6F4 :  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 pistonandcylinder device with 8.4 lb_{m} of air as the working fluid.  


The maximum temperature in the cycle is 80^{o}F and the minimum temperature is 10^{o}F. 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 ft^{3}. 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. 

Read :  Part
(a) 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, V_{4} given, use the ideal gas EOS to determine P_{4}. Use the equation given in the problem statement to determine P_{1}. You can then calaculate V_{1} from the ideal gas EOS. Use an energy balance to determine V_{2} and then the ideal gas EOS to find P_{2}. Finally, use the equation in the problem statement again to determine P_{3}. 

Part
(b) Notice that (U_{2}  U_{1}) = 0 for an ideal gas undergoing an isothermal process because U is a function of T only. Carefully apply an energy balance to step 23 (watch the sign of each term). Here dU = C_{V} dT = R/(g1) dT for the case of constant g. Note that MW_{air} = 28.97 lb_{m}/lbmol. 

Given:  
m  8.4  lb_{m}  g  1.4  
T_{C}  10  ^{o}F  T_{H}  80  ^{o}F  
449.67  ^{o}R  T_{H}  539.67  ^{o}R  
Q_{12}  126  Btu  V_{4}  1.74  ft^{3}  
Find:  a.)  P_{1}, P_{2}, P_{3}, P_{4}  ???  lb_{f}/in^{2}  
b.)  W_{12}, W_{23}, W_{34}, W_{41}  ???  Btu  
c.)  COP_{R}  ???  
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 * ^{o}R)  Conversion Factors:  1 ft^{2} =  144  in^{2}  
R  1.986  Btu/lbmol^{o}R  1 Btu =  778  ftlb_{f}  
MW_{air}  29  lb_{m}/lbmol  1 L =  0.0353  ft^{3}  
T_{ref}  459.67  ^{o}R  
^{}  
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  
P_{4}  964  lb_{f}/in^{2}  
Now, let's apply the equation given in the problem statement to step 14 : 

Eqn 2  
Solving for P_{1 }: 

Eqn 3  P_{1}  509.1  lb_{f}/in^{2}  
P_{2} can also be determined from the ideal gas EOS: 

Eqn 4  
The problem is that we don't know V_{2}. We could determine V_{2} if we knew the value of W_{12} because :  
The work is: 

Eqn 5  
We can then use the IG EOS for isothermal step 12 : 

Eqn 6  
Substituting Eqn 6 into Eqn 5 yields : 

Eqn 7  
Integrating Eqn 7 yields : 

Eqn 8  
Solving for Eqn 8 for V_{2 }: 

Eqn 9  
Now, the issue is that we still don't know W_{12}. Let's write the 1st Law for step 12, 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 12, U_{2} = U_{1} and Eqn 10 reduces to:  

Eqn 11  
W_{12}  126  Btu  
We can determine the number of moles in the system from : 

Eqn 12  
n  0.28966  lbmoles  
Then, we can evaluate V_{1} from the Ideal Gas EOS : 

Eqn 13  
V_{1}  2.7456  ft^{3}  
Now, we can substitute W_{12} and V_{1} into Eqn 9 to evaluate V_{2} :  V_{2}  4.4683  ft^{3}  
At last we can use V_{2} in Eqn 4 to evaluate P_{2} :  P_{2}  312.8  lb_{f}/in^{2}  
That leaves us P_{3} yet to be determined for part (a).  
P_{3} can be determined by applying Eqn 3 to step 23 : 

Eqn 14  
P_{3}  592.4  lb_{f}/in^{2}  
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  
V_{1}  11.38  ft^{3}/lbmol  V_{3}  9.78  ft^{3}/lbmol  
V_{2}  18.51  ft^{3}/lbmol  V_{4}  6.01  ft^{3}/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.  
P_{1}  509  lb_{f}/in^{2}  P_{3}  592  lb_{f}/in^{2}  
P_{2}  313  lb_{f}/in^{2}  P_{4}  964  lb_{f}/in^{2}  
Part b.)  Let's begin by writing the 1st Law, open systems, steadystate 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  
DU_{12} = DU_{34} = 0 because these steps are isothermal processes and the system contains an ideal gas.  
Q_{23} = Q_{41} = 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  
C_{V}  5.0  Btu/lbmole^{o}R  
Now, we can combine Eqn 22 with Eqns 18 & 20 for steps 23 and 41: 
W_{23}  129.432  Btu  
W_{41}  129.432  Btu  
We already know, from part (a), that :  W_{12}  126  Btu  
So, now we need to evaluate W_{34}. Because step 34 is adiabatic (like step 12) we can apply Eqn 8 to step 34 as follows :  

Eqn 26  W_{34}  151.195  Btu  
The assumptions made in this part of the problem cannot be verified with the given information.  
W_{12}  126  Btu  W_{34}  151.19  Btu  
W_{23}  129.43  Btu  W_{41}  129.43  Btu  
Part c.)  The coefficient of performance of a Carnot Cycle is: 

Eqn 27  
COP_{R}  5.0  
Verify:  The assumptions made in this part of the problem cannot be verified with the given information.  
Answers :  a.)  P_{1}  510  lb_{f}/in^{2}  P_{3}  590.0  lb_{f}/in^{2}  
P_{2}  310  lb_{f}/in^{2}  P_{4}  960.0  lb_{f}/in^{2}  
b.)  W_{12}  126  Btu  W_{34}  151  Btu  
W_{23}  129  Btu  W_{41}  129  Btu  
c.)  COP_{R}  5.0 