9F-1 : | Air-Standard Brayton Cycle With and Without Regeneration | 10 pts | |||||
An air-standard Brayton cycle has a compressor pressure ratio of 10. Air enters the compressor at P4 = 14.7 psia, T4 = 540°R, with a mass flow rate of 90,000 lbm/h. The turbine inlet temperature is 2160°R. Calculate the thermal efficiency and the net power developed, in horsepower, if… | |||||||
a.) | the turbine and the compressor isentropic efficiencies are each 100%. | ||||||
b.) | the turbine and compressor isentropic efficiencies are 88 and 84%, respectively. | ||||||
c.) | the turbine and compressor isentropic efficiencies are 88 and 94%, respectively, and a regenerator with an effectiveness of 80% is incorporated. | ||||||
Read: | We will need to know all of the H's in order to determine both Wcycle and h. We can get H4 and H2 immediately from the Ideal Gas Property Table for air. | ||||||
Then, for each part of the problem, use the give isentropic compressor and turbine efficiencies to evaluate H1 and H3. Then, calculate Wcycle and Qin for each part and finally the efficiency. | |||||||
In part (c), renumber the stream carefully so you can easily use most of the H's from part (b). The key to part (c) is to use the regenerator effectiveness to determine the H of the combustor feed. Once you have this, you can compute Qin. Wcycle is the same as in part (b). So, calculate h from its definition. | |||||||
Given: | P1/P4 | 10 | T2 | 2200 | oR | ||
P4 | 14.7 | psia | m | 90,000 | lbm/h | ||
T4 | 540 | oR | |||||
Part a.) | ηS, turb | 1.00 | Part b.) | ηS, turb | 0.88 | ||
ηS, comp | 1.00 | ηS, comp | 0.84 | ||||
Part c.) | ηS, turb | 0.88 | |||||
ηS, comp | 0.84 | ||||||
ηregen | 0.80 | ||||||
Find : | η | ??? | % | Wcycle | ??? | hp | |
Diagrams : |
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Assumptions : | |||||||
- Each component is open system operating at steady-state. | |||||||
- The turbine and compressor processes 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 (oR) |
P (psia) |
Ho (Btu/lbm) |
So (Btu/lbm-oR) |
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1 | 147 |
45.0 |
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1S | 1027.9 |
157.55 |
0.15924 |
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2 | 2200 |
470.7 |
0.3613 |
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3 | 238.8 |
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3S | 1222.2 |
207.2 |
0.20348 |
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4 | 540 |
14.70 |
37.58 |
0.001473 |
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We can calculate the thermal efficiency of the cycle when the compressor and turbine are isentropic using : | |||||||
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Eqn 1 | ||||||
We can determine Q12 and Q34 by applying the 1st Law to HEX #1 and HEX #2 respectively. | |||||||
Each HEX operates at steady-state, involves no shaft work and has negligible changes in kinetic and internal energies. The appropriate forms of the 1st Law are: | |||||||
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Eqn 2 |
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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 T2 and T4, so we can look up H2 and H4 in the Ideal Gas Property Table for air. | |||||||
H2 | 470.670 | Btu / lbm | H4 | 37.580 | Btu / lbm | ||
The two remaining H values depend on the isentropic efficiency of the compressor and the turbine, so they will be diffierent depending on which part of the problem is being considered. | |||||||
We can determine Wcycle by applying the 1st Law to the entire cycle. | |||||||
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Eqn 4 | ||||||
Because the compressor and turbine are assumed to be adiabatic, Eqn 4 simplifies to: | |||||||
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Eqn 5 | ||||||
So, once we determine Q12 and Q34 for each part of the problem, we can use Eqn 5 to evaluate Wcycle. | |||||||
Part a.) | Both the compressor and the turbine are isentropic. | ||||||
This changes Eqns 2 & 3 to: | |||||||
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Eqn 6 |
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Eqn 7 | ||||
We can determine T1S and T3S 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 for the compressor and the turbine are : | |||||||
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Eqn 8 | ||||||
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Eqn 9 | ||||||
We can solve Eqns 8 & 9 for the unknowns SoT1 & SoT3 : | |||||||
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Eqn 10 | ||||||
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Eqn 11 | ||||||
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 10 & 11 to determine SoT3S and SoT1S. We can do this because the HEX's are isobaric. P1 = P2 and P3 = P4. | |||||||
R | 1.987 | Btu/lbmole-oR | |||||
MW | 29.00 | lbm / lbmole | |||||
SoT2 | 0.36125 | Btu / lbm oR | SoT4 | 0.0014732 | Btu / lbm oR | ||
SoT3S | 0.20348 | Btu / lbm oR | SoT1S | 0.15924 | Btu / lbm oR | ||
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 (oR) | Ho (Btu/lbm) | So (Btu/lbmoR) | |||||
1020 | 155.55 | 0.15730 | |||||
T1S | H1S | 0.15924 | T1S | 1027.89 | oR | ||
1030 | 158.08 | 0.15976 | H1S | 157.55 | Btu / lbm | ||
T (oR) | Ho (Btu/lbm) | So (Btu/lbmoR) | |||||
1220 | 206.65 | 0.20302 | |||||
T3S | H3S | 0.20348 | T3S | 1222.20 | oR | ||
1230 | 209.23 | 0.20513 | H3S | 207.22 | Btu / lbm |