7 There's More Than One Kind of Carnot Cycle!

We've looked at the Carnot cycle from the point of view of a steam engine. Steam engines take many forms including jet engines, gas turbines, and the working parts of a nuclear energy plant.

For a steam engine what is important is the heat taken from the hot reservoir, $q_2$ and the work w produced by that heat. So we define the efficiency in terms of those two quantities.

But consider: What if we call the cold reservoir a "refrigerator" and the hot reservoir the "kitchen" ? Then we are interested in pumping heat from the cold reservoir to the hot one. That will take work, but it can be done by running a Carnot cycle in reverse. While only the signs of the various quantities change,²⁹the result is quite different. Here we care about the heat moved from the cold reservior $q_1$ and the work w needed to move it. Now we have:

$$\eta_ = \frac{q_1}{w} \,,$$ (7.1)

where $\eta$ is called the coefficient of performance and the larger it is, the better. The needed quantities are given by equations 3.3 and 3.11. Putting these into equation 7.1 and doing the same manipulations we did above, we get:

$$\eta_ = \frac{q_1}{w} = \frac{-nRT_1 \ln (V_d/V_c)} {-nRT(T_2 - T_1 ) \ln (V_b/V_a)} = \frac{T_1}{T_2 - T_1 } \,.$$ (7.2)

A typical refrigerator at 277 K ( $40^\circ F$) with a room at 295 K ( $72^\circ F$) has a maximum coefficient of performance of 15.4. Of course a real refrigerator will have a lower performance. One way of improving the performance would be to lower the room temperature (since there will then be a smaller temperature difference to work against) or raise the temperature of the refrigerator (which risks spoiling what is inside of it) or both.

Another increasingly important use for a (reverse) Carnot cycle is to use it as a heat pump. A heat pump heats a house, taking in heat from the outside (cold reservoir) and pumping it into the house (hot reservoir). The coefficient of performanc here is:

$$\eta_ = \frac{q_2}{w} = \frac{T_2}{T_2 - T_1} \,,$$ (7.3)

where the manipulations are the same as in equation 7.2. Taking the same numbers as above, but with the house being at 295 K and the outside being at 277 K, we get $\eta = 16.4$. Now this is an amazing figure. If we heated our house by electricity, we'd get one joule into the house for every joule consumed. But if we used an ideal heat pump, we'd get 16.4 joules into the house for every joule spent!³⁰What we are doing is using the energy to pump heat instead of simply converting it to heat. But of course the input electrical energy also turns up as heat in the process. Sadly, real heat pumps are not so efficient, but nevertheless all commercial ones offer an efficiency advantage over a simple electric heater.