Example Problem with Complete Solution

5E-2 : Charging a Tank With R-134a 7 pts
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. 
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
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 :
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 ammonia.
In the initial state : At -20oC : P1 190.08 kPa
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 :
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 -20oC : Usat liq 89.095 kJ/kg Usat vap 1299.9 kJ/kg
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