�
�
�
�
�
�
�
�
�
�
X = C 1 cos (λx) + C 2 sin (λx)
−
λy
Y = C 3 e λy + C 4 e
[
]
−
λy
θ = C 1 cos (λx) + C 2 sin (λx) C 3 e λy + C 4 e
(3.24)
Boundary conditions:
i. x = 0: θ = 0
ii. x = a: θ = 0
iii. y = 0: θ = 0
iv. y = b: θ = cx − T 0
Applying BC (i) into Equation 3.24, C 1 = 0

Applying BC (ii) into Equation 3.24, C 2 sinh (λa) = 0, λ n = nπ/a.

Applying BC (iii) into Equation 3.24, C 3 = −C 4 .

(
)
( nπx ) (
)
λ n y − e −λ n y
θ x, y = C 2 · C 4 sin
e
a
(
)
∞
( nπx )
( nπy )
(3.25)
θ x, y =
C n sin
sinh
a
a
n=1
Applying BC (iv) and using orthogonal functions to evaluate C n ,
a
0 (cx − T 0 ) sin (nπx/a) dx
(3.26)
a
0
C n =
sin 2 (nπx/a) dx
(
)
nπb/a
sinh
a
0
(
)
nπx
(cx − T 0 ) sin
dx
a
(
) 2
(
)
(
) a
(
) a
a
nπx
ax
nπx
T 0 · a
nπx
= c
sin
−
cos
+
cos
.
nπ
a
nπ
a
nπ
a 0
0
a
(
)
2
nπx
ca
T 0 · a
(cx − T 0 ) sin
dx =
(− cos (nπ)) +
(cos (nπ) − 1)
a
nπ
nπ
0
a
( nπx )
ca 2
T 0 · a [
]
n+1 +
n − 1
(cx − T 0 ) sin
dx =
(−1)
(−1)
a
nπ
nπ
0
a
0
(
)
nπx
a
T 0 · a
n
(cx − T 0 ) sin
dx =
(−1) (−ca + T 0 ) −
a
nπ
nπ
57
2-D Steady-State Heat Conduction