## An Elementary Consideration of Heat Losses from Streams of Water in Caves

Our next article is an example of
the scientific approach to caving problems. Members of the B.E.C. are paying increasing attention to this side of
caving, and this article is both appropriate and topical

by Mike Luckwill.

Conditions of heat exchange in a large cave are extremely
complex and will not be unravelled until a large number of temperature
measurements, together with water flow rates and other variables, have been
obtained. The construction of an
extremely simple model enables hypotheses to be tested and may point the
direction in which results are required. I therefore make no apologies for the crudity of the model I shall
investigate. I will, however, attempt to
examine the consequences of the over simplification involved.

First of all, let us look at some facts. The temperature differences concerned are so
small that heat losses due to radiation are negligible and so we need only
consider heat losses due to conduction. Water conducts heat about twice as well as rock and about twenty times
as well as air. We shall assume that the
water loses heat only to the rock and that the rock is capable of absorbing
this heat without increasing in temperature. This assumption is quite reasonable when one considers the vast
quantities of rock in relation to the small amount of heat. Furthermore, we
shall assume that the temperature of the rock remains constant at 8°C.

We shall assume that the greater mass of water has a
temperature t°C and that this
temperature only starts to drop within 1 cm of the water/rock surface, giving a
temperature gradient of (t-8) °C
per cm.

The temperature drop of the water
is then given by T where

°C/sec. where K =
Coefficient of conductivity of water = 0.0015.

A = Water/Rock surface area and V
= volume of water.

Let us consider a semicircular
channel, radius r cm and length l cm.

The A = prl and V = pr^{2}1/2
hence A/V = 2/r.

T therefore = 0.0015(t-8)2/r^{O}C/second
and, if r = 40cms, T = 0.000075(t-8)^{O}C/second.

The correct method of finding the manner in which the
temperature varies with time would be to use the calculus. However, in the interests of simplicity, I
will make an approximation by assuming that the water stays at the same
temperature for ten minutes, and calculate the temperature drop every ten
minutes. Thus,

T = 0.000075(t-8).600^{O}C/10
minutes or 0.045(t-8)^{O}C/10min.

Graph 1 expresses the above result

We now construct a graph showing the change of temperature
of the water with time by the following method. A time O, we shall assume that the water
temperature is 12^{O}C. From
graph 1, we find that the temperature drop at 12^{O}C is 0.18^{O}C. Subtracting 0.18 from 12 we get 10.86, and we
plot this as the temperature for time 10 minutes. We now find the temperature drop for 10.86
and calculate the temperature at 20 minutes, and so on. By this method, we obtain graph 2

Let us now consider water percolating through rock. Typical dimensions for a suitable model would
be a cylinder of water radius 1cm. Once
again, A = 2prl
and V = pr^{2}l,
giving

A/V = 1/r = 1 in this case.

Hence T= 0.0015(t-8)^{O}C/second
= 0.9(t-8)^{O}C/10 minutes.

Thus, at 12^{O}C, a drop of 3.6^{O}C might
occur in ten minutes. Using the same
methods as we did to produce graph 2, we can now produce graph 3, which has
been plotted on the same graph above, again showing the change in temperature
with time.

What conclusions are to be drawn from these results? Firstly, a stream arising from percolating
water will quickly reach the rock temperature. Thus, the temperature of the rock may be measured by measuring the
temperature of a suitable water inlet in the cave e.g. the drinking fountain in
St. Cuthberts. Secondly, the
temperature of such a stream is likely to be 1^{O}C lower than the
temperature of the Main Stream in the cave, as the main stream would take
several hours to reach rock temperature.

To construct a simple model to find air temperature is more difficult. I suggest that deep in a cave, the air
reaches the rock temperature of approximately 8^{O}C. Accepted temperatures of about 11^{O}C
are probably wrong because of the difficulty of measurement. As soon as one approaches a thermometer near
enough to read it, ones breath must quickly cause the thermometer to give a
false reading.

## Book Review

### DOOLIN ST. CATHERINES CAVE. by
Dr. O.C. Lloyd, published by the U.B.S.S. at ten shillings.

This is yet another very worth while publication by the
U.B.S.S. on one of the caves of

The bulk of the text is devoted to a detailed description of
the cave system and an explanation of its origin. This is preceded by a history of the
exploration of the system in which is included a song inspired by the

The cave consists of nearly five miles of passages and any
person proposing to explore the system would be well advised to make reference
to this publication.

Tony Meadon