The GATE Grind

GATE 2016 ME (ME3) – Question 57

Thermal Engineering Applications · Refrigeration and heat pump cycles · 2 marks · Numerical answer

In the vapour compression cycle shown in the figure, the evaporating and condensing temperatures are 260 K and 310 K, respectively. The compressor takes in liquid-vapour mixture (state 1) and isentropically compresses it to a dry saturated vapour condition (state 2). The specific heat of the liquid refrigerant is 4.8 kJ/kg-K and may be treated as constant. The enthalpy of evaporation for the refrigerant at 310 K is 1054 kJ/kg. The difference between the enthalpies at state points 1 and 0 (in kJ/kg) is ____________ State 0 is saturated liquid at 260 K.

Diagram for GATE 2016 ME (ME3) question 57

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: 1095 to 1130

Explanation

Isentropic compression gives s₁=s₂. Along the saturated-liquid approximation, s₃−s₀=c_p ln(310/260), and vaporisation at310K adds1054/310. Thus s₁−s₀=4.8ln(310/260)+1054/310. At260K, h₁−h₀=260(s₁−s₀)=**1103.5116kJ/kg**. This uses the stated constant liquid specific heat.