Isotherm migration method
215
isotherm, in
order
to
find
the
curvatures. The angles 8
i
.
O
=
'Tr/2
on
the
y-axis
and
8
i
.M,
say, =
'Tr/4
on
the diagonal are already known. Thus
on
the
axis, x
2
+ (y -
Yo)2
=
r2,
and we find froro (0,
Yi,O),
(Xj,1>
Yi,l)
that
r = [(Yi,o-
Yi,l)2+
xT.l]/2(Yi,o-
Yi,l)'
On
the
diagonal
the
use of
(x
-
af+
(y-a)2=
r2
yields
r
=
{(Xj.M
-
Xj.M_l)2
+
(Yi.M
-
Yi,M_l)2}/2(Xj.M
+
Yi,M
-
Xj,M-l
-
Yi.M-l),
and
one
or
other
of these expressions for r replaces (S.117).
Crank
and
Crowley (1979) described a corresponding implicit method
for tracking isotherms along orthogonal flow lines, using a linearized form
of equation (S.113).
In
general,
the
implicit scheme offered
the
usual
advantage over
the
explicit method in enabling the use of longer time
steps. However, they choose
to
illustrate their implicit scheme by consid-
ering
the
problem of Bonnerot
and
Jamet (1977) specified in equations
(4.24-8). This problem is a more exacting test of Crank
and
Crowley's
method for two important reasons. Firstly, the radius of curvature is
positive
at
the
left-hand
end
of each isotherm, and negative as the
isotherm approaches
x = 1 (Fig. 4.4(a»). Therefore, proceeding in
the
direction of x increasing along any isotherm the curvature increases from
finite positive values, through infinity,
and
then from negative infinity
to
finite negative values,
the
central
part
of each isotherm being virtually
straight.
In such a situation, care is necessary
to
ensure that
the
finite-
difference equations are written correctly when
rl,m
<0
and
"'.t,m
= -ri,m'
The
second difficulty is associated with
the
initial temperature distribu-
tion.
On
the
left boundary, x = 0,
iJr/iJt
in (S.113) can
be
identified with
iJy/iJt
since
on
x = 0
the
isotherms are concave downwards
and
the
centres
of curvature are
on
x =
O.
When
the
necessary derivatives are obtained
from
the
second of
the
conditions (4.27) and substituted into equation
(S.113) we find that
on
x=O,
iJy/iJt=-'Tr2(1-u)~0,
O<u<1.
However,
from
the
first of (4.28)
on
the
phase-change boundary, u
=0,
we have
iJy/iJt=1!3A
on
x=O.
Thus,
at
t=O, all the isotherms except
u=O
initially move in
the
direction of Y decreasing
at
the
left boundary x = 0,
whereas
the
phase-change isotherm, u = 0, moves upwards. In fact, there
is a discontinuity in
iJy/iJt
at
t = 0
on
x = 0, and it is
not
surprising that
the
finite-difference solution of
Crank
and Crowley (1979) showed a loss of
accuracy there. Accuracy can
be
improved by adding an extra isotherm
with the temperature
~
Bu.
On
the right boundary, x = 1, all isotherms move upwards initially, since
iJy/iJt
=
'Tr
2
(1-
u), 0 <
u.,;;;
1, x = 1,
and
for u = 0,
iJy/iJt
=
1/A,
x =
1.
These
statements are physically consistent with
the
initial temperature distribu-
tion in (4.27).
The
negative temperature gradient from left
to
right along
any line of constant y produces a corresponding sideways heat
flow
which
causes
the
isotherms to move downward
on
x = 0 and upward
on
x =
1.