�
�
�
�
� �
� �
�
�

�
θ dx − θ(δ, t) δ
⎡
⎤
δ
d
dt
⎣
�
0
�� �
� � �
∂θ
∂θ
⎦ = α
−
∂x
∂x
δ
0
δ
�
� � �
d
dt
θ dx = −α ρL
dδ
∂θ
+
dt
∂x 0
0
Assume θ = c 1 (x − δ) + c 2 (x − δ)
2
(4.37)
Boundary conditions:
x = 0, θ s = −c 1 δ + c 2 δ
2
(4.37a)
dθ �
∂θ dδ
∂θ
x = δ,
= 0 =
+
, that is, θ(δ, t) = 0
dt
∂x
dt
∂t
δ
δ
δ
� � 2
� �
∂θ
k
∂θ
= −
+
∂x δ ρL
∂t δ
∂ 2 θ
�
∂θ
� 2 k

∴ α
=
∂x 2
∂x
ρL
δ
δ
α · 2c 2 = c 1
2 k .
(4.37b)
ρL
From Equations 4.37a and 4.37b, we obtain
)
αρL
c 1 =
[1 − 1 + μ]
δk
αδ + θ s
c 2 = δ 2
where μ (2θ s C p /L) is the Stefan number.
=
90
Analytical Heat Transfer
x
T m = T
Moving boundary
δ
δ
Solid
Liquid
0
T s
T i
q s ″
x = (t)
( ,t)
FIGURE 4.16
Slow melting: T m ≡ T i .
where L is the latent heat of melting.