Power output of isentropic steam turbine with 5% feed water diversion

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CptJackWest
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I know the answer is 6328kW, I just not sure on how the book came up with that answer. Can anyone help?

An isentropic steam turbine processes 5 kg/s of steam at 4MPa,which is exhausted at 50 kPa and 100oC. 5% of this flow is diverted for feed water heating at 700kPa. Determine the power produced by this turbine in kW.

Mf=5 kg/s
P1= 4MPa
P2=50 kPa
T2=100oC
x=95%
Pout=700kpa
At final
H=2683.78KJ/kg
S=7.69054 KJ/(kg/K)

The attempt at a solution[/b]
Calculated T at diversion using T2/T1=(P2/P1)^((k-1)/k)
1 at diversion and 2 at the end of process
Which i got to equal 714.7K
Then T at start using same formula
1 at start and 2 at the end of process
Tstart=1098.17K
But not sure where to go from there. Any help would be great
Cheers, Jack
 
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You know S at the exhaust point from the turbine. Since the process in the turbine is isentropic, this should help you determine the H at the inlet pressure of 4 Mpa and also the H at the extraction pressure of 700 kPa. Because an extraction is taking place, make sure you use the correct flow before and after this point. To find H, referring to steam tables would simplify finding the unknown H values.
 
Thanks for the help
I have found the H values and such.
But wondering what formula I would use to put in the values. That is where I am getting stuck. Cheers
 
You don't have any notes from your course about how to determine the work output of a turbine? (Hint: you can always try Google.)