# Example Problem with Complete Solution

9E-2 : 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 air-standard analysis to determine
a.) QC and Wnet, 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 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 (c) 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 (c).
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) & (b).  We repeat these calculations in part (c) using H1S instead of H1 and H3S instead of H3.  Finally, calculate the % Change in each answer for our comparison.
Given: hS, comp 0.80 T2 1800 K
hS, turb 0.85 T4 300 K
P1/P4 15
Find: a.) Wcycle ??? kJ/kg c.) (Wcycle)ideal ??? kJ/kg
Qout ??? kJ/kg (Qout)ideal ??? kJ/kg
b.) h ??? % hideal ??? %
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 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
(K)
Ho
(kJ/kg)
So
(kJ/kg-K)
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 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 1791.5 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 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 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 1.9784 kJ/kg-K SoT4 0.0061681 kJ/kg-K
SoT3S 1.2020 kJ/kg-K SoT1S 0.78254 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-
630 429.25 0.77137
T1S H1S 0.78254 Interpolation yields : T1S 636.61 K
640 439.98 0.78826 H1S 436.34 kJ/kg

T (K) Ho (kJ/kg) So (kJ/kg-
920 747.82 1.1867
T3S H3S 1.2020 Interpolation yields : T3S 932.67 K
940 770.33 1.2109 H3S 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 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) 986.20 Pr(T4) 1.0217
Pr(T3S) 65.747 Pr(T1S) 15.326

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)
630 14.7 429.25
T1S 15.326 H1S Interpolation yields : T1S 637.02 K
640 15.591 439.98 H1S 436.78 kJ/kg

T (K) Pr Ho (kJ/kg)
920 62.489 747.82
T3S 65.747 H3S Interpolation yields : T3S 931.84 K
940 67.990 770.33 H3S 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 H3 and H1, respectively : Eqn 14 Eqn 15
Plugging values into Eqns 14 & 15 gives: H1 523.58 kJ/kg
H3 916.49 kJ/kg
And by interpolation on the Ideal Gas Property Tables:
T (K) Ho (kJ/kg) T (K) Ho (kJ/kg)
710 515.58 1060 906.80
T1 523.58 T3 916.49
720 526.46 1080 929.77
T1 717.35 K T3 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).
WS,turb 875.01 kJ/kg
WS,comp -436.17 kJ/kg Wcycle 438.84 kJ/kg
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 -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 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 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 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) 875.0 1029.4 -15.0%
WS,comp (kJ/kg) -436.2 -348.9 25.0%
Wcycle (kJ/kg) 438.8 680.5 -35.5%
Q34 (kJ/kg) -829.1 -674.7 22.9%
Q12 (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
WS,turb (kJ/kg) 875.0 1029.4 -15.0%
WS,comp (kJ/kg) -436.2 -348.9 25.0%
a.) Wcycle (kJ/kg) 438.8 680.5 -35.5%
Q34 (kJ/kg) -829.1 -674.7 22.9%
Q12 (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. 