fvm

The Finite Volume Method: Basic Principles andExamples

Mainak ChowdhuryIndian Institute of Technology,Kanpur,

KANPUR (UP) - 208016.Email: [email protected]

Under the guidance ofDr. Gautam Biswas

IIT Kanpur

Mainak Chowdhury (IIT Kanpur) December 15, 2009 1 / 49

Overview

1 Introduction

2 Diffusion: The steady case

3 Diffusion: The non-steady case

4 Diffusion and Convection

5 Acknowledgements


Outline

1 Introduction




5 Acknowledgements


What is finite volume method and what does it solve?

A finite volume method (FVM) is a prescription for spatialdiscretization of a continuous system of differential equations.

The basic idea of FVM is that volume integrals of divergence termscan be represented as surface integrals. (Divergence theorem)

The quantities at the surface can be approximated using differentinterpolation techniques from discrete data(control variables).

Ok. . . what kind of equations are we talking about?


The Conservation Equation

Consider a control volume

n

v

Figure: Definition of a control volume

Then for a conserved scalar U,t

RUd +

H

(U(~v .~n) (U).~n)dS = RQV d +

H

( ~QS .~n)dS

This equation is the fundamental equation of many flow problemswhere U is typically the momentum, energy, or density

The problem is steady when the time derivative is zero and nonsteady otherwise.


Outline

1 Introduction




5 Acknowledgements


A 1-D steady diffusion problem

Let us consider a special case of conservation equation in one dimension.

d

dx(

dU

dx) + QV = 0 (1)

Let us see how we can solve this equation.

x

x

Uw UeUp

xx

Node

Control volume

x x

Discretize the domain: Here the points xi divide the domain ofdefinition into several control volumes


A 1-D steady diffusion problem (Contd.)

Integrate the diffusion equation over each control volume: This yields x2x1

(d

dx(

dU

dx) + QV )dx = 0 (2)

This can be simplified as

(dU

dx)x2 (

dU

dx x1) +

x2x1

QV dx = 0 (3)

Assume an interpolation profile: A linear function between nodes willlead to the following equation

x2Ue Upx2

x1Up Uwx1

+ QVx = 0 (4)

where QV represents the average value of QV over the controlvolume.


A 1-D steady diffusion problem (Contd.)

Source term linearization: If we assume

QV = QVc + QVpUp (5)

the equation corresponding to each control volume becomes

apUp = aeUe + awUw + b (6)

where

ap = ae + aw QVpx (7)ae =

x2x2

(8)

aw =x1x1

(9)

b = QVcx (10)


Some issues regarding aforementioned scheme

Choice of the interpolating function

Positivity of coefficients

Boundedness of the iterations: A sufficient condition is that |anb|ap

1

where anb stands for coefficients of neighbouring terms to point p.

Choice of control nodes(structured/unstructured)

But is the above system of equations completely determined?


The Boundary Conditions

They can be specified in a variety of ways:

Dirichlet

Neumann

A combination of the above

The boundary conditions enter the linear equations as source terms.

Together with the boundary conditions, the linear system of equations iscompletely determined and is solvable by sparse matrix or iterativetechniques.


The Boundary Conditions

They can be specified in a variety of ways:

Dirichlet

Neumann

A combination of the above

The boundary conditions enter the linear equations as source terms.Together with the boundary conditions, the linear system of equations iscompletely determined and is solvable by sparse matrix or iterativetechniques.


Extension to 2 dimensionsLet us consider the Laplace equation in two dimensions:

2

x2+2

y 2= 0 (11)

The control volume is now

i+1

i+1/2

i-1/2i-1

i

j

j+1

j-1

j-1/2

j+1/2

A

B

C

D

A'

B'

C'

D'

Figure: The Control Volume ABCD


Solving the Laplace equation

Consider the integral: ABCD

(2

x2+2

y 2)dxdy = 0

By Greens theorem, the above expression is equivalent to(

xdy

ydx) = 0

The line integral is over the boundary of ABCD and can beapproximately evaluated as the length of each segment times thevalue of the partial derivative at the midpoint of the segment.


Solving the Laplace equationThe integral form of the Laplace equation applied to ABCD yields

x|i ,j 1

2yAB

y|i ,j 1

2xAB +

x|i+ 1

2,jyBC

y|i+ 1

2,jxBC

+

x|i ,j+ 1

2yCD

y|i ,j+ 1

2xCD + +

x|i 1

2,jyDA

y|i 1

2,jxDA = 0

Estimating the derivatives at the intermediate points: One way ofdoing this is to assign the average value of the partial derivative overthe control volume to its central point. This yields the followingexpressions

x|i ,j 1

2 1

AreaABC D

(

x)dxdy =

1

AreaABC D

dy

y|i ,j 1

2 1

AreaABC D

(

y)dxdy = 1

AreaABC D

dx


Solving the Laplace equation

The right hand side of the last equation can be approximated asABC D

dy i ,j1yAB + ByBC + i ,jyC D + AyDA

Similarly

ABC D

dx i ,j1xAB + BxBC + i ,jxC D + AxDA

The area AB C D can be approximated as

| ~AB x ~AD | = xAByj1,j yABxj1,j


Solving the Laplace equationAssume

Distance between coordinate lines constant locally

Intermediate value is the average of node values at the corners,i.e.,

A =1

4(i1,j1 + i1,j + i ,j + i ,j1)

The above assumptions give a linear equation in the unknowns at a nodeand its surrounding nodesThus, we have

gi ,ji ,j + gi1,j+1i1,j+1 + gi ,j+1i ,j+1 + gi+1,j+1i+1,j+1 + gi1,ji1,j+ gi+1,ji+1,j + gi1,j1i1,j1 + gi ,j1i ,j1 + gi+1,j1i+1,j1 = 0

The gs depend on the geometry of the mesh, more specifically on thex s and y s.


Some observations

The finite volume method in 2 dimensions presented so far, offers anunifying framework to deal with differential equations, irrespective ofthe grid coordinate system

By allowing flexibility in the choice of the coordinate system, theboundary conditions can be more accurately represented, than infinite difference methods.


Simulations: The 2D Laplace equationConsider the Laplace equation :

2 = 0

defined on the following:

r=0.1

r=1

=0

=/2

Figure: The domain of definition

The boundary condition is given by:

=sin()

r


The exact solution

The exact solution corresponding to the above conditions is given by:

=sin()

r

It is plotted below:

0 0.2

0.4 0.6

0.8 1 0

0.2 0.4

0.6 0.8

1

0

2

4

6

8

10

"EXACT_LAPLACE.OUT" using 1:2:3


A contour plot of the solution

9.5 9

8.5 8

7.5 7

6.5 6

5.5 5

4.5 4

3.5 3

2.5 2

1.5 1

0.5

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1


Numerical SolutionsThe following figures were obtained for a grid obtained from triangulationof the solution region.

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

(solution variable - Analytical Solution)

0.18

0.12

0.06

0.00

0.06

0.12

0.18

0.24

(a) Number of unknowns 448, RMS er-ror 0.0316

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

(solution variable - Analytical Solution)

0.12

0.08

0.04

0.00

0.04

0.08

0.12

0.16

(b) Number of unknowns 3985, RMSerror 0.0134

Figure: The above plots show the error of the numerical solutions on anunstructured grid.


Numerical Solutions

The following figures were obtained for a cylindrical grid in the solutionregion,with uniformly distributed and .

0 0.2

0.4 0.6

0.8 1 0

0.2 0.4

0.6 0.8

1

-0.05-0.045-0.04

-0.035-0.03

-0.025-0.02

-0.015-0.01

-0.005 0

"DIFF21X21.OUT" using 1:2:3

(a) Number of unknowns 400, RMS er-ror 0.0138

0 0.2

0.4 0.6

0.8 1 0

0.2 0.4

0.6 0.8

1

-0.007-0.006-0.005-0.004-0.003-0.002-0.001

0

"DIFF51X51.OUT" using 1:2:3

(b) Number of unknowns 2500, RMSerror 0.0015

Figure: The above plots show the error of the numerical solutions on a cylindricalgrid.


Observations

Structure of the problem is important

Finite Volume method offers a simple way of exploiting this structure.

Requires coordinate transformation in finite difference methods.


Outline

1 Introduction




5 Acknowledgements


The time dependent 1 D diffusion with source term

Retaining the time derivative in the conservation equation we get,

d

dx(

dU

dx) + QV =

U

t(12)

This may be rewritten asU

t= (U, x) (13)

The right hand side includes effects of the source term and the diffusionterm.


Temporal discretization of RHS

Integrating the RHS, we have t+tt

= (f t+t + (1 f )t)t

for0 f 1

I f=1:Fully implicit schemeI f=0.5:Crank-Nicholson schemeI f=0:Fully explicit scheme

Explicit schemes:More stringent restrictions on time steps, noiteration required

Implicit schemes:Less severe time step restrictions, requires iterationsto arrive at the result


Temporal discretization of the transient terms

Consider the following integral t1t0

U

tdtdx

This may be approximated as

(Ut1 Ut0)x

Forward Eulert1 = t + t, t0 = t

Backward Eulert1 = t, t0 = t t


Temporal discretization of the transient terms(contd.)

A second order scheme:

Ut1 =3

2Ut 1

2Utt (14)

Ut0 =3

2Utt 1

2Ut2t (15)


Simulations: 1D Transient Diffusion

Consider the following transient diffusion problem:

t= 2 (16)

with boundary conditions

= 1 at x = 0 (17)

= 0 at x = (18) = 0 at t = 0 (19)

The exact analytical solution to the above is

(x , t) = 1 erf (x/2t)


Explicit scheme with 90% of allowed time step


Explicit scheme with 110% of allowed time step


Implicit scheme with the same time step


Outline

1 Introduction




5 Acknowledgements


The steady state equation

With no source terms the conservation equation reads, for any volume ,

(U(~v .~n) (U).~n)dS = 0

Using divergence theorem,

.(U~v (U)) = 0In 1 D , the above equation reduces to

d(Uv)

dx d

2U

dx2= 0


Solving the steady state equation

x

x

Uw UeUp

xx

Node

Control volume

x x

Integrating the equation over the control volume and using centraldifferencing for the diffusion term, we get

vx2Ux2 vx1Ux1 = ae(Ue Up) aw (Up Uw ) (20)


Interpolation schemes for estimating the LHS

Central Difference

Upwind Differencing

Hybrid Differencing

Power Law

Assume the general equation to be

hpUp = hwUw + heUe (21)

with

hp = hw + he + vx2 vx1 (22)


The coefficient values in the different interpolation schemes

Scheme hw heCentral Differencing aw + vx1/2 ae vx2/2Upwind Differencing aw + max(vx1 , 0) ae + max(0,vx2)Hybrid Differencing max(vx1 , (aw + vx1/2), 0) max(0,vx2 , (ae vx2/2)

Power Law awmax(0, (1 0.1|Pe1 |)5) aemax(0, (1 0.1|Pe2 |)5)+max(vx1 , 0) +max(vx2 , 0)


Some issues in the choice of an interpolation scheme

Conservativeness: The interpolation scheme should give the samevalue of the flux at the same face of adjacent volumes

Boundedness: This means that the value at any node should notblow up.

Transportiveness:This represents the ability of the interpolationscheme to address the relative strengths of diffusion and convection inthe interpolation process.

Accuracy


A closer look at conservativeness

Linear

Interpolation

Control Volume

Control node

Volume Face

(a) A Linear Interpolation

Quadratic

Interpolation

Control Volume

Control node

Volume Face

(b) A Quadratic Interpolation

Figure: A Comparison of Linear vs Quadratic interpolation

However, not all quadratic schemes suffer from this problem. QUICKinterpolation is one of the most widely used today.


A closer look at boundedness

A sufficient condition which will ensure converging solution ([Versteeg et al]) is that |anb|

|ap| 1 for all nodes p (23)< 1 for at least one node (24)

All coefficients should have the same sign to ensure that increase inany neighbouring node leads to increase of value at a particular node.


A closer look at transportiveness

Define the Peclet number as

Pe =vxi L

xi(25)

where L is the length of the domain of interest.

A high |Pe | indicates strong convection over diffusion, low valueindicates the converse.

Central Differencing tends to break down with high |Pe | as it givesequal priority to upstream and downstream flows

Upwind differencing or power law schemes try to accommodate thisinformation


A 2D convection and diffusion problemConsider the problem of diffusion in the annular region of the following:

=0

r=3

y

x

The equation for in the annular region is

2y

x 2x

y=

1

Pe(2

x2+2

y 2) (26)


SolutionThe exact solution to the above problem in the annular region is

= 1 log(x2 + y 2)/(2log3) (27)It is plotted below:

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Analytical Solution

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Analytical Solution

Figure: Exact Solution

Consider the problem in the region R. This becomes a 2D problem .Mainak Chowdhury (IIT Kanpur) December 15, 2009 43 / 49

The numerical errors from the Finite Volume MethodThe following plots were obtained from applying a finite volume method toa Cartesian grid (21X21)

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

(Analytical Solution - solution variable)

0.00030

0.00024

0.00018

0.00012

0.00006

0.00000

0.00006

0.00012

0.00018

0.00024


(a) Pe = 0.05

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4


0.06

0.05

0.04

0.03

0.02

0.01

0.00

0.01

0.02


(b) Pe = 20

Figure: The above plots illustrate the sensitivity of the numerical solutions tohigh Pe. Note the false diffusion


Summary-Why FVM?

Conservativeness:I Conservativeness is explicitly enforcedI In sharp contrast to finite-difference and finite-element methods

Handling unstructured geometries:I Suitable for representing complicated geometries.I Easier incorporation of mixed boundary conditions over complicated

boundariesI Formulation stays the same


Outline

1 Introduction




5 Acknowledgements


Acknowledgements

I would like to thank Dr. G. Biswas, for his insightful comments and helpfulsuggestions. I would also like to thank all members of the organising teamof the 8th Winter Academy for giving us this unique opportunity to learnand interact with our fellow students from India and abroad.


Bibliography

H.K. Versteeg and W. Malalasekera.An introduction to Computational Fluid Dynamics: The Finite VolumeMethodLongman Scientific & Technical

C.A.J. Fletcher.Computational Techniques for Fluid DynamicsSpringer-Verlag

J. Blazek.Computational Fluid Dynamics: Principles and AppicationsElsevier

CFD-Wikiwww.cfd-online.com/Wiki

FiPyhttp://www.ctcms.nist.gov/fipy/


End

Thank you for your kind attention.

Questions?


End

Thank you for your kind attention.Questions?


IntroductionDiffusion: The steady caseDiffusion: The non-steady caseDiffusion and ConvectionAcknowledgements

0.0: 0.1: 0.2: 0.3: 0.4: 0.5: 0.6: 0.7: 0.8: 0.9: 0.10: 0.11: 0.12: 0.13: 0.14: 0.15: 0.16: 0.17: 0.18: 0.19: 0.20: 0.21: 0.22: 0.23: 0.24: 0.25: 0.26: 0.27: 0.28: 0.29: 0.30: 0.31: 0.32: 0.33: 0.34: 0.35: 0.36: 0.37: 0.38: 0.39: 0.40: 0.41: 0.42: 0.43: 0.44: 0.45: 0.46: 0.47: 0.48: 0.49: 0.50: 0.51: 0.52: 0.53: 0.54: 0.55: 0.56: 0.57: 0.58: 0.59: 0.60: 0.61: 0.62: 0.63: 0.64: 0.65: 0.66: 0.67: 0.68: 0.69: 0.70: 0.71: 0.72: 0.73: 0.74: 0.75: 0.76: 0.77: 0.78: 0.79: 0.80: 0.81: 0.82: 0.83: 0.84: 0.85: 0.86: 0.87: 0.88: 0.89: 0.90: 0.91: 0.92: 0.93: 0.94: 0.95: 0.96: 0.97: 0.98: 0.99: anm0: 1.0: 1.1: 1.2: 1.3: 1.4: 1.5: 1.6: 1.7: 1.8: 1.9: 1.10: 1.11: 1.12: 1.13: 1.14: 1.15: 1.16: 1.17: 1.18: 1.19: 1.20: 1.21: 1.22: 1.23: 1.24: 1.25: 1.26: 1.27: 1.28: 1.29: 1.30: 1.31: 1.32: 1.33: 1.34: 1.35: 1.36: 1.37: 1.38: 1.39: 1.40: 1.41: 1.42: 1.43: 1.44: 1.45: 1.46: 1.47: 1.48: 1.49: 1.50: 1.51: 1.52: 1.53: 1.54: 1.55: 1.56: 1.57: 1.58: 1.59: 1.60: 1.61: 1.62: 1.63: 1.64: 1.65: 1.66: 1.67: 1.68: 1.69: 1.70: 1.71: 1.72: 1.73: 1.74: 1.75: 1.76: 1.77: 1.78: 1.79: 1.80: 1.81: 1.82: 1.83: 1.84: 1.85: 1.86: 1.87: 1.88: 1.89: 1.90: 1.91: 1.92: 1.93: 1.94: 1.95: 1.96: 1.97: 1.98: 1.99: anm1: 2.0: 2.1: 2.2: 2.3: 2.4: 2.5: 2.6: 2.7: 2.8: 2.9: 2.10: 2.11: 2.12: 2.13: 2.14: 2.15: 2.16: 2.17: 2.18: 2.19: 2.20: 2.21: 2.22: 2.23: 2.24: 2.25: 2.26: 2.27: 2.28: 2.29: 2.30: 2.31: 2.32: 2.33: 2.34: 2.35: 2.36: 2.37: 2.38: 2.39: 2.40: 2.41: 2.42: 2.43: 2.44: 2.45: 2.46: 2.47: 2.48: 2.49: 2.50: 2.51: 2.52: 2.53: 2.54: 2.55: 2.56: 2.57: 2.58: 2.59: 2.60: 2.61: 2.62: 2.63: 2.64: 2.65: 2.66: 2.67: 2.68: 2.69: 2.70: 2.71: 2.72: 2.73: 2.74: 2.75: 2.76: 2.77: 2.78: 2.79: 2.80: 2.81: 2.82: 2.83: 2.84: 2.85: 2.86: 2.87: 2.88: 2.89: 2.90: 2.91: 2.92: 2.93: 2.94: 2.95: 2.96: 2.97: 2.98: 2.99: anm2:

fvm

Documents