# Example Problem with Complete Solution

6F-4 : Pressure, Work and COP for a Carnot Gas Refrigeration Cycle 8 pts
The PV Diagram, below, shows the process path for a Carnot refrigeration cycle carried out in a piston-and-cylinder device with 8.4 lbm of air as the working fluid. The maximum temperature in the cycle is 80oF and the minimum temperature is -10oF. The isothermal expansion requires 126 Btu of heat transfer and the volume of air in the cylinder at the end of the
isothermal compression is 1.74 ft3. Assume air behaves as an ideal gas with a constant heat capacity ratio of g = 1.4. The following relationship is valid for the adiabatic steps in the Carnot cycle: Calculate
a.) The pressure at states 1 to 4
b.) The work for each of the four processes
c.) The COP for the refrigeration cycle.

Note the direction (
CW or CCW) of the cycle. It will be the opposite direction to that of a power cycle. With the final volume, V4 given, use the ideal gas EOS to determine P4. Use the equation given in the problem statement to determine P1. You can then calaculate V1 from the ideal gas EOS. Use an energy balance to determine V2 and then the ideal gas EOS to find P2. Finally, use the equation in the problem statement again to determine P3.
Part (b)
Notice that
(U2 - U1) = 0 for an ideal gas undergoing an isothermal process because U is a function of T only. Carefully apply an energy balance to step 2-3 (watch the sign of each term). Here dU = CV dT = R/(g-1) dT for the case of constant g. Note that MWair = 28.97 lbm/lbmol.
Given:
m 8.4 lbm g 1.4
TC -10 oF TH 80 oF
449.67 oR TH 539.67 oR
Q12 126 Btu V4 1.74 ft3
Find: a.) P1, P2, P3, P4 ??? lbf/in2
b.) W12, W23, W34, W41 ??? Btu
c.) COPR ???
Diagram: Assumptions: 1 - The system consists of air modeled as an ideal gas with g = 1.4.
2 - Boundary work is the only form of work interaction.
3 - Changes in kinetic and potential energies are negligible.
4 - Because the cycle is a Carnot Cycle, the cycle is reversible and therefore each step in the cycle is a reversible process.
5 - The heat capacities of the ideal gas are constant.
Equations / Data / Solve:
R 1545.35 (ft * lbf)/(lbmol * oR) Conversion Factors: 1 ft2 = 144 in2
R 1.986 Btu/lbmol-oR 1 Btu = 778 ft-lbf
MWair 29 lbm/lbmol 1 L = 0.0353 ft3
Tref 459.67 oR
Part a.) Since air is modeled as an ideal gas, we can determine the pressure at  state 4 from the ideal gas EOS: Eqn 1
P4 964 lbf/in2
Now, let's apply the equation given in the problem statement to step 1-4 : Eqn 2
Solving
for P1 : Eqn 3 P1 509.1 lbf/in2
P2 can also be determined from the ideal gas EOS: Eqn 4
The problem is that we don't know V2.  We could determine V2 if we knew the value of W12 because :
The work is: Eqn 5
We can then use the IG EOS for isothermal step 1-2 : Eqn 6
Substituting Eqn 6 into Eqn 5 yields : Eqn 7
Integrating Eqn 7 yields : Eqn 8
Solving for Eqn 8 for V2 : Eqn 9
Now, the issue is that we still don't know W12.  Let's write the 1st Law for step 1-2, assuming changes in kinetic and potential energies are negligible. Eqn 10
Since the internal energy of an ideal gas depends on temperature only and the temperature is constant along Process 1-2, U2 = U1 and Eqn 10 reduces to: Eqn 11
W12 126 Btu
We can determine the number of moles in the system from : Eqn 12
n 0.28966 lbmoles
Then, we can evaluate V1 from the Ideal Gas EOS : Eqn 13
V1 2.7456 ft3
Now, we can substitute W12 and V1 into Eqn 9 to evaluate V2 : V2 4.4683 ft3
At last we can use V2 in Eqn 4 to evaluate P2 : P2 312.8 lbf/in2
That leaves us P3 yet to be determined for part (a).
P3 can be determined by applying Eqn 3 to step 2-3 : Eqn 14
P3 592.4 lbf/in2
Verify : The ideal gas assumption needs to be verified.
We need to determine the specific volume at each state and check if : Eqn 15
From the Ideal Gas EOS, we obtain : Eqn 16
V1 11.38 ft3/lbmol V3 9.78 ft3/lbmol
V2 18.51 ft3/lbmol V4 6.01 ft3/lbmol
It is NOT accurate to treat the air in this process as an ideal gas !
We were instructed to do so and we did, but we need to keep in mind that the results may not be accurate to 2 significant figures.
P1 509 lbf/in2 P3 592 lbf/in2
P2 313 lbf/in2 P4 964 lbf/in2
Part b.) Let's begin by writing the 1st Law, open systems, steady-state with changes in kinetic and potentail energies negligible : Eqn 16
Apply Eqn 16 to each step in the cycle : or : Eqn 17 or : Eqn 18 or : Eqn 19 or : Eqn 20
DU12 = DU34 = 0 because these steps are isothermal processes and the system contains an ideal gas.
Q23 = Q41 = 0 because these steps are adiabatic.
Integrating the definition of the
constant volume heat capacity we obtain : Eqn 21
For an ideal gas with constant
heat capacities, Eqn 21 becomes : Eqn 22
The heat capacities of ideal gases are related by the following equations : Eqn 23 Eqn 24 Eqn 25
CV 5.0 Btu/lbmole-oR
Now, we can combine Eqn 22 with
Eqns 18 & 20 for steps 2-3 and 4-1:
W23 -129.432 Btu
W41 129.432 Btu
We already know, from part (a), that : W12 126 Btu
So, now we need to evaluate W34.  Because step 3-4 is adiabatic (like step 1-2) we can apply Eqn 8 to step 3-4 as follows : Eqn 26 W34 -151.195 Btu
The assumptions made in this part of the problem cannot be verified with the given information.
W12 126 Btu W34 -151.19 Btu
W23 -129.43 Btu W41 129.43 Btu
Part c.) The coefficient of performance of a Carnot Cycle is: Eqn 27
COPR 5.0
Verify: The assumptions made in this part of the problem cannot be verified with the given information.
Answers : a.) P1 510 lbf/in2 P3 590.0 lbf/in2
P2 310 lbf/in2 P4 960.0 lbf/in2
b.) W12 126 Btu W34 -151 Btu
W23 -129 Btu W41 129 Btu
c.) COPR 5.0 