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7.2 1D PROBLEMS FOR PURE SUBSTANCES
we must have 0 521 85326 5 May 25, 2005 11:14 217
√ √
X i ∝ t or X i = C t, (7.7)
where C can be determined from the interface condition (7.2). The transcendental
equation for determination of C thus becomes
ρλC T m − T w K s 2
= √ √ exp (−C /4α s )
2 erf(C/ 4α s ) πα s
T m − T sup K l 2
+ √ √ exp (−C /4α l ). (7.8)
erfc(C/ 4α l ) πα l
This transcendental equation shows that C = C (T m − T w , T m − T sup , K s , K l ,
α s ,α l ). Thus, C will be different for each initial and boundary condition and for each
specification of physical properties. The value of C must be iteratively determined
to calculate dX i (t)/dt from Equation 7.7 and hence to calculate the temperature as
a function of x and t from Equations 7.5 and 7.6. It can be shown that the system is
governed by a dimensionless number, called the Stefan number, which is defined as
C ps (T m − T w )
St = . (7.9)
λ
The larger the value of St, the faster is the interface movement. A further point to
note is that, although the temperature profiles show discontinuity at the interface,
they are smooth within each phase and the variation of T with t at any x is also
continuous and smooth.
7.2.2 Simple Numerical Solution
It might appear that it is a straightforward matter to discretise Equation 7.4 to ob-
tain a numerical solution. However, there is a difficulty associated with predicting
continuous temperature histories when a numerical solution is obtained. To ap-
preciate the difficulty, we assume uniform and equal properties for both phases
(i.e., ρ s = ρ l = ρ, C ps = C pl = C p , and K s = K l = K). Thus, Equation 7.4 can be
written as
2
∂ ∂ θ
= , (7.10)
∂τ ∂ X 2
where
h − h s
= (dimensionless enthalpy), (7.11)
λ
C p (T − T m )
θ = (dimensionless temperature), (7.12)
λ
α t
τ = (dimensionless time), (7.13)
L 2
x
X = (dimensionless length). (7.14)
L