Answer
$T_{2}$ = 290.6K
$W_{a}$=243.2 kJ/kg
Work Step by Step
Given data:
Inlet temperature =300$^{\circ}$C
Inlet pressure = 2200kPa
outlet pressure = 200kPa
isentropic efficiency($n_{T}$) = 85%
Solution:
For air specific heat at constant pressure $C_{p}$ = 1.013 kJ/kg$^{\circ}$C
k =$C_{p}$$\div$$C_{v}$ = 1.395
The outlet temperature of the air,, can be calculated using p, V, T Relation for isentropic process:
$T_{2}$ = $T_{1}$ $(P_{2}/P_{1})^{(k-1/k)}$
$T_{2}$ = 573 $(200/2200)^{(0.395/1.395)}$
$T_{2}$ = 290.6K
From the definition of the isentropic efficiency:
Actual Work produced by this turbine is given by formula;
$W_{a}$ = ($n_{T}$)*Isentropic Turbine work($W_{s}$)
$W_{a}$=($n_{T}$)*$C_{p}$($T_{1}$-$T_{2}$)
$W_{a}$=( 0.85)( 1.013kJ/kg K)(573-290.6)K
$W_{a}$=243.2 kJ/kg
