In simple gas turbine cycle following data are given: Isentropic efficiency of compressor = 80%. Turbine isentropic efficiency = 100%. All shafts have mechanical efficiency = 90%. Turbine output shaft is coupled to compressor and generator. Pressure ratio of compression = 3 : 1.

Maximum turbine temperature = 1000 K. Ambient condition 1 bar, 327 K. Net specific power output from the turbine is ____ KJ/kgk. Take generator efficiency = 95%, cp = 1.005 kJ/kgk.

87.564

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Solution

The correct option is A 87.564
$P_{1} = 1 b a r, P_{2} = P 3 = 3 b a r, T_{3} = 1000 K$

$\frac{T_{2 s}}{T_{1}} = {(3)}^{0.4 / 1.4}$

$\Rightarrow T_{2 s} = 447.58 K$

$ω_{c} = \frac{C_{p} (T_{2 s} - T_{1})}{n_{c}} = 151.5 k J / k g$

$\Rightarrow n_{m} = 0.9, n_{G} = 0.95$

$\Rightarrow ω_{n e t} = (ω_{T} - \frac{ω_{c}}{n_{m}}) \times n_{m} \times n_{g}$

$\Rightarrow \frac{T_{3}}{T_{1}} = 3^{0.4 / 1.4} \Rightarrow T_{4} = 730.6 K$

$ω_{T} = C_{p} (T_{3} - T_{4}) = 270.75 k J / k g$
$ω_{n e t} = 87.564 k J / k g$

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The correct option is A 87.564P1=1bar,P2=P3=3bar,T3=1000K T2sT1=(3)0.4/1.4 ⇒T2s=447.58K ωc=Cp(T2s−T1)nc=151.5kJ/kg ⇒nm=0.9,nG=0.95 ⇒ωnet=(ωT−ωcnm)×nm×ng ⇒T3T1=30.4/1.4⇒T4=730.6K ωT=Cp(T3−T4)=270.75kJ/kg ωnet=87.564kJ/kg