y = 0, ∂θ/∂y = 0; ⇒ C 4 = 0

y = b, θ 2 = 0; ⇒ λ n = nπ/2b

x = 0, θ 2 = 0; ⇒ C 1 = 0

∞
θ 2 =
C n2 sinh(λ n x) cos(λ n y )
n odd
∞
θ 2 (a, y ) = (T 2 − T 1 ) =
C n2 sinh(λ n a) cos(λ n y ),
n odd
� b
(
)
(T 2 − T 1 ) 0 cos (nπy /2b) dy
C n2 =
(
) � b
sinh (nπa/2b) 0 cos 2 (λ n y ) dy
4(T 2 − T 1 ) sin ((nπ/2))
=
(
)
nπ sinh (nπa/2b)
where sin((nπ/2)) = (−1) (n−1/2) for n odd
∞ 4(T 2 − T 1 )(−1) (n−1)/2
(
)
(
)
nπx
nπy
θ 2 =
(
) sinh
cos
nπ sinh (nπa/2b)
2b
2b
n odd
Finally,
⎧

⎨
∞ [
1 − (−1) n ]
(
)
(
)

2(T 3 − T 1 )
nπy
nπx
T (x, y ) = T 1 +
(
) cosh
sin
⎩
nπ
cosh nπb/a
a
a
n=1
⎫
∞
(−1) (n−1)/2
(
)
(
)⎬
4(T 2 − T 1 )
nπx
nπy
+
(
) sinh
cos
nπ
sinh (nπa/2b)
2b
2b ⎭
n odd
3.4. A long rectangular rod 0 ≤ x ≤ a, 0 ≤ y ≤ b, as shown in Figure 3.10, has the
following thermal BCs:
x = 0, 0 < y < b : T = T o
x = a, 0 < y < b : T = T 0 + T 1 sin(πy /b)
y = 0, 0 < x < a : T = T 0

y = b, 0 < x < a : T = T 0 + T 2 sin(πx/a)

a. Determine the steady-state temperature distribution.
b. Sketch the isotherm and isoflux.
62
Analytical Heat Transfer