9E2 :  Performance of a "Real" Brayton Cycle  10 pts 

A real gas power cycle is similar to the Brayton Cycle, but the compressor and turbine are not isentropic. The compressor and turbine have isentropic efficiencies of 80% and 85%, respectively.  


Use airstandard analysis to determine… a.) Q_{C} and W_{net}, in kJ/kg b.) The thermal efficiency c.) Repeat parts (a) and (b) for an ideal Brayton Cycle (isentropic turbine and compressor). 

Read :  In order to evaluate all of the W_{S} and Q values that we need to answer all the parts of this question, we will need to know the H values at every state. In addition, we will need to know H_{1S} and H_{3S} because part (c) requires that we analyze the ideal cycle as well the actual cycle.  
We can lookup H_{2} and H_{4} immediately, but we need to use the isnetropic efficiency and either the Ideal Gas Entropy Function and the 2nd Gibbs Equation or the Ideal Relative Pressure to determine H_{1} and H_{3}. Whichever method we choose, we will compute H_{1S} and H_{2S} in the process, so when we are done with part (a), we will have all the values we need to complete part (c).  
Once we know all the H values, it is a straightforward process to apply the 1st Law to each process in the cycle in order to answer the questions in parts (a) & (b). We repeat these calculations in part (c) using H_{1S} instead of H_{1} and H_{3S} instead of H_{3}. Finally, calculate the % Change in each answer for our comparison.  
Given:  h_{S, comp}  0.80  ^{}  T_{2}  1800  K  
h_{S, turb}  0.85  T_{4}  300  K  
P_{1}/P_{4}  15  ^{}  
Find:  a.)  W_{cycle}  ???  kJ/kg  c.)  (W_{cycle})_{ideal}  ???  kJ/kg  
Q_{out}  ???  kJ/kg  (Q_{out})_{ideal}  ???  kJ/kg  
b.)  h  ???  %  h_{ideal}  ???  %  
Diagram:  The flow diagram in the problem statement is adequate. A TS Diagram will also be useful.  


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 (K) 
H^{o} (kJ/kg) 
S^{o} (kJ/kgK) 

1  717.35  523.58  
1S  636.61  436.34  0.78254  
2  1800  1791.5  1.9784  
3  1068.4  916.49  
3S  932.67  762.08  1.20203  
4  300  87.410  0.0061681  
Part a.)  Only the compressor and the turbine have shaft work interactions, so the net work for the cycle is given by:  

Eqn 1  
Apply the 1st Law to the turbine and the compressor. They are adiabatic, operate at steadystate and changes in kinetic and potential energies are negligible.  

Eqn 2 

Eqn 3  
We know T_{2} and T_{4}, so we can lookup H_{2} and H_{4}_{ }in the Ideal Gas Properties Table for air.  
H_{2}  1791.5  kJ/kg  H_{4}  87.410  kJ/kg  
We can determine T_{1} and T_{3} 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 is : 

Eqn 4  

Eqn 5  
We can solve Eqns 4 & 5 for the unknowns S^{o}_{T1} & S^{o}_{T3} : 

Eqn 6  

Eqn 7  
We can look up S^{o}_{T2} and S^{o}_{T4} in the Ideal Gas Property Table for air and use it with the known compression ratio in Eqns 6 & 7 to
determine S^{o}_{T3} and S^{o}_{T1}. We can do this because the HEX's are isobaric. P_{1} = P_{2} and P_{3} = P_{4}. 

R  8.314  J/molK  MW  29.00  g/mol  
S^{o}_{T2}  1.9784  kJ/kgK  S^{o}_{T4}  0.0061681  kJ/kgK  
S^{o}_{T3S}  1.2020  kJ/kgK  S^{o}_{T1S}  0.78254  kJ/kgK  
Now, we can use S^{o}_{T1S} and S^{o}_{T3S} and the Ideal Gas Property Table for air to determine T_{1S} and T_{3S} and then H_{1S} and H_{3S} by interpolation :  
T (K)  H^{o} (kJ/kg)  S^{o} (kJ/kg  
630  429.25  0.77137  
T_{1S}  H_{1S}  0.78254  Interpolation yields :  T_{1S}  636.61  K  
640  439.98  0.78826  H_{1S}  436.34  kJ/kg  
T (K)  H^{o} (kJ/kg)  S^{o} (kJ/kg  
920  747.82  1.1867  
T_{3S}  H_{3S}  1.2020  Interpolation yields :  T_{3S}  932.67  K  
940  770.33  1.2109  H_{3S}  762.08  kJ/kg  
Method 2: Use the Ideal Gas Relative Pressure.  
When an ideal gas undergoes an isentropic process :  

Eqn 8 

Eqn 9  
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 8 & 9 For P_{r}(T_{3}) and P_{r}(T_{1}), as follows :  

Eqn 10 

Eqn 11  
Lookup P_{r}(T_{2}) and P_{r}(T_{2}) and use them in Eqns 10 & 11, respectively, To determine P_{r}(T_{3}) and P_{r}(T_{1}):  
P_{r}(T_{2})  986.20  P_{r}(T_{4})  1.0217  
P_{r}(T_{3S})  65.747  P_{r}(T_{1S})  15.326  
We can now determine T_{3S} and T_{1S} by interpolation on the the Ideal Gas Property Table for air.  
Then, we use T_{3S} and T_{1S} to determine H_{3S} and H_{1S} from the Ideal Gas Property Table for air.  
T (K)  P_{r}  H^{o} (kJ/kg)  
630  14.7  429.25  
T_{1S}  15.326  H_{1S}  Interpolation yields :  T_{1S}  637.02  K  
640  15.591  439.98  H_{1S}  436.78  kJ/kg  
T (K)  P_{r}  H^{o} (kJ/kg)  
920  62.489  747.82  
T_{3S}  65.747  H_{3S}  Interpolation yields :  T_{3S}  931.84  K  
940  67.990  770.33  H_{3S}  761.15  kJ/kg  
Since the two methods differ by only about 0.1%, I will use the results from Method 1 in the remaining calculations of this problem.  
Next, we use the isentropic efficiencies of the compressor and the turbine to determine the actual T and H of states 1 and 3.  

Eqn 12 

Eqn 13  
Solve Eqns 12 & 13 for H_{3} and H_{1}, respectively : 

Eqn 14  

Eqn 15  
Plugging values into Eqns 14 & 15 gives:  H_{1}  523.58  kJ/kg  
H_{3}  916.49  kJ/kg  
And by interpolation on the Ideal Gas Property Tables:  
T (K)  H^{o} (kJ/kg)  T (K)  H^{o} (kJ/kg)  
710  515.58  1060  906.80  
T_{1}  523.58  T_{3}  916.49  
720  526.46  1080  929.77  
T_{1}  717.35  K  T_{3}  1068.44  K  
Now that we have fixed all the states and determined the values of all the H's, we can plug values back into Eqns 1  3 and complete part (a).  
W_{S,turb}  875.01  kJ/kg  
W_{S,comp}  436.17  kJ/kg  W_{cycle}  438.84  kJ/kg  
Heat tranfer out of the system occurs in step 34. We can determine Q_{34} by appplying the 1st Law to HEX #2. The HEX operates at steadystate, has no shaft work interaction and changes in kinetic and potential energies are negligible. So, the appropriate form of the 1st Law is:  

Eqn 16  
Plugging values into Eqn 16 gives us:  Q_{34}  829.08  kJ/kg  
Part b.)  We can calculate the thermal efficiency of the cycle from: 

Eqn 17  
Heat tranfer into of the system occurs in step 12. We can determine Q_{12} by appplying the 1st Law to HEX #1. The HEX operates at steadystate, has no shaft work interaction and changes in kinetic and potential energies are negligible. So, the appropriate form of the 1st Law is:  

Eqn 18  
Plugging values into Eqn 18 gives us:  Q_{12}  1267.92  kJ/kg  
Plugging values into Eqn 17 gives us:  h  34.61%  
Part c.)  In the ideal cycle, the compressor and turbine are isentropic. So, all we need to do to complete this part of the problem is use H_{1S} and H_{3S} instead of H_{1} and H_{3} when we calculate W_{S,comp}, W_{S,turb}, W_{cycle}, Q_{34}, Q_{12} and h.  
The equations from parts (a)  (c) become:  

Eqn 19 

Eqn 20  

Eqn 21 

Eqn 22  

Eqn 23 

Eqn 24  
Plugging values into Eqns 19  24 yields the values in the following table. The "% Change" is defined as :  

Eqn 25  
Real Cycle  Ideal Cycle  % Change  
W_{S,turb}  (kJ/kg)  875.0  1029.4  15.0%  
W_{S,comp}  (kJ/kg)  436.2  348.9  25.0%  
W_{cycle}  (kJ/kg)  438.8  680.5  35.5%  
Q_{34}  (kJ/kg)  829.1  674.7  22.9%  
Q_{12}  (kJ/kg)  1267.9  1355.2  6.4%  
h  34.6%  50.2%  31.1%  
Verify:  The assumptions made in the solution of this problem cannot be verified with the given information.  
Answers :  Real Cycle  c.)  Ideal Cycle  %Change  
W_{S,turb}  (kJ/kg)  875.0  1029.4  15.0%  
W_{S,comp}  (kJ/kg)  436.2  348.9  25.0%  
a.)  W_{cycle}  (kJ/kg)  438.8  680.5  35.5%  
Q_{34}  (kJ/kg)  829.1  674.7  22.9%  
Q_{12}  (kJ/kg)  1267.9  1355.2  6.4%  
b.)  h  34.6%  50.2%  31.1%  
Although the isentropic efficiencies of the compressor and turbine are very high, 80% and 85%, they reduce the work output by 35% and reduce the efficiency by 31%. This shows the enormous significance of the these isentropic efficiencies in the overall performance of the power cycle. 