Question

In a vapour compression refrigeration cycle, the refrigerant enters the compressor in saturated vapour state at evaporator pressure with enthalpy 250 kJ/kg. The exit of the compressor is at 300 kJ/kg. COP of the cycle is 3. The refrigerant after condensation is throttled to evaporator pressure. If the enthalpy of saturated liquid at evaporator pressure is 50 kJ/kg, the dryness fraction at entry to evaporator is ________________.

• A. 0.2
• B. 0.25
• C. 0.3
• D. 0.35

Hint:

In throttling, enthalpy remains constant. Use $h = h_f + xh_{fg}$ to find the dryness fraction at evaporator inlet.

Answer 1

The Correct Option is B

For a vapour compression cycle, \[ COP = \frac{\text{Refrigeration Effect}}{\text{Compressor Work}}. \]
Step 1: Compute compressor work. \[ W = h_2 - h_1 = 300 - 250 = 50\ \text{kJ/kg}. \]
Step 2: Use COP = 3. \[ COP = \frac{RE}{W} = 3 \Rightarrow RE = 3W = 150\ \text{kJ/kg}. \]
Step 3: Refrigeration effect definition. \[ RE = h_1 - h_4 = 150. \] Given \(h_1 = 250\): \[ 250 - h_4 = 150 \Rightarrow h_4 = 100. \]
Step 4: Throttled state is a mixture. \[ h_4 = h_f + x\,h_{fg} = 50 + x(250 - 50) = 50 + 200x. \] Set equal to 100: \[ 50 + 200x = 100 \Rightarrow 200x = 50 \Rightarrow x = 0.25. \]
Thus dryness fraction = 0.25.
Final Answer: 0.25