Example Problem with Complete Solution

7B-2 : Work Output of an Adiabatic, Reversible Turbine 5 pts
An adiabatic turbine lets 10 Mpa steam down to 2.1 Mpa. Determine the maximum work output if the inlet temperature is 500oC 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 500 oC Find: WS ??? kJ/kg
P1 10000 kPa
P2 2100 kPa
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 P1 : Tsat 311.00 oC
Since T1 > Tsat, state 1 is a superheated vapor.
The superheated Steam Tables and the NIST Webbook yield :
H1 3375.1 kJ/kg S1 6.5995 kJ/kg-K
Because the process is both reversible and adiabatic, it is isentropic.
Therefore, S2 = S1 : S2 6.5995 kJ/kg-K
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 H2.  First, we need to determine the phases that exist at state 2.
At P2 : Tsat 214.86 oC Ssat vap 6.3210 kJ/kg-K
Ssat liq 2.4699 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 :
At 2.1 MPa : T (oC) S (kJ/kg-K) H (kJ/kg)
265 6.5903 2937.1
T2 6.5995 H2
270 6.6133 2949.4
Interpolation yields : T2 266.99 oC
H2 2941.99 kJ/kg
Now, we can plug values back into Eqn 2 : WS 433.13 kJ/kg
Verify: None of the assumptions made in this problem solution can be verified.
Answers : W 433 kJ/kg