Sonntag, Borgnakke and van Wylen
9.34 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 Solution: air
i
C.V. Air turbine. Adiabatic: q = 0, reversible: sgen = 0
. W
Turbine
Energy Eq.6.13: Entropy Eq.9.8:
e
wT = hi − he , s e = si
hi = 1277.8 kJ/kg, Pr i = 191.17 The constant s process is done using the Pr function from A.7.2
a) Table A.7:
100 ⇒ Pr e = Pr i (Pe / Pi) = 191.17 800 = 23.896 ⇒ Te = 705.7 K, he = 719.7 kJ/kg w = hi - he = 1277.8 – 719.7 = 558.1 kJ/kg
Interpolate in A.7.1
b) Table A.5: CPo = 1.004 kJ/kg K, R = 0.287 kJ/kg K, k = 1.4, then from Eq.8.32 Te = Ti (Pe/Pi)
k-1 k
1000.286 = 1200 800 = 662.1 K
w = CPo(Ti - Te) = 1.004(1200 - 662.1) = 539.8 kJ/kg