8C2 :  Power & Entropy Generation in Turbine With a Flash Drum  8 pts 

A flash drum is a vessel in which gravity is allowed to separate a liquid and a gas. A throttling device can be used to reduce the pressure on a liquid stream to produce a twophase mixture.  
The twophase mixture enters the flash drum where the liquid settles to the bottom and the vapor rises to the top. Because the vapor and liquid phases are allowed to reach equilibrium, the vapor and liquid leaving a flash drum  


Calculate the power output of the turbine and the entropy generation rate for the valve, for the flash drum and for the turbine. Which unit or units generate a large amount of entropy?  
Read :  This is a complicated problem just because it is an ensemble of three processes. Drawing a good flow diagram that includes all the given information is essential. Fortunately, the problem statement includes a good flow diagram.  
Start from State 1 because you know a lot of information about this stream. The valve behaves as a throttling valve. Apply mass and energy balances to the flash drum to determine m_{3} and m_{4}. Use the isentropic efficiency to calculate H_{5} and then use H_{5} to get x_{5}. Application of the 1st Law to the turbine yields W_{turb}.  
To get the correct entropy generated, you must be very accuarate in your calculations. Do not round off until the very end. Save the intermediate results in your calculator memory or use Excel.  
Given:  P_{1}  240  lb_{f}/in^{2}  P_{2}  70  lb_{f }/ in^{2}  
T_{1}  80  ^{o}F  P_{3}  70  lb_{f }/ in^{2}  
m_{1}  5  lb_{m}/s  P_{4}  70  lb_{f }/ in^{2}  
h_{S, turb}  0.88  P_{5}  20  lb_{f }/ in^{2}  
Find:  a.)  W_{turb}  ?  Btu/s  b.)  (S_{gen})_{valve}  ?  Btu / s^{o}R  
(S_{gen})_{flash}  ?  Btu / s^{o}R  
(S_{gen})_{turb}  ?  Btu / s^{o}R  
Diagram:  The necessary flow diagram was provided in the problem statement.  
Assumptions:  1   Each component operates at steadystate with negligible heat transfer between the flowing fluid and the surroundings.  
2   Kinetic and potential energy changes are negligible.  
3   The expansion across the valve is an isenthalpic throttling process.  
Equations / Data / Solve:  
Part a.)  Let's begin by applying the steadystate mass balance equation to the valve, the flash drum and the turbine, one unit at a time.  

Eqn 1 

Eqn 2 

Eqn 3  
Next, apply the 1st Law to the turbine. The turbine is a steadystate, SISO process with negligible changes in kinetic and potential energies. The appropriate form of the 1st Law is:  



Eqn 4  
In Eqn 4, m_{5} was eliminated using Eqn 3. Because the turbine is also assumed to be adiabatic, Q_{turb} = 0 and Eqn 4 becomes:  

Eqn 5  
We can lookup H_{4} in the Ammonia Tables or NIST Webbook because we know it is a saturated vapor at 70 psia.  
H_{4}  622.25  Btu/lb_{m}  
We must use the isentropic efficiency of the turbine to determine H_{5} because we only know the value of one intensive variable at state 5 (P_{5}).  
Isentropic
efficiency applied to our turbine is defined by: 

Eqn 6  
We can solve Eqn 6 for H_{5}, as follows: 

Eqn 7  
H_{5S} is the enthalpy of the effluent (at P_{2}) from an adiabatic, isentropic turbine that has the same feed as the actual turbine. Because this hypothetical turbine is isentropic: S_{5S} = S_{4}  
We can look up S_{4} in the Ammonia Tables or the NIST Webbook :  S_{4}  1.2651  Btu lb_{m}^{o}R  
S_{5S}  1.2651  Btu lb_{m}^{o}R  
Now, we know the values of two intensive properties at state 5S, so we can determine the values of other properties at this state, such as T_{5S} and H_{5S}, by interpolating on the Ammonia Tables or the NIST Webbook. We begin by determining the phases present.  
At P = 20 psia :  S_{sat liq}  0.057608  Btu/lb_{m}^{o}R  Since S_{sat liq} < S_{5} < S_{sat vap}, state 5S is a saturated mixture.  
S_{sat vap}  1.3691  Btu/lb_{m}^{o}R  
Determine x_{5S} from the specific entropy, using: 

Eqn 8  
x_{5S}  0.9207  lb_{m} vap/lb_{m}  
Then, we can use
the quality to determine H_{5S}, using: 

Eqn 9  
At P = 20 psia :  H_{sat liq}  24.887  Btu/lb_{m}  
H_{sat vap}  605.98  Btu/lb_{m}  H_{5S}  559.90  Btu/lb_{m}  
Now, we can use Eqn 7 to evaluate H_{5} :  H_{5}  567.38  Btu/lb_{m}  
Next, we need to evaluate m_{4}. Combine Eqns 1 & 2 to get : 

Eqn 10  
We know m_{1}, so we need to find m_{3} to calculate m_{4}.  
We can determine m_{3} by applying the 1st Law to the flash drum. The flash drum is adiabatic, operates at steadystate, no shaft work crosses it boundaries and changes in kinetic and potential energies are negligible. Therefore, the appropriate form of the 1st Law is:  

Eqn 11  
Use Eqn 10 to eliminate m_{4} from Eqn 11 and use Eqn 1 to eliminate m_{2} from Eqn 11 and you are left with:  

Eqn 12  
We can solve Eqn 12 for m_{3} , as follows: 

Eqn 13  
We can lookup the specific enthalpies in states 1 and 3 in the Ammonia Tables or the NIST Webbook and we already know H_{4}. Then, we can plug these values into Eqn 13 to evaluate m_{3} :  
H_{1}  131.96  Btu/lb_{m}  
H_{3}  84.109  Btu/lb_{m}  m_{3}  4.56  lb_{m}/s  
Next we can evaluate m_{4} from Eqn 10:  m_{4}  0.44  lb_{m}/s  
At last, we can plug values back into Eqn 5 to complete part (a) :  W_{turb}  24.39  Btu/s  
Part b.)  The entropy generation rate is defined in the 2nd Law as: 

Eqn 14  
Since we assumed that
each of our three processes was adiabatic, Eqn 14 simplifies to: 

Eqn 15  
Apply Eqn 15 to each of the three processes: 

Eqn 16  

Eqn 17  

Eqn 18  
At this point, we have determined the state of all five streams in this process, so we can use the Ammonia Tables or the NIST Webbook to evaluate the entropy of each.  
The specific entropies of streams 1, 3 and 4 come straight out of the Ammonia Tables or NIST Webbook.  
S_{1}  0.27391  Btu / lb_{m} ^{o}R  S_{3}  0.18317  Btu / lb_{m} ^{o}R  
S_{4}  1.2651  Btu / lb_{m} ^{o}R  
For S_{5}, we must first determine the quality using: 

Eqn 19  
x_{5}  0.9336  lb_{m} vap/lb_{m}  
Then we can evaluate S_{5} using : 

Eqn 20  
S_{5}  1.2820  Btu / lb_{m} ^{o}R  
In order to determine S_{2}, we must apply the 1st Law to the valve. We assume the valve operates at steadystate, is adiabatic, exhibits negligible changes in kinetic or potential energies and involves no shaft work. Under these conditions, the 1st Law tells us that the valve is an isenthalpic throttling device.  

Eqn 21  H_{2}  131.96  Btu/lb_{m}  
Now, we know the values of two intensive properties at state 2, so we can determine the values of other properties at this state, such as S_{2} , by interpolating on the Ammonia Tables or the NIST Webbook. We begin by determining the phases present.  
At P = 70 psia :  H_{sat liq}  84.109  Btu/lb_{m}  Since H_{sat liq} < H_{2} < H_{sat vap}, state 2 is a saturated mixture.  
H_{sat vap}  622.25  Btu/lb_{m}  
Determine x_{2} from the specific entropy, using: 

Eqn 22  
x_{2}  0.08892  lb_{m} vap/lb_{m}  
Then, we can use
the quality to determine S_{2}, using: 

Eqn 23  
At P = 70 psia :  S_{sat liq}  0.18317  Btu/lb_{m}^{o}R  
S_{sat vap}  1.2651  Btu/lb_{m}^{o}R  S_{2}  0.27938  Btu/lb_{m}^{o}R  
Finally, we have all the values necessary to plug into Eqns 1618 to evaluate the entropy generation in each device and then Eqn 24 to evaluate the lost work in each device.  
(S_{gen})_{valve}  0.02734  Btu / s^{o}R  (S_{gen})_{turb}  0.00751  Btu / s^{o}R  
(S_{gen})_{flash}  0.00000  Btu / s^{o}R  
Verify:  The assumptions made in the solution of this problem cannot be verified with the given information.  
Answers:  a.)  W_{turb}  24.4  Btu/s  
b.)  (S_{gen})_{valve}  0.0273  Btu / s^{o}R  The expansion valve generates the most entropy.  
(S_{gen})_{flash}  0.00000  Btu / s^{o}R  
(S_{gen})_{turb}  0.00751  Btu / s^{o}R 