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Solve C 3 and C 4 for the y-direction equation at y = 0, T = 0, Y = 0 = C 3 +
C 4 ⇒ C 4 = −C 3 .
−
λ n y
−λ n y
− e
λ n y
Y( y) = C 3 e
λ n y
− C 3 e
= C 3 (e
)
With sinh(x) = (e x − e −x )/2, let C 5 = C 3 /2, the above equation can be
written as
Y( y) = C 5 sinh(λ n y)
Then solve for the product equation, and let C 2 C 5 = C n .
nπx
nπy
nπx
nπy
T(x, y) = X(x) · Y(y) = C 2 C 5 sin
sinh
=
C n sin
sinh
a
a
a
a
at y = b, T = T s = C n sin(nπ/a)x sinh(nπ/a)b.
Multiplying both sides by sin(nπx/a) dx, one obtains
mπx
nπx
nπb
mπx
sin
· T s dx =
C n sin
sinh
sin
dx
a
a
a
a
C n can be determined by the integration over x,
T s sin(mπx/a) dx
(3.6)
sin(nπx/a) sinh(nπb/a) sin(mπx/a) dx
C n =
C n = 0, if m = n;
T s sin(nπx/a) dx
, if m = n.
(nπx/a) sinh(nπb/a) dx
C n =
sin
2
one perform integration? From the integration table, one
How does
obtains
a
a
nπx
a
nπx
a
1. sin
dx = −
(1 − cos(nπ))
cos
� =
a
nπ
a 0
nπ
0
=
a
nπ
[
1 − (−1)
n
] =
⎧
⎨
⎩
0, n = even
2a
nπ
, n = odd
a
�
� a
a
nπx 1
2nπx
a
sin
2 nπx dx =
− sin
=
2.
2nπ
2
2
a
a
a
0
0
or
a
a
sin
2 nπx
a
dx =
a
1 − cos(2nπx/a)
a 1 a
2nπx �
a
dx = −
sin
=
2
2 2 2nπ
2
a 0
0
0
48
Analytical Heat Transfer