Air conditioner in real life

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

\displaystyle \mbox{COP}=\frac{H_{1}-H_{3}}{H_{2}-H_{1}} \ \ \ \ \ (1)

where we’ve used the fact that the enthalpy is constant along the throttling edge {3\rightarrow4} so {H_{4}=H_{3}}.

Using Schroeder’s Tables 4.3 and 4.4, a reasonable temperature range for an air conditioner would be a high temperature of {T_{h}=46.3^{\circ}\mbox{ C}} with corresponding pressure {P_{3}=12\mbox{ bars}}. 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 {20^{\circ}\mbox{ C}} so we need the cold temperature of the air conditioner to be less than this so it can absorb heat from the room. Using {T_{c}=8.9^{\circ}\mbox{ C}} 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 {P_{4}=4\mbox{ bars}}.

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

\displaystyle H_{4} \displaystyle = \displaystyle H_{3}=116\mbox{ kJ}\ \ \ \ \ (2)
\displaystyle H_{1} \displaystyle = \displaystyle 252\mbox{ kJ} \ \ \ \ \ (3)

To find {H_{2}}, we use the fact that the path {1\rightarrow2} is adiabatic so entropy is constant. From the table, {S_{1}=0.915\mbox{ kJ K}^{-1}} 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 {S} for {T=50^{\circ}\mbox{ C}} 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 {H_{2}=276\mbox{ kJ}}. This gives a COP of

\displaystyle \mbox{COP}=5.67 \ \ \ \ \ (4)

A Carnot refrigerator operating between {T_{c}=8.9^{\circ}\mbox{ C}} and {T_{h}=46.3^{\circ}\mbox{ C}} has a COP of

\displaystyle \mbox{COP}=\frac{273+8.9}{46.3-8.9}=7.54 \ \ \ \ \ (5)

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