The steady two-dimensional temperature (7) distribution in an isotropic heat conducting materials is given by Laplace equation, H = 6 The side lengths of the domain are L=8 and H=6. Assuming consistent units are used, boundary conditions are shown in Figure 2. Use the grid indicated in Figure 2 to solve for the temperature distribution. L = 8 f'(x) = T=100 f"(x)= T=50 T=50 T=100 T=40 T₁ T3 ²T ²T ax² dy² T=40 Hint: Because of symmetry, T₁-T3 and T2=T4. Central finite difference formula f(x+h)-f(x-h) 2h f(x+h)-2f(x) + f(x-h) h² + =0 T=15 T₂ T4 T=20 ar Əx Figure 2 Finite difference nodal scheme = X

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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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The steady two-dimensional temperature (7) distribution in an isotropic heat conducting materials is
given by Laplace equation,
H = 6
The side lengths of the domain are L=8 and H=6.
Assuming consistent units are used, boundary conditions are shown in Figure 2. Use the grid indicated
in Figure 2 to solve for the temperature distribution.
L = 8
f'(x) =
T=100
f"(x)=
T=50
T=50
T=100
T=40
T₁
T3
²T ²T
ax² dy²
T=40
Hint: Because of symmetry, T₁-T3 and T2=T4.
Central finite difference formula
f(x+h)-f(x-h)
2h
f(x+h)-2f(x) + f(x-h)
h²
+ =0
T=15
T₂
T4
T=20
ar
Əx
Figure 2 Finite difference nodal scheme
=
X
Transcribed Image Text:The steady two-dimensional temperature (7) distribution in an isotropic heat conducting materials is given by Laplace equation, H = 6 The side lengths of the domain are L=8 and H=6. Assuming consistent units are used, boundary conditions are shown in Figure 2. Use the grid indicated in Figure 2 to solve for the temperature distribution. L = 8 f'(x) = T=100 f"(x)= T=50 T=50 T=100 T=40 T₁ T3 ²T ²T ax² dy² T=40 Hint: Because of symmetry, T₁-T3 and T2=T4. Central finite difference formula f(x+h)-f(x-h) 2h f(x+h)-2f(x) + f(x-h) h² + =0 T=15 T₂ T4 T=20 ar Əx Figure 2 Finite difference nodal scheme = X
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