Question 11.12: (a) Why are the back work ratios relatively high in gas turb...
(a) Why are the back work ratios relatively high in gas turbine plants compared to those of steam power plants ?
(b) In a gas turbine plant compression is carried out in two stages with perfect intercooling and expansion in one stage turbine. If the maximum temperature (T _{max} K) and minimum temperature (T _{min} K) in the cycle remain constant, show that for maximum specific output of the plant, the optimum overall pressure ratio is given by
where γ = Adiabatic index ; η_{T} = Isentropic efficiency of the turbine.
η_{C} = Isentropic efficiency of compressor. (AMIE Summer, 2005)
(a) Back work ratio may be defined as the ratio of negative work to the turbine work in a power plant. In gas turbine plants, air is compressed from the turbine exhaust pressure to the combustion chamber pressure. This work is given by – ∫ vdp. As the specific volume of air is very high (even in closed cycle gas turbine plants), the compressor work required is very high, and also bulky compressor is required. In steam power plants, the turbine exhaust is changed to liquid phase in the condenser. The pressure of condensate is raised to boiler pressure by condensate extraction pump and boiler feed pump in series since the specific volume of water is very small as compared to that of air, the pump work (– ∫ vdp), is also very small. From the above reasons, the back work ratio
= \frac{– ∫ vdp}{Turbine work }
for gas turbine plants is relatively high compared to that for steam power plants.
(b) Refer Fig. 29.
Assuming optimum pressure ratio in each stage of the compressors \sqrt{r} ,
\frac{ T_{2} }{ T_{1} } = (\frac{p_{2}}{p_{1}})^{\frac{γ – 1}{γ }}or T_{2} = T_{min} × (r)^{\frac{γ – 1}{2 γ }}
W_{compressor} = 2[c_{p} (T_{2}′ – T_{1})] for both compressors
= 2 c_{p} \frac{ T_{2} – T_{1}}{η_{C}} = \frac{2 c_{p}}{η_{C}} T_{min} [ (r)^{\frac{γ – 1}{2 γ }} – 1] , as T_{1} = T_{min}
Also, \frac{ T_{5} }{ T_{6} } = (\frac{p_{2}}{p_{1}})^{\frac{γ – 1}{γ }} = (r)^{\frac{γ – 1}{γ }}
∴ T_{6} = \frac{ T_{5}}{(r)^{γ – 1/γ } } = \frac{ T_{max}}{(r)^{γ – 1/γ } } , as T_{5} = T_{max}
W_{turbine} = c_{p} (T_{5} – T_{6}′) = c_{p} [ T_{max} – \frac{ T_{max}}{(r)^{γ – 1/γ }}] η_{T} , as η_{T} = \frac{T_{5} – T_{6}′}{T_{5} – T_{6}}= c_{p} T_{max} [ 1 – \frac{1}{(r)^{γ – 1/γ }} ] η_{T}
W_{net} = W_{turbine} – W_{compressor}= c_{p} η_{T} T_{max} [ 1 – \frac{1}{(r)^{γ – 1/γ }} ] – \frac{ 2 c_{p}}{η_{C} } T_{min} [ (r)^{\frac{γ – 1}{2γ }} – 1]
For maximum work output,
\frac{dW_{net} }{dr} = 0
or – c_{p} η_{T} T_{max} (- \frac{γ – 1}{γ } ) (r) ^{- ( \frac{γ – 1}{γ } ) – 1} – \frac{ 2 c_{p}}{η_{C} } T_{min} (\frac{γ – 1}{2 γ }) (r)^{\frac{γ – 1}{2 γ } – 1} = 0
or η_{T} η_{C} \frac{ T_{max}}{ T_{min}} = (r)^{3(γ – 1)/2 γ} , on simplification.
Hence, the optimum pressure ratio is
r_{opt} = [η_{T} . η_{C} . \frac{T _{max}}{T _{min}} ]^{\frac{2 γ}{3 (γ – 1)}} ……. Proved.
