Why does the COP rise from 3.97 to 4.19?

Subcooling raises the refrigerating effect more than superheat raises the work

Practice · 6 marks

VCR cycle with subcooling and superheating

Short answer

Refrigerating effect = 155.0 kJ/kg; mass flow rate = 0.0681 kg/s; compressor power = 2.52 kW; COP = 4.19. Compared to saturated cycle (COP 3.97), subcooling and superheating improved COP by 5.6%.

A vapour compression refrigerator of 3 TR capacity uses R-134a. The vapour leaves the evaporator slightly superheated with $h_1 = 243.5$ kJ/kg and, after isentropic compression, has $h_2 = 280.5$ kJ/kg. The liquid leaves the condenser subcooled with $h_3 = 88.5$ kJ/kg. Calculate (a) the refrigerating effect, (b) the mass flow rate, (c) the compressor power, and (d) the COP. Compare the COP with that of the saturated cycle $h_1 = 239.16$ , $h_2 = 275.39$ , $h_3 = 95.47$ kJ/kg (COP = 3.97) and comment. Take 1 TR = 3.517 kW.

Capacity: $Q_L = 3 \times 3.517 = 10.551$ kW. Throttling gives $h_4 = h_3 = 88.5$ kJ/kg.

(a) Refrigerating effect

q_L = h_1 - h_4 = 243.5 - 88.5 = 155.0\ \text{kJ/kg}

(b) Mass flow rate

\dot m = \frac{10.551}{155.0} = 0.0681\ \text{kg/s}

(c) Compressor power

w = h_2 - h_1 = 280.5 - 243.5 = 37.0\ \text{kJ/kg}

P = 0.0681 \times 37.0 = 2.52\ \text{kW}

(d) COP

\text{COP} = \frac{155.0}{37.0} = 4.19

Comment

The COP rises from 3.97 to 4.19, about 5.6 %. The refrigerating effect increased from 143.7 to 155.0 kJ/kg: subcooling lowered $h_3$ by 6.97 kJ/kg, and the superheat (taken as useful, inside the evaporator) raised $h_1$ by 4.34 kJ/kg. Meanwhile the compressor work rose only slightly, from 36.2 to 37.0 kJ/kg.

Answer: (a) 155 kJ/kg; (b) 0.0681 kg/s; (c) 2.52 kW; (d) 4.19.

✓ Answer checked by Claude Opus 5.5 (high effort)

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