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 : |
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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) |
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1 | 631.8 |
431.2 |
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1S | 599.6 |
396.80 |
0.71857 |
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2 | 1400 |
1305.0 |
1.6730 |
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3 | 818.1 |
634.1 |
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3S | 750.3 |
559.6 |
0.96060 |
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4 | 300 |
87.41 |
0.006168 |
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Part a.) | Only the compressor and the turbine have shaft work interactions, so the net work for the cycle is given by: | ||||||
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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. | |||||||
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Eqn 2 |
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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 : | |||||||
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Eqn 4 | ||||||
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Eqn 5 | ||||||
We can solve Eqns 4 & 5 for the unknowns SoT1 & SoT3 : | |||||||
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Eqn 6 | ||||||
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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 |