Answer:
i) 0.5071 kg/s
ii) -1407.1 kJ/kg
iii) 204.05 kW
iv) 5.881
v) 9.238
Explanation:
Given Data:
evaporation temperature (T) = 4°C = 277.15 K
Condensation Temperature (T) = 34°C = 307.15 K
n (compressor efficiency) = 0.76
refrigeration rate = 1200 kJ/si) to find the circulation rate of the refrigerant
m = =
------- 1
Q = 1200 kJ/s
H2 = entropy at step 2 = 2508.9 (kJ/kg) (obtained from Table F)
H4 = entropy at step 4 = 142.4 (kJ/kg)
back to equation 1
m (circulation rate of refrigerant) = 0.5071 kg/s
ii) heat transfer rate in the condenser
Q = m (H4 - H3)
= 0.5071 (142.4 - 2911.27)
= -1407.1 kJ/kg
where H3 = H2 + ΔH23 = 2911.27 (kJ/kg) (as calculated)
iii) power requirement
w = m * ΔH23
= 0.5071 (kg/s) * 402.37 (kJ/kg) = 204.05 kW
where: ΔH23 = =
= 402.37 (kJ/kg)
iv) coefficient of performance of a cycle
W = Qc / w
= 1200 kJ/s / 204.05 kW
= 5.881
v) coefficient of performance of a Carnot refrigeration cycle
= 277.15 / (307.15 - 277.15)
= 9.238
