9E1 :  Optimal Compressor Outlet Pressure for the Ideal Brayton Power Cycle  8 pts 

Consider the TS Diagram, below, for an ideal Brayton Cycle.  


The working fluid is air and the cold airstandard assumptions apply.  
Using a given pair of thermal reservoirs, T_{1} and T_{3} are fixed. Show that the maximum net work from the cycle is obtained when the compressor effluent temperature is T_{2} = (T_{1} T_{3})^{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 W_{cycle}.  
Differentiate your equation for W_{cycle} with respect to r_{P} 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 W_{cycle} with respect to r_{P} and verify that it is positive at the point determined above so that the extremum you found is a maximum of W_{cycle} and not a minimum.  
Given:  An ideal Brayton Cycle is analyzed on a cold airstandard basis.  
Find:  Show that the compressor effluent temperature that maximizes net work per unit mass of air flow is given by T_{2} = (T_{1}T_{3})^{1/2}.  
Diagram: 


Assumptions:  1   Each component is an open system operating at steadystate.  
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, C_{P}_{ ,} and the specific heat ratio, g , are constant.  
Equations / Data / Solve:  
The net work for the power cycle is: 

Eqn 1  
The 1st Law can be applied to the compressor and to the turbine, assuming they operate adiabatically at steadystate with negligible changes in kinetic and potential energies.  

Eqn 2  
Combining Eqn 1 and Eqn 2 yields: 

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.  

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.  

Eqn 5  
The goal is to determine the maximum W_{cycle}, so we need to set dW_{cycle}/dr_{P} = 0 and solve for T_{2}. This will help us find an extremum, either a maximum or a minimum. Later, we will need to make sure the 2nd derivative, d^{2}W_{cycle}/dr^{2}_{P} , is positive so we can be sure this extremum is a maximum value of W_{cycle}.  

Eqn 6  
We can simplify Eqn 6 because T_{1}, T_{3} and C_{P} are all constant.  

Eqn 7  
A little algebra makes
Eqn 7 easier to work with. 

Eqn 8  
Because T_{2} and T_{4} change as the compression ratio, r_{P} changes. We need to figure out the relationship between T_{2} and T_{4} and r_{P} 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.  

Eqn 9 

Eqn 10  
Solve Eqns 9 and 10 for T_{2} and T_{4}, respectively.  

Eqn 11 

Eqn 12  
Now, we can differentiate Eqns 11 and 12 with respect to r_{P} so we can use the results in Eqn 8.  

Eqn 13 

Eqn 14  
Now, we can
substitute Eqns 13 and 14 into Eqn 8. 

Eqn 15  
When the minus sign is combined with the constant term, it is clear that the two constant terms, (g1)/g, cancel. 

Eqn 16  
This equation is more manageable, but we can simplify it a bit further. 

Eqn 17  
The result is
simple, but there is noT_{2} in Eqn 18! 

Eqn 18  
Here, we need to use Eqn 11 again, but in a slightly different form. 

Eqn 19  
Squaring Eqn 19 yields : 

Eqn 20  
A slight
rearrangement of Eqn 20 yields : 

Eqn 21  
Now, we can use Eqn 21 to eliminate r_{P} from Eqn 18. 

Eqn 22  
Now, solve for T_{2} : 

Eqn 23  

Eqn 24  
Now we need to make sure this is the maximum W_{cycle} and not the minimum. The criterion for a maximum is:  

Eqn 25  
Combining Eqns 5 and 25 yields : 

Eqn 26  
We can simplify Eqn 26 because T_{1}, T_{3} and C_{P} are all constant. 

Eqn 27  
Now, we need to differentiate Eqns 13 and 14, as follows.  

Eqn 28  

Eqn 29  
Combine Eqns 27  29 :  

Eqn 30  
Let the algebra fly :  Eqn 31  
Multiply by g . This does not change the > sign because g > 0.  

Eqn 32  
Combine the r_{P} terms : 

Eqn 33  
Combine Eqn 18 with Eqn 33 to get : 

Eqn 34  
We can divide Eqn 34 by T_{1} without changing the > sign because T_{1} > 0 and then do some more algebra.  

Eqn 35 

Eqn 35  
Finally : 

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 W_{cycle} and NOT the minimum !  
Verify:  The assumptions made in the solution of this problem cannot be verified with the given information.  
Answers : 

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. 