7E4 :  Performance of an Ideal Gas Cycle  10 pts 

An ideal
gas contained in a pistonandcylinder device undergoes a thermodynamic
cycle made up of three quasiequilibrium processes. Step 12: Adiabatic compression from 20^{o}C and 110 kPa to 400 kPa 

Step 23: Isobaric cooling Step 31: Isothermal expansion a.) Carefully draw this process in a traditional pistonandcylinder 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 C_{P} = (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, C_{P} and C_{V} to a cycle on an ideal gas with constant heat capacities. Take advantage of the fact that step 12 is both adiabatic and reversible, so it is isentropic. Power cycles produce a net amount of work and proceed in a clockwise direction on a PV Diagram.  
Given:  T_{1}  20  ^{o}C  P_{1}  110  kPa  
293.15  K  P_{2}  400  kPa  
T_{3}  20  ^{o}C  P_{3}  400  kPa  
293.15  K  R  8.314  J/moleK  
Q_{12}  0  J/mole  C_{P}  20.785  J/moleK  
Find:  For each of the three steps and for the entire cycle:  DU  ???  J/mole  
DH  ???  J/mole  
Q  ???  J/mole  
W  ???  J/mole  
Diagram:  
Part a.) 


Part b.) 


Assumptions:  1   Step 12 is adaibatic, Step 23 is isobaric, Step 31 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 12, 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 T_{2}. So, our next task is to determine T_{2}.  
Since the entire cycle is reversible and this step is also adiabatic, this step is isentropic.  
The fastest way to determine T_{2} is to use one of the PVT relationships for isentropic processes.  

Eqn 5  
Solve Eqn 5 for T_{2} : 

Eqn 6  
Now, we need to evaluate g : 

Eqn 7  
But for ideal gases : 

Eqn 8  
Solving Eqn 8 for C_{V} yields : 

Eqn 9  
Plugging values into Eqn 9 and then Eqn 7 yields :  C_{V}  12.471  J/moleK  
g  1.667  
Now, plug values into Eqn 5 to get T_{2}_{ }and plug that into Eqn 4 to get W_{12} :  T_{2}  491.31  K  
W_{12}  2471.3  J/mole  
Plugging values into Eqn 1 yields :  DU_{12}  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 :  DH_{12}  4118.8  J/mole  
Next, let's analyze step 23, isobaric cooling.  
T_{2}  491.31  K  P_{2}  400  kPa  
T_{3}  293.15  K  P_{3}  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 process: 

Eqn 15  
Because the system contains and ideal gas: 

Eqn 16  
W_{23}  1647.5  J/mole  
Next we can calculate DU by applying Eqn 3 to step 23: 

Eqn 17  
DU_{23}  2471.3  J/mole  
Now, solve Eqn 13 to determine Q_{23} : 

Eqn 18  
Q_{23}  4118.8  J/mole  
Now, we apply Eqn 12 to step 23 to determine DH :  

Eqn 19  
DH_{23}  4118.8  J/mole  
Next, we analyze step 31, isothermal expansion.  
For ideal gases, U and H are functions of T only. Therefore :  DU_{31}  0.0  J/mole  
DH_{31}  0.0  J/mole  
The appropriate form of the 1st Law is: 

Eqn 20  
But since ΔU_{31} = 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 V_{1} and V_{3} 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 :  W_{31}  3146.5  J/mole  
Q_{31}  3146.5  J/mole  
Finally, we can calculate Q_{cycle} and W_{cycle} from :  

Eqn 30 

Eqn 31  
Plugging values into Eqns 29 & 30 yields :  W_{cycle}  972.3  J/mole  
Q_{cycle}  972.3  J/mole  
This result confirms what an application of the 1st Law to the entire cycle tells us: Q_{cycle} = W_{cycle}  
Part d.)  The cycle is a refrigeration cycle because both W_{cycle} and Q_{cycle} are negative.  
The coefficient of performance of
a refrigeration cycle is defined as : 

Eqn 32  
Q_{C} is the heat absorbed by the system during the cycle. In this case, Q_{C} = Q_{31}.  
W is the net work input to the system during the cycle. In this case, W =  W_{cycle}.  
Therefore :  Q_{C}  3146.5  J/mole  
W  972.3  J/mole  
Plug values into Eqn 31 to get :  COP_{R}  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  
V_{1}  22.16  L/mol  
V_{2}  10.21  L/mol  
V_{3}  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.  
COP_{R}  3.2  COP_{HP}  4.2 