Example 9F - 1: Air-Standard Brayton Cycle With and Without Regeneration

9F-1 :	Air-Standard Brayton Cycle With and Without Regeneration	10 pts

Consider three air-standard power cycles operating between the same two thermal reservoirs. All three cycles have the same pressure ratio, 12, and the same maximum and minimum temperatures, 2500^oR and 560^oR, respectively.

In each cycle, the mass flow rate of air is 25,000 lb_m/h and the pressure at the compressor inlet is 14.7 psia. Cycle A is an ideal Brayton Cycle. In Cycle B, the compressor and turbine have isentropic efficiencies

of 85% and 90%, respectively. Cycle C uses the same compressor and turbine as Cycle B, but also incorporates a regenerator with an effectiveness of 75%. Calculate the net power output, in hP, and thermal efficiency of each cycle.

Read :

We will need to know all of the H's in order to determine both W_cycle and h. We can get H₁ and H₃ immediately from the Ideal Gas Property Table for air.

Then, for each part of the problem, use the given isentropic compressor and turbine efficiencies to evaluate H₂ and H₄. Then, calculate W_cycle and Q_in for each part and finally the thermal efficiency.

In Cycle C, re-number the streams carefully so you can easily use most of the H's from Cycle B. The key to Cycle C is to use the regenerator effectiveness to determine the H of the combustor feed. Once you have this, you can compute Q_in. W_cycle is the same as in Cycle B. So, calculate h from its definition.

Given:

P₂/P₁

T₃

2500

^oR

P₁

14.7

psia

25,000

lb_m/h

T₁

560

^oR

Cycle A

h_{S, turb}

1.00

Cycle B

h_{S, turb}

0.90

h_{S, comp}

1.00

h_{S, comp}

0.85

Cycle C

h_{S, turb}

0.90

h_{S, comp}

0.85

h_regen

0.75

Find:

For each cycle :

???

W_cycle

???

Diagram:

Cycle A and Cycle B

Assumptions:

1 -

Each component is an open system operating at steady-state.

2 -

The turbine and compressor are adiabatic.

3 -

There are no pressure drops for flow through the heat exchangers.

4 -

Kinetic and potential energy changes are negligible.

5 -

The working fluid is air modeled as an ideal gas.

Equations / Data / Solve:

Stream

T
(^oR)

P
(psia)

H^o
(Btu/lb_m)

S^o
(Btu/lb_m-^oR)

P_r

560

14.7

42.351

0.010149

1.1596

176.4

204.58

1117.2

176.4

180.24

0.010149

13.915

2500

176.4

555.34

0.39732

329.12

14.7

269.21

1338.5

14.7

237.42

0.39732

27.427

We can calculate the thermal efficiency of the cycle when the compressor and turbine are isentropic using :

Eqn 1

We can determine Q₁₂ and Q₃₄ by applying the 1st Law to HEX #1 and HEX #2, respectively.

Each HEX operates at steady-state, involves no shaft work and has negligible changes in kinetic and potential energies. The appropriate forms of the 1st Law are:

Eqn 2

Eqn 3

In order to use Eqns 1 - 3, we must first evaluate H at each state in the cycle.

Let's begin with states 4 and 2 because they are completely fixed by the given data.

We know T₁ and T₃, so we can look-up H₁ and H₃ in the Ideal Gas Property Table for air.

H₁

42.35

Btu / lb_m

H₃

555.3

Btu / lb_m

The two remaining H values depend on the isentropic efficiency of the compressor and the turbine, so they will be different depending on which part of the problem is being considered.

We can determine W_cycle by applying the 1st Law to the entire cycle.

Eqn 4

Because the compressor and turbine are assumed to be adiabatic, Eqn 4 simplifies to:

Eqn 5

So, once we determine Q₂₃ and Q₄₁ for each part of the problem, we can use Eqn 5 to evaluate W_cycle.

Part a.)

Both the compressor and the turbine are isentropic. This changes Eqns 2 & 3 to:

Eqn 6

Eqn 7

We can determine T_1S and T_3S using either the Ideal Gas Entropy Function and the 2nd Gibbs Equation or we can use Relative Properties. Both methods are presented here.

Method 1: Use the Ideal Gas Entropy Function and the 2nd Gibbs Equation.

The 2nd Gibbs Equation for Ideal Gases in terms of the Ideal Gas Entropy Function for the compressor and the turbine are :

Eqn 8

Eqn 9

We can solve Eqns 8 & 9 for the unknowns S^o_T2S & S^o_T4S :

Eqn 10

Eqn 11

We can look up S^o_T1 and S^o_T3 in the Ideal Gas Property Table for air and use it with the known compression ratio in Eqns 10 & 11 to determine S^o_T2S and S^o_T4S. We can do this because the HEX's are isobaric. P₂ = P₃ and P₄ = P₁.

1.987

Btu/lbmole-^oR

29.00

lb_m / lbmole

S^o_T1

0.010149

Btu / lb_m^oR

S^o_T3

0.39732

Btu / lb_m^oR

S^o_T2S

0.18041

Btu / lb_m^oR

S^o_T4S

0.22706

Btu / lb_m^oR

Now, we can use S^o_T2S and S^o_T4S and the Ideal Gas Property Table for air to determine T_2S and T_4S and then H_2S and H_4S by interpolation :

T (^oR)

H^o (Btu/lb_m)

S^o (Btu/lb_m^oR)

1100

175.86

0.17647

T_2S

H_2S

0.18041

Interpolation yields :

T_2S

1117.16

^oR

1120

180.97

0.18106

H_2S

180.24

Btu / lb_m

T (^oR)

H^o (Btu/lb_m)

S^o (Btu/lb_m^oR)

1300

227.38

0.21948

T_4S

H_4S

0.22706

Interpolation yields :

T_4S

1338.52

^oR

1350

240.41

0.22932

H_4S

237.42

Btu / lb_m

Method 2: Use the Ideal Gas Relative Pressure.

When an ideal gas undergoes an isentropic process :

Eqn 12

Eqn 13

Where P_r is the Ideal Gas Relative Pressure, which is a function of T only and we can look-up in the Ideal Gas Property Table for air.

We can solve Eqns 12 & 13 For P_r(T₃) and P_r(T₁), as follows :

Eqn 14

Eqn 15

Look-up P_r(T₁) and P_r(T₃) and use them in Eqns 14 & 15, respectively, To determine P_r(T_2S) and P_r(T_4S):

P_r(T₁)

1.1596

P_r(T₃)

329.12

P_r(T_2S)

13.915

P_r(T_4S)

27.427

We can now determine T_2S and T_4S by interpolation on the the Ideal Gas Property Table for air.

Then, we use T_2S and T_4S to determine H_2S and H_4S from the Ideal Gas Property Table for air.

T (^oR)

P_r

H^o (Btu/lb_m)

1100

13.124

175.86

T_2S

13.915

H_2S

Interpolation yields :

T_2S

1117.39

^oR

1120

14.034

180.97

H_2S

180.30

Btu / lb_m

T (^oR)

P_r

H^o (Btu/lb_m)

1300

24.581

227.38

T_4S

27.427

H_4S

Interpolation yields :

T_4S

1337.49

^oR

1350

28.376

240.41

H_4S

237.15

Btu / lb_m

Since the two methods differ by less than 0.1%, I will use the results from Method 1 in the remaining calculations of this problem.

Now, that we have values for all of the H's, we can plug values back into Eqns 2, 3, 5 & 1 to complete our analysis of Cycle A.

Q₂₃

375.10

Btu / lb_m

W_cycle

4.501E+06

Btu / h

Q₄₁

-195.07

Btu / lb_m

W_cycle

1769

1 hp

2544.5

Btu/h

48.00%

Cycle B

h_{S, turb}

0.90

h_{S, comp}

0.85

We use the isentropic efficiencies of the compressor and the turbine to determine the actual T and H of states 1 and 3.

Eqn 16

Eqn 17

Solve Eqns 12 & 13 for H₂ and H₄, respectively :

Eqn 18

Eqn 19

Plugging values into Eqns 18 & 19 gives:

H₂

204.58

Btu / lb_m

H₄

269.21

Btu / lb_m

We have all of the H's, so we can plug values back into Eqns 2, 3, 5 & 1 to complete our analysis of Cycle B.

Q₂₃

350.76

Btu / lb_m

W_cycle

3.098E+06

Btu / h

Q₄₁

-226.86

Btu / lb_m

W_cycle

1217

W_cycle

123.90

Btu / lb_m

35.32%

Cycle C

h_{S, turb}

0.90

h_{S, comp}

0.85

e_{S, regen}

0.75

Diagram:

Stream

T
(^oR)

P
(psia)

H^o
(Btu/lb_m)

S^o
(Btu/lb_m-^oR)

P_r

560

14.7

42.351

0.010149

1.1596

176.4

204.58

1117.2

176.4

180.24

0.010149

13.915

2500

176.4

555.34

0.39732

329.12

14.7

269.21

1338.5

14.7

237.42

0.39732

27.427

253.05

We can determine the thermal efficiency of the regenerative cycle using:

Eqn 20

The addition of a regenerator does not effect the value of W_cycle. It is the same as in Cycle B.

W_cycle

123.902

Btu / lb_m

The regenerator does reduce the magnitude of both Q₆₃ and Q₅₁.

We can determine Q₁₂ by applying the 1st Law to the combustor.

The combustor operates at steady-state, involves no shaft work and has negligible changes in kinetic and potential energies. The appropriate form of the 1st Law is:

Eqn 21

From Cycle B the following values of H do not change:

H₂

204.58

Btu / lb_m

H₃

555.34

Btu / lb_m

H₄

269.21

Btu / lb_m

So, we need to use the effectiveness of the regenerator to determine H₆.

The effectiveness of the regenerator is given by:

Eqn 22

We can solve Eqn 22 for H₃ :

Eqn 23

Now, we can plug values back into Eqns 23, 21 & 20 :

H₆

253.05

Btu / lb_m

Q₆₂

302.29

Btu / lb_m

W_cycle

1217

W_cycle

3.098E+06

Btu / h

40.99%

Verify:

The assumptions made in the solution of this problem cannot be verified with the given information.

Answers :

Cycle A

W_cycle

1770

48.0%

Cycle B

W_cycle

1220

Cycle C

W_cycle

1220

35.3%

41.0%

Example Problem with Complete Solution