From: xie@phoenix.Princeton.EDU (Yan-bo Xie)
Subject: Supplement to the ice melting rate problem
Originator: news@nimaster
Organization: Princeton University
Distribution: usa
Date: Thu, 20 Aug 1992 23:32:01 GMT

In a previous posting, the mechanism of the melting rate
problem is given for the ice in a room.  Note that it is
assumed that, however, the contact with other objects is not
important which may be not true for general situation.

For example, if the air is on the ground, the heat exchange
between the ground and ice should be important.  Again, the
melting rate is sensitive to the condition, depends on the
shape of the ground and materials of the ground.  But admitly
this dependence is much weaker than the air mechanism.  And
also the melting rate is much larger (unless a very strange
particular shape of the ground or certain arrangement between
the ice and ground).

In this situation, the calculation can be made more specifically.

There are several important quantities:

The specific heat of the ground, the thermoconductivity of the
ground, the contact area, the latent heat of ice and its volume
and the temperature of the ground far away from the ice (assume
say 60F).  Note that the contact area should contain the
information about how fast the heat is transfered if there is
temperature difference between the ice and the surface of the ground.

If the contact is very good, then the surface of the ground can
be assumed to have temperature 0C.

There is one important length scale (of the ground) in the problem.
If all the heat (needed) to melt the ice is Q, how much volume
of the ground material is needed to lower its temperature from 60F to 0C.
The answer would be V=Q/(60-32)/c. c is the specific heat per volume.
a=V**(1/3) would be the length scale in this problem.

It is important to point out that only the thermoconductivity is related
to time.  If the quantities listed above are the relevant ones,
then the melting rate is directly proportional to the thermoconductivity.

The exact melting rate even in this simple model still requires
a solution of the thermo-transport equation( because the temperature
is a function of position also) and the boundary condition.  However,
an estimation can be made without much calculation.

Suppose the initial ice is cubic with volume Vi.  Then we have two
possibility:

A: Vi>V, take the length to be l=V/(Vi)**(2/3)
B: Vi<V, take the length to be l=V**(1/3)

In case A, one can assume on the ground T=0C, below the ground l, T=60F
(after some melting say half of the ice has been melted).  The heat
flow would be

K*(60-32)*(Vi)**(2/3)/l

K is the thermoconductivity means: the heat flow rate (per second)
per unit area when the temperature gradient is per degree in one
unit length scale.  So in the above formula, (60-32)/l represents
the temperture gradient, (Vi)**(2/3) represents the surface 
contact area.  This number divided by the latent heat (per volume)
gives the rough melting rate (volume loss/per second).

In case B the above formula is still roughly valid with the
substitution of l.  The reason, in case B, V is very large,
and the temperature gradient becomes spheric.  A geometric factor
may be mutiplied (between 0.3 and 3), but this estimation
is already very crude.


Another limit case is that the contact between the ice and the ground
is not good.  In this case, one may assume the surface of the ground
having the temperature 60F.  But the melting rate would be directly
related to the contact condition which could be varied greatly.

If the contact is extremely inefficient, then the contact between
the air and ice becomes important.
Newsgroups: sci.physics