Example Problem with Complete Solution

9E-1 : Performance of a "Real" Brayton Cycle 10 pts
The compressor and turbine of a simple gas power cycle each have isentropic efficiencies of 90%. The compressor pressure ratio is 12. The minimum and maximum temperatures are 300K and 1400K, respectively. On the basis of an air-standard analysis, determine the following:
               
a.) the net work per unit mass of air flowing, in kJ/kg,
b.) the heat rejected per unit mass of air flowing, in kJ/kg,
c.) the thermal efficiency,
d.) Compare your answers for parts (a) - (c) to the values you would expect for an ideal cycle (isentropic turbine and compressor).
 
               
Read : In order to evaluate all of the WS 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 H1S and H3S because part (d) requires that we analyze the ideal cycle as well the actual cycle.
We can lookup H2 and H4 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 H1 and H3.  Whichever method we choose, we will compute H1S and H2S in the process, so when we are done with part (a), we will have all the values we need to complete part (d).
Once we know all the H values, it is a straight-forward process to apply the 1st Law to each process in the cycle in order to answer the questions in parts (a) - (c).  We repeat these calculations in part (d) using H1S instead of H1 and H3S instead of H3.  Finally, calculate the % Change in each answer.
Given: ηS, comp 0.90 T2 1400 K
ηS, turb 0.90 T4 300 K
P1/P4 12
Find : a.) Wcycle ? kJ/kg
b.) Qout ? kJ/kg
c.) η ? %
d.) (Wcycle)id ? kJ/kg
(Qout)ideal ? kJ/kg
ηideal ? %
Diagrams :
Assumptions :
- Each component is an open system operating at steady-state.
- The turbine and compressor 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
(K)
P
(kPa)
Ho
(kJ/kg)
So
(kJ/kg-K)
1
631.8
431.2
1S
599.6
396.80
0.71857
2
1400
1305.0
1.6730
3
818.1
634.1
3S
750.3
559.6
0.96060
4
300
87.41
0.006168
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 steady-state and changes in kinetic and potential energies are negligible.
Eqn 2
Eqn 3
We know T2 and T4, so we can look up H2 and H4 in the Ideal Gas Properties Table for air.
H2 1305.0 kJ/kg H4 87.410 kJ/kg
We can determine T1 and T3 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 is :
   
Eqn 4
   
Eqn 5
   
We can solve Eqns 4 & 5  for the unknowns SoT1 & SoT3 :  
   
Eqn 6
   
Eqn 7
   
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 6 & 7 to determine SoT3 and SoT1.  We can do this because the HEX's are isobaric.  P1 = P2 and P3 = P4.
   
R 8.314 J/mol-K MW 29.00 g/mol
   
SoT2 1.6730 kJ/kg-K SoT4 0.0061681 kJ/kg-K
SoT3S 0.96060 kJ/kg-K SoT1S 0.71857 kJ/kg-K
   
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 (K) Ho (kJ/kg) So (kJ/kg-K)  
590 386.57 0.70137  
T1S H1S 0.71857 T1S 599.61 K
600 397.21 0.71926 H1S 396.80 kJ/kg
   
T (K) Ho (kJ/kg) So (kJ/kg-K)  
750 559.20 0.96012  
T3S H3S 0.96060 T3S 750.33 K
760 570.15 0.9746   H3S 559.56 kJ/kg

Method 2: Use the Ideal Gas Relative Pressure.
When an ideal gas undergoes an isentropic process :    
   
Eqn 8
Eqn 9
   
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 8 & 9 For Pr(T3) and Pr(T1), as follows :  
   
Eqn 10
Eqn 11
   
Look-up Pr(T2) and Pr(T2) and use them in Eqns 10 & 11, respectively, To determine Pr(T3) and Pr(T1):
Pr(T2) 340.20 Pr(T4) 1.022
Pr(T3S) 28.350 Pr(T1S) 12.260
   
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 (K) Pr Ho (kJ/kg)  
600 12.259 397.21  
T1S 12.260 H1S T1S 600.02 K
610 13.035 407.87 H1S 397.23 kJ/kg
   
T (K) Pr Ho (kJ/kg)  
750 28.376 559.20  
T3S 28.350 H3S T3S 749.82 K
760 29.847 570.15   H3S 559.01 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 H3 and H1, respectively :
Eqn 14
Eqn 15
Plugging values into Eqns 14 & 15 gives: H1 431.17 kJ/kg
H3 634.11 kJ/kg
And by interpolation on the Ideal Gas Property Tables:
T (K) Ho (kJ/kg)
630 429.25
T1 431.17
640 439.98 T1 631.79 K
T (K) Ho (kJ/kg)
810 625.19
T3 634.11
820 636.25 T3 818.06 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).
WS,turb 670.89 kJ/kg
WS,comp -343.76 kJ/kg Wcycle 327.13 kJ/kg
Part b.) Heat tranfer out of the system occurs in step 3-4.  We can determine Q34 by appplying the 1st Law to HEX #2.  The HEX operates at steady-state, 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: Q34 -546.70 kJ/kg
Part c.) We can calculate the thermal efficiency of the cycle from:
Eqn 17
Heat tranfer into the system occurs in step 1-2.  We can determine Q12 by appplying the 1st Law to HEX #1.  The HEX operates at steady-state, 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: Q12 873.83 kJ/kg
Plugging values into Eqn 17 gives us: η 37.44%  
Part d.) 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 H1S and H3S instead of H1 and H3 when we calculate WS,comp, WS,turb, Wcycle, Q34, Q12 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
WS,turb (kJ/kg) 670.9 745.4 -10.0%
WS,comp (kJ/kg) -343.8 -309.4 11.1%
Wcycle (kJ/kg) 327.1 436.0 -25.0%
Q34 (kJ/kg) -546.7 -472.2 15.8%
Q12 (kJ/kg) 873.8 908.2 -3.8%
η   37.4% 48.0% -22.0%
Verify : The assumptions made in the solution of this problem cannot be verified with the given information.
Answers :     Real Cycle
WS,turb (kJ/kg) 670.9
WS,comp (kJ/kg) -343.8
a.) Wcycle (kJ/kg) 327.1
b.) Q34 (kJ/kg) -546.7
Q12 (kJ/kg) 873.8
c.) η   37.4%
d.) Ideal Cycle %Change
745.4 -10.0%
-309.4 11.1%
436.0 -25.0%
-472.2 15.8%
908.2 -3.8%
48.0% -22.0%
Although the isentropic efficiencies of the turbine and compressor are very high, 90%,  they reduce the work output by 25% and reduce the efficiency by 22%.  This shows the enormous significance of the these isentropic efficiencies in the overall performance of the power cycle.