Gas Refrigeration Cycle (Reverse Brayton): 04-BS-10, December 2015, Q3

Question 3: Gas Refrigeration Cycle (Reverse Brayton) (20 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): $P_1=100$ kPa, $T_1=270$ K. Pressure ratio $=3$ ($P_2=300$ kPa). Turbine inlet (state 3): $T_3=310$ K at 300 kPa, expanding to 100 kPa. $\eta_c=0.85$, $\eta_t=0.88$. Variable specific heats.

Find. (a) $w_{net,in}$ [kJ/kg]; (b) $q_L$ [kJ/kg]; (c) COP; (d) $s_{gen}$ in compressor and turbine [kJ/kg·K].

Fig. Q3 — T–s state points for the gas (reverse-Brayton) refrigeration cycle (1→2 actual compressor, 2→3 constant-P heat rejection, 3→4 actual turbine/expander, 4→1 constant-P heat absorption — refrigeration effect). No saturation dome for an ideal-gas working fluid.

Approach

This is the reverse (refrigeration) counterpart of the Brayton cycle: the compressor raises pressure from the cold space, an ambient-temperature heat exchanger rejects heat at constant pressure down to the given turbine-inlet temperature, an expander (turbine) drops the pressure and temperature further than a throttle would, and the resulting cold air absorbs the refrigeration load before returning to the compressor inlet. Solve the compressor and turbine independently with the same variable-cp root-finding method as the power-cycle version, then combine for net work, cooling load, and COP.

Compressor (1→2), 100→300 kPa. Isentropic exit $T_{2s}=369.22$ K gives $w_{c,s}=99.97$ kJ/kg. Actual: $$w_{c,a}=\frac{99.97}{0.85}=\boxed{117.61\ \text{kJ/kg}},\qquad T_2=386.66\ \text{K}\ (113.51\ ^\circ\text{C}).$$
Turbine/expander (3→4), 300→100 kPa. Isentropic exit $T_{4s}=226.58$ K gives $w_{t,s}=83.91$ kJ/kg. Actual: $$w_{t,a}=0.88\times83.91=\boxed{73.84\ \text{kJ/kg}},\qquad T_4=236.59\ \text{K}\ (-36.56\ ^\circ\text{C}).$$
Net work input (part a). $$w_{net,in}=w_{c,a}-w_{t,a}=117.61-73.84=\boxed{43.78\ \text{kJ/kg}}.$$
Refrigeration effect (part b). Cold air warms from $T_4$ back to $T_1$ at constant pressure (100 kPa) in the refrigerated space: $$q_L=h(T_1)-h(T_4)=\boxed{33.60\ \text{kJ/kg}}.$$
Coefficient of performance (part c). $$\text{COP}=\frac{q_L}{w_{net,in}}=\frac{33.60}{43.78}=\boxed{0.7674}.$$
Entropy generation, compressor and turbine (part d). Both are adiabatic, so $s_{gen}$ equals the actual entropy rise across each: $$s_{gen,c}=s(T_2,P_2)-s(T_1,P_1)=\boxed{0.04669\ \text{kJ/kg}\cdot\text{K}}$$ $$s_{gen,t}=s(T_4,P_1)-s(T_3,P_2)=\boxed{0.04349\ \text{kJ/kg}\cdot\text{K}}$$ Both positive, consistent with irreversible (non-isentropic) adiabatic devices.

Quantity	Result
(a) $w_{net,in}$	43.78 kJ/kg
(b) $q_L$	33.60 kJ/kg
(c) COP	0.7674
(d) $s_{gen,c}$	0.04669 kJ/kg·K
(d) $s_{gen,t}$	0.04349 kJ/kg·K

Concept & Theory

Topic. Simple gas (reverse-Brayton) refrigeration cycle using air as the working fluid, distinct from every vapor-compression (R-134a) refrigeration cycle elsewhere in this subject.

Key relations. Same variable-cp isentropic root-finding as the power-cycle Brayton, applied to both the compressor (pressure rising) and the turbine/expander (pressure falling): $s^\circ(T_2)-s^\circ(T_1)=R\ln(P_2/P_1)$. COP $=q_L/w_{net,in}$.

Why this works. Unlike a vapor-compression cycle's throttle valve, the gas cycle's low-pressure device is a work-producing turbine (not an isenthalpic throttle) — its work output partially offsets the compressor's work input, which is why $w_{net,in}=w_{c,a}-w_{t,a}$ rather than just $w_{c,a}$ alone. This is also why the cycle needs an isentropic efficiency for *both* the compressor and the "expansion valve" (here, an actual turbine).

Common pitfall. Treating this like the vapor-compression cycles elsewhere in the subject and looking for saturation properties or a two-phase state — air here never leaves the single-phase (ideal-gas) region, so there is no saturation dome on the T–s diagram.

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

Question 3 of 9: Gas Refrigeration Cycle (Reverse Brayton)

Question 3: Gas Refrigeration Cycle (Reverse Brayton) (20 marks)

Approach