Ideal-Gas Refrigeration Cycle: 04-BS-10, December 2019, Q4

Question 4: Ideal-Gas Refrigeration Cycle (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. $T_1=300$ K, $P_1=100$ kPa, $r_p=3.75$, turbine inlet $T_3=350$ K. No "variable specific heats" instruction is given for this question (unlike Q1/Q2), so cold-air-standard properties apply: $c_p=1.005$ kJ/kg·K, $k=1.4$. Both compressor and turbine ideal (isentropic); no regenerator.

Find. (a) $w_{net,in}$ [kJ/kg]; (b) $q_L$ [kJ/kg]; (c) COP.

Fig. Q4 — T–s, reversed (refrigeration) Brayton cycle: 1→2 isentropic compression, 2→3 constant-$P$ heat rejection, 3→4 isentropic expansion (turbine), 4→1 constant-$P$ refrigeration load.

Approach

Both legs are isentropic, so the cold-air-standard power law $T_2/T_1=r_p^{(k-1)/k}$ applies directly at both the compressor and the turbine (using the SAME pressure ratio, since $P_2/P_1= P_3/P_4=r_p$). Compressor work, turbine work, and the refrigeration load are then simple $c_p\Delta T$ terms.

Compressor exit temperature and work. $$T_2=T_1\,r_p^{(k-1)/k}=300\times3.75^{0.2857}=437.65\ \text{K}$$ $$w_c=c_p(T_2-T_1)=1.005(437.65-300)=138.34\ \text{kJ/kg}.$$
Turbine exit temperature and work. $$T_4=\frac{T_3}{r_p^{(k-1)/k}}=\frac{350}{1.4589}=239.92\ \text{K}$$ $$w_t=c_p(T_3-T_4)=1.005(350-239.92)=110.63\ \text{kJ/kg}.$$
Net work input (part a). $$w_{net,in}=w_c-w_t=138.34-110.63=\boxed{27.71\ \text{kJ/kg}}.$$
Refrigeration capacity (part b). Heat absorbed in the cold space, process 4→1: $$q_L=c_p(T_1-T_4)=1.005(300-239.92)=\boxed{60.38\ \text{kJ/kg}}.$$
COP (part c). $$\text{COP}=\frac{q_L}{w_{net,in}}=\frac{60.38}{27.71}=\boxed{2.179}.$$

Quantity	Result
(a) $w_{net,in}$	27.71 kJ/kg
(b) $q_L$	60.38 kJ/kg
(c) COP	2.179

Concept & Theory

Topic. Ideal-gas (reversed Brayton / Bell-Coleman) refrigeration cycle, cold-air-standard.

Key relations. $T_2/T_1=T_3/T_4=r_p^{(k-1)/k}$ (same pressure ratio spans both the compressor and the turbine); $w_{net,in}=w_c-w_t=q_H-q_L$ by cycle energy balance.

Why this works. Because the cycle is a steady-flow, open-system arrangement (no piston-cylinder boundary work terms to track separately), each device's work is simply $c_p\Delta T$ and the whole-cycle energy balance $w_{net,in}=q_H-q_L$ holds automatically — no extra constant-pressure boundary-work term is hidden anywhere, unlike a closed piston-cylinder cycle.

Common pitfall. Using different pressure ratios for the compressor and turbine leg — both spans use the SAME $r_p=3.75$, since $P_2=P_3$ and $P_1=P_4$ in this simple (non-regenerative) arrangement.

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

Question 4 of 9: Ideal-Gas Refrigeration Cycle

Question 4: Ideal-Gas Refrigeration Cycle (15 marks)

Approach