At steadystate, a steam turbine produces 1,050 MW by letting the pressure down from 50 bar to 1 bar.
The steam enters the turbine at 320^{o}C with a velocity of 8 m/s
and a mass flow rate
of 150 kg/min. 
The steam leaves the turbine with a quality of 0.94 kg vapor/kg and a velocity of 65 m/s.
Determine the rate of heat loss from the turbine to the surroundings in kW. 


Read : 
Apply the steadystate form of the 1st Law for open systems and solve for Q.
Assume changes
in potential energy
are negligible. We know the values of two intensive
variables for state 1, so we can look up H_{1}. We know the pressure and quality for state 2, so we can also determine H_{2}. Then, just plug back into the 1st Law to get Q ! 










Given: 
m 
150 
kg/min 



v_{1} 
8 
m/s 


2.500 
kg/s 



P_{2} 
100 
kPa 

W_{s} 
1050 
kW 



x_{2} 
0.94 


P_{1} 
5000 
kPa 



v_{2} 
65 
m/s 

T_{1} 
320 
^{o}C 
















Find: 
Q 
??? 
kW 
















Diagram: 























Assumptions: 
1  
The turbine operates at steadystate. 


2  
The change in the potential energy of the fluid from
the inlet to the outlet is negligible. 










Equations
/ Data / Solve: 


















Let's begin by writing
the steadystate form of the 1st Law for open systems. 















Eqn 1 











Solve Eqn 1 for
Q : 



Eqn 2 











We must use the Steam Tables to determine H_{2} and H_{1} : 
















Eqn 3 











H_{1} 
2986.2 
kJ/kg 

















At P_{2} = 100 kPa : 



H_{sat liq} 
417.50 
kJ/kg 







H_{sat vap} 
2674.95 
kJ/kg 







H_{2} 
2539.5 
kJ/kg 











Now, we can plug
values into Eqn 2 to evaluate Q : 

Q 
61.62 
kW 










Verify: 
None of the assumptions
made in this problem solution can be verified. 













Answers : 
Q 
61.6 
kW 















