An adiabatic
turbine lets 10 Mpa steam
down to 2.1 Mpa.
Determine the maximum work output if the inlet temperature is 500^{o}C and changes in kinetic
and potential energies
are negligible. 











Read : 
The key to solving this problem is to
recognize that any
process that is both adiabatic and reversible is ISENTROPIC. This means that S_{2} = S_{1} and this allows you to
fix state 2 and evaluate H_{2}. Use H_{2} in the 1st Law to evaluate W_{S}. 










Given: 
T_{1} 
500 
^{o}C 


Find: 
W_{S} 
??? 
kJ/kg 

P_{1} 
10000 
kPa 







P_{2} 
2100 
kPa 
















Diagram: 











Assumptions: 
1  
The turbine is both adiabatic and reversible. 


2  
Changes in kinetic and potential energies are negligible. 


3  
Shaft
work is the only form of work that crosses the system boundary. 










Equations
/ Data / Solve: 


















Begin by applying the 1st Law to the process, assuming changes in kinetic and potential energies are negligible: 


















Eqn 1 











The process is adiabatic so Eqn 1 can be simplified to : 




















Eqn 2 











Use the NIST Webbook to obtain properties
for state 1. First we have to determine the phases present. 











At P_{1} : 

T_{sat} 
311.00 
^{o}C 















Since T_{1} > T_{sat}, state 1
is a superheated vapor. 

The superheated Steam Tables and the NIST Webbook yield : 











H_{1} 
3375.1 
kJ/kg 



S_{1} 
6.5995 
kJ/kgK 











Because the process is
both reversible and adiabatic, it is isentropic. 











Therefore, S_{2} = S_{1} : 




S_{2} 
6.5995 
kJ/kgK 











At this point we know
values of two intensive variables for state 2, so we can use the NIST Webbook to determine the value
of any other property. In this case, we need H_{2}. First, we need to
determine the phases
that exist at state 2. 











At P_{2} : 
T_{sat} 
214.86 
^{o}C 


S_{sat vap} 
6.3210 
kJ/kgK 







S_{sat liq} 
2.4699 
kJ/kgK 











Because S_{2} > S_{sat vap} at P_{2}, we can conclude that state 2 is a superheated vapor. We could have reached
the same conclusion after careful consideration of a TS Diagram. 











We can get the
following data from the superheated Steam Tables or from the NIST Webbook : 











At 2.1
MPa : 
T (^{o}C) 
S (kJ/kgK) 
H (kJ/kg) 







265 
6.5903 
2937.1 







T_{2} 
6.5995 
H_{2} 







270 
6.6133 
2949.4 















Interpolation yields : 



T_{2} 
266.99 
^{o}C 







H_{2} 
2941.99 
kJ/kg 











Now, we can plug
values back into Eqn 2 : 

W_{S} 
433.13 
kJ/kg 










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










Answers : 
W 
433 
kJ/kg 















