Heat pumps

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 4.14.

A heat pump is essentially a backwards air conditioner, in that it extracts an amount of heat {Q_{c}} from the cold outside air using an amount {W} of electrical energy and dumps an amount {Q_{h}=Q_{c}+W} of heat into an inside room in order to heat it. We can analyze its performance using the same logic as for a refrigerator. The coefficient of performance in this case is the amount of heat dumped into the room divided by the energy (work) needed to achieve this, so

\displaystyle \mbox{COP}=\frac{Q_{h}}{W} \ \ \ \ \ (1)

From conservation of energy

\displaystyle \mbox{COP}=\frac{Q_{h}}{Q_{h}-Q_{c}}=\frac{1}{1-Q_{c}/Q_{h}} \ \ \ \ \ (2)

Thus the COP is always greater than 1.

In an ideal refrigerator (e.g. one working on a reversed Carnot cycle) the entropy gained in absorbing {Q_{c}} is equal to the entropy lost in expelling {Q_{h}}, so

\displaystyle \frac{Q_{c}}{T_{c}} \displaystyle = \displaystyle \frac{Q_{h}}{T_{h}}\ \ \ \ \ (3)
\displaystyle Q_{c} \displaystyle = \displaystyle \frac{T_{c}}{T_{h}}Q_{h} \ \ \ \ \ (4)

The upper limit on the COP is therefore

\displaystyle \mbox{COP}_{max}=\frac{1}{1-T_{c}/T_{h}}=\frac{T_{h}}{T_{h}-T_{c}} \ \ \ \ \ (5)

A heat pump is more efficient than a purely electric heater (that is, a heater that generates heat by electrical resistance rather than by transferring heat from outside to inside) since in an electric heater, the work {W} is equal to {Q_{h}}, while in a heat pump, this work can be used to acquire an additional amount {Q_{c}} of heat which is expelled into the room along with the electrical work {W}. In other words, with a heat pump you have {Q_{h}=W+Q_{c}} while with an electric heater, you have only {Q_{h}=W}.

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