| 4F-4 : | Heat & Work for a Closed Cycle Containing R-134a | 8 pts |
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| A piston-and-cylinder device contains R-134a which is initially at -10°C and 200 kPa. The R-134a undergoes the following 3-step cycle : | |||||||||
| Step 1-2: | Isochoric heating to 80°C. | ||||||||
| Step 2-3: | Isothermal compression to a state in which the quality is 0.700. Q23 = -85.6 kJ. | ||||||||
| Step 3-1: | Adiabatic expansion. | ||||||||
| a.) | Carefully sketch the process path for this cycle on a PV Diagram. Put a point on the diagram for each state. Be sure to include and label all the important features for a complete PV diagram for this system. | ||||||||
| b.) | Determine Qcycle and Wcycle in kJ/kg. | ||||||||
| c.) | Is this cycle a power cycle or a refrigeration cycle ? Explain. | ||||||||
| Read : | We are given T1 and P1, so we can determine any and all properties of the system using the R-134a Tables. In particular, we can evaluate the specific volume and we know that this does not chnge in step 1-2. This gives us a 2nd intensive property for state 2 and allows us to evaluate all of the properties of state 2. We expect T2 >T1. Step 2-3 is an isothermal compression to a quality of x3 = 0.700. Because T3 = T2, we will be able to evaluate all of the properties of state 3, again using the R-134a Tables. In each of the three steps, we know the value of either the heat or the work. W12 = 0 because the process is isochoric. Q23 is given and Q31 = 0 because the process is adiabatic. So, when we apply the 1st Law to each step, there is just one unknown and we can evaluate it. Once we know Q and W for each step, we can determine Qcycle and Wcycle because they are the sum of the Q's and W's for the steps that make up the cycle, respectively. | ||||||||
| Given : | T1 | -10 |
°C | Find : | Q12 | ??? |
kJ/kg | ||
| P1 | 200 |
kPa | W23 | ??? |
kJ/kg | ||||
| T2 | 80 |
°C | W31 | ??? |
kJ/kg | ||||
| T3 | 80 |
°C | Qcycle | ??? |
kJ/kg | ||||
| x3 | 0.700 |
kg vap/kg | Wcycle | ??? |
kJ/kg | ||||
| Q23 | -85.6 |
kJ/kg | |||||||
| Q31 | 0 |
kJ | |||||||
| Diagrams : |  |
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| Part a.) |  |
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| Assumptions : | |||||||||
| 1 - | Changes in kinetic and potential energies are negligible. | ||||||||
| 2 - | Boundary work is the only form of work that crosses the system boundary. | ||||||||
| Equations / Data / Solve : | |||||||||
| Let's begin by writing the 1st Law for each of the three steps that make up the cycle, assuming that changes in potential and kinetic energy are negligible. | |||||||||
| Step 1-2 : |  |
Eqn 1 | |||||||
| Step 2-3 : |  |
Eqn 2 | |||||||
| Step 3-1 : |  |
Eqn 3 | |||||||
| Step 1-2 is isochoric, so no boundary work occurs. If we assume that boundary work is the only form of work interaction in this cycle, then W12 = 0. Q31 = 0 because step 3-1 is adiabatic. | |||||||||
| We can solve Eqns 1 - 3 to evaluate the unknowns Q12, W23 and W31. | |||||||||
| Step 1-2 : |  |
Eqn 4 | |||||||
| Step 2-3 : |  |
Eqn 5 | |||||||
| Step 3-1 : |  |
Eqn 6 | |||||||
| Our next step must be to determine the value of the specific internal energy at states 1, 2 and 3 because, once we know these, we can use Eqns 4 - 6 to evaluate the unknowns Q12, W23 and W31. | |||||||||
| Let's begin with state 1. First, we must determine the phase or phases that exist in state 1. We can accomplish this by comparing P1 to Psat(T1). | |||||||||
| Psat(T1) | 2633.2 |
kPa | |||||||
| Since P1 < Psat(T1), we conclude that a superheated vapor exists in the cylinder at state 1. | |||||||||
| We can determine U1 from the Superheated R-134a Tables. We can also determine V1 because we know V2 = V1 and the knowledge of this 2nd intensive variable for state 2 will allow us to evaluate U2. | |||||||||
| V1 = V2 = | 0.099915 |
m3/kg | U1 | 224.56 |
kJ/kg | ||||
| Next, let's work on state 2. We know the value of 2 intensive variables, T2 and V2, and we know that if a superheated vapor expands at constant volume, it must still be a superheated vapor. Consequently, we can use the Superheated R-134a Tables to determine U2 (and any other properties at state 2 that we want. | |||||||||
| At T2 = 80°C, it turns out that V2 = 0.099915 m3/kg falls between 240 kPa and 280 kPa, so we must interpolate to determine U2. | |||||||||
| At T2 = 80°C : | V (m3/kg) |
U (kJ/kg) |
P (kPa) |
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0.11675 |
295.91 |
240 |
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0.099915 |
U2 |
P2 |
U2 | 295.58 |
kJ/kg | ||||
0.099612 |
295.57 |
280 |
P2 | 279.29 |
kPa | ||||
| Now, let's work on state 3. We know the temperature and the quality, so we can determine U3 using : | |||||||||
 |
Eqn 7 | ||||||||
| We can use the Saturated R-134a Tables to determine Usat vap and Usat liq at 80°C and then we can plug numbers into Eqn 7 to evaluate U3. | |||||||||
| Usat liq | 171.41 |
kJ/kg | |||||||
| Usat vap | 263.69 |
kJ/kg | U3 |
236.0 |
kJ/kg | ||||
| Now, we can go back and plug the values of the specific internal energies into Eqns 4 - 6 to evaluate the unknowns Q12, W23, and W31. | |||||||||
| Q12 | 71.02 |
kJ/kg | W23 | -26.03 |
kJ/kg | ||||
W31 |
11.45 |
kJ/kg | |||||||
| Next, we need to evaluate the specific work and specific heat for the entire cycle. | |||||||||
| The specific work for the cycle is the sum of the specific work for each step. | |||||||||
| The specific amount of heat transfer for the cycle is the sum of the specific amount of heat transfer for each step. | |||||||||
| Qcycle | -14.58 |
kJ/kg | Wcycle | -14.58 |
kJ/kg | ||||
| Verify: | The assumptions made in this problem solution cannot be verified. | ||||||||
| Answers: | Q (kJ/kg) |
W (kJ/kg) |
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| Step 1 - 2 | 71.0 |
0 |
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| Step 1 - 3 | -85.6 |
-26.0 |
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| Step 1 - 4 | 0 |
11.4 |
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| Cycle | -14.6 |
-14.6 |
Because Wcycle < 0, this is a refrigeration cycle ! |
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