Example Problem with Complete Solution

7D-3 : Work, Efficiency and the T-S Diagram for an Ideal Gas Power Cycle 8 pts
Air contained in a piston-and-cylinder device undergoes a power cycle made up of three internally reversible processes.
Step 1-2: Adiabatic compression from 20 psia and 570oR to 125 psia
Step 2-3: Isothermal expansion to 20 psia
Step 3-1: Isobaric compression
a.) Sketch the process path for this power cycle on both PV and TS diagrams
b.) Calculate T3 in oF
c.) Calculate the boundary work in Btu/lbm
d.) Calculate the thermal efficiency of the power cycle
Read : For part (a) sketch the cycle first to get a better understanding of the processes.
For part (b) recall that the Process 2-3 is isothermal and therefore T3 = T2.  Determine So(T2) and look it up in the Ideal Gas Entropy Table for air to determine T2.
For part (c) determine the net work by determining the work for each process and then adding them together. 
For part (d) determine the thermal efficiency as the ratio of the net work to the heat going into the system.
Given: T1 570 oR P2 125 psia
P1 20 psia P3 20 psia
Find: Part (a) Sketch PV and TS diagrams Part (c) Wcycle ? Btu/lbm
Part (b) T3 ? oR Part (d) h ?
Assumptions: 1 - The system is the air inside the cylinder.
2 - The air is modeled as an ideal gas.
3 - Each process is internally reversible.
4 - Boundary work is the only form of work that crosses the system boundary.
5 - There is no change in kinetic or potential energy for either of the two processes.
Equations / Data / Solve:
Part b.) Since Process 2-3 is isothermal:
Eqn 1
So we should work on determining T2.
Let's apply the 2nd  Gibbs Equation
for ideal gases to Process 1-2 :
Eqn 2
We can determine T2 and thus T3 from:
Eqn 3
Lookup So(T1) in the Ideal Gas Entropy Tables: So(T1) 0.014381 Btu/lbm-oR
Now, plug values into Eqn 3 to determine So(T2):
R 1.987 Btu/lbmol-oR So(T2) 0.14007 Btu/lbm-oR
MW 28.97 lbm/lbmol
Now that we know the value of So at T2, we can interpolate on the air Ideal Gas Property Table to determine T2.
T (oR) So (Btu/lbm-oR)
950 0.13939
T2 0.14007 Interpolation yields : T2 = T3 = 952.60 oR
960 0.14202 492.93 oF
Part c.) The net work is the sum of the
work done during each process:
Eqn 4
We need to determine the work involved in each process.  Begin with Process 1-2.
Since Process 1-2 is adiabatic, and changes in internal and kinetic energies are negligible, the appropriate form of the 1st Law is :
Eqn 5
Eqn 6
Because we know both T1 and T2, we can look up the U's in the Ideal Gas Property Table:
At T1 = 570 oR, no interpolation is required: U1 5.6705 Btu/lbm
T (oR) Uo (Btu/lbm)
950 72.806
952.60 U2 Interpolation yields : U2 73.281 Btu/lbm
960 74.631
Now, we can plug values back into Eqn 6 to determine W12 : W12 -67.610 Btu / lbm
Boundary work done by the system during Process 2-3 can be calculated from the definition of boundary work :
Eqn 7
For an ideal gas substitute P = nRT/V into Eqn 7 :
Eqn 8
Integrate Eqn 2
(the process is isothermal, T2 = T3 ) :
Eqn 9
Since P2V2 = nRT2 and P3V3 = nRT3 and
T2 = T3, we conclude that P2 V2 = P3 V3 , or :
Eqn 10
Combining Eqn 10 and Eqn 9 yields :
Eqn 11
We can plug numbers into Eqn 11 to evaluate W23 : W23 119.74 Btu / lbm
Boundary work done by the system during Process 3-1 can be calculated from the definition of boundary work :
Eqn 12
Since Process 3-1 is isobaric
(P=constant), Eqn 12 simplifies to :
Eqn 13
Since :
Eqn 14
Eqn 15
And since P1 = P3, Eqns 13, 14 & 15 can be combined to obtain :
Eqn 16
Now, we can plug values into Eqn 16 to evaluate W31 : W31 -26.24 Btu / lbm
Now, we can calculate Wcycle from the sum of the work terms for each step, using Eqn 4 :
Wcycle 25.88 Btu / lbm
Part d.) The thermal efficiency of the cycle is defined by :
Eqn 17
We know Wcycle, so we need to determine Qin.  We also know that Q12 = 0 (adiabatic process) and from the TS Diagram it can be concluded that Q23 > 0 and Q31 < 0. Therefore, Qin = Q23.
Eqn 18
Now we need to determine the heat transferred into the cycle during Process 2-3.  Start from the definition of entropy :
Eqn 19
Because Process 2-3 is internally reversible, we can integrate Eqn 19 to get:
Eqn 20
Now, because Process 2-3 is isothermal, the T
pops out of the integral and Eqn 20 is easy to integrate:
Eqn 21
Now, we can again apply the 2nd Gibbs Equation for ideal gases to Process 2-3 to evaluate DS:
Eqn 22
Since the process is isothermal: So(T2) = So(T3)
Eqn 22 simplifies to:
Eqn 23
When we substitute Eqn 23 into Eqn 21 we get :
Eqn 24
DS23 0.12569 Btu/lbm-oR Q23 119.74 Btu / lbm
Finally, we plug values back into Eqn 18 to evaluate the thermal efficiency of the cycle :
h 21.62%
Verify: The ideal gas assumption needs to be verified.
We need to determine the specific volume
at each state and check if:
Air can be considered a diatomic gas.
Solving the Ideal Gas EOS for molar volume yields :
Use : R 10.7316 psia-ft3 / lbmol-oR
V1 305.85 ft3/lbmol
V2 81.78 ft3/lbmol V3 511.15 ft3/lbmol
The specific volume is greater than 80 ft3/lbmol for all states so the ideal gas assumption is valid.
Answers : a.) See diagrams above. c.) Wcycle 25.9 Btu/lbm
b.) T3 953 oR d.) h 21.6%