Question

In a vapour compression refrigeration cycle, the refrigerant enters the compressor in saturated vapour state at evaporator pressure, with specific enthalpy equal to 250 kJ/kg. The exit of the compressor is superheated at condenser pressure with specific enthalpy equal to 300 kJ/kg. At the condenser exit, the refrigerant is throttled to the evaporator pressure. The coefficient of performance (COP) of the cycle is 3. If the specific enthalpy of the saturated liquid at evaporator pressure is 50 kJ/kg, then the dryness fraction of the refrigerant at entry to evaporator is _________.

Accepted Answer

The correct answer is (B) 0.25. COP = (refrigeration effect)/(work input) = (h1−h4)/(h2−h1). 3 = (250−h4)/(300−250) = (250−h4)/50, so h4=100 kJ/kg. Throttling is isenthalpic, so h4=h3 (state entering the evaporator). hfg at evaporator pressure = h1(sat. vapour)−hf(sat. liquid) = 250−50 = 200 kJ/kg. Dryness fraction x4 = (h4−hf)/hfg = (100−50)/200 = 0.25.

Explanation