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# Re: a thermodynamics question

• From: Leigh Palmer <palmer@sfu.ca>
• Date: Thu, 23 Oct 1997 11:49:44 -0700

Here is a new (maybe trivial) question for all you thermodynamics buffs.
Our building just got the heat turned on. Unfortunately, due to recent
construction, none of the thermostats are working, so it's getting very
hot. My colleague in the next office can't open his window, so he is
cooling the room by running his air conditioner. Now the textbook air
conditioners remove "Qc" from indoors and exhaust "Qh" outdoors. Trouble
is the temperatures are Th = 50 and Tc = 80. What happens in this case?
Does the air conditioner generate electricity rather than consume it?

I don't believe Martha's question got answered. I'm letting questions
compost these days for several reasons. There are often multiple answers
to the questions and I would have nothing to contribute. There are often
multiple postings of questions (I saw a triple today, posted to three
lists). There are fewer than twenty-four hours in my waking day.

sentence is rhetorical, but the answer to it, of course, is "No". The
problem here arises from interpreting the term Th as the ambient
outdoor temperature; that is not what is meant. Th is the temperature
of the heat exchanger which is dissipating Qh convectively in the
outside air. To do so the air conditioner must always exhaust heat to
the atmosphere at a temperature higher than the outside ambient, and
always at a temperature higher than the inside ambient as well, since
Tc must be *below* the inside ambient temperature. The formulae which
employ qair temperatures as though they are the reservoir temperatures
do not take into account the considerable temperature differences that
must be maintained at the heat exchangers for them to operate at
sufficient capacity to perform their function. The formulae imply only
ideal limits to efficiency, coefficient of performance, etc. when the
ambient temperatures are used.

Leigh