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 350^{o}C. 
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 T_{in} and P_{in}. 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 T_{2} using U_{2} and the given value of P_{2}. 










Given: 
P_{in} 
1800 
kPa 



Q 
0 
kJ 

T_{in} 
350 
^{o}C 



W_{S} 
0 
kJ 

P_{2} 
1800 
kPa 
















Find: 
T_{2} 
??? 
^{o}C 
















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 singleinput, singleoutput 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 H_{in}. First, we need to
determine the state of
the system. 











In the NIST Webbook, we can find the P_{sat}(T_{in}) for steam: 
P_{sat}(T_{in}) 
16.529 
MPa 











Because P_{in} < P_{sat}(T_{in}), we must consult the Superheated Steam Tables to
evaluate H_{in}. 











From the NIST Webbook, we can obtain : 


H_{in} 
3141.8 
kJ/kg 











Plugging
values into Eqn 5 gives
us : 


U_{2} 
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 T_{2}. 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 : 















U_{sat liq} 
882.37 
kJ/kg 



U_{sat vap} 
2597.2 
kJ/kg 











Because U_{2} > U_{sat vap}(P_{2}), we must consult the Superheated Steam Tables to
evaluate T_{2}. 











At P
= 1.8 MPa, the NIST
Webbook tells us that : 














T (^{o}C) 
U (kJ/kg) 








510 
3135.7 








T_{2} 
3141.8 








520 
3152.9 

Interpolation yields : 

T_{2} 
513.56 
^{o}C 










Verify: 
None of the assumptions
made in this problem solution can be verified. 













Answers : 
T_{2} 
513.6 
^{o}C 

















The temperature of the steam in the tank in the final state is greater than the temperature of the steam in the feed line because the surroundings did flow work on the system as it filled with steam. This flow work caused the internal energy of the steam in the tank to exceed
the internal energy of
the steam in the feed line. Consequently, the temperature of the steam in the tank in the final state must be greater than the temperature of feed. 









