Finite-Difference Method Example

ME 350, Heat TransferKim Shollenberger

Steps for Finite-Difference Method

1.  Define geometry, domain (including mesh and elements), and properties

2.  Simplify (or model) by making assumptions3.  Define boundary (and initial) conditions4.  Use energy balance to develop system of finite-

difference equations to solve for temperatures5.  Use Fourier’s Law and known temperatures to

solve for heat fluxes

1) Define Geometry and Properties

qx = 0

T1

Lx = Ly = 2.0 cm

T1 = 100 ˚C

h = 100 W/m2•K

T∞ = 20 ˚C

k = 0.72 W/m•K

Note: qx = 0 on line of thermal symmetry at x = 0

y

x

h, T∞

h, T∞

Lx

Ly

Define Domain, Mesh, and Elements

qx = 0

T1

0 ≤ x ≤ Lx

0 ≤ y ≤ Ly

x = (i -1) ∆x

y = (j -1) ∆y

i = 1, 2, 3,…, Nx + 1

j = 1, 2, 3,…, Ny + 1

∆x = Lx / Nx

∆y = Ly / Ny

y, j

x, ii = 1 2 3 Nx Nx+1

3

2

j = 1

h, T∞

h, T∞

Ny +1

Ny

Lx

Ly

Mesh Definitions

Differential Elements - discrete areas over which temperature and properties are assumed constant

Node - center of each elementMesh - combination of all

nodes (or elements)

i,j i+1,j i-1,j

i,j+1

i,j-1

node

element

Mesh Definitions, cont.

Structured Mesh - mesh with regularly spaced nodes that align with axes

Mesh Refinement - decreasing size of mesh elements resulting in more nodes and higher accuracy

Boundary Elements - elements on boundary of domain that can be full, half, or quarter size. Nodes located on the boundary are more accurate and required for symmetry boundary conditions.

2) Simplify (or model) by �Making Assumptions

•  Steady state•  2-D conduction in brick•  Constant properties•  Negligible radiation

3) Define Boundary Conditions

•  Symmetry plane at x = 0: �qx (x = 0) = 0

•  Fixed temperature at y = 0:�T(y = 0) = T1

•  Convection x = Lx and y = Ly�

4) Develop Finite-Difference Equations

Interior Nodes:

i, j i+1, ji-1, j

i, j+1

i, j-1

∆x

∆y€

Ti, j =Ti+1, j +Ti−1, j +Ti, j+1 +Ti, j−1

4

Develop Equations Continued…

Boundary at j = 1 (y = 0) and i = 1 to Nx +1:Temperatures are all fixed at T1

Ti, 1 = T1

Develop Equations Continued…

Boundary at i = 1 (x = 0) and j = 2 to Ny :

1, j 2, j

1, j - 1

1, j + 1

∆x/2

∆y

€

T1, j =2 T2, j +T1, j+1 +T1, j−1

4

Develop Equations Continued…

Boundary at i = 1 (x = 0) and j = Ny + 1:

1, Ny + 1 2, Ny +1

1, Ny

h, T∞

∆y/2

∆x/2€

T1,Ny+1 =T2,Ny+1 +T1,Ny

+Bi T∞2+Bi

Develop Equations Continued…

Boundary at i = Nx + 1 (x = Lx) and j = 2 to Ny:

Nx, j Nx + 1, j

Nx + 1, j - 1

Nx + 1, j + 1

∆x/2

∆yh, T∞

€

TNx+1, j =2 TNx , j +TNx+1, j+1 +TNx+1, j−1 +2 Bi T∞

4 +2 Bi

Develop Equations Continued…

Boundary at i = Nx + 1 (x = Lx) and j = Ny + 1:

Nx, Ny + 1

Nx + 1, Ny + 1

Nx + 1, Ny

∆x/2

∆y/2 h, T∞

h, T∞

€

TNx+1,Ny+1 =TNx ,Ny+1 +TNx+1,Ny

+2 Bi T∞2+2 Bi

Develop Equations Continued…

Boundary at j = Ny + 1 (y = Ly) and i = 2 to Nx:

i, Ny + 1 i + 1, Ny + 1i - 1, Ny + 1

i, Ny

∆x

∆y/2

h, T∞

€

Ti,Ny+1 =2 Ti,Ny

+Ti+1,Ny+1 +Ti−1,Ny+1 +2 Bi T∞4 +2 Bi

Solve System of Linear Equations

Procedure for Direct Matrix Inversion:a.  Renumber each node: k = (j - 1)(Nx + 1) + i�

for a total of N = (Nx + 1)(Ny + 1) nodes.b.  Use equations derived above to write an equation

for each node. Example for interior nodes:

k k+1k-1

k + (Nx+1)

k - (Nx +1)

€

Tk −14Tk+1 +Tk−1 +Tk+(Nx+1) +Tk−(Nx+1)[ ] = 0

Solve System Continued …

c. Express equations as follows:

d. Calculate the A matrix and C vector.e. Calculate temperature: [T] = [A]-1 [C]

€

a1,1 a1,2 ! a1,k ! a1,Na2,1 a2,2 ! a2,k ! a2,N" " # "ak,1 ak,2 ak,k ak,N" " " # "

aN ,1 aN ,2 ! aN ,k ! aN ,N

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

T1T2"Tk"TN

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

=

C1C2"Ck"CN

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

Solve System Continued …

Procedure for Gauss-Seidel Iteration:a. Assume initial temperature distributionPick something reasonable: • For this case either T1 or T∞ works well• Alternatively, T = 0 ˚C would work• Picking good initial guess for temperature

will speed up convergence

Solve System Continued …

b. Solve for Ti, j at Each Node:• Use equations derived above for each node• Always use most recently calculated

temperatures for each new calculation• Perform calculations in systematic order to

improve efficiency (i.e. T1, 1, T2, 1, T3, 1, …)

Solve System Continued …

c. Check for Convergence• Calculate |Tnew- Told| after updating all of the

temperature data• When difference falls below desired

tolerance at every node solution has converged

• Repeat steps b. and c. until convergence

5) Solve for Heat Fluxes

Use known temperatures and Fourier’s Law:

where gradient for 2-D Cartesian is:

thus, components and magnitude are:

€

ʹ ́ ! q = −k ∇T

€

∇T =∂T∂x

ˆ i +∂T∂y

ˆ j

€

ʹ ́ q x = −k ∂T∂x,

€

ʹ ́ q y = −k ∂T∂y,

€

ʹ ́ ! q = ʹ ́ q x( )2 + ʹ ́ q y( )2

Use Finite Differences for Gradient

For 2nd order accurate approximation:

Central differences:

Forward differences:

Backwards differences:

€

∂T∂x

≈Ti+1 −Ti−1

2 Δx

€

∂T∂x

≈−3Ti + 4 Ti+1 −Ti+2

2 Δx

€

∂T∂x

≈3Ti − 4 Ti−1 +Ti−2

2 Δx

Calculating Heat Transfer at y = 0�(j = 1) per Unit Length (z-direction)

Must integrate over x-direction:

Approximate integral using:

€

ʹ q y = 0( ) = ʹ ́ q y y = 0( )dxx=0Lx∫

€

ʹ q y = 0( ) ≈ ʹ ́ q y,1Δx2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + ʹ ́ q y, i Δx

i=2

Nx

∑ + ʹ ́ q y, Nx+1Δx2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

MATLAB Programs

Two programs in MATLAB to solve Example #8:1.  conduction_matrix_inv.m2.  conduction_Gauss_Seidel.m

Both are available online on my webpage at www.calpoly.edu/~kshollen.

All system parameters, such as Nx, h, and k, can be changed to test their effect on the solution.

Mesh for Solution with Nx = 5

q = 0

T1

y, j

x, i1 2 3 4 5

1

h, T∞

h, T∞

Lx

Ly

2

3

4

5

Temperature in ˚C with Nx = 5

Mesh for Solution with Nx = 20

q = 0

T1

y, j

x, i1 2 3 20 21

321

h, T∞

h, T∞

2120

Lx

Ly

Temperature in ˚C with Nx = 20

Temperature in ˚C �with h = 10,000 W/m2•K

Temperature in ˚C with k = 10