EXAMPLE 1
One of the fastest ways to solve simple cycle problems like this one is to use a pressure-enthalpy chart. The problem can�t be solved unless there are at least two properties known at a given point so start wherever these two point are known. Using the assumptions given on the vapor compression discussion page, we can start at point one at the inlet to the compressor. This point is assumed to be a saturated, dry vapor. Also, the problem statement gives you the lower operating temperature of -15 which we can use for point 1. These two points are enough to start your solution.
The problem statement asks for the coefficient of performance (COP). Per the equation below your after Qin and Qout.
Per the appropriate chart for R134a, we can locate point 1 by noting the temperature and the fact the refrigerant is a dry vapor:
At this point from the chart we can determine that h1 is roughly 100 btu/lb. For an ideal refrigeration cycle, S1 is assumed to be equal to S2. Also, the upper operating temperature for the cycle is given as 100F which means that both points 2 and 3 are operating at this temperature. Therefore, following a constant entropy line from point 1 to the temperature line of 100F give the second operating point:
From the chart, the enthalpy at point 2 is roughly 119 btu/lb.
Since the condenser operation from point 2 to 3 is at constant temperature, you can follow the temperature line of 100F at point 2 to the saturated liquid point at 3:
From the chart, h3 is approximately 45 btu/lb. Now that the enthalpy at points 2 and 3 are known, Qout can be calculated:
The enthalpy from point 3 to 4 across the throttle valve is equal. Therefore h4 = h3 = 45 btu/lb. Qin is simply the difference between the enthalpy going into and out of the evaporator:
Therefore the COP is:
