5E-2 : | Charging a Tank With R-134a | 7 pts |
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A rigid tank contains ammonia at -20oC with a quality of 0.50 kg vap/kg. Superheated ammonia vapor at 800 kPa and 80oC slowly and steadily enters the tank from a supply line through a pipe with a valve. | ||||||||||||||||
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When the pressure in the tank reaches 600 kPa, the valve is closed. At this point, the last drop of liquid ammonia in the tank vaporizes and the tank contains only saturated ammonia vapor. | ||||||||||||||||
The total volume
of the tank is 350 L.
Determine... a.) The final temperature of the ammonia in the tank b.) The mass of ammonia that has entered the tank |
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c.) The heat transfer between the tank and the surroundings 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 ammonia 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 ammonia 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 | 350 | L | Pin | 800 | kPa | ||||||||||
0.35 | m3 | Tin | 40 | oC | ||||||||||||
T1 | -20 | oC | P2 | 600 | kPa | |||||||||||
x1 | 0.50 | kg vap/kg total | x2 | 1.00 | kg vap/kg total | |||||||||||
Find: | T2 | ??? | oC | Q | ??? | kJ | ||||||||||
min | ??? | kg | ||||||||||||||
Diagram: | The diagram in the problem statement is adequate. | |||||||||||||||
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 ammonia at a pressure of 600 kPa. We can obtain this temperature from the NIST Webbook. | |||||||||||||||
T2 | 9.285 | oC | ||||||||||||||
Part b.) | The integral form of the transient mass balance on the tank is : |
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Eqn 1 | |||||||||||||
We can determine the intial and final mass of R-134a in the system using : |
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Eqn 2 | ||||||||||||||
Our next step is to determine the intial and final specific volume of the ammonia. | ||||||||||||||||
In the initial state : | At -20oC : | P1 | 190.08 | kPa | ||||||||||||
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Eqn 3 | |||||||||||||||
Vsat liq | 0.0015035 | m3/kg | ||||||||||||||
Vsat vap | 0.62373 | m3/kg | V1 | 0.31262 | m3/kg | |||||||||||
The final state is simpler because the ammonia in the tank is a saturated vapor. | ||||||||||||||||
At 600 kPa : | Vsat vap | 0.21035 | m3/kg | V2 | 0.21035 | m3/kg | ||||||||||
Now, we can use Eqn 2 to to determine the initial and final mass of ammonia in the tank. | ||||||||||||||||
m1 | 1.12 | kg | m2 | 1.66 | kg | |||||||||||
Plug these values back into Eqn 1 to determine the mass of ammonia that was added to the tank during this process. | ||||||||||||||||
min | 0.544 | 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 : | |||||||||||||||
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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 : | ||||||||||||||||
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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 -20oC : | Usat liq | 89.095 | kJ/kg | Usat vap | 1299.9 | kJ/kg | ||||||||||
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Eqn 6 | |||||||||||||||
U1 | 694.52 | kJ/kg | ||||||||||||||
The final state is simpler because the ammonia in the tank is a saturated vapor. | ||||||||||||||||
At 600 kPa : | Usat vap | 1326.1 | kJ/kg | U2 | 1326.1 | kJ/kg | ||||||||||
Next, we need to determine Hin. First, we need to determine the state of the system. | ||||||||||||||||
Tsat(Pin) | 17.848 | oC | Tin > Tsat, therefore we must consult the Superheated Ammonia Tables. | |||||||||||||
From the NIST Webbook, we can obtain : | Hin | 1521.3 | kJ/kg | |||||||||||||
Now, we can plug values back into Eqn 5 to evaluate Q : | Q | 600.83 | kJ | |||||||||||||
Verify: | None of the assumptions made in this problem solution can be verified. | |||||||||||||||
Answers : | T2 | 9.3 | oC | Q | 601 | kJ | ||||||||||
min | 0.544 | kg |