Actual VCR cycle with compressor efficiency

Refrigerating effect

Only the heat absorbed in the evaporator is useful cooling. The suction-line heat gain comes from outside the cold space, so:

$$q_L = h_{evap\ exit} - h_4 = 239.16 - 95.47 = 143.69\ \text{kJ/kg}$$

(a) Compressor work per kg

$$\begin{aligned} w_s &= h_{2s} - h_1 = 278.6 - 242.2 = 36.4\ \text{kJ/kg}\\ w_a &= \frac{w_s}{\eta_s} = \frac{36.4}{0.75} = 48.53\ \text{kJ/kg} \end{aligned}$$

Actual discharge enthalpy: $h_2 = 242.2 + 48.53 = 290.7$ kJ/kg.

(b) Mass flow rate

$$\dot m = \frac{10}{143.69} = 0.0696\ \text{kg/s}$$

(c) Power

$$\begin{aligned} P_{comp} &= \dot m\,w_a = 0.0696 \times 48.53 = 3.38\ \text{kW}\\ P_{elec} &= \frac{P_{comp}}{\eta_{motor}} = \frac{3.38}{0.9} = 3.75\ \text{kW} \end{aligned}$$

(d) Actual COP

Based on shaft power: $\text{COP} = 10/3.38 = 2.96$. Based on electrical input: $\text{COP} = 10/3.75 = 2.66$.

(For comparison, the ideal COP with no losses is $143.69/36.23 = 3.97$.)

(e) Condenser heat rejection

$$Q_H = \dot m (h_2 - h_3) = 0.0696 \times (290.7 - 95.47) = 13.59\ \text{kW}$$

(Energy check: $10 + 0.21$ (suction-line gain) $+ 3.38 = 13.59$ kW.)

Answer: (a) 48.5 kJ/kg; (b) 0.0696 kg/s; (c) 3.38 kW at the shaft, 3.75 kW electrical; (d) 2.96 (shaft) or 2.66 (electrical); (e) 13.6 kW.