Answer to Question #244741 in Molecular Physics | Thermodynamics for Mwalimu

Question

Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa,
450

Answer 1

$P_1= 10 \;MPa \\ T_1 = 450 \;°C \\ h_1 = 3240.9 \;kJ/kg$

At $P_2=10 \;kPa$ and x=0.92, from table

$h_f = 191.83 \;kJ/kg \\ h_{fg} = 2392.8 \;kJ/kg \\ h_2 = h_p + h_{fg} = 191.83 +0.92(2392.8) = 2393.206 \;kJ/kg \\ Q-W = m[(h_2-h_1) + \frac{v^2_2 -v^2_1}{2000}] \\ Q=0 \\ W = m[(h_1-h_2) + \frac{v^2_1 -v^2_2}{2000}] \\ = 14 [(3240.9 -2393.206) + \frac{85^2 -54^2}{2000}] \\ = 11897.88 \;kJ/s \\ = 11.90 \;MW$