In a gas turbine unit air is drawn in at 1.02 bar and 15°C, and is compressed to 6.12 bar. Calculate the thermal efficiency and the work ratio of the ideal constant pressure cycle, when the maximum cycle temperature is 800°C.
η=1−1rpγ−1γ=1−(1.026.12)1.4−11.4=0.401=40.1%,\eta=1-\frac 1{r_p^{\frac{\gamma-1}{\gamma}}}=1-(\frac{1.02}{6.12})^{\frac{1.4-1}{1.4}}=0.401=40.1\%,η=1−rpγγ−11=1−(6.121.02)1.41.4−1=0.401=40.1%,
T2T1=T3T4=(p2p1)γ−1γ=1.669,\frac{T_2}{T_1}=\frac{T_3}{T_4}=(\frac{p_2}{p_1})^{\frac{\gamma-1}{\gamma}}=1.669,T1T2=T4T3=(p1p2)γγ−1=1.669,
T2=1.669T1=480.5 K,T_2=1.669T_1=480.5~K,T2=1.669T1=480.5 K,
T4=T31.669=642.9 K,T_4=\frac{T_3}{1.669}=642.9~K,T4=1.669T3=642.9 K,
NWO=cp(T3−T4)−cp(T2−T1)=238.8 kJkg,NWO=c_p(T_3-T_4)-c_p(T_2-T_1)=238.8~\frac{kJ}{kg},NWO=cp(T3−T4)−cp(T2−T1)=238.8 kgkJ,
GWO=cp(T3−T4)=432.3 kJkg,GWO=c_p(T_3-T_4)=432.3~\frac{kJ}{kg},GWO=cp(T3−T4)=432.3 kgkJ,
WR=NWOGWO=0.553=55.3 %.WR=\frac{NWO}{GWO}=0.553=55.3~\%.WR=GWONWO=0.553=55.3 %.
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Dear DENNIS TONUI, 1.4 is the adiabatic index for air
How did you arrive at isentropic index of 1.4?
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Dear DENNIS TONUI, 1.4 is the adiabatic index for air
How did you arrive at isentropic index of 1.4?
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