Freon-12 Window Air Conditioner — Flow Rate, Power, COP: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, May 2013, Q4

Question 4: Freon-12 Window Air Conditioner — Flow Rate, Power, COP (Part A, 20 marks)

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.

Check: the printed question has "saturated liquid" and "saturated vapour" the wrong way round (an evaporator in a vapour-compression cycle cannot discharge liquid, nor a condenser vapour). The physically consistent reading — sat. VAPOUR leaves the evaporator at 15°C (below the 20°C room, so heat flows in) and sat. LIQUID leaves the condenser at 50°C (above the 40°C outside air, so heat flows out) — is used below, per Note 1. All refrigerant properties are taken from the paper's own Freon-12 tables (Appendices 1 and 3, reference state $h=s=0$ for saturated liquid at $-40^{\circ}\text{C}$).

Given. $h=585$ W/m²·K, $A=0.5$ m² at the evaporator; $T_{room}=20^{\circ}\text{C}$; evaporator (sat. vapour exit) $T_{evap}=15^{\circ}\text{C}$; condenser (sat. liquid exit) $T_{cond}=50^{\circ}\text{C}$; compressor isentropic efficiency $\eta_c=80\%$.

Find. (a) refrigerant mass flow rate, (b) compressor power, (c) COP.

Figure — T–s diagram of the vapour-compression cycle: 1→2 compression (dashed 2s = isentropic, solid 2 = actual, $\eta_c=80\%$), 2→3 condensation at 50°C, 3→4 throttling (constant $h$), 4→1 evaporation at 15°C.

Approach. The evaporator's convective heat-transfer data is the only route to the refrigeration load $\dot Q_{evap}$ (it is not given directly); once $\dot Q_{evap}$ is known, the R-12 saturation-table enthalpies at the two given temperatures give the flow rate, and an isentropic compression from state 1 to the condenser pressure gives the compressor work.

Evaporator load, from the coil's own convective duty. The evaporator coil runs essentially isothermal (phase change at $T_{evap}$), so simple convection applies directly: $$\dot Q_{evap}=hA(T_{room}-T_{evap})=585\times0.5\times(20-15)=\boxed{1462.5\ \text{W}}.$$
State properties (R-12 saturation tables). Sat. vapour at $15^{\circ}\text{C}$: $h_1=h_g=193.644$ kJ/kg, $s_1=s_g=0.6897$ kJ/kg·K, $p_1=0.4914$ MPa (Appendix 1). Sat. liquid at $50^{\circ}\text{C}$: $h_3=h_f=84.868$ kJ/kg, $p_3=1.2193$ MPa. Throttling is isenthalpic, so $h_4=h_3=84.868$ kJ/kg.
Refrigerant mass flow rate (a). $$\dot m=\frac{\dot Q_{evap}}{h_1-h_4}=\frac{1.4625}{193.644-84.868}=\frac{1.4625}{108.776}=\boxed{0.01345\ \text{kg/s}} \ (48.4\ \text{kg/hr}).$$
Isentropic and actual compressor work. Isentropic compression from state 1 to $p_3$ ($s_{2s}=s_1=0.6897$). Interpolating the superheated table (Appendix 3) at this entropy gives $h=209.45$ kJ/kg at 1.20 MPa (between the 50 and 60°C rows) and $h=212.17$ kJ/kg at 1.40 MPa (between 60 and 70°C), so at 1.2193 MPa $h_{2s}=209.71$ kJ/kg ($T_{2s}\approx54.1^{\circ}\text{C}$) and $w_{s}=h_{2s}-h_1=16.07$ kJ/kg. Applying the 80% isentropic efficiency, $$w_{actual}=\frac{w_s}{\eta_c}=\frac{16.07}{0.80}=\boxed{20.09\ \text{kJ/kg}}$$ (actual discharge $h_2=h_1+w_{actual}=213.73$ kJ/kg, $T_2\approx59.1^{\circ}\text{C}$ superheated).
Compressor power (b) and COP (c). $$\dot W=\dot m\,w_{actual}=0.01345\times20.09=\boxed{0.270\ \text{kW}},\qquad \text{COP}=\frac{\dot Q_{evap}}{\dot W}=\frac{q_L}{w_{actual}}=\frac{108.776}{20.09}=\boxed{5.41}.$$

Final results — Question 4
Quantity	Value
(a) Refrigerant mass flow rate	$0.01345$ kg/s ($48.4$ kg/hr)
(b) Compressor power	$0.270$ kW
(c) COP	$5.41$

Topic: Ideal vapour-compression refrigeration cycle sized from a convective evaporator duty rather than a stated cooling load.

Key relations: $\dot Q=hA\Delta T$ (isothermal coil convection); $\dot m=\dot Q_{evap}/(h_1-h_4)$; $\eta_c=w_s/w_{actual}$ for a compressor; $\text{COP}=\dot Q_{evap}/\dot W_{compressor}$.

Why this works: a saturated-vapour-in/saturated-liquid-out cycle boundary condition fixes BOTH state points' enthalpies from tables alone (no quality calculation needed at 1 or 3); the only "sizing" datum in the whole problem is the evaporator's own $hA\Delta T$, which is why it is given even though it plays no role in the state-point property lookups.

Common pitfall: using $T_{room}-T_{cond}$ or $T_{outside}-T_{evap}$ (mixing the two coils) in the convection duty; or applying $\eta_c$ multiplicatively the wrong way ($w_{actual}=\eta_c\,w_s$ instead of $w_s/\eta_c$), which understates the real compressor power.

Source: Çengel & Boles, Thermodynamics: An Engineering Approach, Ch. 11 (vapour-compression refrigeration cycle, compressor isentropic efficiency).

Question 4 of 8: Freon-12 Window Air Conditioner — Flow Rate, Power, COP

Question 4: Freon-12 Window Air Conditioner — Flow Rate, Power, COP (Part A, 20 marks)