Neumann's solution; generalizations; volume changes
103
be
written down from (3.7), (3,8), and (3.9). They are
U
l
X
Ul
= U
1
-
erf(a/2kt) erf 2(k
1
t)!'
(3.12)
U
2
x
~
= - U
2
+ erfc(a/2k!) erfc
2(k2t)~
.
(3.13)
It
is
now possible to see what initial conditions
are
satisfied by (3.10) and
(3.13).
At
t = 0 they give s = 0 and
~
= - U
2
,
i.e. the whole region
x>
0
is
solid
at
uniform
temperatur~
- U
2
•
The
special case in which the solid
is
initially at its melting temperature, so that U
2
= 0, is
the
single-phase
problem introduced in §1.2.1 and (3.11) reduces to
Ae
A2
erf A = U1cl/(L-rr!), (3.14)
where A =
a/(2kl).
Carslaw and Jaeger (1959, p. 287) show a graph of the
left-hand side of (3.14)
as
a function of A from which values of A can be
read for any given value
of
UlCl/(L~).
They also point
out
that for
small values of
A,
and hence of the right-hand side of (3.14), use of the
first
term
in the series expansion for erf A gives approximately
(3.15)
The
solutions (3.12), (3.13) and
the
equation (3.11) are quoted by
Carslaw and Jaeger (1959, p. 288). Their solution of
the
corresponding
solidification of a liquid in
x > 0 initially
at
a uniform temperature above
freezing
is
deducible by suitable changes of nomenclature.
Carslaw
and
Jaeger (1959, Chapter XI) also present Neumann-type
solutions for
other
physically important problems including the region
x>
0 initially liquid and x < 0 solid, the case in which melting occurs over
a temperature range, and three-phase problems. They solve the commonly
neglected problem in which motion of the liquid results from a change of
volume
on
solidification, due
to
a difference between the densities of solid
and liquid. As an example they consider the freezing of a semi-infinite
liquid in the region
x>
0 when the density of the solid,
PI'
exceeds that of
the liquid,
P2,
and both phases are incompressible. Thus
the
problem
is
defined by equation (3.1) in
the
solid phase but in the liquid phase (3.2)
is
replaced by
x>s(t),
(see equations (1.32) and (1.33».
On
the solidification boundary, x = s(t),
they take
U1
=
~
=
UM,
together with (3.3).
For
their solidification prob-
lem, instead
of
(3.4), (3.5), they have
U1
= 0, X = 0,
U2
~
U
as
x
~
00.
The