Question 9.PS.30: Air enters a turbine at 800 kPa, 1200 K, and expands in a re......

Air enters a turbine at 800 kPa, 1200 K, and expands in a reversible adiabatic process to 100 kPa. Calculate the exit temperature and the work output per kilogram of air, using
a. The ideal gas tables, Table A.7
b. Constant specific heat, value at 300 K from table A.5

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C.V. Air turbine.
Adiabatic: q = 0, reversible: s _{ gen }=0
Energy Eq.6.13: w _{ T }= h _{ i }- h _{ e } ,
Entropy Eq.9.8: s _{ e }= s _{ i }

a) Table A.7:    h _{ i }=1277.8 \,kJ / kg , \quad P _{ r\,i }=191.17

The constant s process is done using the P_r function from A.7.2

\Rightarrow P _{ r\,e }= P _{ r \,i }\left( P _{ e } / P _{ i }\right)=191.17\left\lgroup\frac{100}{800} \right\rgroup=23.896

Interpolate in A.7.1    \Rightarrow T _{ e }=705.7 \,K , \quad h _{ e }=719.7 \,kJ / kg

w = h _{ i }- h _{ e }=1277.8-719.7=558.1 \,k J / k g

b) Table A.5:    C _{\text {Po }}=1.004 \,kJ / kg K , \quad R =0.287 \,kJ / kg K , \quad k =1.4 then from Eq.8.32

\begin{gathered}T _{ e }= T _{ i }\left( P _{ e } / P _{ i }\right)^{\frac{ k -1}{ k }}=1200\left\lgroup\frac{100}{800} \right\rgroup^{0.286}= 6 6 2 . 1 \,K \\w = C _{ P o}\left( T _{ i }- T _{ e }\right)=1.004(1200-662.1)= 5 3 9 . 8 \, k J / k g\end{gathered}
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