| 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. Check your work using 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 | % | ||||||||||||||
