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Introduction to Finite Difference and Finite VolumeMethods for CFD

By

Prof. B.V.S.S.S Prasad

Thermal Turbo Machines Lab,Department of Mechanical Engineering,

IIT Madras, Chennai.

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Pre-Processor Definition ofgeometry/computational domain

Grid generation Selection of

phenomena to be

modeled andselectinggoverningequations

Definition of fluidproperties

Specification ofinitial/boundaryconditions

Solver Numerical solution of

governing equations Approximation of

the unknownvariables bysimple functions

Discretization bysubstitution oftheapproximationsinto thegoverningequations

Solution ofalgebraicequations

Post-Processor

Presentation ofresults in terms of

Vector plots

Contour plots

Particle tracking View

manipulations

Printouts

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SolverApproximation ofthe unknownvariables by simplefunctions

Discretization by

substitution of theapproximations intothe governingequations

Solution of

algebraic equations

Finite Difference Method

Finite Element Method

Spectral Method

Finite Volume Method

Others!

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Finite Differences

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A Typical Finite Difference Grid

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Fundamental of Finite-Difference Methods

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Difference Representation of Partial Differential Equations

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Simple Theoretical Basis

Consider an ODE L()=0 Assume the solution as

Substituting in the above ODE

The residue is (R)=L( )

Ideally, we would like to have this residue to be zero

Based on the principles of calculus of variations, wepropose to minimize the residue such that:

where W is a weighing function and the integrationsperformed over the domain of interest.

2 n0 1 2 n=a +a x+a x +....+a x

W.Rdx 0

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In the approximate (numerical) solution, the number ofweighing functions is equal to the number of unknowncoefficients.

Choices for weighing functions

The residue is forced to be zero at selected points, i.e.,R(x1)=0, R(x2)=0 . R (xn+1)=0.

(n+1) equations for the (n+1) unknownsCollocation method

The residue itself is used as weighing function i.e., W=R

R2 dx=0 Least Squares Method

The function used for approximating the solution is taken asthe weighing function

Galerkin MethodR dx 0

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Numerical Algorithm of FVM

Formal Integration of the governing equations of fluidflow over all the control volumes of the solutiondomain.

Discretisation i.e. conversion of integral equationsinto a system of algebraic equations.

Solution of algebraic equations by an iterative

method.

Si l Th i l B i

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Simple Theoretical Basis

Consider an ODE L()=0 Assume the solution as

Substituting in the above ODE

The residue is (R)=L( )

Ideally, we would like to have this residue to be zero Based on the principles of calculus of variations, we

propose to minimize the residue such that:

where W is a weighing function and the integrationsperformed over the domain of interest.

2 n0 1 2 n=a +a x+a x +....+a x

W.Rdx 0

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In the approximate (numerical) solution, the number ofweighing functions is equal to the number of unknowncoefficients.

Choices for weighing functions

The residue is forced to be zero at selected points, i.e.,R(x1)=0, R(x2)=0 . R (xn+1)=0.

(n+1) equations for the (n+1) unknownsCollocation method

The residue itself is used as weighing function i.e., W=R

R2 dx=0 Least Squares Method

The function used for approximating the solution is taken asthe weighing function

Galerkin MethodR dx 0

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Weighing function W=1 is chosen. The number of

weighted residual equations are obtained by dividing the

computational domain into several sub domains and

setting the weighing function to be unity over onechosen sub-domain and zero everywhere else.

Finite Volume Method

One-Dimensional Steady State Diffusion Equation

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One Dimensional Steady State Diffusion Equation

The diffusion process is governed by

Step 1: Discretisation of domain (Grid generation)

Propertyd d

S 0dx dx

Control Volume, vol A x

vol vol

d ddvol S dvol 0

dx dx

dvol A dx

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Step 2: Discretization:

(i) Integration of given equation over a control volume

(ii) Obtain discretized equation at the nodal point

Taking W=1 from w to e and zero elsewhere

Equation for one sub-domain (w-e)

d

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Considering as the flux

J J S xe w 0

J kd

dx

The principal task is to evaluate the fluxes crossing the control

surface

Evaluation of fluxes is done using any finite difference scheme,

Presently using central difference

andd

dx xw

P W

w

d

dx xe

E P

e

Physically, Je diffusion flux of leaving east face

Jw diffusion flux of entering west face

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Kx

Kx

S xeE P

e

w

P W

w

0

K

x

K

x

K

x

K

xS x

e

e

w

w

P

e

e

E

w

w

W

a a a b p P E E W W

a a b p P nb nb

Fi it Diff F l ti

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Finite Difference Formulation

2

2

E P W

2

P P E E W W

P P nb nb

d s 0dx

2s 0

xa a a b

a a b

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In general, the source term S may be non-linear

function of the dependant variable

Linearise the source term,

such thatSp is negative

( )a S a S x p p P nb nb c

The discretisation equation becomes

u pS S S

Example Problem

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Example Problem

A fin has a base temperature of 100C (TB) and the end

insulated. It is exposed to an ambient temperature of 20 C.One dimensional heat transfer in this situation is governed by

where h is the convective heat transfer coefficient, P is theperimeter, k the thermal conductivity of the material and the

ambient temperature. Calculate the temperature distributionalong the fin and compare the results with the analyticalsolution given by

where n2=hP/(kA), L is the length of the fin and x the distancealong the fin. Data L=1m, hP/(kA)=25m-2

( ) 0d dT

kA hP T T dx dx

T

cosh[ ( )]cosh( )

T T n L xnLT T

B

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Solution:The governing equation in the example contains a sink term, th

convective heat loss, which is a function of the local temperature T. Thefirst step in solving the problem by the finite volume method is to set up

grid. We assume a uniform grid and divide the length into five controlvolumes so that x=0.2m.

( )hP T T

When kA=constant, the governing equation can be written as

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, g g q

where n2 =hp/(kA)

Integration of the above equation over a control volume gives

The second integral due to the source term in the equation is evaluatedby assuming that the integral is locally constant within each controlvolume.

First we develop a formula valid for nodal points 2,3 and 4 byintroducing the usual linear approximations for the temperaturegradient. Subsequent division by cross sectional area A gives

This can be rearranged as

2( ) 0d dT

n T Tdx dx

2( ) 0

V V

d dTdV n T T dV

dx dx

2[ ( ) ] 0Pe w

dT dT A A n T T A x

dx dx

2[ ( ) ] 0P wE P

P

T TT Tn T T x

x x

2 21 1 1 1P w E P

T T T n xT n xT x x x x

For interior nodal points 2,3 and 4 we write

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For interior nodal points 2,3 and 4 we write

with

Next we apply the boundary conditions at node points 1 and 5. At node 1 thewest control volume boundary is kept at a specified temperature.

The coefficients of the discretised equation at boundary node 1 are

At node 5 the flux across the east boundary is zero since the east side of thecontrol volume is an insulated boundary.

P P W W E E ua T a T a T S

aW aE aP Sp Su1/(x) 1/(x) aW+aE-Sp -n

2x n

2xT

2[ ( ) ] 0/ 2

E P P BP

T T T T n T T x

x x

aW aE aP Sp Su0 1/(x) aW+aE-Sp -n2x-2/x n2xT+2/xTB

20 0P W PT T

n T T xx

Hence the east coefficient is set to zero. There are no additional

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source terms associated with the zero flux boundary conditions. Thecoefficients at the boundary node 5 are given by

Substituting numerical values gives the coefficients

aW aE aP Sp Su

1/(x) 0 aW+aE-Sp -n2x n2xT

Node aW

aE

Su SP

aP=aW

+aE-Sp

1 0 5 100+10TB

-15 20

2 5 5 100 -5 15

3 5 5 100 -5 15

4 5 5 100 -5 15

5 5 0 100 -5 10

Th t i f f th ti t i

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The matrix form of the equations set is

The solution to the above system is

Comparison with the analytical solution:

The maximum percentage error ((analytical solution-finite volumesolution)/analytical solution) is around 6%. Given the coarseness of the grid

used in the calculation the numerical solution is reasonably close to theexact solution

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The numerical solution can be improved by employing a finer grid. Letus consider the same problem with the rod length subdivided into 10control volumes. The derivation of the discretized equations is the

same as before, but the numerical values of the coefficients and sourceterms are different owing to the smaller grid spacing of x=0.1m. Thecomparison of results of the second calculation with the analyticalsolution is given. The second numerical results show better agreementwith the analytical solution with maximum deviation of 2% only.

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Finite Volume Method for

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Two-dimensional diffusion problems

S 0x x y y

Two-dimensional steady state diffusion equation

Step 1 : Discretization using 2 D grid

vol vol vol

dx dy dx dy S dvol 0x x y y

Integrating over a control volume and then obtaining the

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g g g

discretized equation

Ae=Aw= y and An=As= x

[Je-Jw]+[Jn-Js]+ Where

Generalized discretized equation form for interior nodes

aPP = aPP+ aPP+ aPP+ aPP+Su

aW+aE+aS+aN-Sp

aPaNaSaEaW

w w

WP

A

x

e e

PE

A

x

s s

SP

A

y

n n

PN

A

y

S V 0

W W P Ww

WP

AJ

x

E E W P

e PE

A

J x

N N P Nn

NP

AJ

x

S S S Ps SP

A

J x

Finite Volume Method for

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Three-dimensional steady state diffusion equation

Three-dimensional diffusion problems

Typical ControlVolume is

S 0x x y y z z

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aNaSaEaW

w w

WP

A

x

e e

PE

A

x

s s

SP

A

y

n n

PN

A

y

aB aT aP

aW+aE+aS+aN+aB+

aT-SP

b b

BP

A

z

t t

Pt

A

z

aPP = aPP+ aPP+ aPP+ aPP+Su

The discretized equation for interior nodes is

where

SUMMARY

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SUMMARY

1. CFD overview

2. Basics of important CFD solvers FDM,FVM and FEM

3. Finite Volume Method for 1D diffusion problems

importance of fluxes4. Example problem of heat conduction in a fin importance

of grid size

5. Extensions to 2D and 3D diffusion problems

fdm-fvm

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