Steam at 350^{o}C and 650
kPa is mixed with subcooled water at 30^{o}C and 650
kPa in an open
feedwater heater (FWH) as a way to produce saturated
liquid water at the same pressure. 

Assuming the open FWH is adiabatic, determine the mass flow rate of steam
required per kilogram of subcooled liquid water fed to the open FWH. 














Read : 
The feedwater heater is just a fancy mixer. When we write the MIMO form of the 1st Law at steadystate, there are three unknowns: the three mass flow rates. The states of all three streams are fixed, so we can determine the specific enthalpy of each of
them.
Mass conservation
tells us that m_{3} = m_{1} + m_{2}. We can use this to
eliminate m_{3} from the 1st Law. Then we can solve the 1st Law for m_{1} / m_{2} ! 












Given: 
T_{1} 
30 
^{o}C 



P_{3} 
650 
kPa 


P_{1} 
650 
kPa 



x_{3} 
0 
kg vap/kg 

T_{2} 
350 
^{o}C 



Q 
0 



P_{2} 
650 
kPa 


















Find: 

m_{dot1}
/ m_{dot,2}
= 
??? 


















Diagram: 













Assumptions: 
1  
The feedwater heater operates at steadystate. 



2  
Changes in potential and kinetic energies are negligible. 



3  
Heat
transfer is negligible. 



4  
No shaft work crosses the system boundary in this process. 












Equations
/ Data / Solve: 




















An open
feedwater heater is essentially a mixer in which superheated vapor is used to raise
the temperature of a subcooled liquid. We can begin our analysis with the steadystate form of the 1st Law. 














Eqn 1 













The assumptions in the list above allow
us to simplify the 1st Law considerably: 



















Eqn 2 













Conservation
of mass on the feedwater
heater operating at steadystate tells us that : 




















Eqn 3 













We can solve Eqn 3 for m_{dot,3} and use the result
to eliminate m_{dot,3} from Eqn 2. The result is: 


















Eqn 4 













The easiest way to
determine m_{dot,1} / m_{dot,2} is to divide Eqn 4 by m_{dot,2}. 



















Eqn 5 













Now, we can solve Eqn 5 for m_{dot,1} / m_{dot,2} : 



Eqn 6 













Now, all we need to do
is to determine the specific enthalpy of all three streams and plug these values into Eqn
6 to complete the problem. 













First we must
determine the phase(s)
present in each stream. 

T_{sat}(650kPa) = 
161.98 
^{o}C 













Therefore: 

Stream
1 is a subcooled liquid because T_{1} < T_{sat} 




Stream
2 is a superheated
vapor because T_{2} > T_{sat} 




T_{3} = T_{sat} because it is a saturated liquid. 













Data from the Steam Tables of the NIST Webbook (using the default
reference state) : 













H_{1} 
126.32 
kJ/kg 



H_{2} 
3165.1 
kJ/kg 








H_{3} 
684.1 
kJ/kg 













Now, plug these values
into Eqn 6 to obtain : 

m_{dot1}
/ m_{dot,2}
= 
4.448 













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















Answers : 
m_{dot1} / m_{dot,2} = 
4.45 

















