Question 11.13: In a gas turbine the compressor takes in air at a temperatur...
In a gas turbine the compressor takes in air at a temperature of 15°C and compresses it to four times the initial pressure with an isentropic efficiency of 82%. The air is then passed through a heat exchanger heated by the turbine exhaust before reaching the combustion chamber. In the heat exchanger 78% of the available heat is given to the air. The maximum temperature after constant pressure combustion is 600°C, and the efficiency of the turbine is 70%. Neglecting all losses except those mentioned, and assuming the working fluid throughout the cycle to have the characteristic of air find the efficiency of the cycle.
Assume R = 0.287 kJ/kg K and γ = 1.4 for air and constant specific heats throughout.
Given : T_{1} = 15 + 273 = 288 K, Pressure ratio, \frac{p_{2} }{p_{1} } = \frac{p_{3} }{p_{4} } = 4 , η_{compressor} = 82%.
Effectiveness of the heat exchanger, ε = 0.78,
η_{ turbine} = 70%, Maximum temperature, T_{3} = 600 + 273 = 873 K.
Efficiency of the cycle η_{cycle} :
Considering the isentropic compression 1–2, we have
\frac{ T_{2} }{ T_{1} } = (\frac{p_{2}}{p_{1}})^{\frac{γ – 1}{γ }} = (4) ^{\frac{1.4 – 1}{1.4 }} = 1.486
∴ T_{2} = 288 × 1.486 = 428 K
Now, η_{compressor} = \frac{T_{2} – T_{1}}{T_{2}′ – T_{1}}
i.e., 0.82 = \frac{428 – 288}{T_{2}′ – 288}
∴ T_{2}′ = \frac{428 – 288}{0.82} + 288 = 459 K
Considering the isentropic expansion process 3–4, we have
\frac{ T_{3} }{ T_{4} } = (\frac{p_{3}}{p_{4}})^{\frac{γ – 1}{γ }} = (4) ^{\frac{1.4 – 1}{1.4 }} = 1.486
∴ T_{4} = \frac{ T_{3} }{1.486} = \frac{ 873 }{1.486} = 587.5 K.
Again, η_{turbine } = \frac{T_{3} – T_{4}′}{T_{3} – T_{4}} = \frac{873 – T_{4}′}{873 – 587.5}
i.e., 0.70 = \frac{873 – T_{4}′}{873 – 587.5}
∴ T_{4}′ = 873 – 0.7 (873 – 587.5) = 673 K
W_{compressor} = c_{p} (T_{2}′ – T_{1})
But c_{p} = R × \frac{γ }{γ – 1} = 0.287 × \frac{1.4 }{1.4 – 1} = 1.0045 kJ/kg K
∴ W_{compressor } = 1.0045 (459 – 288) = 171.7 kJ/kg
W_{turbine} = c_{p} (T_{3} – T_{4}′) = 1.0045 (873 – 673) = 200.9 kJ/kg
∴ Network = W_{turbine} – W_{compressor} = 200.9 – 171.7 = 29.2 kJ/kg.
Effectiveness for heat exchanger, ε = \frac{T_{5} – T_{2}′}{T_{4}′ – T_{2}′}
i.e., 0.78 = \frac{T_{5} – 459}{673 – 459}
∴ T_{5} = (673 – 459) × 0.78 + 459 = 626 K
∴ Heat supplied by fuel per kg
= c_{p} (T_{3} – T_{5}) = 1.0045 (873 – 626) = 248.1 kJ/kg
∴ η_{cycle} = \frac{Network done }{Heat supplied by the fuel} = \frac{29.2}{248.1} = 0.117 or 11.7%.
