9E-1 : | Optimal Compressor Outlet Pressure for the Ideal Brayton Power Cycle | 8 pts |
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Consider the TS Diagram, below, for an ideal Brayton Cycle. | |||||||||||||||||||||||||||
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The working fluid is air and the cold air-standard assumptions apply. | |||||||||||||||||||||||||||
Using a given pair of thermal reservoirs, T1 and T3 are fixed. Show that the maximum net work from the cycle is obtained when the compressor effluent temperature is T2 = (T1 T3)1/2. | |||||||||||||||||||||||||||
Read : | Use the 1st Law to write an equation for the net work produced by the cycle. | ||||||||||||||||||||||||||
Use the constant heat capacity to eliminate enthalpy from your equation for Wcycle. | |||||||||||||||||||||||||||
Differentiate your equation for Wcycle with respect to rP and set the result equal to zero. | |||||||||||||||||||||||||||
Simplify the result using the fact that the compressor and turbine are isentropic and the heat capacity ratio, g, is constant. | |||||||||||||||||||||||||||
Finally, calculate the 2nd derivative of Wcycle with respect to rP and verify that it is positive at the point determined above so that the extremum you found is a maximum of Wcycle and not a minimum. | |||||||||||||||||||||||||||
Given: | An ideal Brayton Cycle is analyzed on a cold air-standard basis. | ||||||||||||||||||||||||||
Find: | Show that the compressor effluent temperature that maximizes net work per unit mass of air flow is given by T2 = (T1T3)1/2. | ||||||||||||||||||||||||||
Diagram: |
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Assumptions: | 1 - | Each component is an open system operating at steady-state. | |||||||||||||||||||||||||
2 - | The turbine and compressor are isentropic. | ||||||||||||||||||||||||||
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. | ||||||||||||||||||||||||||
6 - | The specific heat, CP , and the specific heat ratio, g , are constant. | ||||||||||||||||||||||||||
Equations / Data / Solve: | |||||||||||||||||||||||||||
The net work for the power cycle is: |
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Eqn 1 | |||||||||||||||||||||||||
The 1st Law can be applied to the compressor and to the turbine, assuming they operate adiabatically at steady-state with negligible changes in kinetic and potential energies. | |||||||||||||||||||||||||||
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Eqn 2 | ||||||||||||||||||||||||||
Combining Eqn 1 and Eqn 2 yields: |
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Eqn 3 | |||||||||||||||||||||||||
The change in enthalpy of the air in the turbine and in the compressor can be determined as follows because we have assumed that the heat capacity is constant. | |||||||||||||||||||||||||||
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Eqn 4 | ||||||||||||||||||||||||||
Now we can apply Eqn 4 to the turbine and to the compressor and use the results to eliminate enthalpy from Eqn 3. | |||||||||||||||||||||||||||
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Eqn 5 | ||||||||||||||||||||||||||
The goal is to determine the maximum Wcycle, so we need to set dWcycle/drP = 0 and solve for T2. This will help us find an extremum, either a maximum or a minimum. Later, we will need to make sure the 2nd derivative, d2Wcycle/dr2P , is positive so we can be sure this extremum is a maximum value of Wcycle. | |||||||||||||||||||||||||||
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Eqn 6 | ||||||||||||||||||||||||||
We can simplify Eqn 6 because T1, T3 and CP are all constant. | |||||||||||||||||||||||||||
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Eqn 7 | ||||||||||||||||||||||||||
A little algebra makes
Eqn 7 easier to work with. |
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Eqn 8 | |||||||||||||||||||||||||
Because T2 and T4 change as the compression ratio, rP changes. We need to figure out the relationship between T2 and T4 and rP in order to solve Eqn 8. | |||||||||||||||||||||||||||
Next, we can take advantage of the fact that the compressor and the turbine are isentropic and use the following relationships from Lesson 7E, page 6. | |||||||||||||||||||||||||||
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Eqn 9 |
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Eqn 10 | ||||||||||||||||||||||||
Solve Eqns 9 and 10 for T2 and T4, respectively. | |||||||||||||||||||||||||||
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Eqn 11 |
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Eqn 12 | ||||||||||||||||||||||||
Now, we can differentiate Eqns 11 and 12 with respect to rP so we can use the results in Eqn 8. | |||||||||||||||||||||||||||
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Eqn 13 |
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Eqn 14 | ||||||||||||||||||||||||
Now, we can
substitute Eqns 13 and 14 into Eqn 8. |
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Eqn 15 | |||||||||||||||||||||||||
When the minus sign is combined with the constant term, it is clear that the two constant terms, (g-1)/g, cancel. |
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Eqn 16 | |||||||||||||||||||||||||
This equation is more manageable, but we can simplify it a bit further. |
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Eqn 17 | |||||||||||||||||||||||||
The result is
simple, but there is noT2 in Eqn 18! |
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Eqn 18 | |||||||||||||||||||||||||
Here, we need to use Eqn 11 again, but in a slightly different form. |
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Eqn 19 | |||||||||||||||||||||||||
Squaring Eqn 19 yields : |
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Eqn 20 | |||||||||||||||||||||||||
A slight
rearrangement of Eqn 20 yields : |
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Eqn 21 | |||||||||||||||||||||||||
Now, we can use Eqn 21 to eliminate rP from Eqn 18. |
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Eqn 22 | |||||||||||||||||||||||||
Now, solve for T2 : |
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Eqn 23 | |||||||||||||||||||||||||
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Eqn 24 | ||||||||||||||||||||||||||
Now we need to make sure this is the maximum Wcycle and not the minimum. The criterion for a maximum is: | |||||||||||||||||||||||||||
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Eqn 25 | ||||||||||||||||||||||||||
Combining Eqns 5 and 25 yields : |
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Eqn 26 | |||||||||||||||||||||||||
We can simplify Eqn 26 because T1, T3 and CP are all constant. |
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Eqn 27 | |||||||||||||||||||||||||
Now, we need to differentiate Eqns 13 and 14, as follows. | |||||||||||||||||||||||||||
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Eqn 28 | ||||||||||||||||||||||||||
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Eqn 29 | ||||||||||||||||||||||||||
Combine Eqns 27 - 29 : | |||||||||||||||||||||||||||
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Eqn 30 | ||||||||||||||||||||||||||
Let the algebra fly : | Eqn 31 | ||||||||||||||||||||||||||
Multiply by g . This does not change the > sign because g > 0. | |||||||||||||||||||||||||||
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Eqn 32 | ||||||||||||||||||||||||||
Combine the rP terms : |
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Eqn 33 | |||||||||||||||||||||||||
Combine Eqn 18 with Eqn 33 to get : |
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Eqn 34 | |||||||||||||||||||||||||
We can divide Eqn 34 by T1 without changing the > sign because T1 > 0 and then do some more algebra. | |||||||||||||||||||||||||||
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Eqn 35 |
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Eqn 35 | ||||||||||||||||||||||||
Finally : |
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Eqn 36 | |||||||||||||||||||||||||
The heat capacity ratio, g, is always greater than 1. So the 2nd derivative is positive and we have indeed found the maximum Wcycle and NOT the minimum ! | |||||||||||||||||||||||||||
Verify: | The assumptions made in the solution of this problem cannot be verified with the given information. | ||||||||||||||||||||||||||
Answers : |
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The optimal temperature of the compressor effluent in a Brayton Cycle is the geometric average of the temperatures of the compressor and turbine feed streams. |