Question 9.CSGP.136: A turbine receives air at 1500 K, 1000 kPa and expands it to......
A turbine receives air at 1500 K, 1000 kPa and expands it to 100 kPa. The turbine has an isentropic efficiency of 85%. Find the actual turbine exit air temperature and the specific entropy increase in the actual turbine.
C.V. Turbine. steady single inlet and exit flow.
To analyze the actual turbine we must first do the ideal one (the reference).
Energy Eq.6.13: w _{ T }= h _1- h _2 \text {; }
Entropy Eq.9.8: s _2= s _1+ s _{ gen }= s _1
Entropy change in Eq.8.19 and Table A.7.1:
s _{ T 2}^{\circ}= s _{ T 1}^{\circ}+ R \ln \left( P _2 / P _1\right)=8.61208+0.287 \ln (100 / 1000)=7.95124
Interpolate in A.7 ⇒ T _{2 s }=849.2, \quad h _{2 s }=876.56 \Rightarrow
w _{ T }=1635.8-876.56=759.24 \,kJ / kg
Now we can consider the actual turbine from Eq.9.26 and Eq.6.13:
The entropy balance equation is solved for the generation term
s _{ gen }= s _{2 ac }- s _1=8.078-8.6121-0.287 \ln (100 / 1000)= 0 . 1 2 6 8 \, k J / k g ~ K
