Let θ = T − T 1 .

The above governing equation and the associated BCs become

∂ 2 θ
∂ 2 θ ∂ 2 θ
+
+
= 0
∂x 2
∂y 2
∂z 2
x = 0, θ = 0, or ∂θ/∂x = 0; x = a, θ = 0, two homogeneous BCs,
y = 0, θ = 0, or ∂θ/∂y = 0; y = b, θ = 0, two homogeneous BCs,
z = 0, θ = θ 0 ; z = c, θ = 0, one nonhomogeneous BC.
Let θ = X(x)Y(y)Z(z). Put its derivatives in the above 3-D heat conduction
equation and obtain
−
1
X
d 2 X
dx 2 =
1
Y
d 2 Y
dy 2 +
1
Z
d 2 Z
dz 2 = λ
2
(3.16)
This implies
d 2 X
dx 2 + λ
2 X = 0
(3.17)
−
1
Y
d 2 Y
dy 2 =
1
Z
d 2 Z
dz 2 − λ
2
= μ
2
The last equation can be further written as
⎧ d 2 Y
⎪ ⎪
+ μ 2 Y = 0
⎨ dy 2
(3.18)
⎪ ⎪ d 2 Z
⎩
− (λ 2 + μ 2 )Z = 0
dz 2
Therefore, we need to solve two eigenvalue equations. The x-direction solution will be sine and cosine, the y-direction solution is sine and cosine, and
the z-direction solution is sinh and cosh. The only nonhomogeneous BC
in the z-direction will be used to determine the final unknown integrated
value C n .
From the previous discussion, the solutions for X, Y, and Z follow:
⎧ X ∼ = c 1 cos λx + c 2 sin λx
⎪ ⎪ ⎨
Y ∼
sin μy
= c 3 cos μy + c 4
⎪
√
√
⎪ ⎩ Z ∼
− λ 2 +μ 2 z + c 6 e λ 2 +μ 2 z
= c 5 e
55
2-D Steady-State Heat Conduction