R-134a Refrigerator, Subcooled Condenser Exit: 04-BS-10, Undated paper, Q6

Question 6: R-134a Refrigerator, Subcooled Condenser Exit (Part B – 15 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.

Given. Compressor inlet (state 1): $P_1=0.14$ MPa, $T_1=-10\ ^\circ\text{C}$ (superheated), $\dot m=0.12$ kg/s. Compressor exit (state 2, actual): $P_2=0.7$ MPa, $T_2=50\ ^\circ\text{C}$. Condenser exit (state 3): $T_3=24\ ^\circ\text{C}$ at 0.65 MPa — note $T_{sat}(0.65\ \text{MPa})\approx24.2\ ^\circ\text{C}$, so this is a very slightly subcooled liquid, essentially at saturation. Throttle exit (state 4): $P_4=0.15$ MPa, $h_4=h_3$.

State	Description	P	T	h (kJ/kg)
1	Compressor inlet	0.14 MPa	−10.0°C	394.50
2	Compressor exit (actual)	0.7 MPa	50.0°C	436.67
2s	Compressor exit (isentropic)	0.7 MPa	—	429.31
3	Condenser exit (subcooled)	0.65 MPa	24.0°C	233.12
4	Evaporator inlet	0.15 MPa	—	233.12

Find. (a) $\dot Q_L$, $\dot W_c$; (b) $\eta_c$; (c) COP.

Approach

All four states are directly fixed from the given (P,T) pairs; the only unknown is the isentropic compressor exit enthalpy $h_{2s}$, obtained at the same exit pressure and the inlet entropy $s_1$.

Refrigeration capacity and compressor power — part (a). The evaporator absorbs heat from $h_4$ to $h_1$; the compressor is the only work input: $$\dot Q_L=\dot m(h_1-h_4)=0.12\times(394.50-233.12)=\boxed{19.37\ \text{kW}}.$$ $$\dot W_c=\dot m(h_2-h_1)=0.12\times(436.67-394.50)=\boxed{5.061\ \text{kW}}.$$
Isentropic efficiency — part (b). $$\eta_c=\frac{h_{2s}-h_1}{h_2-h_1}=\frac{429.31-394.50}{436.67-394.50}=\frac{34.80}{42.17}=\boxed{0.8253\ (82.5\%)}.$$
COP — part (c). $$COP=\frac{\dot Q_L}{\dot W_c}=\frac{19.37}{5.061}=\boxed{3.827}.$$

Quantity	Value
(a) $\dot Q_L$	19.37 kW
(a) $\dot W_c$	5.061 kW
(b) $\eta_c$	82.53%
(c) COP	3.827

Topic. Actual (non-ideal) vapor-compression refrigeration cycle with a subcooled condenser exit.

Key relations. $\eta_c=(h_{2s}-h_1)/(h_2-h_1)$; $\dot Q_L=\dot m(h_1-h_4)$; $COP=\dot Q_L/\dot W_c$; throttle: $h_4=h_3$.

Why this works. Because the compressor exit state is given directly (0.7 MPa, 50°C) rather than assumed isentropic, its ACTUAL enthalpy rise fixes both the real compressor work and (via the isentropic-efficiency definition) lets the efficiency be back-calculated instead of assumed — this is the reverse of the usual "given $\eta_c$, find $h_2$" problem shape.

Common pitfall. Assuming $T_{sat}$ at 0.65 MPa for the condenser exit instead of using the given 24°C directly — the problem specifies a subcooled liquid state, about 2.7°C below saturation, and the correct enthalpy must come from the compressed-liquid (or subcooled) region, not the saturation table.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., §11-3 (actual vapor-compression refrigeration cycle).

Question 6 of 9: R-134a Refrigerator, Subcooled Condenser Exit

Question 6: R-134a Refrigerator, Subcooled Condenser Exit (Part B – 15 marks)

Approach