# Example Problem with Complete Solution

8D-3 : Isentropic and 2nd Law Efficiencies of a Steam Turbine 6 pts
A steam turbine lets 2000 psia steam down to 60 psia. The inlet steam temperature is 1500oF and the isentropic efficiency is 88%.
a.) Calculate WS,act in Btu/lbm
b.) Calculate the 2nd Law Efficiency of the turbine. Assume Tsurr = 75oF.

Read : The key to this problem is to assume that the turbine is adiabatic.
We can calculate the isentropic work of the turbine because S2 = S1 gives us the additional intensive variable value that we need to fix the state of the outlet stream.  Then we can calculate the actual work from the isentropic work and the isentropic efficiency.
The 2nd Law Efficiency is the ratio of the actual work that we found in part (a) to the reversible work.  We need to know the actual entropy of the outlet stream in order to determine the reversible work for the turbine.  We can use the actual work and the 1st Law to determine the actual enthalpy of the effluent.  This gives us the second intensive property we need in order to use the Steam Tables to evaluate S2.
Given: hs 88% Find: Ws,act ??? Btu/lbm
P1 2000 psia hii ??? %
T1 1500 oF
P2 60 psia
Tsurr 75 oF
Diagram: Assumptions: 1 - The turbine is assumed to be adiabatic.
2 - Changes in kinetic and potential energies are negligible.
Equations / Data / Solve:
Part a.) The isentropic efficiency of an Eqn 1
We can solve Eqn 1 for WS,act : Eqn 2
Because we know the values of two intensive properties at state 1, we can use the Steam Tables or the NIST Webbook to look-up H1.  Because T1 > Tcritical (1165.3oR), we need to look in the superheated vapor table for properties at state 1.
H1 1779.2 Btu/lbm
The key to determining H2S is the fact that S2S = S1 and we can determine S1 from the Steam Tables or the NIST Webbook.
S1 1.7406 Btu/lbm-oR
S2S 1.7406 Btu/lbm-oR
Now, we know the values of two intensive properties at state 2S, so we can determine the values of other properties at this state, such as T2S and H2S, by interpolating on the Steam Tables or the NIST Webbook.  We begin by determining the phases present.
At P = 60 psia : Ssat liq 0.4276 Btu/lbm-oR Since S2 > Ssat vap, state 2S is a superheated vapor.
Ssat vap 1.6454 Btu/lbm-oR
Interpolation within the 60 psia superheated steam table is required.
At P = 60 psia : T (oF) H (Btu/lbm) S (Btu/lbm-°R)
400 1234.5 1.7149
T2S H2S 1.7406 T2S 449.9 oF
600 1333.1 1.8181 H2S 1259.1 Btu/lbm
Now, we can plug values back into Eqn 2 to evaluate WS,act : Ws,act 457.7 Btu/lbm
Part b.) The 2nd Law Efficiency of a turbine is defined as: Eqn 3
The reversible work can be determined from : Eqn 4
We could determine S2 if we knew H2.  We can determine H2 from and WS,act by applying the 1st Law to the actual, adiabatic process where changes in kinetic and potential energies are negligible. Eqn 5 or: Eqn 6
Plugging values into Eqn 6 yields: H2 1321.5 Btu/lbm
Now that we know the values of two intensive variables at state 2, P2 and H2, we can interpolate on the Steam Tables or NIST Webbook data to determine S2.  Because H2 > H2S , and state 2S is a superheated vapor, we know that the actual state 2 is a superheated vapor.
At P = 60 psia : T (oF) H (Btu/lbm) S (Btu/lbm-°R)
400 1234.5 1.7149
T2 1321.51 S2 T2 576.5 oF
600 1333.1 1.8181 S2 1.8059 Btu/lbm-oR
Now, using Tsurr = 75oF, we can use Eqn 4 to calculate WS,rev and then use Eqn 3 to evaluate the 2nd Law Efficiency of the turbine.
Tsurr 534.67 oR WS,rev 492.6 Btu/lbm
hii 92.9%
Verify: The assumptions made in the solution of this problem cannot be verified with the given information.
Answers : Part a.) Ws,act 458 Btu/lbm Part b.) hii 92.9% 