Question 12.24: A vapour compression heat pump is driven by a power cycle ha...
A vapour compression heat pump is driven by a power cycle having a thermal efficiency of 25%. For the heat pump, refrigerant-12 is compressed from saturated vapour at 2.0 bar to the condenser pressure of 12 bar. The isentropic efficiency of the compressor is 80%. Saturated liquid enters the expansion valve at 12 bar. For the power cycle 80% of the heat rejected by it is transferred to the heated space which has a total heating requirement of 500 kJ/min. Determine the power input to the heat pump compressor. The following data for refrigerant-12 may be used :
| Pressure, bar | Temperature, °C |
Enthalpy, kJ/kg | Entropy, kJ/kg K | ||
| Liquid | Vapour | Liquid | Vapour | ||
| 2.0 12.0 |
– 12.53 49.31 |
24.57 84.21 |
182.07 206.24 |
0.0992 0.3015 |
0.7035 0.6799 |
Vapour specific heat at constant pressure = 0.7 kJ/kg K. (Anna University)
Heat rejected by the cycle = \frac{500}{0.8} = 625 kJ/min.
Assuming isentropic compression of refrigerant, we have
Entropy of dry saturated vapour at 2 bar
= Entropy of superheated vapour at 12 bar
0.7035 = 0.6799 + c_{p} \ln \frac{ T}{(49.31 + 273)} = 0.6799 + 0.7 × \ln (\frac{ T}{322.31})
or \ln (\frac{ T}{322.31}) = \frac{0.7035 – 0.6799 }{0.7} = 0.03371
or T = 322.31 (e)^{0.03371} = 333.4 K
∴ Enthalpy of superheated vapour at 12 bar
= 206.24 + 0.7(333.4 – 322.31) = 214 kJ/kg
Heat rejected per cycle = 214 – 84.21 = 129.88 kJ/kg
Mass flow rate of refrigerant = \frac{625}{129.88} = 4.812 kg/min
Work done on compressor = 4.812 (214 – 182.07)
= 153.65 kJ/min = 2.56 kW
Actual work of compresson = \frac{2.56}{η_{compressor}} = \frac{2.56}{0.8} = 3.2 kW
Hence power input to the heat pump compressor = 3.2 kW.