Answer to Question #297673 in Thermal Power Engineering for Harsh

Question

A turbine operating under steady flow conditions receives steam at the

following state : pressure 13.8 bar ; specific volume 0.143 m3/kg ;

internal energy 2590 kJ/kg ; velocity 30 m/s. The state of the steam

leaving the turbine is : pressure 0.35 bar ; specific volume 4.37 m3/kg

; internal energy 2360

kJ/kg ; velocity 90 m/s. Heat is lost to the surroundings at the rate of

0.25 kJ/s. If the rate of steam flow is 0.38 kg/s, what is the power

developed by the turbine ?

Answer 1

$\dot{m}(u_1+P_1V_1+\frac{C_1^2}{2})-Q=\dot{m}(u_2+P_2V_2+{\frac{C_2^2}{2}})+\dot{w}$

$0.38 \begin{bmatrix} 2590×10^3+(13.8×10^5)(0.143)+\frac{30}{2}^2 \\ \end{bmatrix}-0.25×10^3$

$=1059110.2$

$0.38\begin{bmatrix} 2360×10^3+(0.35×10^5)(4.37)+\frac{90}{2}^2 \\ \end{bmatrix}+\dot{w}$

$=956460+\dot{w}$

$1059110.2=956460+\dot{w}$

$\dot{w}=102650.2$

$=102.65kW$