Question

A refrigerator operates on an ideal vapor compression cycle between the pressure limits of 140 kPa and 800 kPa. The working fluid is refrigerant R-134a. The refrigerant enters the compressor as saturated vapor at 140 kPa and exits at 800 kPa. It leaves the condenser as a saturated liquid at 800 kPa. The coefficient of performance (COP) of the refrigerator is $____________$ (rounded off to two decimal places).
Given property data for R-134a:
At 140 kPa: $h_f = 27.06 \, kJ/kg, \, h_g = 239.19 \, kJ/kg$
At 800 kPa: $h_f = 95.48 \, kJ/kg, \, h_g = 267.34 \, kJ/kg$
At 800 kPa and 60$^\circ$C: $h = 296.82 \, kJ/kg$

Hint:

For vapor compression cycles, always identify enthalpies at 4 key states (saturated vapor, superheated after compression, saturated liquid, and throttling). COP is just refrigerating effect divided by compressor work.

Answer 1

Correct Answer: 2.45

Step 1: State points enthalpies. - State 1 (compressor inlet): Saturated vapor at 140 kPa. \[ h_1 = h_g @ 140 \, kPa = 239.19 \, kJ/kg \] - State 2 (compressor outlet, isentropic): At 800 kPa. Given superheated $h = 296.82 \, kJ/kg$. \[ h_2 = 296.82 \, kJ/kg \] - State 3 (after condenser, saturated liquid at 800 kPa): \[ h_3 = h_f @ 800 \, kPa = 95.48 \, kJ/kg \] - State 4 (throttling to 140 kPa, isenthalpic): \[ h_4 = h_3 = 95.48 \, kJ/kg \]
Step 2: Work and heat interactions. - Compressor work: \[ W_{comp} = h_2 - h_1 = 296.82 - 239.19 = 57.63 \, kJ/kg \] - Refrigerating effect: \[ Q_{in} = h_1 - h_4 = 239.19 - 95.48 = 143.71 \, kJ/kg \]
Step 3: COP of refrigerator. \[ COP = \frac{Q_{in}}{W_{comp}} = \frac{143.71}{57.63} \approx 2.92 \] Final Answer: \[ \boxed{2.92} \] % Quicktip