Question 11.16: In a closed cycle gas turbine there is two-stage compressor ...
In a closed cycle gas turbine there is two-stage compressor and a two-stage turbine. All the components are mounted on the same shaft. The pressure and temperature at the inlet of the first-stage compressor are 1.5 bar and 20°C. The maximum cycle temperature and pressure are limited to 750°C and 6 bar. A perfect intercooler is used between the two-stage compressors and a reheater is used between the two turbines. Gases are heated in the reheater to 750°C before entering into the L.P. turbine. Assuming the compressor and turbine efficiencies as 0.82, calculate :
(i) The efficiency of the cycle without regenerator.
(ii) The efficiency of the cycle with a regenerator whose effectiveness is 0.70.
(iii) The mass of the fluid circulated if the power developed by the plant is 350 kW,
The working fluid used in the cycle is air. For air : γ = 1.4 and c_{p} = 1.005 kJ/kg K.
Given : T _{1} = 20 + 273 = 293 K, T_{5} = T_{7} = 750 + 273 = 1023 K, p_{1} = 1.5 bar,
p_{2} = 6 bar, η _{compressor} = η_{ turbine} = 0.82.
Effectiveness of regenerator, ε = 0.70, Power developed, P = 350 kW.
For air : c_{p} = 1.005 kJ/kg, K, γ = 1.4
As per given conditions : T _{1} = T_{3} , T_{2}′ = T_{4}′
\frac{ T_{2} }{ T_{1} } = (\frac{p_{2}}{p_{1}})^{\frac{γ – 1}{γ }} and p_{x} = \sqrt{p_{1} p_{2}} = \sqrt{1.5 × 6} = 3 bar
Now, T_{2} = T_{1} × (\frac{p_{2}}{T_{1}})^{\frac{γ – 1}{γ }} = 293 × (\frac{3}{1.5})^{\frac{1.4 – 1}{1.4}} = 357 K
η_{compressor (L.P.)} = \frac{T_{2} – T_{1}}{T_{2}′ – T_{1}}0.82 = \frac{357 – 293}{T_{2}′ – 293}
i.e., T_{2}′ = \frac{357 – 293}{0.82} + 293 = 371 K i.e., T_{2}′ = T_{4}′ = 371 K
Now, \frac{ T_{5} }{ T_{6} } = (\frac{p_{5}}{p_{6}})^{\frac{γ – 1}{γ }} = (\frac{p_{2}}{p_{x}})^{\frac{1.4 – 1}{1.4}} [∵ p_{5} = p_{2}
p_{6} = p_{x} ]
\frac{1023}{T_{6}} = (\frac{6}{3})^{0.286} = 1.219
∴ T_{6} = \frac{1023}{1.219} = 839 K
η _{turbine (H.P.)} = \frac{ T_{5} – T_{6}′ }{T_{5} – T_{6}}0.82 = \frac{ 1023 – T_{6}′ }{1023 – 839}
∴ T_{6}′ = 1023 – 0.82 (1023 – 839) = 872 K
T_{8}′ = T_{6}′ = 872 K as η_{ turbine (H.P.)} = η_{turbine (L.P.)}
and T_{7} = T_{5} = 1023 K
Effectiveness of regenerator, ε = \frac{ T′ – T_{4}′ }{T_{8}′ – T_{4}′}
where T′ is the temperature of air coming out of regenerator
∴ 0.70 = \frac{ T′ – 371 }{872 – 371} i.e., T′ = 0.70 (872 – 371) + 371 = 722 K
Network available, W_{net} = [W_{T(L.P.)} + W_{T(L.P.)} ] – [W_{C(H.P.)} + W_{C(L.P.)} ]
= 2 [W_{T(L.P.)} – W_{C(L.P.)}] as the work developed by each turbine is same and work absorbed by each compressor is same.
∴ W_{net} = 2 c_{p} [(T_{5} – T_{6}′) – (T_{2}′ – T_{1})]
= 2 × 1.005 [(1023 – 872) – (371 – 293)] = 146.73 kJ/kg of air
Heat supplied per kg of air without regenerator
= c_{p} (T_{5} – T_{4}′) + c_{p} (T_{7} – T_{6}′)
= 1.005 [(1023 – 371) + (1023 – 872)] = 807 kJ/kg of air
Heat supplied per kg of air with regenerator
= c_{p} (T_{5} – T′) + c_{p} (T_{7} – T_{6}′)
= 1.005 [(1023 – 722) + (1023 – 872)]
= 454.3 kJ/kg
(i) η_{thermal (without regenerator)} = \frac{146.73 }{807} = 0.182 or 18.2%.
(ii) η_{thermal (with regenerator)} = \frac{146.73 }{454.3} = 0.323 or 32.3%.
(iii) Mass of fluid circulated, \dot{m} :
Power developed, P = 146.73 × \dot{m} kW
∴ 350 = 146.73 × \dot{m}
i.e., \dot{m} = \frac{350}{146.73} = 2.38 kg/s
i.e., Mass of fluid circulated = 2.38 kg/s.
