Question 15.2: Compute the COP and refrigerating efficiency for a theoretic......
Compute the COP and refrigerating efficiency for a theoretical single-stage cycle operating with a condenser pressure of 1253 kPa and an evaporator pressure of 201 kPa using R-134a as the refrigerant. The work for the cycle is 31 kJ per kilogram of refrigerant.
From Fig. 15-2, the COP may be expressed as
COP = \frac{i_{3} – i_{1}}{i_{4} – i_{3}} = \frac{i_{3} – i_{1}}{w_{34}}
And from Table A-2b
i_{3} = 392.8 kJ/kg, i_{1} = 268.5 kJ/kg
Then
COP = \frac{392.8 – 268.5}{31} = 4.0
A Carnot cycle operating between the saturation temperatures corresponding to the given condenser and evaporator pressures has a COP given by Eq. 15-5, where from Table A-2b
Carnot cycle COP_{c} = \frac{T_{c}}{T_{c} – T_{e}} (15-5)
T_{e} = −10 + 273 = 263 K
T_{c} = 48 + 273 = 321 K
we have
COP_{c} = \frac{263}{321 – 263} = 4.5
The refrigerating efficiency is then
\eta_{R} = \frac{COP}{COP_{c}} = \frac{4.0}{4.5} = 0.89 = 89 percent
