# Example Problem with Complete Solution

5E-1 : Charging an Evacuated Vessel From a Steam Line 6 pts
An evacuated tank will be filled with steam that flows into the tank from a steam supply line where the pressure is 2.8 MPa and the temperature is 350oC.
The steam from the supply line passes through a valve before it enters the tank. When the valve is opened, the tank fills with steam and the pressure increases until it reaches 2.8 MPa.
At this point, steam flow stops and the valve is closed. If the process is adiabatic and changes in kinetic and potential energies are negligible, determine the final temperature of the steam in the tank.

Read : Use the contents of the tank as the system.  Most of the key assumptions for this problem are given in the problem statement. One additional assumption is that no shaft work crosses the boundary of the system during the process.  Another crucial assumption is that this is a uniform flow, uniform state process.  These assumptions allow us to simplify the 1st Law dramatically.  We can use a transient mass balance to show that the mass in the tank in the final state is equal to the mass that was added to the tank.  This probably seems obvious since the tank was initially empty.  We can determine the specific enthalpy of the steam entering the tank from the Steam Tables because we know both Tin and Pin.  We will be able to determine the internal energy of the steam in the tank in the final state from the 1st Law.  Then, we can use the Steam Tables to determine the value of T2 using U2 and the given value of P2.
Given: Pin 1800 kPa Q 0 kJ
Tin 350 oC WS 0 kJ
P2 1800 kPa
Find: T2 ??? oC
Diagram: Initial : Final :  Assumptions: 1 - Although this is a transient process, it can be analyzed as a uniform flow, uniform state problem because the properties of the steam entering the tank are constant.
2 - Changes in kinetic and potential energies are negligible.
3 - No shaft work crosses the boundary of the system, which consists of the contents of the tank.
4 - The process is adiabatic.
Equations / Data / Solve:
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 1
In our process, no shaft work occurs, no heat transfer occurs, there is no mass leaving the system and there is no mass inside the system initially, so Eqn 1 can be simplified a great deal. Eqn 2
The integral form of the transient mass balance on the tank is : Eqn 3
Eqn 3 can be simplified because there is no mass leaving the sytstem and there is no mass initially inside the system. Eqn 4
We can use Eqn 4 to further simplify Eqn 2 : Eqn 5
Next, we need to determine Hin.  First, we need to determine the state of the system.
In the NIST Webbook, we can find the Psat(Tin) for steam: Psat(Tin) 16.529 MPa
Because Pin < Psat(Tin), we must consult the Superheated Steam Tables to evaluate Hin.
From the NIST Webbook, we can obtain : Hin 3141.8 kJ/kg
Plugging values into Eqn 5 gives us : U2 3141.8 kJ/kg
Now, we know the values of two intensive properties at state 2: pressure and specific internal energy.  The state is completely determined and we can use the Steam Tables to evaluate any other intensive property, such as T2.  To do so, we must first detemine which phase or phases are present in state 2.
At P = 1.4 MPa, the NIST Webbook tells us that :
Usat liq 882.37 kJ/kg Usat vap 2597.2 kJ/kg
Because U2 > Usat vap(P2), we must consult the Superheated Steam Tables to evaluate T2.
At P = 1.8 MPa, the NIST Webbook tells us that :
T (oC) U (kJ/kg)
510 3135.7
T2 3141.8
520 3152.9 Interpolation yields : T2 513.56 oC
Verify: None of the assumptions made in this problem solution can be verified. 