Answer to Question #297393 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

Here,

u - specific internal energy at inlet and exit

P- Pressure at inlet and exit.

V - specific volume at inlet and e

M- mass flow rate through turbine

C - velocity at inlet and exit

Q - Rate of heat transfer between turbine and surroundings

W - rate of work transfer

inlet power

$=m(u+pv+\frac{c_2}{^2})-Q$

$=0.38(2 .590×10^6+13.8×10^5×0.143+450)-250$

$=1059110$

Outlet power

$0.38(2.36×10^6+0.35×10^5×4.37+450)+w$

$=956460+w$

$1059110=956460+w$

$w=1059110-956460$

$w=102650$ W