Question

Air at 1600 K, 30 bar enters a turbine operating at steady state and expands adiabatically to the exit, where the pressure is 2.05 bar. The isentropic turbine efficiency is 90%. Assume ideal gas behavior for the air and ignore kinetic and potential energy effects. Determine the temperature at the exit, in K, and the work developed, in kJ per kg of air flowing.

Answer 1

Solution :

The isentropic efficiency of the turbine is given as :

$\eta = \frac{\text{actual work done}}{\text{isentropic work done}}$

  $=\frac{m(h_1-h_2)}{m(h_1-h_{2s})}$

  $=\frac{h_1-h_2}{h_1-h_{2s}}$

The entropy relation for the isentropic process is given by :

$0=s^\circ_2-s^\circ_1-R \ln \left(\frac{P_2}{P_1}\right)$

$\ln \left(\frac{P_2}{P_1}\right)=\frac{s^\circ_2-s^\circ_1}{R}$

$\frac{P_2}{P_1}=exp\left(\frac{s^\circ_2-s^\circ_1}{R}\right)$

$\left(\frac{P_2}{P_1}\right)_{s=constant}=\frac{P_{r2}}{P_{r1}}$

Now obtaining the properties from the ideal gas properties of air table :

At $T_1 = 1600 \ K,$

$P_{r1}=791.2$

$h_1=1757.57 \ kJ/kg$

Calculating the relative pressure at state 2s :

$\frac{P_{r2}}{P_{r1}}=\frac{P_2}{P_1}$

$\frac{P_{r2}}{791.2}=\frac{2.4}{30}$

$P_{r2}=63.296$

Obtaining the properties from Ideal gas properties of air table :

At $P_{r2}=63.296$,  $T_{2s}\approx 860 \ K$

Considering the isentropic relation to calculate the actual temperature at the turbine exit, we get:

  $\eta=\frac{h_1-h_2}{h_1-h_{2s}}$

$\eta=\frac{c_p(T_1-T_2)}{c_p(T_1-T_{2s})}$

$\eta=\frac{T_1-T_2}{T_1-T_{2s}}$

$0.9=\frac{1600-T_2}{1600-860}$

$T_2= 938 \ K$

So, at $T_2= 938 \ K$, $h_2=975.66 \ kJ/kg$

Now calculating the work developed per kg of air is :

$w=h_1-h_2$

  = 1757.57 - 975.66

  = 781 kJ/kg

Therefore, the temperature at the exit is 938 K and work developed is 781 kJ/kg.