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GATE 2022 ME (ME2) – Question 45

Thermal Engineering Applications · Refrigeration and heat pump cycles · 2 marks · Multiple choice

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 _________.

  1. 0.2
  2. 0.25
  3. 0.3
  4. 0.35

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Correct answer: (B) 0.25

Explanation

**Cycle states.** 1 = compressor inlet (saturated vapour), 2 = compressor exit, 3 = condenser exit, 4 = evaporator inlet after throttling.

**Step 1: compressor work.**
$$w=h_2-h_1=300-250=50\text{ kJ/kg}.$$

**Step 2: refrigerating effect from the COP.**
$$COP=\frac{q_{evap}}{w}=3\;\Rightarrow\;q_{evap}=h_1-h_4=150\text{ kJ/kg}.$$

**Step 3: enthalpy after throttling.**
$$h_4=h_1-150=250-150=100\text{ kJ/kg}.$$

**Step 4: dryness fraction.** Throttling is isenthalpic, so $h_4$ is the enthalpy at the evaporator inlet. With $h_f=50$ and $h_g=250$ kJ/kg at the evaporator pressure:
$$x=\frac{h_4-h_f}{h_g-h_f}=\frac{100-50}{250-50}=\mathbf{0.25}.$$