| 5E-2 : | Charging a Tank With Compresed Air | 7 pts |
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| A 0.2 m3 rigid tank initially contains R-134a at 8°C. In this state, 60% of the mass is in the vapor phase, and the rest is in the liquid phase. The tank is connected by a valve to a supply line where refrigerant at 1 MPa and 120°C flows steadily. Now the valve is opened slightly and the R134a is allowed to enter the tank from the supply line. When the pressure in the tank reaches 800 kPa, the last drop of liquid refrigerant in the tank enters the vapor phase. At this point, the valve is closed. Determine... | ||||||||||
| a.) | … the final tmperature of the refrigerant in the tank | |||||||||
| b.) | … the mass of R-134a that has entered the tank | |||||||||
| c.) | ... the amount of heat transfer that has taken place during this process | |||||||||
| Read : | Part (a) is straightforward because the vapor in the system is saturated at 800 kPa. We can then determine the initial and final mass of R-134a in the tank. A mass balance between the initial and final states of the system (the tank) tells us that the mass added to the tank is just the final mass minus the initial mass. This process is a transient process because the mass of R-134a inside the system (the tank) changes with time. We will need to use the integral form of the transient 1st Law Equation to answer part (c). | |||||||||
| Given : | V | 0.2 |
m3 | Pin | 1000 |
kPa | ||||
| T1 | 8 |
°C | Tin | 120 |
°C | |||||
| x1 | 0.60 |
kg vap/kg total | P2 | 800 |
kPa | |||||
| x1 | 1.00 |
kg vap/kg total | ||||||||
| Find : | T2 | ??? |
°C | Q | ??? |
kJ | ||||
| min | ??? |
kg | ||||||||
| Diagram : |  |
 |
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| Assumptions : | 1 - | Although this is a transient process, it can be analyzed as a uniform flow problem because the properties of the R-134a entering the tank are constant. | ||||||||
| 2 - | Kinetic and potential energies are negligible. | |||||||||
| 3 - | No shaft work crosses the boundary of the system, which consists of the content of the tank. | |||||||||
| Equations / Data / Solve : | ||||||||||
| Part a.) | The vapor inside the tank in the final state is saturated. Therefore, it is at the saturation temperature of R-134a at a pressure of 800 kPa. We can obtain this temperature from the NIST Webbook. | |||||||||
| T2 | 31.327 | °C | ||||||||
| Part b.) | The integral form of the transient mass balance on the tank is : | |||||||||
 |
Eqn 1 | |||||||||
| We can determine the intial and final mass of R-134a in the system using : |  |
Eqn 2 | ||||||||
| Our next step is to determine the intial and final specific volume of the R-134a. | ||||||||||
| In the initial state : | ||||||||||
| At 8oC : | P1 | 387.61 |
kPa |  |
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| Vsat liq | 0.00078873 |
m3/kg | ||||||||
| Vsat vap | 0.052804 |
m3/kg | Eqn 3 | |||||||
| V1 | 0.031998 |
m3/kg | ||||||||
| The final state is simpler because the R-134a in the tank is a saturated vapor. | ||||||||||
| At 800 kPa : | Vsat vap | 0.025625 |
m3/kg | V2 | 0.025625 |
m3/kg | ||||
| Now, we can use Eqn 2 to to determine the initial and final mass of R-134a in the tank. | ||||||||||
| m1 | 6.25 |
kg | m2 | 7.80 |
kg | |||||
| Plug these values back into Eqn 1 to determine the mass of R-134a that was added to the tank during this process. | ||||||||||
| min | 1.554 |
kg | ||||||||
| part c.) | The integral form of the transient energy balance equation for a single-input, single-output system in which kinetic and potential energies are negligible is : | |||||||||
 |
Eqn 4 | |||||||||
| In our process, no shaft work occurs and there is no mass leaving the system, so Eqn 4 can be simplified and solved for Q : | ||||||||||
 |
Eqn 5 | |||||||||
| We can determine U2 and U1 much as we determined V2 and V1 in part (b). | ||||||||||
| The NIST Webbook, using the default reference state, tells that in the initial state : | ||||||||||
| At 8oC : | Usat liq | 210.53 |
kJ/kg | |||||||
| Usat vap | 382.73 |
kJ/kg | Eqn 6 | |||||||
| U1 | 313.85 |
kJ/kg | ||||||||
| The final state is simpler because the R-134a in the tank is a saturated vapor. | ||||||||||
| At 800 kPa : | Usat vap | 394.96 |
kJ/kg | U2 | 394.96 |
kJ/kg | ||||
| Next, we need to determine Hin. First, we need to determine the state of the system. | ||||||||||
| In the NIST Webbook, we cannot find the Psat(Tin) because the R-134a is supercritical at Tin. | ||||||||||
| TC | 101.06 |
°C | ||||||||
| This is enough information to tell us that we need to look in the Superheated Steam Tables to find Hin. We can verify this fact from : | ||||||||||
| Tsat(Pin) | 60.058 |
°C | Tin > Tsat, therefore we must consult the Superheated R-134a Tables. | |||||||
| From the NIST Webbook, we can obtain : | Hin | 504.21 |
kJ/kg | |||||||
| Now, we can plug values back into Eqn 5 to evaluate Q : | Q | 337.15 |
kJ | |||||||
| Verify : | None of the assumptions made in this problem solution can be verified. | |||||||||
| Answers : | T2 | 31.3 |
°C | |||||||
| min | 1.55 |
kg | ||||||||
| Q | 337 |
kJ | ||||||||
