Example Problem with Complete Solution

7B-2 : Work Output of an Adiabatic, Reversible Turbine 5 pts
Steam enters an adiabatic turbine at 5 MPa and 450oC and leaves at a pressure of 1.4 MPa.  Determine the work output of the turbine per kg of steam flowing through the turbine if the process is reversible 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 S2 = S1 and this allows you to fix state 2 and evaluate H2.  Use H2 in the 1st Law to evaluate WS.
Given : T1 450 oC Find:
P1 5000 kPa WS ??? kJ/kg
P2 1400 kPa
Diagram : Figure 1
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:
Equation 1
Eqn 1
The process is adiabatic so Eqn 1 can be simplified to :
Equation 2
Eqn 2
Use the NIST Webbook to obtain properties for state 1.  First we have to determine the phases present.
At P1 : Tsat 263.94 oC
Since T1 > Tsat, state 1 is a superheated vapor.
The superheated steam tables and the NIST Webbook yield :
H1 3317.2 kJ/kg S1 6.8210 kJ/kg-K
Because the process is both reversible and adiabatic, it is isentropic.
Therefore : S2 6.8210 kJ/kg-K
At this point we know values of two intensive variable for state 2, so we can use the NIST Webbook to determine the value of any other property.  In this case, we need H2.  First, we need to determine the phases that exist at state 2.
At P2 : Tsat 195.04 oC
Ssat vap 6.4675 kJ/kg-K
Ssat liq 2.2835 kJ/kg-K
Because S2 > Ssat vap at P2, 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 :
T (oC) S (kj/kg-K) H (kJ/kg)
265 6.8140 2962.5
270 6.8350 2973.8
Interpolation yields : T2 266.67 oC
H2 2966.27 kJ/kg
Now, we can plug values back into Eqn 2 : WS 350.93 kJ/kg
Verify: None of the assumptions made in this problem solution can be verified.
Answers: W 351 kJ/kg