Ideal gas turbine 100 kW: pressure ratio, work, efficiency

Short answer

From T3/T4 = r_p^((γ-1)/γ), r_p = (1023.15/573.15)^3.5 = 7.60. Compressor work w_c = cp(T2 - T1) = 239.2 kJ/kg. Net work 213.0 kJ/kg gives mass flow rate ṁ = 100/213 = 0.469 kg/s. Efficiency η = 44.0%.

An ideal gas turbine has a net power output 100 kW. The working medium is air. The minimum and maximum temperatures of the cycle are 30°C and 750°C respectively. If the turbine outlet temperature is 300°C, determine a) Compressor pressure ratio. b) Compressor work c) Mass flow rate of air d) Thermal efficiency [Take Cp = 1005 J/kgK and γ = 1.4]

Ideal Brayton cycle: 1-2 isentropic compression, 2-3 constant-pressure heat addition, 3-4 isentropic expansion, 4-1 constant-pressure rejection.

Given: W_net = 100 kW, T₁ = 30 + 273.15 = 303.15 K, T₃ = 750 + 273.15 = 1023.15 K, T₄ = 300 + 273.15 = 573.15 K, cp = 1.005 kJ/kg·K, γ = 1.4.

(a) Compressor pressure ratio

For the turbine: T₃/T₄ = r_p^((γ−1)/γ)

r_p^0.2857 = 1023.15/573.15 = 1.7851

r_p = 1.7851^3.5 = 7.60

Compressor outlet temperature

T₂ = T₁ r_p^0.2857 = 303.15 × 1.7851 = 541.2 K

(b) Compressor work

w_c = cp (T₂ − T₁) = 1.005 × (541.2 − 303.15) = 239.2 kJ/kg

(c) Mass flow rate

Turbine work: w_t = cp (T₃ − T₄) = 1.005 × (1023.15 − 573.15) = 452.2 kJ/kg Net work: w_net = 452.2 − 239.2 = 213.0 kJ/kg

ṁ = Ẇ_net/w_net = 100/213.0 = 0.469 kg/s

(d) Thermal efficiency

q_in = cp (T₃ − T₂) = 1.005 × (1023.15 − 541.2) = 484.4 kJ/kg

η = w_net/q_in = 213.0/484.4 = 44.0 %

Check: η = 1 − 1/r_p^0.2857 = 1 − 1/1.7851 = 44.0 % ✓.

Answer: r_p = 7.60; w_c = 239.2 kJ/kg; ṁ = 0.469 kg/s; η = 44.0 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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