2068 Magh · 8 marks

Brayton cycle 80 MW: mass flow, efficiency, compressor fraction

Short answer

Pressure ratio 14, r_p^((γ−1)/γ)=2.1247; T2=637.4 K, T4=706.0 K. Net work per kg = 458.9 kJ/kg. Mass flow = 80 000 kW / 458.9 = 174.3 kg/s. Efficiency = 52.9%. Turbine power = 139.1 MW. Fraction of turbine power to compressor = wc/wt = 0.425 (42.5%).

A power plant operating on an ideal Brayton cycle delivers a power output of 80 MW. The minimum and maximum temperatures during the cycle are 300 K and 1500 K respectively. The pressure at the compressor inlet and outlet are 100 kPa and 1400 kPa respectively. (a) Determine the mass flow rate of air. (b) Determine the thermal efficiency of the cycle. (c) Determine the power output from the turbine. (d) What fraction of the turbine power output is required to drive the compressor? [R = 287 J/kg K, Cp = 1005 J/kg K]

Given: Ẇ_net = 80 MW, T₁ = 300 K, T₃ = 1500 K, P₁ = 100 kPa, P₂ = 1400 kPa, R = 0.287, cp = 1.005 kJ/kg·K. Here γ = cp/(cp − R) = 1.4, so (γ−1)/γ = R/cp = 0.2856.

r_p = 1400/100 = 14; r_p^0.2856 = 2.1247

Temperatures

T₂ = T₁ r_p^0.2856 = 300 × 2.1247 = 637.4 K T₄ = T₃/r_p^0.2856 = 1500/2.1247 = 706.0 K

Specific work and heat (per kg)

Compressor: w_c = cp (T₂ − T₁) = 1.005 × 337.4 = 339.1 kJ/kg
Turbine: w_t = cp (T₃ − T₄) = 1.005 × 794.0 = 798.0 kJ/kg
Net: w_net = 798.0 − 339.1 = 458.9 kJ/kg
Heat input: q_in = cp (T₃ − T₂) = 1.005 × 862.6 = 866.9 kJ/kg

(a) Mass flow rate

ṁ = Ẇ_net/w_net = 80 000/458.9 = 174.3 kg/s

(b) Thermal efficiency

η = w_net/q_in = 458.9/866.9 = 52.9 % (= 1 − 1/2.1247)

(c) Turbine power

Ẇ_t = ṁ w_t = 174.3 × 798.0 = 139117 kW = 139.1 MW

(d) Fraction of turbine work used by the compressor

Back-work ratio = w_c/w_t = 339.1/798.0 = 0.425 (about 42.5 %)

The compressor takes about 42.5 % of the turbine output; the remaining 57.5 % is the net output.

✓ Answer checked by Claude Opus 5.5 (high effort)

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