Answer:
a) Ql = 7.18kW and Win = 1.81kW.
b) Qh = 8.99kW.
c) COPr = 3.96.
Explanation:
This is an ideal vapor-compression refrigeration cycle, and thus compressor is isotropic and the refrigerant leaves the condenser as a saturated liquid and enters the compressor as saturated vapor
From the refrigerant-134a tables, enthalpies are determined as:
P1 = 0.14MPa => h1 = 239.16kJ/kg
P2 = 0.8MPa => h2 = 275.39kJ/kg
P3 = 0.8MPa => h3 = 95.47kJ/kg
h4 ≅ h3 =95.47kJ/kg
a) Rate of heat removal from refrigerated space and the power input to compressor:
Ql = m(h1 - h4) = 0.05[239.16 - 95.47]
Ql = 7.18kW
Win = m(h2 - h1) = 0.05[275.39 - 239.16]
Win = 1.81kW
b) Rate of heat rejection from the refrigerant to the environment:
Qh = m(h2 - h3) = 0.05[275.39 - 95.47] = 8.99kW
And, It can also be determined from the formula:
Qh = Ql + Win = 7.18 + 1.81
Qh = 8.99kW
c) Coefficient of performance of the refrigerator is:
COPr = Ql / Win
COPr = 7.18 / 1.81
COPr = 3.96.
There is no any unit of the coefficient of the performance of the refrigerator.