4F1 :  Heat and Work for a Cycle Carried Out in a Closed System  6 pts 

A gas in a piston and cylinder device undergoes three quasiequilibrium processes to complete a thermodynamic cycle. The following information is known about the three steps that make up the cycle. 

Process 12: constant volume, V = 37 L, ΔU_{12} = 31.6 kJ Process 23: expansion with PV = constant and ΔU_{23} = 0 Process 31: constant pressure, P = 155 kPa, W_{31} = 15.1 kJ 

Assume changes in kinetic
and potential energies
are negligible. a.) Sketch the path for the cycle on a PV Diagram b.) Calculate the total boundary work for the cycle in kJ c.) Calculate Q_{23} in kJ 

d.) Calculate Q_{31} in kJ e.) Determine whether this cycle is a power cycle or a refrigeration/heatpump cycle and calculate the COP or thermal efficiency. 

Read :  Work your way around
the cycle, step by step. Sum the boundary work
for the three steps to determine W_{cycle}._{
}Write and solve the 1st Law for steps 23 and 31 to determine Q_{23} and Q_{31}. Write and solve the 1st Law for steps 23 and 31 to determine Q_{12} and sum the Q's to evaluate Q_{cycle}. Check your work using Q_{cycle} = W_{cycle} because ΔU_{cycle} = 0. Power cycle is W_{cycle} > 0. Refrigeration or HP cycle of W_{cycle} < 0. 

Given:  Step 12:  V_{1} = V_{2}  0.037  m^{3}  Step 31:  P_{3} = P_{1}  155  kPa  
U_{2}  U_{1}  31.6  kJ  W_{31}  15.1  kJ  
Step 23  P_{2} V_{2} = P_{3} V_{3}  
U_{3} = U_{2}  
Diagram:  See the answer to part (a), below.  
Find:  a.)  Sketch the cycle on a PV Diagram.  d.)  Q_{31}  ???  kJ  
b.)  W_{cycle}  ???  kJ  e.)  Power or Refrigeration Cycle ?  
c.)  Q_{23}  ???  kJ  
Assumptions:  
1   The gas is a closed system  
2   Boundary work is the only form of work interaction  
3   Changes in kinetic and potential energies are negligible.  
Equations / Data / Solve:  
Part a.) 


Part b.)  Since W_{cycle} = W_{12} + W_{23} + W_{31}, we will work our way around the cycle and calculate each work term along the way.  
Because the volume is constant in step 12:  W_{12}  0  kJ  
In step 23: P V = C , therefore, the definition of boundary work becomes:  

Eqn 1  
But, we don't know V_{3} !  Perhaps we can use W_{31} to detemine V_{3}.  
Step 31 is isobaric, therefore, the definition of boundary work becomes:  

Eqn 2  
Solve this equation for V_{3} : 

Eqn 3  
V_{3}  0.1344  m^{3}  
Now, plug V_{3} and C = P_{3}V_{3} into Eqn 1 to determine W_{23}:  
W_{23}  26.9  kJ  
Sum the work terms for the three steps to get W_{cycle}:  W_{cycle}  11.78  kJ  
Part c.)  Write the 1st Law for step 23:  Q_{23}  W_{23} = U_{3}  U_{2} = 0  Eqn 4  
Q_{23} = W_{23}  26.88  kJ  
Part d.)  Write the 1st Law for step 31:  Q_{31}  W_{31} = U_{1}  U_{3}  Eqn 5  
But, U_{2} = U_{3} :  Q_{31}  W_{31} = U_{1}  U_{3} = U_{1}  U_{2} =  ( U_{2}  U_{1} )  Eqn 6  
Solve for Q_{31} :  Q_{31} = W_{31}  ( U_{2}  U_{1} )  Eqn 7  
Plug in the given values :  Q_{31}  46.70  kJ  
Part e.)  First, we should determine Q_{12} from the 1st Law:  
Q_{12}  W_{12} = U_{2}  U_{1} = 0  Eqn 8  
Q_{12} = U_{2}  U_{1}  Eqn 9  
Q_{12}  31.6  kJ  
Define:  Q_{cycle} = Q_{12} + Q_{23} + Q_{31}  Eqn 10  
Q_{cycle}  11.78  kJ  
Since Q_{cycle} > 0 and W_{cycle} > 0, this is a power cycle !  
Notice that Q_{cycle} = W_{cycle} because DU_{cycle} = 0.  
Thermal Efficiency is defined by : 

Eqn 11  
h_{th}  20.14  %  
Verify:  The assumptions made in this problem solution cannot be verified. But all of these assumptions are pretty solid.  
Answers :  a.)  See the sketch, above.  d.)  Q_{31}  46.7  kJ  
b.)  W_{cycle}  11.8  kJ  e.)  This is a Power Cycle.  
c.)  Q_{23}  26.9  kJ  h_{th}  20.1  % 