8D3 :  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 1500^{o}F and the isentropic efficiency is 88%. 

a.) Calculate W_{S,act} in Btu/lb_{m} b.) Calculate the 2nd Law Efficiency of the turbine. Assume T_{surr} = 75^{o}F. 

Read :  The key to this problem is to assume that the turbine is adiabatic.  
We can calculate the isentropic work of the turbine because S_{2} = S_{1} 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 S_{2}.  
Given:  h_{s}  88%  Find:  W_{s,act}  ???  Btu/lb_{m}  
P_{1}  2000  psia  h_{ii}  ???  %  
T_{1}  1500  ^{o}F  
P_{2}  60  psia  
T_{surr}  75  ^{o}F  
^{}  
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 adiabatic turbine is defined by: 

Eqn 1  
We can solve Eqn 1 for W_{S,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 lookup H_{1}. Because T_{1} > T_{critical} (1165.3^{o}R), we need to look in the superheated vapor table for properties at state 1.  
H_{1}  1779.2  Btu/lb_{m}  
The key to determining H_{2S} is the fact that S_{2S} = S_{1} and we can determine S_{1} from the Steam Tables or the NIST Webbook.  
S_{1}  1.7406  Btu/lb_{m}^{o}R  
S_{2S}  1.7406  Btu/lb_{m}^{o}R  
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 T_{2S} and H_{2S}, by interpolating on the Steam Tables or the NIST Webbook. We begin by determining the phases present.  
At P = 60 psia :  S_{sat liq}  0.4276  Btu/lb_{m}^{o}R  Since S_{2} > S_{sat vap}, state 2S is a superheated vapor.  
S_{sat vap}  1.6454  Btu/lb_{m}^{o}R  
Interpolation within the 60 psia superheated steam table is required.  
At P = 60 psia :  T (^{o}F)  H (Btu/lb_{m})  S (Btu/lb_{m}°R)  
400  1234.5  1.7149  
T_{2S}  H_{2S}  1.7406  T_{2S}  449.9  ^{o}F  
600  1333.1  1.8181  H_{2S}  1259.1  Btu/lb_{m}  
Now, we can plug values back into Eqn 2 to evaluate W_{S,act }:  W_{s,act}  457.7  Btu/lb_{m}  
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 S_{2} if we knew H_{2}. We can determine H_{2} from and W_{S,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:  H_{2}  1321.5  Btu/lb_{m}  
Now that we know the values of two intensive variables at state 2, P_{2} and H_{2}, we can interpolate on the Steam Tables or NIST Webbook data to determine S_{2}. Because H_{2} > H_{2S} , and state 2S is a superheated vapor, we know that the actual state 2 is a superheated vapor.  
At P = 60 psia :  T (^{o}F)  H (Btu/lb_{m})  S (Btu/lb_{m}°R)  
400  1234.5  1.7149  
T_{2}  1321.51  S_{2}  T_{2}  576.5  ^{o}F  
600  1333.1  1.8181  S_{2}  1.8059  Btu/lb_{m}^{o}R  
Now, using T_{surr} = 75^{o}F, we can use Eqn 4 to calculate W_{S,rev} and then use Eqn 3 to evaluate the 2nd Law Efficiency of the turbine.  
T_{surr}  534.67  ^{o}R  W_{S,rev}  492.6  Btu/lb_{m}  
^{}  
h_{ii}  92.9%  
Verify:  The assumptions made in the solution of this problem cannot be verified with the given information.  
Answers :  Part a.)  W_{s,act}  458  Btu/lb_{m}  Part b.)  h_{ii}  92.9% 