So
√
θ s
¯
− s/αx
θ = e
s
Applying the inverse Laplace transform from a given table, we obtain the
following results as sketched in Figure 4.9.
θ
T − T i
x
=
= erfc √
(4.25)
θ s
T s − T i
4αt
Case 2: Constant surface heat flux BC:
T(x, 0) = T i
T(∞, t) = T i
∂T(0, t)
""
−k
= q s
∂x
Let θ = T − T i , then
θ(x, 0) = 0
θ(∞, 0) = 0
∂θ(0, t)
""
−k
= q s
∂x
The Laplace transform solution is
q "" 1
√
s
− s/αx
θ ¯ =
√
− e
k s s/α
83
Transient Heat Conduction
t
x
0
T s
T i
T (x,t)
FIGURE 4.9
Solution of 1-D transient heat conduction with given surface temperature boundary condition.
√
θ s
¯
− s/αx
θ = e
s
Applying the inverse Laplace transform from a given table, we obtain the
following results as sketched in Figure 4.9.
θ
T − T i
x
=
= erfc √
(4.25)
θ s
T s − T i
4αt
Case 2: Constant surface heat flux BC:
T(x, 0) = T i
T(∞, t) = T i
∂T(0, t)
""
−k
= q s
∂x
Let θ = T − T i , then
θ(x, 0) = 0
θ(∞, 0) = 0
∂θ(0, t)
""
−k
= q s
∂x
The Laplace transform solution is
q "" 1
√
s
− s/αx
θ ¯ =
√
− e
k s s/α
83
Transient Heat Conduction
t
x
0
T s
T i
T (x,t)
FIGURE 4.9
Solution of 1-D transient heat conduction with given surface temperature boundary condition.