Freon-12 Refrigeration Cycle (Coefficient of Performance): 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2014, Q4

Question 4: Freon-12 Refrigeration Cycle (Coefficient of Performance) (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A real (irreversible, with line pressure drops) vapour-compression cycle. The compressor runs 1→2 and rejects $q_\text{out}=4\ \text{kJ/kg}$; the condenser exit / throttle inlet is state 5; throttling holds enthalpy so $h_7=h_6=h_5$; the evaporator absorbs heat 7→8. Enthalpies from the supplied Freon-12 tables:

State	Condition	$h\ (\text{kJ/kg})$
1 — compressor inlet	125 kPa, −10 °C (superheated)	185.2
2 — compressor exit	1.20 MPa, 100 °C (superheated)	245.5
5 — throttle inlet	≈ sat. liquid, 40 °C	74.5
7 — evaporator inlet	140 kPa, $h_7=h_5$	74.5
8 — evaporator exit	100 kPa, −20 °C (superheated)	179.9

Find. The cycle coefficient of performance, $\text{COP}=q_L/w_\text{in}$.

Figure 3 — T–s schematic of the Freon-12 cycle: 1→2 compression (with 4 kJ/kg heat loss), 2→5 condensation/de-superheating to saturated liquid at 40 °C, 5→7 throttling (constant enthalpy), 7→8 evaporation absorbing the refrigeration effect. Irreversibilities and line pressure drops make the actual path non-ideal.

Check

State 5 (1.15 MPa, 40 °C) is a subcooled liquid; its enthalpy is taken as $h_f$ at 40 °C = 74.5 kJ/kg (compressed-liquid enthalpy ≈ saturated-liquid value at the same temperature). The evaporator exit (state 8) at 100 kPa, −20 °C is superheated vapour ($h=179.9$ kJ/kg). The stated point-by-point pressure drops confirm this is a real, not idealised, cycle.

Approach. Write the first law over the compressor (heat loss increases the work input), take the refrigeration effect as the evaporator enthalpy rise with $h_7=h_5$ from the throttle, then form the COP.

Compressor work with heat loss. Energy balance $h_1+w_\text{in}=h_2+q_\text{out}$ gives$$w_\text{in}=(h_2-h_1)+q_\text{out}=(245.5-185.2)+4.0=\boxed{64.4\ \text{kJ/kg}}.$$
Refrigeration effect (evaporator). Throttling keeps $h_7=h_6=h_5=74.5$ kJ/kg, so$$q_L=h_8-h_7=179.9-74.5=\boxed{105.3\ \text{kJ/kg}}.$$
Coefficient of performance. $$\text{COP}=\frac{q_L}{w_\text{in}}=\frac{105.3}{64.4}=\boxed{1.64}.$$

Quantity	Result
Compressor work input	64.4 kJ/kg
Refrigeration effect $q_L$	105.3 kJ/kg
Coefficient of performance	≈ 1.64

Concept & Theory

Topic. Non-ideal vapour-compression refrigeration cycle with a non-adiabatic compressor.

Key relations. Compressor first law $w_\text{in}=(h_2-h_1)+q_\text{out}$; throttle $h_\text{out}=h_\text{in}$; refrigeration effect $q_L=h_\text{evap,out}-h_\text{evap,in}$; cooling COP $=q_L/w_\text{in}$.

Why this works. Only the compressor consumes work, so the COP is the ratio of the useful cooling delivered in the evaporator to that work. Because the compressor loses 4 kJ/kg to its surroundings, the shaft must supply more than the enthalpy rise alone — heat rejected during compression penalises the COP. Throttling is isenthalpic, so the evaporator inlet enthalpy equals the condenser-exit enthalpy.

Common pitfall. Subtracting the compressor heat loss from the work (it adds to the required work), or using the isentropic discharge enthalpy instead of the actual measured state 2.

Source. Çengel & Boles, Thermodynamics §11 (refrigeration cycles); Freon-12 property tables (exam appendix).

Question 4 of 8: Freon-12 Refrigeration Cycle (Coefficient of Performance)

Question 4: Freon-12 Refrigeration Cycle (Coefficient of Performance) (equal value)