Answer to Question #296754 in Mechanical 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

Solution;

Apply the steady flow energy equation;

$\dot{m}(h_1+\frac{c_1^2}{2})-Q=\dot{m}(h_2+\frac{c_2^2}{2})+W$

But;

$h=u+PV$

$W=\dot{m}[(u_1-u_2)+(P_1v_1-P_2v_2)+(\frac{c_1^2-c_2^2}{2})]-Q$

By direct substitution of values;

$W=0.38[(2590-2360)+(13.8×0.143-0.35×4.37)×10^5+(\frac{30^2-90^2}{2})]-0.25kW$

$W=0.38[230+(197.34-152.95)-3.6]-0.25$

$W=102.9-0.25=102.65kW$