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109
Numerical Analysis in Heat Conduction
The solution may be expressed as
[T] = [A]
−1 [C]
(5.10)
where [A]
−1 is the inverse of [A] that is defined as
⎡
⎤
b 11
b 12 · · · b 1N
⎢
⎥
b 21
b 22 · · · b 2N
⎢
⎥
[A]
−1
= ⎢ .
⎥
⎣ . .
⎦
b N1 b N2 · · · b NN
Therefore, temperature can be determined by
T 1 = b 11 C 1 + b 12 C 2 + · · · + b 1N C N
T 2 = b 21 C 1 + b 22 C 2 + · · · + b 2N C N
. .
(5.11)
.
T N = b N1 C 1 + b N2 C 2 + · · · + b NN C N
Example 5.1
We use Figure 5.3 as an example to demonstrate how to solve the 2-D heat
conduction problem by using the finite-difference method. Let Δx = Δy , q ˙ = 0.
Rearrange the temperatures from the energy balance on node 1, 2, 3, ….
1
T 1 = (T s + T s + T 2 + T 3 ) ⇒
4
−4T 1 + T 2 + T 3 + 0 + 0 + 0 + 0 + 0 = −2T s
1
T 2 = (T 1 + T 4 + T 1 + T s )
4
2T 1 − 4T 2 + 0 + T 4 + 0 + 0 + 0 + 0 = −T s
1
T 3 = (T s + T 5 + T 4 + T 1 )
4

1

T 4 = (T 3 + T 6 + T 3 + T 2 )
4

1

T 5 = (T s + T 7 + T 6 + T 3 )
4

1

T 6 = (T 5 + T 8 + T 5 + T 4 )
4
1
hΔx
T 7 = (
) 2T 5 + T 8 + T s + 2
T ∞
4 + 2(hΔx/k )
k
1
hΔx
T 8 = (
) T 6 + T 7 +
T ∞
2 + (hΔx/k )
k