Freon-12 Vapour-Compression Refrigeration: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2015, Q4

Question 4: Freon-12 Vapour-Compression Refrigeration (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. Freon-12 cycle: evaporator (saturation) temperature 15 °C; condenser pressure 1.4 MPa (saturation ≈ 56 °C); condenser liquid subcooled to 48 °C; $\dot m=0.008\ \text{kg/s}$; compressor isentropic efficiency $\eta_c=0.80$. Enthalpies from the exam Freon-12 tables (saturated) with a superheat estimate at the discharge:

State	Condition	$h$ (kJ/kg)
1 — compressor inlet	sat. vapour, 15 °C	193.6
2s — isentropic discharge	1.4 MPa, superheated	212.1
3 — condenser exit	subcooled liquid, 48 °C	82.8
4 — evaporator inlet	throttled, $h_4=h_3$	82.8

Find. (a) compressor power, (b) refrigeration capacity, (c) COP.

Figure 3 — T–s schematic of the Freon-12 cycle: 1→2 compression (80 % efficient), 2→3 condensation and subcooling to 48 °C, 3→4 throttling at constant enthalpy, 4→1 evaporation at 15 °C absorbing the refrigeration effect.

Approach. Find the isentropic discharge enthalpy at the condenser pressure, apply the 80 % efficiency for the real work, take the throttle as isenthalpic to fix the evaporator inlet, then form the capacity and COP from the evaporator enthalpy rise.

Compressor work. The ideal (isentropic) lift is $h_{2s}-h_1=212.1-193.6=18.5\ \text{kJ/kg}$; with $\eta_c=0.80$ the real specific work is$$w_\text{in}=\frac{h_{2s}-h_1}{\eta_c}=\frac{18.5}{0.80}=\boxed{23.1\ \text{kJ/kg}}.$$
(a) Compressor power. $$\dot W=\dot m\,w_\text{in}=0.008(23.1)=\boxed{0.185\ \text{kW}\approx185\ \text{W}}.$$
Refrigeration effect. Throttling holds $h_4=h_3=82.8\ \text{kJ/kg}$, so the evaporator absorbs$$q_L=h_1-h_4=193.6-82.8=110.9\ \text{kJ/kg}.$$
(b) Refrigerator capacity. $$\dot Q_L=\dot m\,q_L=0.008(110.9)=\boxed{0.887\ \text{kW}\ (\approx0.25\ \text{ton})}.$$
(c) Coefficient of performance. $$\text{COP}=\frac{q_L}{w_\text{in}}=\frac{110.9}{23.1}=\boxed{4.8}.$$

Check

The discharge state 2s is only mildly superheated (≈ 62 °C at 1.4 MPa), so $h_{2s}$ is estimated from the saturated-vapour enthalpy plus a small $c_p\,\Delta T_\text{sup}$ term using a reconstructed superheated Freon-12 table. A ±1 kJ/kg uncertainty in $h_{2s}$ moves the COP by about ±0.2 and the power by ±5 W; the capacity is unaffected. Subcooling to 48 °C (below the 56 °C saturation) raises the refrigeration effect over a saturated-liquid exit.

Quantity	Result
Compressor specific work	≈ 23.1 kJ/kg
Compressor power	≈ 185 W
Refrigeration capacity	≈ 0.887 kW
Coefficient of performance	≈ 4.8

Concept & Theory

Topic. Vapour-compression refrigeration cycle with an irreversible compressor and condenser subcooling.

Key relations. Compressor $w_\text{in}=(h_{2s}-h_1)/\eta_c$; throttle $h_4=h_3$; refrigeration effect $q_L=h_1-h_4$; cooling $\text{COP}=q_L/w_\text{in}$.

Why this works. Only the compressor consumes work, so the COP is the ratio of the useful evaporator cooling to that work. The isentropic efficiency inflates the real work above the ideal lift, while subcooling the condensate pushes the throttle inlet to a lower enthalpy, widening the evaporator enthalpy rise and improving the effect.

Common pitfall. Multiplying (rather than dividing) the ideal work by the efficiency, or taking the condenser exit as saturated liquid when the question specifies 48 °C subcooling.

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

Question 4 of 8: Freon-12 Vapour-Compression Refrigeration

Question 4: Freon-12 Vapour-Compression Refrigeration (equal value)