R-134a Refrigerator — Non-Ideal Compressor with a Suction-Line Pressure Drop: 04-BS-10, December 2014, Q9

Question 9: R-134a Refrigerator — Non-Ideal Compressor with a Suction-Line Pressure Drop (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): $0.14$ MPa, $-10\ ^\circ$C, $\dot m=0.12$ kg/s. Compressor exit (state 2): $0.7$ MPa, $50\ ^\circ$C. Condenser exit (state 3): $24\ ^\circ$C, $0.65$ MPa (subcooled liquid). Throttled to $0.15$ MPa (state 4, evaporator inlet) — slightly above the 0.14 MPa compressor-inlet pressure, reflecting a small suction-line pressure drop and superheat gain between the evaporator exit and the compressor inlet, exactly as stated in the source.

Find. (a) $\dot Q_L$ and $\dot W_c$ [kW]; (b) $\eta_c$; (c) COP.

Fig. Q9 — P–h state points for the R-134a refrigerator cycle (1→2 compressor, 2→3 condenser, 3→4 throttle valve, 4→1 evaporator).

Approach

Evaluate all four named states directly from the given $(P,T)$ or $(P,x)$ pairs, using the compressor-inlet entropy to find the isentropic exit enthalpy at the same exit pressure as the actual state. The evaporator and compressor energy balances then give the two rates directly, and their ratio (plus the isentropic-vs-actual compressor work ratio) gives COP and $\eta_c$.

Compressor states. Inlet (0.14 MPa, $-10\ ^\circ$C): $h_1=394.50$ kJ/kg, $s_1=1.7680$ kJ/kg·K. Actual exit (0.7 MPa, 50°C): $h_2=436.67$ kJ/kg. Isentropic exit (0.7 MPa, $s=s_1$): $h_{2s}=429.31$ kJ/kg.
Condenser exit / throttle. Subcooled liquid at 0.65 MPa, 24°C: $h_3=233.12$ kJ/kg; throttled isenthalpically to 0.15 MPa, $h_4=h_3=233.12$ kJ/kg.
Heat removal rate and compressor power (part a). $$\dot Q_L=\dot m(h_1-h_4)=0.12\times(394.50-233.12)=\boxed{19.366\ \text{kW}}.$$ $$\dot W_c=\dot m(h_2-h_1)=0.12\times(436.67-394.50)=\boxed{5.061\ \text{kW}}.$$
Isentropic efficiency (part b). $$\eta_c=\frac{h_{2s}-h_1}{h_2-h_1}=\frac{429.31-394.50}{436.67-394.50}=\boxed{0.8253\ (82.5\%)}.$$
COP of the refrigerator (part c). $$\text{COP}=\frac{\dot Q_L}{\dot W_c}=\frac{19.366}{5.061}=\boxed{3.827}.$$

Quantity	Result
(a) $\dot Q_L$	19.366 kW
(a) $\dot W_c$	5.061 kW
(b) $\eta_c$	0.8253 (82.5%)
(c) COP	3.827

Concept & Theory

Topic. Actual (non-ideal) vapor-compression refrigeration cycle with all four states given directly by measurement, rather than derived from an assumed isentropic efficiency.

Key relations. Evaporator energy balance $\dot Q_L=\dot m(h_1-h_4)$; compressor energy balance $\dot W_c=\dot m(h_2-h_1)$; isentropic efficiency $\eta_c=(h_{2s}-h_1)/(h_2-h_1)$; $\text{COP}=\dot Q_L/\dot W_c$.

Why this works. Every state here is pinned down by a direct, independent measurement (inlet/exit temperature and pressure), so no assumed efficiency or approximation is needed anywhere except to back out $\eta_c$ itself as a derived check — the isentropic state at the same exit pressure is only a reference point for that one calculation, not an input to $\dot Q_L$ or $\dot W_c$.

Common pitfall. Assuming the throttle-valve exit pressure equals the compressor- inlet pressure exactly — the source data show a small but real difference (0.15 MPa vs. 0.14 MPa), reflecting a suction-line pressure drop between the evaporator and the compressor, so $h_4$ must be evaluated at the throttle's own exit pressure (0.15 MPa, matching $h_3$ from the isenthalpic throttle) even though the compressor inlet is quoted at 0.14 MPa.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 11 (Refrigeration Cycles).

Question 9 of 9: R-134a Refrigerator — Non-Ideal Compressor with a Suction-Line Pressure Drop

Question 9: R-134a Refrigerator — Non-Ideal Compressor with a Suction-Line Pressure Drop (15 marks)

Approach