Discussion
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 _________.
Answer: B
The dryness fraction at entry to the evaporator is 0.25.
Number the states: 1 evaporator exit (saturated vapour, \(h_{1} = 250\) kJ/kg), 2 compressor exit (\(h_{2} = 300\) kJ/kg), 3 condenser exit, 4 after throttling. Throttling keeps enthalpy constant, so \(h_{4} = h_{3}\).
\[ \begin{aligned} COP &= \frac{h_{1} - h_{4}}{h_{2} - h_{1}} = \frac{250 - h_{4}}{50} = 3 \\ h_{4} &= 250 - 150 = 100\ \text{kJ/kg} \end{aligned} \]
State 4 is a wet mixture at evaporator pressure, between saturated liquid (\(h_{f} = 50\)) and saturated vapour (\(h_{g} = 250\)):
\[ \begin{aligned} x_{4} &= \frac{h_{4} - h_{f}}{h_{g} - h_{f}} \\ &= \frac{100 - 50}{250 - 50} = 0.25 \end{aligned} \]
0.2 divides by 250 instead of by the latent enthalpy 200 (50/250). 0.3 and 0.35 do not follow from the given enthalpies; any error in the COP step changes \(h_{4}\) and every later figure.
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