# Example Problem with Complete Solution

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, 2500oR and 560oR, respectively.
In each cycle, the mass flow rate of air is 25,000 lbm/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 Wcycle 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 Wcycle and Qin 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 Qin.  Wcycle is the same as in Cycle B.  So, calculate h from its definition.
Given: P2/P1 12 T3 2500 oR
P1 14.7 psia m 25,000 lbm/h
T1 560 oR
Cycle A hS, turb 1.00 Cycle B hS, turb 0.90
hS, comp 1.00 hS, comp 0.85
Cycle C hS, turb 0.90
hS, comp 0.85
hregen 0.75
Find: For each cycle : h ??? % Wcycle ??? hp
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)
Ho
(Btu/lbm)
So
(Btu/lbm-oR)
Pr
1 560 14.7 42.351 0.010149 1.1596
2   176.4 204.58
2S 1117.2 176.4 180.24 0.010149 13.915
3 2500 176.4 555.34 0.39732 329.12
4   14.7 269.21
4S 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 / lbm H3 555.3 Btu / lbm
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 Wcycle 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 Wcycle.
Part a.) Both the compressor and the turbine are isentropic. This changes Eqns 2 & 3 to: Eqn 6 Eqn 7
We can determine T1S and T3S 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 SoT2S & SoT4S : Eqn 10 Eqn 11

We can look up SoT1 and SoT3 in the Ideal Gas Property Table for air and use it with the known compression ratio in Eqns 10 & 11 to determine SoT2S and SoT4S.  We can do this because the HEX's are isobaric.  P2 = P3 and P4 = P1.

R 1.987 Btu/lbmole-oR MW 29.00 lbm / lbmole

SoT1 0.010149 Btu / lbm oR SoT3 0.39732 Btu / lbm oR
SoT2S 0.18041 Btu / lbm oR SoT4S 0.22706 Btu / lbm oR

Now, we can use SoT2S and SoT4S and the Ideal Gas Property Table for air to determine T2S and T4S and then H2S and H4S by interpolation :

T (oR) Ho (Btu/lbm) So (Btu/lbmoR)
1100 175.86 0.17647
T2S H2S 0.18041 Interpolation yields : T2S 1117.16 oR
1120 180.97 0.18106 H2S 180.24 Btu / lbm

T (oR) Ho (Btu/lbm) So (Btu/lbmoR)
1300 227.38 0.21948
T4S H4S 0.22706 Interpolation yields : T4S 1338.52 oR
1350 240.41 0.22932 H4S 237.42 Btu / lbm
Method 2: Use the Ideal Gas Relative Pressure.

When an ideal gas undergoes an isentropic process : Eqn 12 Eqn 13

Where Pr 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 Pr(T3) and Pr(T1), as follows : Eqn 14 Eqn 15

Look-up Pr(T1) and Pr(T3) and use them in Eqns 14 & 15, respectively, To determine Pr(T2S) and Pr(T4S):

Pr(T1) 1.1596 Pr(T3) 329.12
Pr(T2S) 13.915 Pr(T4S) 27.427

We can now determine T2S and T4S by interpolation on the the Ideal Gas Property Table for air.

Then, we use T2S and T4S to determine H2S and H4S from the Ideal Gas Property Table for air.

T (oR) Pr Ho (Btu/lbm)
1100 13.124 175.86
T2S 13.915 H2S Interpolation yields : T2S 1117.39 oR
1120 14.034 180.97 H2S 180.30 Btu / lbm

T (oR) Pr Ho (Btu/lbm)
1300 24.581 227.38
T4S 27.427 H4S Interpolation yields : T4S 1337.49 oR
1350 28.376 240.41 H4S 237.15 Btu / lbm
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 / lbm Wcycle 4.501E+06 Btu / h
Q41 -195.07 Btu / lbm Wcycle 1769 hP
1 hp 2544.5 Btu/h h 48.00%
Cycle B hS, turb 0.90 hS, 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 / lbm
H4 269.21 Btu / lbm
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 / lbm Wcycle 3.098E+06 Btu / h
Q41 -226.86 Btu / lbm Wcycle 1217 hP
Wcycle 123.90 Btu / lbm h 35.32%
Cycle C hS, turb 0.90 hS, comp 0.85 eS, regen 0.75
Diagram:  Stream T
(oR)
P
(psia)
Ho
(Btu/lbm)
So
(Btu/lbm-oR)
Pr
1 560 14.7 42.351 0.010149 1.1596
2   176.4 204.58
2S 1117.2 176.4 180.24 0.010149 13.915
3 2500 176.4 555.34 0.39732 329.12
4   14.7 269.21
4S 1338.5 14.7 237.42 0.39732 27.427
5
6     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 Wcycle.  It is the same as in Cycle B.
Wcycle 123.902 Btu / lbm
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 / lbm
H3 555.34 Btu / lbm H4 269.21 Btu / lbm
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 / lbm
Q62 302.29 Btu / lbm Wcycle 1217 hP
Wcycle 3.098E+06 Btu / h h 40.99%
Verify: The assumptions made in the solution of this problem cannot be verified with the given information.
Answers : Cycle A Wcycle 1770 hP
h 48.0%
Cycle B Wcycle 1220 hP Cycle C Wcycle 1220 hP
h 35.3%   h 41.0% 