5E4 :  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 200^{o}C.  


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 H_{2}O in the tank is 20% of what it was initially, determine the quality of the vaporliquid mixture 

in the tank and the amount of heat
transfer required up to that point. Note: this process can be considered a uniformflow, 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 200^{o}C. 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:  V_{1,liq}  20  L  T_{1} = T_{2} =  200  ^{o}C  
0.020  m^{3}  473.15  K  
V_{1,vap}  45  L  
0.045  m^{3}  
V  0.065  m^{3}  b.)  f  0.2  kg final/kg init  
Find:  a.)  m_{1,vap}  ???  kg  
m_{1,liq}  ???  kg  
m_{1}  ???  kg  
b.)  x_{2}  ???  kg vap/kg total  
when: 


c.)  Q_{12}  ???  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 crosssectional 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 200^{o}C :  V_{sat vap}  0.12721  m^{3}/kg  m_{1,vap}  0.3537  kg  
V_{sat liq}  0.0011565  m^{3}/kg  m_{1,liq}  17.29  kg  
We can determine the total mass of water in the system initially from: 

Eqn 3  
m_{1}  17.65  kg  
Part b.)  We know that: 

Eqn 4  m_{2}  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 :  V_{2}  0.018417  m^{3}/kg  
x_{2}  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 W_{S} = 0, DE_{kin} = DE_{pot} = 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 200^{o}C:  H_{out}  852.27  kJ/kg  
A mass balance allows us to determine m_{out} : 

Eqn 9  
m_{out}  14.12  kg  
Next, we can determine U_{1} and U_{2} : 

Eqn 10  

Eqn 11  
where: 

Eqn 12  
Plugging values into Eqns 10  12 yields:  x_{1}  0.02005  kg vap/kg  
At 200^{o}C :  U_{sat vap}  2594.20  kJ/kg  U_{1}  885.43  kJ/kg  
U_{sat liq}  850.47  kJ/kg  U_{2}  1089.24  kJ/kg  
We are finally ready to put numbers into Eqn 8 to complete this problem.  
m_{2} U_{2}  m_{1} U_{1}  m_{out} H_{out}  
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.)  m_{1,vap}  0.354  kg  b.)  x_{2}  0.137  kg vap/kg total  
m_{1,liq}  17.3  kg  
m_{1}  17.6  kg  c.)  Q_{12}  251  kJ 