Example Problem with Complete Solution

5E-4 : Discharging a Tank Containing Water and Steam 6 pts
The rigid tank, shown below, contains 20 L of liquid water and 45 L of water vapor in equilibrium at 200oC.
   
 
             
When the valve in the drain line is opened slightly, liquid water flows slowly out of the tank at a constant rate. Heat transfer into the tank keeps the temperature within the tank uniform and constant.
a.) Determine the mass of liquid water and the mass of water vapor initially in the tank
b.) When the total mass of H2O in the tank is 20% of what it was initially, determine the quality of the vapor-liquid mixture 
in the tank and the amount of heat transfer required up to that point.
Note: this process can be considered a uniform-flow, unsteady process.
 
Read : The key to this problem is that the process is an isothermal process.  As a result, the properties of the liquid inside this system and leaving the system are always the properties of saturated liquid at 200oC.  As a result, this is a uniform state process.  We can also assume it is a uniform flow process.  If we further assume that changes in kinetic and potential energies are negligible and that no shaft work occurs, we can use the 1st Law to determine Q.  Parts (a) and (b) require use of the Steam Tables or NIST Webbook and a working knowledge of the relationship between mass, volume and specific volume, but should not be difficult.
Diagram:
Given: V1,liq 20 L T1 = T2 = 200 oC
0.020 m3 473.15 K
V1,vap 45 L
0.045 m3
V 0.065 m3 b.) f 0.2 kg final/kg init
Find: a.) m1,vap ??? kg
m1,liq ??? kg
m1 ??? kg
b.) x2 ??? kg vap/kg total
when:
c.) Q12 ??? kJ
Assumptions: 1 - Only saturated liquid water leaves the tank.
2 - The process is isothermal.
3 - Only flow work (no shaft work) crosses the system boundary.
4 - Changes in kinetic and potential energies are negligible.
5 - Uniform Flow: The properties and flow rate of the outlet stream are constant over the cross-sectional area of the pipe and with respect to time.
6 - Uniform State: At all times, the properties of the outlet stream are the same as the properties of the contents of the system at that point in time.
Equations / Data / Solve:
Part a.) Because the water liquid and vapor are in equilibrium with each other at all times throughout the process, they are always saturated.  Therfore, we can determine the mass of liquid and vapor initially in the tank by looking up their specific volumes in the Saturated Steam Tables or the NIST Webbook.  
Eqn 1
Eqn 2
At 200oC : Vsat vap 0.12721 m3/kg m1,vap 0.3537 kg
Vsat liq 0.0011565 m3/kg m1,liq 17.29 kg
We can determine the total mass of water in the system initially from:
Eqn 3
m1 17.65 kg
Part b.) We know that:
Eqn 4 m2 3.53 kg
The key here is that we know both the volume and the total mass in the tank, so we can calculate the specific volume and use it to determine the quality from :
Eqn 5
The real key to this problem is that the process is isothermal.  As a result, the properties of the saturated vapor in the tank remain constant and the properties of the saturated liquid inside the tank and flowing out of the tank also remain constant.
We determine the overall specific volume at state 2 from:
Eqn 6
Now, we can plug values into Eqns 6 & 5 : V2 0.018417 m3/kg
x2 0.1369 kg vap/kg tot
Part c.) To determine Q, we need to apply the 1st Law for transient processes and open systems.
Here, we assume that WS = 0, DEkin = DEpot = 0.  The appropriate form of the 1st Law under these conditions is:
Eqn 7
Solving Eqn 7 for Q yields:
Eqn 8
The specific enthalpy of the water leaving the system is the enthalpy of saturated liquid water at 200oC: Hout 852.27 kJ/kg
A mass balance allows us to determine mout :
Eqn 9
mout 14.12 kg
Next, we can determine U1 and U2 :
Eqn 10
Eqn 11
where:
Eqn 12
Plugging values into Eqns 10 - 12 yields: x1 0.02005 kg vap/kg
At 200oC : Usat vap 2594.20 kJ/kg U1 885.43 kJ/kg
Usat liq 850.47 kJ/kg U2 1089.24 kJ/kg
We are finally ready to put numbers into Eqn 8 to complete this problem.
m2 U2 m1 U1 mout Hout
Q = 3844 - 15625 + 12032 kJ
Q 251.2 kJ
Verify: The assumptions made in this solution cannot be verified with the given information.
Answers : a.) m1,vap 0.354 kg b.) x2 0.137 kg vap/kg total  
m1,liq 17.3 kg
m1 17.6 kg c.) Q12 251 kJ