Gas-Turbine Air Refrigeration Machine: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2016, Q4

Question 4: Gas-Turbine Air Refrigeration Machine (equal value)

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.

Quantity	Value
Compressor inlet / bleed cool-to	15 °C (288.15 K)
Main-turbine inlet	760 °C (1033.15 K)
Cold-space exhaust escape	−5 °C (268.15 K)
Pressure ratio / efficiencies	$r_p=5$; $\eta_c=0.80$, $\eta_t=0.90$
Air (cold-air standard)	$\gamma=1.4$, $c_p=1.005$ kJ/kg·K

Figure 4 — Top: flow schematic (compressor → bleed split → combustor/main turbine and cooler/refrigeration turbine, all on one shaft). Bottom: $T$–$s$ sketch — 1→2 compression, 2→3 combustion, 3→4 main-turbine expansion; the bled stream is cooled at constant pressure 2→5 to 15 °C (entropy falls), expanded 5→6 to −80.6 °C in the refrigeration turbine (entropy rises, as in any real expansion), then warmed 6→7 by the cold-space load to the −5 °C escape temperature.

Approach. Compute the compressor, main-turbine and refrigeration-turbine temperatures from the pressure ratio and component efficiencies; the single-shaft power balance (turbine work = compressor work) fixes the bleed fraction $f$; then form the cold-space heat and combustion heat per unit total flow.

Compressor exit. With $r_p^{(\gamma-1)/\gamma}=5^{0.2857}=1.584$, $$T_2=T_1+\frac{T_1(r_p^{0.2857}-1)}{\eta_c}=288.15+\frac{288.15(0.584)}{0.80}=498.4\ \text{K}$$
Turbine exits. Both turbines drop through the same ratio with $\eta_t=0.9$: $$T_4=T_3\Big[1-\eta_t\big(1-1/1.584\big)\Big]=1033.15(0.668)=690.4\ \text{K}$$ $$T_6=T_5(0.668)=288.15(0.668)=192.6\ \text{K}\;(-80.6\text{ °C})$$
Single-shaft balance ⇒ bleed fraction. Per unit total mass, $w_c=(1-f)w_{t,\text{main}}+f\,w_{t,\text{ref}}$ (all $\div c_p$): $$210.3=(1-f)(342.8)+f(95.6)\ \Rightarrow\ f=\frac{342.8-210.3}{342.8-95.6}$$ ==**$f\approx0.536$ of the air is bled to refrigeration.**==
Cold-space heat. The cold stream enters the cold space at $T_6=192.6$ K and leaves it at the stated escape temperature $T_7=-5$ °C $=268.15$ K: $$\frac{\dot Q_L}{c_p}=f(T_7-T_6)=0.536(268.15-192.6)=40.5\ \text{K (per unit total flow)}$$
Combustion heat and ratio. $\dfrac{\dot Q_H}{c_p}=(1-f)(T_3-T_2)=0.464(534.7)=248.1$ K, so $$\frac{\dot Q_L}{\dot Q_H}=\frac{40.5}{248.1}$$ ==**$\dot Q_L/\dot Q_H\approx0.163$.**==

Check

The machine is taken as self-driving (no net shaft power): the two turbines exactly supply the compressor, which sets the bleed fraction. Cold-air-standard constant $c_p$ is used, so $c_p$ cancels from the ratio. Because the bled air is chilled to 15 °C at high pressure before expanding, its turbine exit reaches −81 °C — the source of the refrigeration effect. Note the sign on the escape temperature: both the question text and state 7 of the printed schematic give $p_7=1$ atm and $T_7=-5$ °C, i.e. below zero. Reading it as $+5$ °C inflates the cold-space rise from 75.6 to 85.6 K and the ratio from 0.163 to 0.185 — a 13 % overstatement of the refrigeration duty.

Quantity	Result
Compressor exit $T_2$	498.4 K (225 °C)
Main-turbine exit $T_4$	690.4 K (417 °C)
Refrigeration-turbine exit $T_6$	192.6 K (−81 °C)
Bleed fraction $f$	≈ 0.536
Cold-space heat / combustion heat (per unit total flow, ÷$c_p$)	40.5 / 248.1 K
Heat-extracted / heat-added ratio	≈ 0.163

Concept & Theory

Topic. Combined Brayton power / reverse-Brayton (air-cycle) refrigeration on a common shaft.

Key relations. Isentropic $T_s/T=r_p^{(\gamma-1)/\gamma}$; component efficiencies $\eta_c=(T_{2s}-T_1)/(T_2-T_1)$, $\eta_t=(T_3-T_4)/(T_3-T_{4s})$; shaft balance $w_c=\sum w_t$; heat terms $q=c_p\Delta T$.

Why this works. Each turbine's exit temperature depends only on its inlet temperature and the (common) expansion ratio and efficiency, so both share the factor 0.668. The refrigeration branch works because pre-cooling the compressed air to 15 °C before expansion carries it to a much lower temperature than the ambient — the classic air-cycle (bootstrap) refrigeration principle used in aircraft. The self-driving constraint replaces an external work input with the excess of main-turbine work over compressor demand.

Common pitfall. Forgetting to pre-cool the bled air before the second turbine (its low exit temperature depends on entering at 15 °C, not 225 °C), or leaving net shaft work unbalanced so $f$ is undetermined.

Source. Çengel & Boles, Thermodynamics §9 (Brayton cycle) and §11 (gas/air-cycle refrigeration).

Question 4 of 8: Gas-Turbine Air Refrigeration Machine

Question 4: Gas-Turbine Air Refrigeration Machine (equal value)