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## Q. 2.19

A steam turbine operates under steady flow conditions. It receives 7500 kg/h of steam from the boiler. The steam enters the turbine at 2800 kJ/kg enthalpy, 70 m/s velocity, and an elevation of 4 m. The steam leaves the turbine at 2000 kJ/kg enthalpy, 140 m/s velocity, and an elevation of 1.5 m. Heat losses from the turbine to surroundings amount to 1600 kJ/h. Calculate the output of the turbine.

## Verified Solution

\begin{aligned}m &=\frac{7500}{3600}=2.083 \ kg / s \\q &=\frac{-1600}{2.083 \times 3600}=-0.213 \ kJ / kg\end{aligned}

The steady flow energy equation per kg mass is :

$\begin{gathered}h_1+½ . c_1^2+g z_1+q=h_2+½ . c_2^2+g z_2+w \\2800+½.70^2 \times 10^{-3}+9.81 \times 4 \times 10^{-3}-0.213=2000+½.140^2 \times 10^{-3}+9.81 \times 1.5 \times 10^{-3}+w\end{gathered}$

w = 791.46 kJ/kg

Turbine output, $\dot{W}=\dot{m} w=2.083 \times 791.46=1648.6 \ kW$