Example Problem with Complete Solution

7E-4 : Performance of an Ideal Gas Cycle 10 pts
An ideal gas contained in a piston-and-cylinder device undergoes a thermodynamic cycle made up of three quasi-equilibrium processes.
Step 1-2: Adiabatic compression from 20oC and 110 kPa to 400 kPa
Step 2-3: Isobaric cooling
Step 3-1: Isothermal expansion
a.) Carefully draw this process in a traditional piston-and-cylinder schematic
b.) Sketch the process path for this cycle on a PV Diagram. 
Put a point on the diagram for each state and label it. Be sure to include and label all the important features for a complete PV Δiagram for this system
c.) Calculate Q, W, ΔU and ΔH, in J/mole, 
for each step in the process and for the entire cycle. Assume that CP = (5/2) R.
d.) Is this cycle a power cycle or a refrigeration cycle?  Explain. Calculate the thermal efficiency or COP of the cycle, whichever is appropriate.
Read : Sketch carefully.  Understanding what is going on in the problem is half the battle.  Apply the 1st Law, the definitions of boundary work, CP and CV to a cycle on an ideal gas with constant heat capacities. Take advantage of the fact that step 1-2 is both adiabatic and reversible, so it is isentropic. Power cycles produce a net amount of work and proceed in a clock-wise direction on a PV Diagram.  
Given: T1 20 oC P1 110 kPa
293.15 K P2 400 kPa
T3 20 oC P3 400 kPa
293.15 K R 8.314 J/mole-K
Q12 0 J/mole CP 20.785 J/mole-K
Find: For each of the three steps and for the entire cycle: DU ??? J/mole
DH ??? J/mole
Q ??? J/mole
W ??? J/mole
Part a.)
Part b.)
Assumptions: 1 - Step 1-2 is adaibatic, Step 2-3 is isobaric, Step 3-1 is isothermal.
2 - The entire cycle and all of the steps in the cycle are internally reversible.
3 - Changes in kinetic and potential energies are negligible.
4 - Boundary work is the only form of work interaction during the cycle.
5 - The PVT behavior of the system is accurately described by the ideal gas EOS.
Equations / Data / Solve:
Part c.) Let's begin by analyzing step 1-2, the adiabatic compression.
Begin by applying the 1st Law for closed systems to each step in the Carnot Cycle.  Assume that changes in kinetic and potential energies are negligible.
Eqn 1
Because internal energy is not a function of pressure for an ideal gas, we can determine DU by integrating the equation which defines the constant volume heat capacity.  The integration is simplified by the fact that the heat capacity for the gas in this problem has a constant value.
Eqn 2
Eqn 3
Combining Eqns 1 & 3 yields:
Eqn 4
The problem is that we do not know T2.  So, our next task is to determine T2.
Since the entire cycle is reversible and this step is also adiabatic, this step is isentropic.
The fastest way to determine T2 is to use one of the PVT relationships for isentropic processes.
Eqn 5
Solve Eqn 5 for T2 :
Eqn 6
Now, we need to evaluate g :
Eqn 7
But for ideal gases :
Eqn 8
Solving Eqn 8 for CV yields :
Eqn 9
Plugging values into Eqn 9 and then Eqn 7 yields : CV 12.471 J/mole-K
g 1.667
Now, plug values into Eqn 5 to get T2 and plug that into Eqn 4 to get W12 : T2 491.31 K
W12 -2471.3 J/mole
Plugging values into Eqn 1 yields : DU12 2471.3 J/mole
Now, we can get DH from its definition :
Eqn 10
But, the gas is an ideal gas:
Eqn 11
Combining Eqns 10 & 11 gives us :
Eqn 12
Now, we can plug values into Eqn 12 : DH12 4118.8 J/mole
Next, let's analyze step 2-3, isobaric cooling.
T2 491.31 K P2 400 kPa
T3 293.15 K P3 400 kPa
The appropriate form of the 1st Law is:
Eqn 13
Because we assumed that boundary work is the only form of work that crosses the system boundary, we can determine work from its definition.
Eqn 14 Isobaric
Eqn 15
Because the system contains and ideal gas:
Eqn 16
W23 -1647.5 J/mole
Next we can calculate DU by applying Eqn 3 to step 2-3:
Eqn 17
DU23 -2471.3 J/mole
Now, solve Eqn 13 to determine Q23 :
Eqn 18
Q23 -4118.8 J/mole
Now, we apply Eqn 12 to step 2-3 to determine DH :
Eqn 19
DH23 -4118.8 J/mole
Next, we analyze step 3-1, isothermal expansion.
For ideal gases, U and H are functions of T only.  Therefore : DU31 0.0 J/mole
DH31 0.0 J/mole
The appropriate form of the 1st Law is:
Eqn 20
But since ΔU31 = 0, Eqn 20 becomes :
Eqn 21
Again, because we assumed that boundary work is the only form of work that crosses the system boundary, we can determine work from its definition.
Eqn 22 Ideal Gas EOS :
Eqn 23
Solve Eqn 23 for P and substitute the result into Eqn 22 to get :
Eqn 24
Eqn 25
Integrating Eqn 25 yields :
Eqn 26
We can use the Ideal Gas EOS to avoid calculating V1 and V3 as follows:
Apply Eqn 23 to both states 3 and 1 :
Eqn 27
Cancelling terms and rearranging leaves :
Eqn 28
Use Eqn 27 to eliminate the V's from Eqn 25 :
Eqn 29
Now, plug values into Eqn 28 and then Eqn 20 : W31 3146.5 J/mole
Q31 3146.5 J/mole
Finally, we can calculate Qcycle and Wcycle from :
Eqn 30
Eqn 31
Plugging values into Eqns 29 & 30 yields : Wcycle -972.3 J/mole
Qcycle -972.3 J/mole
This result confirms what an application of the 1st Law to the entire cycle tells us:  Qcycle = Wcycle
Part d.) The cycle is a refrigeration cycle because both Wcycle and Qcycle  are negative.
The coefficient of performance of a
refrigeration cycle is defined as :
Eqn 32
QC is the heat absorbed by the system during the cycle.  In this case, QC = Q31.
W is the net work input to the system during the cycle.  In this case, W = - Wcycle.
Therefore : QC 3146.5 J/mole
W 972.3 J/mole
Plug values into Eqn 31 to get : COPR 3.236
Verify: The ideal gas assumption needs to be verified.
Eqn 33
We need to determine the specific volume at each state and check if :
Eqn 34
V1 22.16 L/mol
V2 10.21 L/mol
V3 6.09 L/mol
The specific volume at each state is greater than 5 L/mol for all states and the working fluid is a diatomic gas, so the ideal gas assumption is valid.
Answers : a.) See diagram above. b.) See diagram above.
c.) Step DU DH Q W
1 - 2 2471 4119 0 -2471 All values in this table are in J/mole.
2 - 3 -2471 -4119 -4119 -1648
3 - 1 0 0 3146 3146
Cycle 0 0 -972 -972
d.) Refrigeration or Heat Pump Cycle.
COPR 3.2 COPHP 4.2