Example Problem with Complete Solution

9F-1 : Air-Standard Brayton Cycle With and Without Regeneration 10 pts
An air-standard Brayton cycle has a compressor pressure ratio of 10. Air enters the compressor at P4 = 14.7 psia, T4 = 540°R, with a mass flow rate of 90,000 lbm/h. The turbine inlet temperature is 2160°R. Calculate the thermal efficiency and the net power developed, in horsepower, if…
a.) the turbine and the compressor isentropic efficiencies are each 100%.
b.) the turbine and compressor isentropic efficiencies are 88 and 84%, respectively.
c.) the turbine and compressor isentropic efficiencies are 88 and 94%, respectively, and a regenerator with an effectiveness of 80% is incorporated.
Read: We will need to know all of the H's in order to determine both Wcycle and h.  We can get H4 and H2 immediately from the Ideal Gas Property Table for air.
Then, for each part of the problem, use the give isentropic compressor and turbine efficiencies to evaluate H1 and H3.  Then, calculate Wcycle and Qin for each part and finally the efficiency.
In part (c), renumber the stream carefully so you can easily use most of the H's from part (b).  The key to part (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 part (b).  So, calculate h from its definition.
Given: P1/P4 10 T2 2200 oR
P4 14.7 psia m 90,000 lbm/h
T4 540 oR
Part a.) ηS, turb 1.00 Part b.) ηS, turb 0.88
ηS, comp 1.00 ηS, comp 0.84
Part c.) ηS, turb 0.88
ηS, comp 0.84
ηregen 0.80
Find : η ??? % Wcycle ??? hp
Diagrams :
Assumptions :
- Each component is open system operating at steady-state.
- The turbine and compressor processes are adiabatic.
- There are no pressure drops for flow through the heat exchangers.
- Kinetic and potential energy changes are negligible.
- 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)
1
147
45.0
1S
1027.9
157.55
0.15924
2
2200
470.7
0.3613
3
238.8
3S
1222.2
207.2
0.20348
4
540
14.70
37.58
0.001473
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 internal 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 T2 and T4, so we can look up H2 and H4 in the Ideal Gas Property Table for air.
H2 470.670 Btu / lbm H4 37.580 Btu / lbm
The two remaining H values depend on the isentropic efficiency of the compressor and the turbine, so they will be diffierent 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 Q12 and Q34 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 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 SoT1 & SoT3 :  
   
Eqn 10
   
Eqn 11
   
We can look up SoT2 and SoT4 in the Ideal Gas Property Table for air and use it with the known compression ratio in Eqns 10 & 11 to determine SoT3S and SoT1S.  We can do this because the HEX's are isobaric.  P1 = P2 and P3 = P4.
   
R 1.987 Btu/lbmole-oR  
MW 29.00 lbm / lbmole  
   
SoT2 0.36125 Btu / lbm oR SoT4 0.0014732 Btu / lbm oR
SoT3S 0.20348 Btu / lbm oR SoT1S 0.15924 Btu / lbm oR
   
Now, we can use SoT1S and SoT3S and the Ideal Gas Property Table for air to determine T1S and T3S and then H1S and H3S by interpolation :
   
T (oR) Ho (Btu/lbm) So (Btu/lbmoR)  
1020 155.55 0.15730  
T1S H1S 0.15924 T1S 1027.89 oR
1030 158.08 0.15976 H1S 157.55 Btu / lbm
   
T (oR) Ho (Btu/lbm) So (Btu/lbmoR)  
1220 206.65 0.20302  
T3S H3S 0.20348 T3S 1222.20 oR
1230 209.23 0.20513   H3S 207.22 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(T2) and Pr(T2) and use them in Eqns 14 & 15, respectively, To determine Pr(T3) and Pr(T1):
Pr(T2) 194.46 Pr(T4) 1.0217
Pr(T3S) 19.446 Pr(T1S) 10.217
   
We can now determine T3S and T1S by interpolation on the the Ideal Gas Property Table for air.
Then, we use T3S and T1S to determine H3S and H1S from the Ideal Gas Property Table for air.
T (oR) Pr Ho (Btu/lbm)  
1020 9.9221 155.55  
T1S 10.217 H1S T1S 1028.10 oR
1030 10.286 158.08 H1S 157.60 Btu / lbm
   
T (oR) Pr Ho (Btu/lbm)  
1220 19.334 206.65  
T3S 19.446 H3S T3S 1221.85 oR
1230 19.938 209.23   H3S 207.13 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 4, 5 & 1 to complete part (a).
Q12 313.12 Btu / lbm Wcycle 1.291E+07 Btu / h
Q34 -169.64 Btu / lbm Wcycle 5075 hP
1 hp 2544.5 Btu/h h 45.82%
Part b.) ηS, turb 0.88 ηS, comp 0.84
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 H3 and H1, respectively :
Eqn 18
Eqn 19
Plugging values into Eqns 18 & 19 gives: H1 180.40 Btu / lbm
H3 238.83 Btu / lbm
We have all of the H's, so we can plug values back into Eqns 4, 5 & 1 to complete part (b).
Q12 290.27 Btu / lbm Wcycle 8.012E+06 Btu / h
Q34 -201.25 Btu / lbm Wcycle 3149 hP
Wcycle 89.02 Btu / lbm η 30.67%
Part c.) ηS, turb 0.88
ηS, comp 0.84 εS, reg 0.80
Diagram:
Stream T
(oR)
P
(psia)
Ho
(Btu/lbm)
So
(Btu/lbm-oR)
1   147 180.4 0.0000
1S 1027.9   157.55 0.15924
2 2200   470.7 0.3613
3     238.8  
3S 1222.2   207.2 0.20348
4 540 14.70 37.58 0.001473
5        
6     227.14  
We can determine the thermal efficiency
of the regenerative cycles using:
Eqn 20
The addition of a regenerator does not effect the value of Wcycle.  It is the same as in part (b).
Wcycle 89.023 Btu / lbm
The regenerator does reduce the magnitude of both Q62 and Q54.
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 internal energies.  The appropriate form of the 1st Law is:
Eqn 21
From part (b) the following values of H do not change:
H1 180.40 Btu / lbm
H2 470.67 Btu / lbm H3 238.83 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 227.14 Btu / lbm
Q62 243.53 Btu / lbm Wcycle 3148.8 hP
Wcycle 8.012E+06 Btu / h η 36.56%
Verify : The assumptions made in the solution of this problem cannot be verified with the given information.
Answers: a.) Wcycle 5075 hP
η 45.8%  
b.) Wcycle 3149 hP
η 30.7%  
c.) Wcycle 3149 hP
η 36.6%