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√
√

At x = δ, δ ∼ t to obtain θ 1 = 0, δ = b t.

Insert Equations 4.33a through 4.33d into Equations 4.33f and 4.33g
c 1 = θ s
−θ s
c 2 =
√
erf (b/ 4α 1 )
θ ∞
"
c =
√
2
erfc(b/ 4α 2 )
θ ∞
"
c = θ ∞ −
√
1
erfc(b/ 4α 2 )
From Equation 4.33e,
−k 1 θ s exp(−b 2 /4α 1 )
k 2 θ ∞ exp(−b 2 /4α 2 )
b
√
√
− √
√
= Lρ 1
(4.33h)
πα 1 erf (b/ 4α 1 )
πα 2 erfc(b/ 4α 2 )
2
η
2
e
−η 2
∂
∂
x
∵
erf √
dη
√
=
∂x
4α 1 t
π ∂x
0
η
2 ∂η d
−η 2 dη
√
=
e
π ∂x dη
0
2
2
1
−x
= √ √
exp
π 4α 1 t
4α 1 t
∴ b can be obtained from Equation 4.33h:
√
δ = b t
θ 1 can be obtained from Equation 4.33f, θ 2 can be obtained from Equation 4.33g
∂θ 1
q 1 = k 1
can be determined
∂x 0
Special case: slow freezing, as sketched in Figure 4.14, approximately linear.
⇒ θ ∞ = 0:
√
θ 1
erf (x/ 4α 1 t)
= 1 −
√
(4.34)
θ s
erf (b/ 4α 1 )
b can be obtained
√
δ = b t
88
Analytical Heat Transfer