

ji

p

ji AAAAA 1,1,,1,1,

xvvyuu

t

yxb jijijiji

o

ppp

ji

1,

*

,

*

,1

*

,

*

, )()()()(

The source term bp

i,jis the continuity imbalance arising from incorrect velocity field u*and v

*. In

the final iteration, the value of bp

will come out to be practically zero for all the control volume.

Then P=0 at all grid points will be an acceptable solution of equation (42) and the starred

velocities and pressure will themselves be the correct values (i.e. P*= P, u

*= u, v

*= v).

The source term bp

thus serves as a useful indicator of the convergence of the fluid flow solution.

Assembly of a complete SIMPLE algorithm.

1. Guess the pressure field P*2. Solve the momentum equations (37) and (38) to find u*and v*.3. Solve P equation (42).4. Calculate P from PPP * 5. Calculate u and v from u*and v*using equations (39) and (40)6. Check convergence7. If no convergence set P*= P, u*= u and v*= v and return to step 2 and return to step 2 and

repeat the whole procedure until a converged solution is obtained.

Note that the solution of the pressure correction equation (42) is expected to diverge unless someunder-relaxation is used during the iteration, i.e. we add only a fraction of P to P

*, or instead

of equation (34) we employ

PPP p *

where pis the pressure under-relaxation factor. Patankar recommend p0.8.

Implementation of Boundary Conditions

The flows inside a CFD solution domain are driven by the boundary conditions.


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The process of solving a field problem (e.g. fluid flow or heat and mass transfer) is nothing more

than the extrapolation of a set of data defined the boundary surface in the domain interior.

The most common boundary conditions are:

1. Inlet2.

Outlet3. Wall

4. Prescribed pressure5. Symmetry

The above boundary conditions are easy to implement for the velocities, temperatures, ., etc.

But the new pressure correction P equation (42) needs boundary conditions to be solved.

There are two kinds of conditions at the boundary. Either (i) the pressure at the boundary isgiven, or (ii) the velocity component normal to the wall is specified.

(i)

If the guessed pressure field P

*

is arranged such that at a boundary P

*

= Pgiven, then thevalue of P at that boundary will be zero.

(ii) If the grid is designed such that the boundary is on the face of the of the controlvolume as shown in Figure 10 below.

N

S

EWew

s

nny

sy

y

x

eu

boundary

Given velocity

component

Figure 10 Pressure face of control volume

The velocity ueis given as a boundary condition. In the derivation of P equation for the above

control volume from the continuity equation, the flow rate across the boundary should be

expressed in terms of a given ueitself. Then PE(Pi+1,j) will not appear will not appear (Ap

i+1,j) in

the P equation for this case.

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Notes-cfd Viscous Flow