Each of two vessels contains
of steam at a different temperature and pressure. Vessel A
is a rigid tank with a volume of 0.9 m^{3} and vessel B is a pistonandcylinder device that holds 0.7 m^{3} of steam. 
They are connected by a pipe with a closed valve in the line. Initially, tank A contains saturated steam at 150 kPa while cylinder B
contains superheated steam at 350^{o}C and 400
kPa. 






When the valve is opened,
the steam in the two vessels is allowed to come to equilibrium.
a.) Determine the mass of steam in each vessel before the valve
is opened: m_{A1} and m_{B1}.

b.) If the equilibrium temperature is T_{2} = 240^{o}C, calculate Q and W for the equilibration process. 


Read : 
The key aspect of this
problem is whether ANY water remains in the cylinder, B, at equilibrium. If there is water
left in B at the final state, it will exist at T_{2} and P_{2} = P_{B1} because the piston would still be "floating". The other key is that this is a closed system, so the mass of water in the entire system remains constant. We can use the Steam Tables and the given initial volumes to answer part (a). In part (b), the work is done at constant pressure as the piston descends. So it is not difficult to compute. Finally, solve the 1st Law to determine Q.
This is possible because we know the initial and final
states and the work ! 










Diagram: 












Given: 
P_{A1} 
150 
kPa 



V_{B1} 
0.7 
m^{3} 

V_{A} 
0.9 
m^{3} 



T_{B1} 
350 
^{o}C 

x_{A1} 
1 
kg vap/kg 



P_{B1} 
400 
kPa 

T_{2} 
240 
^{o}C 
















Find: 
a.) 
m_{A1} 
??? 
kg 

b.) 
Q 
??? 
kJ 


m_{B1} 
??? 
kg 


W 
??? 
kJ 










Assumptions: 
1  
 The initial and final states are equilibrium states. 


2  
 The process is a quasiequilibrium process. 










Equations
/ Data / Solve: 

















Part a.) 
We can determine m_{A1} because we know the volume of the tank and we can look up the specific
volume of the saturated
vapor that it contains. 












Eqn 1 

NIST WebBook: 
V_{A1} 
1.1594 
m^{3}/kg 





m_{A1} 
0.7762 
kg 











We can use the same
approach to determine m_{B1}, but first we must determine its state. 











At 400
kPa, T_{sat} = 133.52^{o}C. Since T_{B1} > T_{sat}, tank
B initially contains superheated vapor. 













NIST WebBook: 


V_{B1} 
0.71396 
m^{3}/kg 

















m_{B1} 
0.9805 
kg 










Part b.) 
In part
b, there are two possibilities. At equilibrium, either B contains some water or it is completely empty. 











Case
1  B is not
empty: P_{B2} = P_{B1} because the piston is still
floating. 














Case
2  B is empty: V_{B2} = 0 and ALL
of the water is in tank A. 














Let's test Case 1 first. T_{1} = T_{sat} at 150 kPa, so T_{1} = 111.3^{o}C. T_{sat} at 400 kPa is 143.6^{o}C. Since T_{2} > T_{sat} at 400 kPa, the water would still be superheated vapor and the specific volume would be: 







V 
0.58314 
m^{3}/kg 











Therefore the total volume occupied by this superheated vapor would be: 














Eqn 2 


V_{2} 
1.0244 
m^{3} 









Since this volume, which the total mass of water
in the system occupies at P_{B1}, is greater than the volume of tank
A, we can conclude
that all of the water could not fit into tank A at P_{B1}. If P_{2} were less
than P_{B1}, the water would occupy even more volume and again would not fit into tank
A. Some water must remain in the cylinder and, therefore, P_{B2} = P_{B1}. 











Therefore: 



V_{B2} = V_{2}  V_{A1} = 
0.1244 
m^{3} 











Calculate the PV or boundary
work from: 



Eqn 3 











But, since this
process is isobaric: 


Eqn 4 















W 
230.2 
kJ 











Finally, we need to
apply the 1st Law to
determine Q. Use all of the water in both vessels as the system. 


















Eqn 5 











Or: 


Eqn 6 











Use the NIST WebBook and the ASHRAE Convention to determine all
of the specific internal energies. 











m_{tot} 
1.7567 
kg 



U_{A1} 
2519.2 
kJ/kg 

U_{2} 
2710.6 
kJ/kg 



U_{B1} 
2884.4 
kJ/kg 







DU 
21.8 
kJ 

















Q 
252.0 
kJ 

Verify: 
None of the
assumptions can be verified from the data given in the problem statement. 










Answers : 
a.) 
m_{A1} 
0.776 
kg 

b.) 
W 
230 
kJ 












m_{B1} 
0.980 
kg 


Q 
252 
kJ 









