Example Problem with Complete Solution

8D-2 : Sgen and Wlost for a Compressor with Heat Losses 7 pts
An air compressor increases the pressure from 100 kPa to 1 MPa.  The inlet temperature is 300 K and the outlet temperature is 420 K.  The mass flow rate is 0.50 kg/s and the power requirement for the compressor is 155 kW.  The average surface temperature of the compressor is 400 K.  The temperature of the surroundings is 300 K.  Determine...
               
a.) The internal, external and total entropy generation rate.
b.) The internal, external and total lost work.    
c.) Show that the total lost work determined in part (b) is the difference between the power requirement of a completely reversible compressor that accomplishes the same change in state of the air stream and the actual work for the real compressor.
 
 
 
               
Read: Equations for evaluating total and external entropy generation can be obtained for the text.  You could just take the difference betwee these to determine the internal entropy generation.  But it is  useful to understand how the position of the system boundary can be manipulated to directly yield an equation for the internal entropy generation.
The keys are to use the 1st Law to determine Q and to use the Gibbs Equation and the Ideal Gas Entropy Function to evaluate DS.
Once the entropy generation rates are known, it is easy to determine the lost work rates.
In order to verify the definition of lost work, we must determine the amount of work a reversible compressor would require to accomplis the same compression process.  Qrev must be different from Qact.  The key to this part of the problem is that total entropy generation rate for the reversible compressor must be zero.
Given: P1 100 kPa T1 300 K
P2 1000 kPa T2 420 K
m 0.50 kg/s Tsurf 400 K
Ws,act -155 kW Tsurr 300 K
Find: a.) b.)
Sgen,int ??? kW/K Wlost,int ??? kW
Sgen,ext ??? kW/K Wlost,ext ??? kW
Sgen,tot ??? kW/K Wlost,tot ??? kW
c.) Show that : Figure 1a
Eqn 1
Assumptions:
1- The surface temperature of the compressor remains constant.  Therefore, it can be treated as a thermal reservoir.
2- The compressor operates at steady-state with negligible changes in kinetic and potential energies.
3- Air behaves as an ideal gas.
Diagram : Figure 1b
Equations / Data / Solve :
Part a.) We can use the following equations to evaluate the internal and total entropy generation rates.
Figure 2
Eqn 2
If we place the system boundary far from the surface of the compressor, then ALL of the irreversibilities are inside the system because the temperature at the system boundary is the same as the temperature of the surroundings.  So, heat exchange between this system and the surroundings is reversible.
Figure 3
This explains why using Tsurr in Eqn 2 tells us that we are computing the TOTAL entropy generation rate.  Because with this system, there are no irreversibilities OUTSIDE the system boundary.  Notice that heat transfer from this sytem does occur, but it occurs with zero temperature driving force, so it is reversible !
Equation 2
Eqn 3
Using our usual system boundary, right at the surface of the compressor, heat exchange with the surroundings is irreversible.  So, by using Eqn 3, with Tsurf instead of Tsurr, we have excluded the external irreversibility due to heat transfer through a finite temperature difference.  So, this entropy generation equation gives us the INTERNAL entropy generation only !
Equation 3
The external entropy generation rate is equal to the rate at which the entropy of the universe (compressor and surroundings) changes due only to the heat transfer from the compressor to the surroundings.
Equation 4
Eqn 4
The minus sign appears in Eqn 4 because the sign of Q from them the perspective of the system is negative, but from the perspective of the surroundings, Q > 0 because heat is entering the surroundings !
Eqn 4 can be rearranged to give us :
Equation 5
Eqn 5
Let's begin by determing DS using the 2nd Gibbs Equation and the Ideal Gas Property Table for air.
Equation 6
Eqn 6
So1 0.34649 kJ/kg-K So2 0.0061681 kJ/kg-K
R 8.314 J/mole-K
MW 28.97 g/mole DS -0.32049 kJ/kg-K
Next, we need to evaluate Q, the rate of heat loss from the compressor.  We can do this using the 1st Law for open systems operating at steady-state with negligible changes in kinetic and potential energies.
Equation 7
Eqn 7
We can solve Eqn 7 for Q : Equation 8
Eqn 8
Now, we can use the Ideal Gas Property Table for air to evaluate H1 and H2.
Ho1 87.41 kJ/kg Ho2 208.88 kJ/kg
Plug values back into Eqn 7 to evaluate Q : Q -94.27 kW
Now, we are able to plug values back into Eqns 2, 3 & 5 to evaluate the entropy generation rates.
Sgen,int 0.075418 kW/K
Sgen,ext 0.078554 kW/K Sgen,tot 0.153972 kW/K
Double check your calculations using :
Equation 9
Eqn 9
Part b.) Once we have completed part (a), part (b) is a straightforward application of the relationship between lost work and entropy generation.
Equation 10
Eqn 10
Eqn 10 applies for internal, external and total entropy generation and lost work.
Wlost,int 22.63 kW
Wlost,ext 23.57 kW Wlost,tot 46.19 kW
Part c.) Here we must verify that our answer from part (b) is consistent with the defintion of lost work.
Equation 11
Eqn 11
In order to do this, we must evaluate WS,rev.
The key to doing this is to understand that the reversible process still operates between states 1 and 2.
The values of Q and WS are different from those for the real compressor.
The fact that ties this part of the problem together and allows us to determine Q and WS for the reversible compressor is that :
Equation 12
Eqn 12
We can use this fact and solve Eqn 3 for Qrev :
Equation 13
Eqn 13
Qrev -48.07 kW
Next, we can apply the 1st Law to the reversible compressor to evaluate WS,rev.
We can solve Eqn 7 for WS,rev.  The result is :
Equation 14
Eqn 14
Plugging values into Eqn 14 yields : WS,rev -108.81 kW
Finally, put values into the right-hand side of Eqn 11 :
Equation 15
46.19 kW
This matches our result for lost work from part (b).  So, we have confirmed the relationship between reversible work, actual work and lost work.
Verify : None of the assumptions made in this problem solution can be verified.
Answers : a.) Sgen,int 0.075418 kW/K
Sgen,ext 0.078554 kW/K
Sgen,tot 0.153972 kW/K
b.) Wlost,int 22.63 kW
Wlost,ext 23.57 kW
Wlost,tot 46.19 kW
c.) WS,rev - WS,act = 46.19 kW