9F1 :  AirStandard Brayton Cycle With and Without Regeneration  10 pts 

Consider three airstandard 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^{o}R and 560^{o}R, 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 H_{1} and H_{3} 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 H_{2} and H_{4}. Then, calculate W_{cycle} and Q_{in} for each part and finally the thermal efficiency.  
In Cycle C, renumber 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:  P_{2}/P_{1}  12  ^{}  T_{3}  2500  ^{o}R  
P_{1}  14.7  psia  m  25,000  lb_{m}/h  
T_{1}  560  ^{o}R  
^{}  
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 :  h  ???  %  W_{cycle}  ???  hp  
Diagram:  Cycle A and Cycle B  



Assumptions:  1   Each component is an open system operating at steadystate.  
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 (^{o}R) 
P (psia) 
H^{o} (Btu/lb_{m}) 
S^{o} (Btu/lb_{m}^{o}R) 
P_{r}  
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 Q_{12} and Q_{34} by applying the 1st Law to HEX #1 and HEX #2, respectively.  
Each HEX operates at steadystate, 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 T_{1} and T_{3}, so we can lookup H_{1} and H_{3} in the Ideal Gas Property Table for air.  
H_{1}  42.35  Btu / lb_{m}  H_{3}  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 Q_{23} and Q_{41} 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. P_{2} = P_{3} and P_{4} = P_{1}.  
R  1.987  Btu/lbmole^{o}R  MW  29.00  lb_{m} / lbmole  
S^{o}_{T1}  0.010149  Btu / lb_{m }^{o}R  S^{o}_{T3}  0.39732  Btu / lb_{m }^{o}R  
S^{o}_{T2S}  0.18041  Btu / lb_{m }^{o}R  S^{o}_{T4S}  0.22706  Btu / lb_{m }^{o}R  
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 (^{o}R)  H^{o} (Btu/lb_{m})  S^{o} (Btu/lb_{m}^{o}R)  
1100  175.86  0.17647  
T_{2S}  H_{2S}  0.18041  Interpolation yields :  T_{2S}  1117.16  ^{o}R  
1120  180.97  0.18106  H_{2S}  180.24  Btu / lb_{m}  
T (^{o}R)  H^{o} (Btu/lb_{m})  S^{o} (Btu/lb_{m}^{o}R)  
1300  227.38  0.21948  
T_{4S}  H_{4S}  0.22706  Interpolation yields :  T_{4S}  1338.52  ^{o}R  
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 lookup in the Ideal Gas Property Table for air.  
We can solve Eqns 12 & 13 For P_{r}(T_{3}) and P_{r}(T_{1}), as follows :  

Eqn 14 

Eqn 15  
Lookup P_{r}(T_{1}) and P_{r}(T_{3}) and use them in Eqns 14 & 15, respectively, To determine P_{r}(T_{2S}) and P_{r}(T_{4S}):  
P_{r}(T_{1})  1.1596  P_{r}(T_{3})  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 (^{o}R)  P_{r}  H^{o} (Btu/lb_{m})  
1100  13.124  175.86  
T_{2S}  13.915  H_{2S}  Interpolation yields :  T_{2S}  1117.39  ^{o}R  
1120  14.034  180.97  H_{2S}  180.30  Btu / lb_{m}  
T (^{o}R)  P_{r}  H^{o} (Btu/lb_{m})  
1300  24.581  227.38  
T_{4S}  27.427  H_{4S}  Interpolation yields :  T_{4S}  1337.49  ^{o}R  
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.  
Q_{23}  375.10  Btu / lb_{m}  W_{cycle}  4.501E+06  Btu / h  
Q_{41}  195.07  Btu / lb_{m}  W_{cycle}  1769  hP  
1 hp  2544.5  Btu/h  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 H_{2} and H_{4}, respectively : 

Eqn 18  

Eqn 19  
Plugging values into Eqns 18 & 19 gives:  H_{2}  204.58  Btu / lb_{m}  
H_{4}  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.  
Q_{23}  350.76  Btu / lb_{m}  W_{cycle}  3.098E+06  Btu / h  
Q_{41}  226.86  Btu / lb_{m}  W_{cycle}  1217  hP  
W_{cycle}  123.90  Btu / lb_{m}  h  35.32%  
Cycle C  h_{S, turb}  0.90  h_{S, comp}  0.85  e_{S, regen}  0.75  
Diagram:  



Stream  T (^{o}R) 
P (psia) 
H^{o} (Btu/lb_{m}) 
S^{o} (Btu/lb_{m}^{o}R) 
P_{r}  
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 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 Q_{63} and Q_{51}.  
We can determine Q_{12} by applying the 1st Law to the combustor.  
The combustor operates at steadystate, 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:  
H_{2}  204.58  Btu / lb_{m}  
H_{3}  555.34  Btu / lb_{m}  H_{4}  269.21  Btu / lb_{m}  
So, we need to use the effectiveness of the regenerator to determine H_{6}.  
The effectiveness of the regenerator is given by: 

Eqn 22  
We can solve Eqn 22 for H_{3} : 

Eqn 23  
Now, we can plug values back into Eqns 23, 21 & 20 :  
H_{6}  253.05  Btu / lb_{m}  
^{}  
Q_{62}  302.29  Btu / lb_{m}  W_{cycle}  1217  hP  
W_{cycle}  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  W_{cycle}  1770  hP  
h  48.0%  
Cycle B  W_{cycle}  1220  hP  Cycle C  W_{cycle}  1220  hP  
h  35.3%  h  41.0%  