Example Problem with Complete Solution

8A-4 : Entropy Production for the Adiabatic Compression of Air
6 pts
One kilogram of air in a piston-cylinder assembly is compressed adiabatically from 300 K, 1 bar to 5 bar.
                   
a.) If the air is compressed without internal irreversibilities, determine the temperature at the final state, in K, and the work required, in kJ.
 
b.) If the compression requires 20% more work than found in (a) determine the temperature at the final state, the amount of entropy produced in kJ/K and the amount of lost work, assuming Tsurr = 300 K.
 
                   
Read : Assume ideal gas behavoir for air. Apply an energy balance and an entropy balance.
Notice that in part (a) the problem asks for the "work required", therefore our answer will be positive.
To get S and U data, use the Gibbs Equation in terms of the Ideal Gas Entropy Function.
Given: T1 300 K m 1 kg
P1 1 bar Q 0 KJ
P2 5 bar Part (b) Wpart (b) 20% greater than Wpart (a)
Tsurr 300 K
Find : Part (a) T2S ? K Part (b) T2 ? K
-Wb ? kJ Sgen ? kJ/K
Wlost ? kJ
 
Diagram:
Assumptions : - As shown in the diagram, the system is the air inside the cylinder.
- Air is modeled as an ideal gas.
- No heat transfer occurs.
- Boundary work is the only form of work that crosses the system boundary.
- Changes in kinetic and potential energies are negligible.
- For Part (a), there is no entropy generated.
Equations / Data / Solve :
Part a.) To determine the work required we need to apply the 1st Law for closed systems:
Eqn 1
Because the process is adiabatic and we assume that changes in kinetic and potential energies are negligible and we assume boundary work is the only form of work, Eqn 1 becomes:
 
Eqn 2 or: Eqn 3
So, in order to answer part (a), we need to determine U for both the initial and final states.
Use the Ideal Gas Property Table for air to evaluate U1 and U2, but first we must know T1 and T2.
Because this process is both adiabatic and internally reversible, the process is isentropic.
In this problem we have air and we assume it behaves as an ideal gas.
We can solve this problem using the ideal gas entropy function.
The 2nd Gibbs Eqn in
terms of the So is:		 Eqn 4
Since part (a) is an isentropic process, Eqn 4 becomes:
Eqn 5
The final temperature, T2, can be determined from determining So(T2) and then interpolating on the Ideal Gas Properties Table for air.
Solving Eqn 5 for So(T2) yields: Eqn 6
Properties for state are determined from Ideal Gas Properties Table for air. So(T1) 0.00617 kJ/kg-K
U(T1) 1.314 kJ/kg
Now, we can plug values into Eqn 3: R 8.314 kJ/kmol K
MW 29.00 kg/kmol
So(T2) 0.468 kJ/kg-K
Now, we can go back to the Ideal Gas Properties Table for air and determine T2 and U2 by interpolation.
T
(K)		 U kJ/kg So
kJ/kg-K
470 125.61 0.463
T2 U2 0.468 T2s 472.28 K
480 133.12 0.484 U2 127.33 kJ/kg
Put values into Eqn 3 to finish this part of the problem: -Wb 126.01 kJ
Part b.) The actual work is 20% greater than the work determined in Part (a):
-Wb 151.21 kJ
In this part of the problem, we know the actual work, but we don't know T2 or U2.
We can solve Eqn 3 for U2 in terms
of the known variables m, Wb and U1 :
Eqn 7
Plugging values into Eqn 7 yields : U2 152.53 kJ/kg
Next, we can determine T2 and So(T2) by interpolating on the Ideal Gas Properties Table for air.
U
kJ/kg		 T
(K)		 So
kJ/kg-K
148.23 500 0.5270
152.53 T2 So(T2) T2 505.7 K
155.81 510 0.5477 So(T2) 0.5387 kJ/kg-K
 
The 2nd Law in terms of the
entropy generated is:		 Eqn 8
Since there is no heat transfer
in this problem: 		 Eqn 9
The entropy change can be
determined from Eqn 4:		 Eqn 10
Now, we can plug values into Eqns 9 & 10
 to determine the entropy generated:
Sgen 0.071157 kJ/K
Lost work can be determined using: Eqn 11
Wlost 21.35 kJ
Verify : The assumptions made in the solution of this problem cannot be verified with the given information.
Answers: Part a.) T2S 472 K Part b.) T2 506 K
-Wb 126.0 kJ Sgen 0.0712 kJ/K
Wlost 21.3 kJ
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