# Example Problem with Complete Solution

4F-1 : Heat and Work for a Cycle Carried Out in a Closed System 6 pts
A gas in a piston and cylinder device undergoes three quasi-equilibrium processes to complete a thermodynamic cycle. The following information is known about the three steps that make up the cycle.
Process 1-2: constant volume, V = 37 L, ΔU12 = 31.6 kJ
Process 2-3: expansion with PV = constant and ΔU23 = 0
Process 3-1: constant pressure, P = 155 kPa, W31 = -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 Q23 in kJ
d.) Calculate Q31 in kJ
e.) Determine whether this cycle is a power cycle or a refrigeration/heat-pump 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 Wcycle.
Write and solve the 1st Law for steps 2-3 and 3-1 to determine Q23 and Q31.
Write and solve the
1st Law for steps 2-3 and 3-1 to determine Q12 and sum the Q's to evaluate Qcycle.
Qcycle = Wcycle because ΔUcycle = 0.
Power cycle is Wcycle > 0. Refrigeration or HP cycle of Wcycle < 0.
Given: Step 1-2: V1 = V2 0.037 m3 Step 3-1: P3 = P1 155 kPa
U2 - U1 31.6 kJ W31 -15.1 kJ
Step 2-3 P2 V2 = P3 V3
U3 = U2
Diagram: See the answer to part (a), below.
Find: a.) Sketch the cycle on a PV Diagram. d.) Q31 ??? kJ
b.) Wcycle ??? kJ e.) Power or Refrigeration Cycle ?
c.) Q23 ??? 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 Wcycle = W12 + W23 + W31, we will work our way around the cycle and calculate each work term along the way.
Because the volume is constant in step 1-2: W12 0 kJ
In step 2-3:   P V = C , therefore, the definition of boundary work becomes: Eqn 1
But, we don't know V3 ! Perhaps we can use W31 to detemine V3.
Step 3-1 is isobaric, therefore, the definition of boundary work becomes: Eqn 2
Solve this equation for V3 : Eqn 3
V3 0.1344 m3
Now, plug V3 and C = P3V3 into Eqn 1 to determine W23:
W23 26.9 kJ
Sum the work terms for the three steps to get Wcycle: Wcycle 11.78 kJ
Part c.) Write the 1st Law for step 2-3: Q23 - W23 = U3 - U2 = 0 Eqn 4
Q23 = W23 26.88 kJ
Part d.) Write the 1st Law for step 3-1: Q31 - W31 = U1 - U3 Eqn 5
But, U2 = U3 : Q31 - W31 = U1 - U3 = U1 - U2 = - ( U2 - U1 ) Eqn 6
Solve for Q31 : Q31 = W31  - ( U2 - U1 ) Eqn 7
Plug in the given values : Q31 -46.70 kJ
Part e.) First, we should determine Q12 from the 1st Law:
Q12 - W12 = U2 - U1 = 0 Eqn 8
Q12 = U2 - U1 Eqn 9
Q12 31.6 kJ
Define: Qcycle = Q12 + Q23 + Q31 Eqn 10
Qcycle 11.78 kJ
Since Qcycle > 0 and Wcycle > 0, this is a power cycle !
Notice that Qcycle = Wcycle because DUcycle = 0.
Thermal Efficiency is defined by : Eqn 11
hth 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.) Q31 -46.7 kJ
b.) Wcycle 11.8 kJ e.) This is a Power Cycle.
c.) Q23 26.9 kJ hth 20.1 % 