# Example Problem with Complete Solution

8C-2 : 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 two-phase mixture.
The two-phase 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 are both saturated. Consider the system in the diagram shown below.

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 m3 and m4. Use the isentropic efficiency to calculate H5 and then use H5 to get x5. Application of the 1st Law to the turbine yields Wturb.
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: P1 240 lbf/in2 P2 70 lbf / in2
T1 80 oF P3 70 lbf / in2
m1 5 lbm/s P4 70 lbf / in2
hS, turb 0.88 P5 20 lbf / in2
Find: a.) Wturb ? Btu/s b.) (Sgen)valve ? Btu / s-oR
(Sgen)flash ? Btu / s-oR
(Sgen)turb ? Btu / s-oR
Diagram: The necessary flow diagram was provided in the problem statement.
Assumptions: 1 - Each component operates at steady-state 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 steady-state 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 steady-state, SISO process with negligible changes in kinetic and potential energies.  The appropriate form of the 1st Law is:  Eqn 4
In Eqn 4, m5 was eliminated using Eqn 3.  Because the turbine is also assumed to be adiabatic, Qturb = 0 and Eqn 4 becomes: Eqn 5
We can lookup H4 in the Ammonia Tables or NIST Webbook because we know it is a saturated vapor at 70 psia.
H4 622.25 Btu/lbm
We must use the isentropic efficiency of the turbine to determine H5 because we only know the value of one intensive variable at state 5 (P5).
Isentropic efficiency applied to
our
turbine is defined by: Eqn 6
We can solve Eqn 6 for H5, as follows: Eqn 7
H5S is the enthalpy of the effluent (at P2) from an adiabatic, isentropic turbine that has the same feed as the actual turbine.  Because this hypothetical turbine is isentropic:      S5S = S4
We can look up S4 in the Ammonia Tables or the NIST Webbook : S4 1.2651 Btu lbm-oR
S5S 1.2651 Btu lbm-oR
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 T5S and H5S, by interpolating on the Ammonia Tables or the NIST Webbook.  We begin by determining the phases present.
At P = 20 psia : Ssat liq 0.057608 Btu/lbm-oR Since Ssat liq < S5 < Ssat vap, state 5S is a saturated mixture.
Ssat vap 1.3691 Btu/lbm-oR
Determine x5S from the specific entropy, using: Eqn 8
x5S 0.9207 lbm vap/lbm
Then, we can use the
quality to determine H5S, using: Eqn 9
At P = 20 psia : Hsat liq 24.887 Btu/lbm
Hsat vap 605.98 Btu/lbm H5S 559.90 Btu/lbm
Now, we can use Eqn 7 to evaluate H5 : H5 567.38 Btu/lbm
Next, we need to evaluate m4.  Combine Eqns 1 & 2 to get : Eqn 10
We know m1, so we need to find m3 to calculate m4.
We can determine m3 by applying the 1st Law to the flash drum.  The flash drum is adiabatic, operates at steady-state, 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 m4 from Eqn 11 and use Eqn 1 to eliminate m2 from Eqn 11 and you are left with: Eqn 12
We can solve Eqn 12 for m3 , 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 H4.  Then, we can plug these values into Eqn 13 to evaluate m3 :
H1 131.96 Btu/lbm
H3 84.109 Btu/lbm m3 4.56 lbm/s
Next we can evaluate m4 from Eqn 10: m4 0.44 lbm/s
At last, we can plug values back into Eqn 5 to complete part (a) : Wturb 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.
S1 0.27391 Btu / lbm -oR S3 0.18317 Btu / lbm -oR
S4 1.2651 Btu / lbm -oR
For S5, we must first determine the quality using: Eqn 19
x5 0.9336 lbm vap/lbm
Then we can evaluate S5 using : Eqn 20
S5 1.2820 Btu / lbm -oR
In order to determine S2, we must apply the 1st Law to the valve.  We assume the valve operates at steady-state, 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 H2 131.96 Btu/lbm
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 S2 , by interpolating on the Ammonia Tables or the NIST Webbook.  We begin by determining the phases present.
At P = 70 psia : Hsat liq 84.109 Btu/lbm Since Hsat liq < H2 < Hsat vap, state 2 is a saturated mixture.
Hsat vap 622.25 Btu/lbm
Determine x2 from the specific entropy, using: Eqn 22
x2 0.08892 lbm vap/lbm
Then, we can use the
quality to determine S2, using: Eqn 23
At P = 70 psia : Ssat liq 0.18317 Btu/lbm-oR
Ssat vap 1.2651 Btu/lbm-oR S2 0.27938 Btu/lbm-oR
Finally, we have all the values necessary to plug into Eqns 16-18 to evaluate the entropy generation in each device and then Eqn 24 to evaluate the lost work in each device.
(Sgen)valve 0.02734 Btu / s-oR (Sgen)turb 0.00751 Btu / s-oR
(Sgen)flash 0.00000 Btu / s-oR
Verify: The assumptions made in the solution of this problem cannot be verified with the given information. 