Ideal Reverse-Brayton (Gas) Refrigeration Cycle: 04-BS-10, May 2017, Q6

Question 6: Ideal Reverse-Brayton (Gas) Refrigeration Cycle (15 marks)

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Given. Cold-air-standard reverse-Brayton (gas-refrigeration) cycle, $c_p=1.005$ kJ/kg·K, $k=1.4$. Compressor: $P_1=100$ kPa, $T_1=260$ K $\to P_2=300$ kPa (isentropic, $\eta_c=100\%$). Turbine: $P_3=300$ kPa, $T_3=300$ K $\to P_4=100$ kPa (isentropic, $\eta_t=100\%$).

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

Fig. Q6 — T–s diagram for the ideal reverse-Brayton refrigeration cycle (1→2 compressor, 2→3 high-pressure cooler, 3→4 turbine/expander, 4→1 cold-space heat absorption; constant-$c_p$ entropy shown relative to an arbitrary reference, isentropic legs vertical by construction).

Approach

With both isentropic efficiencies at 100%, the compressor and turbine states follow directly from the isentropic $T$-$P$ relation with $r_p=3$ on each leg. Net work is the compressor input minus the turbine output; refrigeration load is the heat absorbed warming the cold air from the turbine exit ($T_4$) back up to the compressor inlet ($T_1$).

Compressor (1→2, isentropic, $r_p=3$). $$T_2=T_1\,r_p^{(k-1)/k}=260\times3^{0.2857}=355.87\ \text{K}\ (82.72\ ^\circ\text{C}).$$ $$w_c=c_p(T_2-T_1)=1.005\times(355.87-260)=96.35\ \text{kJ/kg}.$$
Turbine/expander (3→4, isentropic, $r_p=3$). $$T_4=\frac{T_3}{r_p^{(k-1)/k}}=\frac{300}{3^{0.2857}}=219.18\ \text{K}\ (-53.97\ ^\circ\text{C}).$$ $$w_t=c_p(T_3-T_4)=1.005\times(300-219.18)=81.22\ \text{kJ/kg}.$$
Net work (part a). $$w_{net}=w_c-w_t=96.35-81.22=\boxed{15.13\ \text{kJ/kg}}\ \text{(net work INPUT to the cycle)}.$$
Refrigeration load and COP (part b). The cold air returns to the compressor inlet state by absorbing heat from the refrigerated space between states 4 and 1: $$q_L=c_p(T_1-T_4)=1.005\times(260-219.18)=41.02\ \text{kJ/kg}.$$ $$\text{COP}=\frac{q_L}{w_{net}}=\frac{41.024}{15.127}=\boxed{2.712}.$$

Quantity	Result
(a) $w_{net}$	15.13 kJ/kg
(b) COP	2.712

Concept & Theory

Topic. Ideal (reverse) Brayton gas-refrigeration cycle, the air-cycle counterpart of vapor-compression refrigeration — used where a phase-change refrigerant is impractical (e.g. aircraft cabin cooling, cryogenics).

Key relations. Isentropic ideal-gas: $T_2/T_1=r_p^{(k-1)/k}$. Net work $=w_c-w_t$.

Why this works. Both legs share the same pressure ratio, so the work each does scales with its own temperature span: $T_2-T_1=95.87$ K for the compressor versus $T_3-T_4=80.82$ K for the turbine. Because the compressor's inlet (260 K) is colder than the turbine's inlet (300 K), the compressor span is larger even though $T_1w_t$ and net work is a genuine INPUT to the cycle — always compute both legs explicitly rather than assuming which dominates from the inlet temperatures alone. Unlike vapor-compression refrigeration, the reverse-Brayton cycle also uses a real turbine/expander (recovering shaft work) instead of an irreversible throttle, which is why $w_t$ appears explicitly in the net-work balance rather than being discarded.

Common pitfall. Confusing $q_L=c_p(T_1-T_4)$ (heat absorbed by the circulating air between the turbine exit and the compressor inlet) with $c_p(T_3-T_2)$ or any other state pair — the refrigeration effect is specifically the leg that closes the cycle back to the compressor inlet.

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

Question 6 of 9: Ideal Reverse-Brayton (Gas) Refrigeration Cycle

Question 6: Ideal Reverse-Brayton (Gas) Refrigeration Cycle (15 marks)

Approach