Yes, the temperature will rise until the rate of heat leaving the
container equals the power (rate of energy) entering the box. As the
interior of the box gets hotter (more exactly, as the inside-outside
temperature differential increases), heat exiting the box will increase.
So at some internal temperature an equilibrium will be reached. You'd
want to use a power level high enough to make this temperature well
above ambient so that small variations in ambient temperature won't
badly skew the results, but low enough that you don't cook the box's
innards.

The method you propose seems workable, except I didn't see any mention
of heat leaving the box via thermal conduction along the wires.
Depending on the box's insulating property and the wires, this could be
a significant contributor to the total heat loss from the box. Therefore
it's very important to either insure that this loss is negligible
compared to the loss through the styrofoam, or else to manage it (most
easily by using exactly the same wires and wire orientation during
calibration and test).

As I mentioned earlier, it's probably within the means of an amateur to
make useful measurements, but it would take a lot of care and attention
to detail. There may well be other factors none of us have considered
(as examples, the presence of variable drafts in the vicinity of the
box, which could impact the heat transfer rate in a major way, or
nonlinearity of convective cooling), so measurements of various bodies
of known dissipation would have to be measured to confirm that all
significant factors are accounted for before there will be any
confidence in the results.

Roy Lewallen, W7EL

Paul Keinanen wrote:

On Tue, 17 Aug 2004 15:50:05 GMT, Richard Clark
wrote:


However, as to the statement above, and the presumptions that follow,
there is no equilibrium to observe as the test implementation has been
described. You are pouring calories into an insulated environment
which can only raise temperature without end (short of destruction, of
course).

As I described elsewhere, this is called a caloric bomb, and as such
should be limited in time for all comparisons.


You seem to assume that no thermal power will flow through the
styrofoam walls of the box.

Your statement is true if you insert a huge amount of power into the
box, in which case the temperature would climb, until the styrofoam
would melt. However, the measurement can be done at much lower power
levels that are compatible with the thermal conductivity of the
styrofoam.

If styrofoam would be an ideal isolator, you could put some deep
freeze food into a styrofoam box, move to your summer cottage and hope
that the food would still be eatable after a week. Unfortunately this
is not true :-(.

When doing antenna efficiency measurements, the power levels should be
set to a level, in which the temperature increase is manageable (below
the melting point of the styrofoam).

To give an example of the power levels required, assume that
you insert a 1 kg (1 liter) bottle of drink at 0 C into the test
container and after 10000 seconds (about 3 hours) the drink
temperature is at +10 C, the energy needed is about 40 kJ, thus the
leakage through the container walls would be about 4 W. With the
outside temperature of 20 C, the average temperature difference would
be 15 C and the thermal conductivity about 4 C/W. This is what you
usually get with some power transistors with a heat sink :-).

If you build an igloo around the "miracle" antenna, you should be able
to measure the heat generated (and thus efficiency) with power levels
well below the legal limit in most countries and with bad antennas
10-100 W should be enough.

Of course, you should make sure that the styrofoam own RF dissipation
is sufficiently small at the test frequency.

Paul OH3LWR