Ideal Brayton Refrigeration (Gas Refrigeration) Cycle: 04-BS-10, Undated paper, Q5

Question 5: Ideal Brayton Refrigeration (Gas Refrigeration) Cycle (Part B – 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. Ideal (isentropic compressor and turbine) Brayton refrigeration cycle, cold-air-standard properties. Compressor inlet $T_1=270$ K, $P_1=100$ kPa; pressure ratio $r_p=3$; turbine inlet $T_3=310$ K. Reference reservoirs for part (d): $T_c=270$ K, $T_h=310$ K.

Find. (a) $w_{net,in}$; (b) $q_L$; (c) COP; (d) $COP_{rev}$.

Approach

The cycle is a reversed Brayton cycle: isentropic compression $1\to2$, constant-pressure cooling $2\to3'$ (external, not asked), isentropic expansion through a turbine $3\to4$ that produces the refrigeration effect as the cold, low-pressure air absorbs heat $4\to1$.

Compressor exit (state 2). $$T_2=T_1\,r_p^{(k-1)/k}=270\times3^{0.2857}=270\times1.3687=369.56\ \text{K}.$$
Turbine exit (state 4). $$T_4=T_3\left(\frac{1}{r_p}\right)^{(k-1)/k}=310\times3^{-0.2857}=\frac{310}{1.3687}=226.49\ \text{K}.$$
Net work input — part (a). $$w_{comp}=c_p(T_2-T_1)=1.005\times(369.56-270)=100.06\ \text{kJ/kg},\quad w_{turb}=c_p(T_3-T_4)=1.005\times(310-226.49)=83.93\ \text{kJ/kg}.$$ $$w_{net,in}=w_{comp}-w_{turb}=100.06-83.93=\boxed{16.13\ \text{kJ/kg}}.$$
Refrigeration capacity — part (b). The refrigerated space is warmer than the turbine-exit air ($T_4=226.5$ K $
Coefficient of performance — part (c). $$COP=\frac{q_L}{w_{net,in}}=\frac{43.73}{16.13}=\boxed{2.712}.$$
Reversible (Carnot) COP — part (d). $$COP_{rev}=\frac{T_c}{T_h-T_c}=\frac{270}{310-270}=\boxed{6.75}.$$

Quantity	Value
(a) $w_{net,in}$	16.13 kJ/kg
(b) $q_L$	43.73 kJ/kg
(c) COP	2.712
(d) $COP_{rev}$	6.75

Topic. Reversed (gas-refrigeration) Brayton cycle.

Key relations. Isentropic $T_2/T_1=r_p^{(k-1)/k}$; $COP=q_L/w_{net,in}$; Carnot bound $COP_{rev}=T_c/(T_h-T_c)$.

Why this works. Unlike a vapor-compression cycle, the working fluid never changes phase — the turbine's isentropic expansion is what drives the air temperature well below the refrigerated-space temperature, and the "refrigeration effect" is simply constant-pressure heating of that cold air back up to $T_1$. The COP is intrinsically far below the vapor-compression cycle's (and here, well below the Carnot bound of 6.75) because expansion work is only partially recovered compared to a throttle, which recovers none but also destroys none of the compression work directly as turbine loss. Note how sensitive the cycle is to the pressure ratio: at $r_p=3$ the two large gross works (100.1 and 83.9 kJ/kg) nearly cancel, so a small proportional error in either one moves $w_{net,in}=16.1$ kJ/kg — and hence the COP — by a much larger proportion.

Common pitfall. Computing $q_L$ from $T_3$ instead of $T_1$ — the refrigerated space interacts with the LOW-pressure air between states 4 and 1 (after the turbine, before the compressor), not between 2 and 3.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., §11-8 (gas refrigeration cycles).

Question 5 of 9: Ideal Brayton Refrigeration (Gas Refrigeration) Cycle

Question 5: Ideal Brayton Refrigeration (Gas Refrigeration) Cycle (Part B – 15 marks)

Approach