Window Air Conditioner Using Freon-12: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2013, Q4

Question 4: Window Air Conditioner Using Freon-12 (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. Room air 20 °C, $hA = 585\times0.5$; evaporator (state 1, saturated vapour) 15 °C; condenser (state 3, saturated liquid) 50 °C; compressor efficiency 80 %. From the Freon-12 appendix:

State	Condition	h (kJ/kg)	s (kJ/kg·K)
1 — evaporator exit	sat. vapour, 15 °C (0.4914 MPa)	193.64	0.6897
2s — isentropic comp.	1.22 MPa	209.45	0.6897
3 — condenser exit	sat. liquid, 50 °C (1.2193 MPa)	84.87	—
4 — after throttle	0.4914 MPa (wet)	84.87	—

Find. (a) refrigerant mass flow; (b) compressor power at 80 % efficiency; (c) cooling COP.

Figure 4 — P–h diagram of the vapour-compression cycle: 1→2 compression from 0.49 to 1.22 MPa, 2→3 condensation to saturated liquid at 50 °C, 3→4 throttle, 4→1 evaporation at 15 °C. The refrigeration effect is h₁ − h₄.

Check

The printed question labels the evaporator exit "saturated liquid" and the condenser exit "saturated vapour"; these phase words are evidently transposed. Physically the evaporator delivers saturated vapour at the low temperature (15 °C, below the 20 °C room) and the condenser delivers saturated liquid at the high temperature (50 °C, above the 40 °C outdoor air). The solution uses the physically consistent reading; the stated temperatures are used as printed.

Approach. The evaporator duty is set by the room-side convection; dividing by the refrigeration effect gives the flow, and the isentropic compression (corrected for efficiency) gives the power and COP.

Evaporator duty and refrigerant flow (a). Heat drawn from the room is $\dot Q_L=hA(T_\text{room}-T_\text{evap})=585(0.5)(20-15)=1462.5$ W. With refrigeration effect $q_L=h_1-h_4=193.64-84.87=108.78$ kJ/kg,$$\dot m=\frac{\dot Q_L}{q_L}=\frac{1462.5}{108{,}780}=\boxed{0.01345\ \text{kg/s}}\;(48.4\ \text{kg/hr}).$$
Compressor power (b). Isentropic work to 1.22 MPa is $w_s=h_{2s}-h_1=209.45-193.64=15.81$ kJ/kg; at 80 % efficiency $w=w_s/0.80=19.76$ kJ/kg, so$$\dot W=\dot m\,w=0.01345(19.76)=\boxed{265.7\ \text{W}}.$$
Coefficient of performance (c). For cooling,$$\text{COP}=\frac{q_L}{w}=\frac{108.78}{19.76}=\boxed{5.50}.$$

Quantity	Result
Evaporator duty	1462.5 W
(a) Refrigerant flow	0.01345 kg/s (48.4 kg/hr)
(b) Compressor power	265.7 W
(c) COP (cooling)	5.50

Concept & Theory

Topic. Ideal vapour-compression refrigeration cycle with compressor efficiency.

Key relations. Refrigeration effect $q_L=h_1-h_4$ (with $h_4=h_3$ across the throttle); compressor work $w=(h_{2s}-h_1)/\eta_c$; $\text{COP}_R=q_L/w$; evaporator heat load $\dot Q_L=hA\,\Delta T$.

Why this works. The throttle is isenthalpic, so the condenser-liquid enthalpy carries straight to the evaporator inlet; the refrigeration effect is then simply the evaporator enthalpy rise. The room convection independently fixes the cooling load, so the flow follows immediately, and the compressor state 2s is found by following the constant-entropy line from state 1 to the condenser pressure.

Common pitfall. Reading the transposed phase labels literally, or dividing the isentropic work by the flow before correcting for the 80 % efficiency.

Source. Çengel & Boles, Thermodynamics §11 (vapour-compression refrigeration cycles).

Question 4 of 8: Window Air Conditioner Using Freon-12

Question 4: Window Air Conditioner Using Freon-12 (equal value)