Example 8C - 2: Power & Entropy Generation in Turbine With a Flash Drum

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 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:

P₁

240

lb_f/in²

P₂

lb_f/ in²

T₁

^oF

P₃

lb_f/ in²

m₁

lb_m/s

P₄

lb_f/ in²

h_{S, turb}

0.88

P₅

lb_f/ in²

Find:

a.)

W_turb

Btu/s

b.)

(S_gen)_valve

Btu / s-^oR

(S_gen)_flash

Btu / s-^oR

(S_gen)_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, Q_turb = 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/lb_m

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

H_5S 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: S_5S = S₄

We can look up S₄ in the Ammonia Tables or the NIST Webbook :

S₄

1.2651

Btu lb_m-^oR

S_5S

1.2651

Btu lb_m-^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 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-^oR

Since S_{sat liq} < S₅ < S_{sat vap}, state 5S is a saturated mixture.

S_{sat vap}

1.3691

Btu/lb_m-^oR

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 H₅ :

H₅

567.38

Btu/lb_m

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/lb_m

H₃

84.109

Btu/lb_m

m₃

4.56

lb_m/s

Next we can evaluate m₄ from Eqn 10:

m₄

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.

S₁

0.27391

Btu / lb_m -^oR

S₃

0.18317

Btu / lb_m -^oR

S₄

1.2651

Btu / lb_m -^oR

For S₅, we must first determine the quality using:

Eqn 19

x₅

0.9336

lb_m vap/lb_m

Then we can evaluate S₅ using :

Eqn 20

S₅

1.2820

Btu / lb_m -^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/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 S₂ , 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} < H₂ < H_{sat vap}, state 2 is a saturated mixture.

H_{sat vap}

622.25

Btu/lb_m

Determine x₂ from the specific entropy, using:

Eqn 22

x₂

0.08892

lb_m vap/lb_m

Then, we can use the
quality to determine S₂, using:

Eqn 23

At P = 70 psia :

S_{sat liq}

0.18317

Btu/lb_m-^oR

S_{sat vap}

1.2651

Btu/lb_m-^oR

S₂

0.27938

Btu/lb_m-^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.

(S_gen)_valve

0.02734

Btu / s-^oR

(S_gen)_turb

0.00751

Btu / s-^oR

(S_gen)_flash

0.00000

Btu / s-^oR

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-^oR

The expansion valve generates the most entropy.

(S_gen)_flash

0.00000

Btu / s-^oR

(S_gen)_turb

0.00751

Btu / s-^oR

Example Problem with Complete Solution