y
θ = T−T 1
T ∞ , h
∂θ b
h (θ− θ ∞ ) = −k ∂y
b
T 1
T 1
θ = 0
θ = 0
x
T 1 θ = 0
a
""
−(q /k)(a/nπ)[1 − (−1) n ]
s
=
(nπ/a) cosh(nπb/a)[(a/2)]
""
−(q /k)(2/nπ)[1 − (−1) n ]
s
=
(nπ/a) cosh(nπb/a)
Therefore, we obtain
∞
""
2q a[1 − (−1) n ]
nπx
nπy
s
T(x, y) = T 1 −
sin
sinh
(3.11)
kn 2 π 2 cosh(nπb/a)
a
a
n=1
3.2.2 Given Surface Convection BC
A similar procedure can be applied for the only nonhomogeneous convection
BC problem [2] shown in Figure 3.3. Again, the solution will be a product of
sine and cosine in the x-direction and sinh and cosh in the y-direction, and
the nonhomogeneous convection BC will be used to solve the final unknown
integrated value of C n .
A long rectangular bar with one side cooled by convection and the others
maintained at a constant temperature T 1 .
Define θ = T − T 1 . Laplace’s equation applies to this steady 2-D conduction
problem.
∂ 2 θ
∂ 2 θ
+
= 0
∂x 2
∂y 2
which must be solved subject to BCs
x = 0, 0 < y < b : θ = 0,
y = 0, 0 < x < a : θ = 0,
x = a, 0 < y < b : θ = 0,
51
2-D Steady-State Heat Conduction
FIGURE 3.3
2-D heat conduction with one convective boundary conditions.