Geology Reference
In-Depth Information
T
m
T
z
Figure 5.5. Sketch of the actual temperature profile of a cooling body.
loss is only half that for the thinner layer, because the temperature gradient is only
half, so that would double the time. However, there is also twice as much heat to
be removed, because a larger volume of material has to cool. That also contributes
a factor of 2. Thus there is twice as much heat to be lost, but it is lost at only half
the rate, so it takes four times as long.
The time
t
s
in Eq. (5.13) is an upper bound, but it still gives us an understanding
of how the cooling time
scales
, i.e. how the time changes if the depth is changed,
or if the material properties are changed. This is because the actual cooling time is
a constant fraction of the upper bound
t
s
. This makes some sense if you note that
the logic for the layer of thickness 2
d
is the same as for the thinner layer, so you
might expect everything to scale in the same way as
d
is changed. Thus
t
s
is not the
actual time, but it is a useful
timescale
for this problem.
This expectation is borne out by a rigorous mathematical analysis, which yields
an actual cooling time of
d
2
/
4
κ
[1]. As expected, the actual time is less than our
upper bound, but it scales in the same way, so the actual time is always half of
our upper bound. The temperature profile that results from a rigorous analysis is
sketched in Figure 5.5. Mathematically, the curve is an
error function
, which I will
spare you the details of. However, you will notice that the temperature approaches
the initial temperature asymptotically with depth. This makes it harder to measure
the thickness of the cooled layer. So as to be specific, the depth
d
is defined as the
depth at which the temperature reaches 84% of the asymptotic temperature. (The
84% makes simple sense in the context of the error function.)
The result we have obtained is quite powerful. Let us write it in two different
forms:
t
∼
d
2
/κ,
(5.15a)
(
κt
)
1
/
2
.
d
∼
(5.15b)
∼
The symbol '
' means 'is of the order of', or 'is roughly equal to'. Thus if
we know a length scale in a thermal diffusion problem, we can simply estimate a
corresponding timescale, as we have done above. You can as easily apply Eq. (5.15a)
to a cooling magmatic sill or dyke, to a cooling pluton, or to the cooling crust after
a thermo-tectonic event, for example.
