Ideal vapour compression cycle for 5 TR

Cycle states

• State 1: dry saturated vapour, $h_1 = 239.16$ kJ/kg.
• State 2: after isentropic compression, $h_2 = 275.39$ kJ/kg.
• State 3: saturated liquid, $h_3 = 95.47$ kJ/kg.
• State 4: after throttling, $h_4 = h_3 = 95.47$ kJ/kg.

Capacity: $Q_L = 5 \times 3.517 = 17.585$ kW.

(a) Refrigerating effect

$$q_L = h_1 - h_4 = 239.16 - 95.47 = 143.69\ \text{kJ/kg}$$

(b) Mass flow rate

$$\dot m = \frac{Q_L}{q_L} = \frac{17.585}{143.69} = 0.1224\ \text{kg/s}$$

(c) Compressor power

$$w = h_2 - h_1 = 275.39 - 239.16 = 36.23\ \text{kJ/kg}$$ $$P = \dot m\,w = 0.1224 \times 36.23 = 4.43\ \text{kW}$$

(d) COP

$$\text{COP} = \frac{q_L}{w} = \frac{143.69}{36.23} = 3.97$$

(e) Heat rejected in condenser

$$Q_H = \dot m (h_2 - h_3) = 0.1224 \times (275.39 - 95.47) = 22.02\ \text{kW}$$

Check: $Q_L + P = 17.585 + 4.43 = 22.02$ kW, equal to $Q_H$.

(f) Volume flow rate at compressor inlet

$$\dot V_1 = \dot m\,v_1 = 0.1224 \times 0.1402 = 0.0172\ \text{m}^3/\text{s}$$

Answer: (a) 143.7 kJ/kg; (b) 0.1224 kg/s; (c) 4.43 kW; (d) 3.97; (e) 22.0 kW; (f) 0.0172 m³/s.