Example Problem with Complete Solution

4C-2 : Equilibration of a Tank and a Piston-and-Cylinder Device 5 pts
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 m3 and vessel B is a piston-and-cylinder device that holds 0.7 m3 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 350oC 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: mA1 and mB1.
b.) If the equilibrium temperature is T2 = 240oC, 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 T2 and P2 = PB1 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: PA1 150 kPa VB1 0.7 m3
VA 0.9 m3 TB1 350 oC
xA1 1 kg vap/kg PB1 400 kPa
T2 240 oC
Find: a.) mA1 ??? kg b.) Q ??? kJ
mB1 ??? kg W ??? kJ
Assumptions: 1 - - The initial and final states are equilibrium states.
2 - - The process is a quasi-equilibrium process.
Equations / Data / Solve:
Part a.) We can determine mA1 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: VA1 1.1593 m3/kg
mA1 0.7763 kg
We can use the same approach to determine mB1, but first we must determine its state.
At 350 kPa, Tsat = 133.52oC.  Since TB1 > Tsat, tank B initially contains superheated vapor.
NIST WebBook: VB1 0.71396 m3/kg
mB1 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: PB2 = PB1 because the piston is still floating.
Case 2 - B is empty: VB2 = 0 and ALL of the water is in tank A.
Let's test Case 1 first.  Since T2 > Tsat at 500 kPa, the water would still be superheated vapor and the specific volume would be:
V 0.58314 m3/kg
Therefore the total volume occupied by this superheated vapor would be:
Eqn 2 V2 1.0245 m3
Since this volume, which the total mass of water in the system occupies at PB1, is greater than the volume of tank A, we can conclude that all of the water could not fit into tank A at PB1.  If P2 were less than PB1, the water would occupy even more volume and again would not fit into tank A.  Some water must remain in the cylinder and, therefore, PB2 = PB1.
Therefore: VB2 = V2 - VA1 =   0.1245 m3
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.
mtot 1.7568 kg UA1 2519.2 kJ/kg
U2  2651.0 kJ/kg UB1 2884.4 kJ/kg
DU -126.6 kJ
Q -356.8 kJ
Verify: None of the assumptions can be verified from the data given in the problem statement.
Answers : a.) mA1 0.776 kg b.) W -230 kJ
mB1 0.980 kg Q -357 kJ