Example Problem with Complete Solution

9E-1 : 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 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:
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:
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.
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 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.
Eqn 6
We can simplify Eqn 6 because T1, T3 and CP are all constant.
Eqn 7
A little algebra makes Eqn 7
easier to work with.
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.
Eqn 9
Eqn 10
Solve Eqns 9 and 10 for T2 and T4, respectively.
Eqn 11
Eqn 12
Now, we can differentiate Eqns 11 and 12 with respect to rP 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, (g-1)/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 noT2 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 rP from Eqn 18.
Eqn 22
Now, solve for T2 :
Eqn 23
Eqn 24
Now we need to make sure this is the maximum Wcycle 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 T1, T3 and CP 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 rP terms :
Eqn 33
Combine Eqn 18 with Eqn 33 to get :
Eqn 34
We can divide Eqn 34 by T1 without changing the > sign because T1 > 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 Wcycle 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.