Practice · 6 marks

Ideal Brayton cycle work and efficiency

Short answer

Using isentropic relations: T2=300×8^0.2857=543.4 K, T4=1200/8^0.2857=662.5 K. Compressor work 244.7 kJ/kg, turbine work 540.2 kJ/kg, net work 295.6 kJ/kg, power 1478 kW, thermal efficiency 44.8%, back work ratio 0.453.

An ideal Brayton cycle takes in air at 100 kPa and 300 K. The pressure ratio is 8 and the maximum cycle temperature is 1200 K. The mass flow rate is 5 kg/s. Taking $c_p = 1.005$ kJ/kg K and $\gamma = 1.4$ , find (a) the temperatures at all cycle points, (b) the compressor and turbine work, (c) the net work and power, (d) the thermal efficiency and (e) the back work ratio.

$(\gamma-1)/\gamma = 0.2857$ and $r_p^{0.2857} = 8^{0.2857} = 1.8114$ .

(a) Temperatures

T_2 = T_1 r_p^{0.2857} = 300 \times 1.8114 = 543.4\ \text{K}, \qquad T_4 = \frac{T_3}{r_p^{0.2857}} = \frac{1200}{1.8114} = 662.5\ \text{K}

Point	1	2	3	4
T (K)	300	543.4	1200	662.5

(b) Compressor and turbine work

w_c = c_p (T_2 - T_1) = 1.005 \times 243.4 = 244.7\ \text{kJ/kg}

w_t = c_p (T_3 - T_4) = 1.005 \times 537.5 = 540.2\ \text{kJ/kg}

(c) Net work and power

w_{net} = w_t - w_c = 540.2 - 244.7 = 295.6\ \text{kJ/kg}, \qquad P = 5 \times 295.6 = 1478\ \text{kW}

(d) Thermal efficiency

q_{in} = c_p (T_3 - T_2) = 1.005 \times 656.6 = 659.8\ \text{kJ/kg}, \qquad \eta = \frac{w_{net}}{q_{in}} = 0.448

Check: $1 - 1/8^{0.2857} = 0.448$ .

(e) Back work ratio

\text{BWR} = \frac{w_c}{w_t} = \frac{244.7}{540.2} = 0.453

Answer: $T_2 = 543.4$ K, $T_4 = 662.5$ K; $w_c = 244.7$ , $w_t = 540.2$ , $w_{net} = 295.6$ kJ/kg; $P = 1478$ kW; $\eta = 44.8\%$ ; BWR = 0.453.

✓ Answer checked by Claude Opus 5.5 (high effort)

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