Example Problem with Complete Solution

8A-5 : Entropy Production for the Adiabatic Compression of Air 6 pts
Air is compressed in an adiabatic piston-and-cylinder device, as shown below, from 200 kPa and 360 K to 800 kPa.
a.) Calculate the final temperature, T2, and the boundary work if the process is internally reversible.
b.) Calculate T2 and the entropy generation if a real piston-and-cylinder device requires 15% more work than the internally reversible device.
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 2nd Gibbs Equation in terms of the Ideal Gas Entropy Function.
Given: T1 360 K m 2.9 kg
P1 200 kPa Q 0 KJ
P2 800 kPa Part (b) Wpart (b) = 15% > Wpart (a)
Find: Part (a) T2S ??? K Part (b) T2 ??? K
-Wb ??? kJ Sgen ??? kJ/K
Wlost ??? kJ
Assumptions: 1 - As shown in the diagram, the system is the air inside the cylinder.
2 - Air is modeled as an ideal gas.
3 - No heat transfer occurs.
4 - Boundary work is the only form of work that crosses the system boundary.
5 - Changes in kinetic and potential energies are negligible.
6 - 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 Equation 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 1 are determined from Ideal Gas Properties Table for air.
So(T1) 0.189360 kJ/kg-K
U(T1) 44.3940 kJ/kg
Now, we can plug values into Eqn 3: R 8.314 kJ/kmol K
MW 28.97 kg/kmol
So(T2) 0.58721 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) Uo kJ/kg So kJ/kg-K
520 163.42 0.56803
T2 U2 0.58721 T2 529.59 K
530 171.04 0.58802 U2 170.73 kJ/kg
Put values into Eqn 3 to finish this part of the problem: -Wb 366.38 kJ
Part b.) The actual work is 15% greater than the work determined in Part (a):
Eqn 7
-Wb 421.33 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 interms of
known variables m, Wb and U1 :
Eqn 8
Plugging values into Eqn 7 yields : U2 189.68 kJ/kg
Next, we can determine T2 and So(T2) by interpolating on the Ideal Gas Properties Table for air.
Uo kJ/kg T (K) So kJ/kg-K
186.36 550 0.62702
189.68 T2 So(T2) T2 554.3 K
194.05 560 0.64605 So(T2) 0.63524 kJ/kg-K
The 2nd Law in terms of the
entropy generated is:
Eqn 9
Since there is no heat transfer in this problem: 
Eqn 10
The entropy change can be determined from Eqn 4:
Eqn 11
Now, we can plug values into Eqns 9 & 10 to determine the entropy generated:
Sgen 0.04803 kJ/K
Verify: The ideal gas assumption needs to be verified.
Eqn 12
We need to determine the specific volume at each state and check if :
Eqn 13
V1 14.97 L/mol
V2A 5.50 L/mol V2B 5.76 L/mol
The specific volume at each state is greater than 5 L/mol for all states and the working fluid can be treated as a diatomic gas, so the ideal gas assumption is valid.
Answers : Part a.) T2S 530 K Part b.) T2 554 K
-Wb 366 kJ Sgen 0.0480 kJ/K