13.5 Applications of thermodynamics: heat pumps and refrigerators  (Page 3/7)

The best COP _hp Of a heat pump for home use

A heat pump used to warm a home must employ a cycle that produces a working fluid at temperatures greater than typical indoor temperature so that heat transfer to the inside can take place. Similarly, it must produce a working fluid at temperatures that are colder than the outdoor temperature so that heat transfer occurs from outside. Its hot and cold reservoir temperatures therefore cannot be too close, placing a limit on its ${\text{COP}}_{\text{hp}}$ . (See [link] .) What is the best coefficient of performance possible for such a heat pump, if it has a hot reservoir temperature of $\text{45}\text{.}0\text{º}\text{C}$ and a cold reservoir temperature of $-\text{15}\text{.}0\text{º}\text{C}$ ?

Strategy

A Carnot engine reversed will give the best possible performance as a heat pump. As noted above, ${\text{COP}}_{\text{hp}}=1/\text{Eff}$ , so that we need to first calculate the Carnot efficiency to solve this problem.

Solution

Carnot efficiency in terms of absolute temperature is given by :

The temperatures in kelvins are $T_{\text{h}}=\text{318 K}$ and $T_{\text{c}}=\text{258 K}$ , so that

Thus, from the discussion above,

or

so that

Discussion

This result means that the heat transfer by the heat pump is 5.30 times as much as the work put into it. It would cost 5.30 times as much for the same heat transfer by an electric room heater as it does for that produced by this heat pump. This is not a violation of conservation of energy. Cold ambient air provides 4.3 J per 1 J of work from the electrical outlet.