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Gas turbine 6:1: power output and efficiency

Short answer

Compressor work = 241.87 kJ/kg, turbine work = 290.01 kJ/kg, net work 48.14 kJ/kg. Power output = 16 kg/s × 48.14 = 770.3 kW. Thermal efficiency = 12.24%.

A gas turbine unit has a pressure ratio of 6:1 and maximum cycle temperature of 610°C. The isentropic efficiencies of the compressor and turbine are 80% and 82% respectively. Calculate the power output in kilowatts of an electric generator geared to the turbine when the air enters the compressor at 15°C at the rate of 16 kg/s. Find the thermal efficiency of the plant as well. Take Cp = 1.005 kJ/kg K and γ = 1.4 for the compression process, Cp = 1.11 kJ/kg K and γ = 1.333 for the expansion process and Cp = 1.11 kJ/kg K for the combustion process.

Data: rp=6r_p = 6, T3=610+273=883T_3 = 610 + 273 = 883 K, ηc=0.80\eta_c = 0.80, ηt=0.82\eta_t = 0.82, T1=15+273=288T_1 = 15 + 273 = 288 K, m˙=16\dot m = 16 kg/s. Compression: cp=1.005c_p = 1.005, γ=1.4\gamma = 1.4 (γ−1γ=0.2857\frac{\gamma-1}{\gamma} = 0.2857). Expansion: cp=1.11c_p = 1.11, γ=1.333\gamma = 1.333 (γ−1γ=0.2498\frac{\gamma-1}{\gamma} = 0.2498). Combustion: cp=1.11c_p = 1.11 kJ/kg K. Mass of fuel neglected.

 1        2          3        4
 --> [C] ---> [CC] ---> [T] ---> exhaust
      \================/  ==> Generator

Compressor

T2s=288×60.2857=480.53 KT2=288+480.53−2880.80=528.66 KWc=1.005(528.66−288)=241.87 kJ/kg\begin{aligned} T_{2s} &= 288 \times 6^{0.2857} = 480.53\ \text{K} \\ T_2 &= 288 + \frac{480.53 - 288}{0.80} = 528.66\ \text{K} \\ W_c &= 1.005(528.66 - 288) = 241.87\ \text{kJ/kg} \end{aligned}

Turbine

T4s=88360.2498=8831.5646=564.38 KT4=883−0.82(883−564.38)=621.73 KWt=1.11(883−621.73)=290.01 kJ/kg\begin{aligned} T_{4s} &= \frac{883}{6^{0.2498}} = \frac{883}{1.5646} = 564.38\ \text{K} \\ T_4 &= 883 - 0.82(883 - 564.38) = 621.73\ \text{K} \\ W_t &= 1.11(883 - 621.73) = 290.01\ \text{kJ/kg} \end{aligned}

Power output

Wnet=290.01−241.87=48.14 kJ/kgP=m˙Wnet=16×48.14=770.3 kW\begin{aligned} W_{net} &= 290.01 - 241.87 = 48.14\ \text{kJ/kg} \\ P &= \dot m W_{net} = 16 \times 48.14 = 770.3\ \text{kW} \end{aligned}

Thermal efficiency

Qs=cp(T3−T2)=1.11(883−528.66)=393.31 kJ/kgηth=48.14393.31=0.1224=12.24%\begin{aligned} Q_s &= c_p (T_3 - T_2) = 1.11(883 - 528.66) = 393.31\ \text{kJ/kg} \\ \eta_{th} &= \frac{48.14}{393.31} = 0.1224 = 12.24\% \end{aligned}

Answer: Generator power output ≈ 770.3 kW (assuming no mechanical/generator losses); thermal efficiency = 12.24 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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