Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 4.32.
Suppose we want to use the refrigeration cycle discussed in an earlier post to design an air conditioner using the coolant fluid HFC-134a as before. The cycle looks like this:
The COP for this cycle is
where we’ve used the fact that the enthalpy is constant along the throttling edge so .
Using Schroeder’s Tables 4.3 and 4.4, a reasonable temperature range for an air conditioner would be a high temperature of with corresponding pressure . This temperature is hotter than any outdoor temperature (at least in populated areas), so heat will be able to flow from the air conditioner into the outdoor environment. A typical indoor temperature is around so we need the cold temperature of the air conditioner to be less than this so it can absorb heat from the room. Using is probably a bit too much, but it’s the best we can get from Table 4.3. This corresponds to a low pressure of .
The fluid is 100% liquid at point 3, 100% saturated gas at point 1, and superheated gas at point 2. Point 4 is a mixture of saturated liquid and gas. From the tables we have
To find , we use the fact that the path is adiabatic so entropy is constant. From the table, so we need to find an entry in Table 4.4 with the same entropy at a pressure of 12 bars. This value is slightly less than for but since no value is given for any lower temperature and the values are almost the same we can use this value for the superheated temperature. The corresponding enthalpy is . This gives a COP of
A Carnot refrigerator operating between and has a COP of