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1/17

The Control-Volume Finite-Di ff erenceApproximation to the Di ff usion Equation

ME 448/548 Notes

Gerald RecktenwaldPortland State University

Department of Mechanical & Materials [email protected]

21 January 2014

ME 448/548: 2D Diff usion Equation

Motivation

Numerical solution of the 2D Poisson equation is the next step in developing ourknowledge of CFD technique.

Introduce the Finite Volume Method Naturally deal with material discontinuity Can naturally enforce conservation of mass and energy Core idea in many (not all) commercial CFD codes

Extend analysis to two spatial dimensions More interesting practical applications More complex data structures More complex procedures to solve the Ax = b problem

ME 448/548: 2D Diff usion Equation 1


2/17

Overview

Goals for this unit

Introduce the Poisson equation A model of steady heat conduction with a source term

Form of the pressure equation for incompressible ow Precursor to the generalized advection diff usion equation

Use the nite volume method to obtain discrete equations Allow for non-uniform mesh, diff usion coeffi cient, and source term. Introduce a set of Matlab codes for 2D Control Volume Finite Diff erence (CVFD) Demonstrate solutions for three model problems. Measure truncation error for a model problem with a simple solution Apply to fully-developed ow in rectangular ducts


Model Problem

The two dimensional diff usion equation in Cartesian coordinates is

x

x

+ y

y

+ S = 0 (1)

where is the scalar eld, is the diff usion coeffi cient, and S is the source term.

Example: Heat conduction in a rectangular domain

= k, thermal conductivity or = kc p

=

S = volumetric heat source



3/17

2D Cartesian Finite Volume Mesh

x

y

1 2

i = 0 2 j = 0

2

3

3

1

1

+1 +2

Interior node

Boundary node

Ambiguous corner node


Finite Volume Mesh: Compass Point Notation

Nodes are labelled with their positionrelative to the typical point P.

N

S

EW

x

y

Capital letters designate node locations.

N, S, E, W are the names of north,south, east and west neighbors.

N , S , E , W are the values of atthe north, south, east and west neighbor nodes .

Use of compass point names simpliesthe algebra. For example the value of at point N is N or i,j +1 , but N issimpler to write.

Compass point notation is an historicconvention that does not extend tounstructured meshes.



4/17

Finite Volume Mesh

Detailed notation for a typical 2D controlvolume

x w x e

S

PW E

N

x

y

xe xw

yn

y s y s

yn

Lowercase letters designate the interfacebetween the nodes.

x e and x w are the x coordinates of theeast and west faces of the controlvolume around P

yn and ys are the y coordinates of thenorth and south faces of the controlvolume around P

Similarly, n , s , e , w are the valuesof at the north, south, east and westcontrol volume interfaces .


Finite Volume Mesh

Detailed notation for a typical 2D controlvolume

x w x e

S

PW E

N

x

y

xe xw

yn

y s y s

yn

Regardless of whether the controlvolumes sizes are uniform, node P isalways located in the geometric center of each control volume.

Thus,

x P

x w = xe

x P =

x

2yP ys = yn yP =

y2

for uniform or non-uniform meshes



5/17

Finite Volume Mesh

Detailed notation for a typical 2D control

volume

x w x e

S

PW E

N

x

y

xe xw

yn

y s y s

yn

In contrast, the distances between the

nodes is not assumed to be uniform

x e = x E x P 6= x w = x P x W yn = yN yP 6= ys = yP yS

x e = x E x P 6= x w = x P x W yn = yN yP 6= ys = yP yS Note the use of upper and lower casesubscripts: Upper case refers to nodes;

lower case refers to interfaces.


Convert the Di ff erential Equation to a Discrete Equation

Integrate over the control volume

Z yn

ys Z xe

x w

x

x

dxdy

= Z yn

ys

x e

x w

dy

x e

x w y

e E P

x e w P W

x w y

xw xe

S

PW E

N

x

y

xe x w

yn

y s ys

yn

The nal step is obtained by using central diff erence approximations for the derivatives atthe interfaces.



6/17


Integrate the source term

Z xe

x w Z yn

ysS dy dx S P x y (2)

Note: The Control Volume Finite Di ff erence (CVFD) method treats the source term anddiff usion coeffi cients as piecewise constants. This is a rather crude approximation, say,compared to allowing the source term and diff usion coeffi cient to vary linearly within thecontrol volume.

However, piecewise constant proles of S and allow the method to be conservative,i.e., conserving mass or energy, automatically. The conservative nature of the CVFDmethod is one of its primary strengths.



Putting pieces back into the model equation gives the discrete system of equations

a S S a W W + a P P a E E a N N = b (3)where

a E = e

x x e, a W =

w

x x w, a N =

n

y yn, a S =

s

y y s

a P = a E + a W + a N + a S (4)

b = S P (5)

This is not a tridiagonal system of equations.



7/17

Non-uniform

Continuity of uxes at the interface requires

P

x x e =

E

x x e + = e

x x e

Use central diff erence approximations

e E P

x e= P

e P x e

(6)

e E P

x e= E

E ex e+

(7) x

e

material 1 material 2

x e + x e -

P E


Non-uniform

Equations 6 and 7 can be rearranged as

e P = x e

P

e

x e( E P ) (8)

E e = x e+

E

e

x e( E P ) (9)

Add Equation 8 and Equation 9

E P = e

x e( E P ) x

e P

+ x e+ E

.

x e

material 1 material 2

x e + x e -

P E

Cancel the factor of ( E P ) and solve for e / x e to get e

x e=

x e P

+ x e+

E

1=

E P

x e E + x e+ P .



8/17

Non-uniform

Thus, the diff usion coeffi cient at the interface that results in ux continuity is

e = E P

E + (1 ) P (10)where

x e

x e=

x e x P x E x P

(11)

An analogous derivation gives formulas for w , n , and s .


Solving the System of Equations

Regardless of uniform or variable , the discrete equation has a ve-point stencil, and thediscrete equation for any interior node can be written.

a S S a W W + a P P a E E a N N = b (12)

To set up the matrix for this system of equations, we need to re-number the unknowns.

Important: i and j subscripts for the mesh are not the same as the row andcolumn indices in the system Ax = b.



9/17


Order the nodes:

n = i + ( j 1) nxn is the node number (row number)in Ax = b. i and j are the meshindices corresponding to x i and y j .

With natural ordering the neighborsin the compass point notation havethese indices

np = i + (j-1)*nxn e = n p + 1n w = n p - 1nn = np + nxns = np - nx

x

y

1 2

i = 0 2 j = 0

2

3

3

1

1

+1 +2

Interior node

Boundary node

Ambiguous corner node



This leads to a vector of unknowns

1,1 2,1

... nx, 1 1,2 2,2

... i,j

... nx,ny

1 2...

nx nx +1 nx +2

... n...

N

(13)



10/17


The coeffi cient matrix has 5 non-zero diagonals

A = a S a pa W a E a N


Algorithm for obtaining the numerical solution

1. Dene physical parameters: L x , L y , (x, y ) and boundary conditions2. Dene the mesh: n x and n y if uniform3. Compute the coeffi cient matrix4. Solve the system of equations A = b, where is the vector of unknowns.5. Post-process to visualize the solution

The Poisson equation is steady. Each step is performed only once.



11/17

Structured Mesh

The CVFD Matlab codes use structured meshes. in size.

The cells in the domain are topologically equivalent to a rectangular array

The cells need not be uniform in size. Each cell not on a boundary touches four other cells Each row has the same number of cells Each column has the same number of cells

Uniform Mesh Block-Uniform Mesh

Lx1, nx 1 Lx2, nx 2

Ly1, ny1

Ly2, ny2

Ly3, ny3

x

y

x

y


Boundary Conditions (part 1)

Boundarytype

BoundaryCondition Post-processing in fvpost

1 Specied T Compute q00 from discrete approximation to Fouriers law.q00 = k T b T i

x b x iwhere T i and T b are interior and boundary temperatures, respectively.

2 Specied q00 Compute T b from discrete approximation to Fouriers law.T b = T i + q00

x b x ik

where T i and T b are interior and boundary temperatures, respectively.



12/17

Boundary Conditions (part 2)

Boundarytype

BoundaryCondition Post-processing in fvpost

3 Convection From specied h and T , compute boundary temperature and heat uxthrough the cell face on the boundary. Continuity of heat ux requires

kT b T ix b x i

= h (T b T amb )

which can be solved for T b to give

T b = hT amb + ( k/ x e )T i

h + ( k/ x e )

where x e = x b x i4 Symmetry q00 = 0 . Set boundary T b equal to adjacent interior T i .


Matlab codes for obtaining the numerical solution

A set of general purpose codes has been written to facilitate experimentation with theCVFD method.

Algorithm Tasks Core Routines

Dene the mesh fvUniformMesh orfvUniBlockMesh

Dene boundary conditions

Compute nite-volume coeffi cients for interior cells fvcoef

Adjust coeffi cients for boundary conditions fvbcSolve system of equations

Assemble coeffi cient matrix fvAmatrix

Solve

Compute boundary values and/or uxes fvpost

Plot results



13/17

Model Problem 1

Choose a source term that may be physically unrealistic, but one that gives an exact

solution that is easy to evaluate

S = " L x2

+2L y

2#sin xL x sin 2 yL yThe exact solution is

= sin xL x

sin2 yL y

Main code to solve this problem is in demoModel1.m


Model Problem 1

The exact solution is

= sin xL x

sin2 yL y

0

0.5

1

00.20.40.60.81

0

0.5

1

1.5

2

yx



14/17

Solutions to Model Problem 1 Show Correct Truncation Error

The local truncation error at each node ise i O ( x 2) .Since the exact solution is known we can

compute

kek2N

= q Pe2i

N

N e2N

= e N .

where N = n x n y is the total number of interior nodes in the domain, and e is theaverage truncation error per node.

103 10

2 101 10 0

108

107

106

105

104

103

102

101

x

M e a s u r e

d e r r o r

MeasuredTheoretical error ~ ( x)3

Since e i O ( x 2) , N n 2x , and x = L x / (n x + 1) , we can estimate

kek2N

e

N =

O ( x 2)

n x=

O L 2x / (n x + 1)2

n xO

1

n x

3

= O ( x 3) .


Model Problem 2

Uniform source term: S = 1 .

Analytical solution is an inniteseries

Code in demoModel2.m

0

0.5

1

00.20.40.60.81

0

0.010.02

0.03

0.04

0.05

0.06

0.07

0.08

yx



15/17

Model Problem 3

Heat conduction in a rectangle consisting of two material regions.

Inner rectangle with high conductivity Outer rectangle with low conductivity Inner region has uniform heat source

0.25 L x

0.25 L y

0.5 L y

0.5 L x

L x

L y 1 = 1

S 1 = 0

2

= 1

S 2 = 1000

The discontinuity and diff erence in material properties can be used to stress the solutionalgorithm.

= 2

1

The analytical solution does not exist. Code in demoModel3.m


Model Problem 3

0.25 L x

0.25 L y

0.5 L y

0.5 L x

L x

L y

1 = 1

S 1 = 0

2

= 1

S 2 = 1000

Solution with = 100

0

0.5

1

00.20.40.60.81

0

5

10

15

20

25

30

35

yx



16/17

Model Problem 4: Fully Developed Flow in a Rectangular Duct

For simple fully-developed ow the governing equation for the axial velocity w is

" 2w x 2 + 2w y2 # dpdz = 0This corresponds to the generic model equation with

= w, = (= constant ) , S = dpdz

.



The symmetry in the problem allows alternative ways of dening the numerical model

Full Duct

x

y

L x

L y

Quarter Duct

x y

L x

L y

For the full duct simulation depicted on the left hand side, the boundary conditions are noslip conditions on all four walls.

w(x, 0) = w(x, L y ) = w(0 , y ) = w (L x , y ) = 0 . (full duct)



17/17


The symmetry in the problem allows alternative ways of dening the numerical model

Full Duct

x

y

L x

L y

Quarter Duct

x y

L x

L y

For the quarter duct simulation depicted on the right hand side, the boundary conditionsare no slip conditions on the solid walls (x = L x and y = L y )

w(L y , y ) = w(x, L x ) = 0 (quarter duct)

and symmetry conditions on the other two planes

u x x =0

= u y y=0

= 0 .


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