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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 vapor 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, barTemperature,°CEnthalpy, kJ/kgEntropy, kJ/kg K2.0– 12.53LiquidVapourLiquidVapour12.049.3124.5784.21182.07206.240.09920.3015 0.70350.6799Vapour specific heat at constant pressure = 0.7 kJ/kg K. |
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Answer» Heat rejected by the cycle = \(\cfrac{500}{0.8}\) = 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 + cp ln \(\cfrac{T}{(49.31+273)}\) = 0.6799 + 0.7 × ln \(\left(\cfrac{T}{322.31}\right)\) ln \(\left(\cfrac{T}{322.31}\right)\) = \(\cfrac{0.7035-0.6799}{0.7}\) = 0.03371 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 = \(\cfrac{625}{129.88}\) = 4.812 kg/min Work done on compressor = 4.812 (214 – 182.07) = 153.65 kJ/min = 2.56 kW Actual work of compresson = \(\cfrac{2.56}{η_{compressor}}\) = \(\cfrac{2.56}{0.8}\) = 3.2 kW Hence power input to the heat pump compressor = 3.2 kW. |
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