Here is a problem having to do with the filling of a tank that initially contains water at
1 bar and then the final state contains still water at 4 bar.
Please read this problem and then attempt to sketch out the solution on your own before
viewing this screencast.
I have drawn out the situation here based on the problem statement.
We have a tank that initially is at a pressure of 1 bar and you are given information about
the mass inside of the tank which is 1 kg.
The total volume which is 0.10 m^3.
So we can get the per mass volume of the water that is initially in the tank which is 0.10
m^3/kg.
This allows us to have two state variables that are specified.
The pressure and the per mass volume so that we can determine all the other properties
including the internal energy of the stream.
We will start off by going ahead and doing that.
The first thing we can do is determine the quality of the stream which again is the mass
fraction that is in the vapor.
We can do this by the lever rule which says that we can take the per mass volume that
we just determined of 0.1 subtract off the volume of the liquid and then divide that
through by the volume of the vapor divided by the volume of the liquid.
Which is equal to 0.059.
Again these numbers 1.69 and 0.001 they come from the steam tables at a pressure of 1.0
bar.
Actually the saturated steam tables.
If it was unsaturated steam then we would find that the per mass volume would not be
in between the volume of the liquid and the volume of the vapor.
But because it is between the volume of the liquid and the volume of the vapor at a pressure
of 1.0 bar we know that we have a saturated two phase system with a quality of 5.9%.
Now we can evaluate any of the thermodynamic properties.
For example the internal energy of the tank which I am calling condition 1 and multiply
by 0.059 kg of the vapor.
This is the internal energy of the vapor according to the steam tables at a pressure of 1.0 bar
plus the mass of the liquid which is just 1 kg minus 0.059 kg multiply that by 417 kJ/kg
which is the internal energy of the liquid as determined from the steam tables, saturated
steam tables at a pressure of 1.0 bar.
When you add that together you get 540 kJ as the initial internal energy.
Now we can look at the stream that is entering which is saturated steam at 5 bar.
The way I am going to set up this problem, you can actually solve this problem in a couple
of different ways.
The way I am going to show you is not necessarily the easiest but hopefully it will still be
illuminating.
What I am going to say is that my system in this case, I am going to operate on a closed
system, is going to be all the steam and all the water that ends up in the tank at the
final state of 4.0 bar after we have filled the vessel by opening and then closing this
valve.
That means that my system consists of all the steam that is originally in the tank plus
some steam that enters through these lines.
We do not know how much enters.
We are not given that in the problem statement.
So that is something we are going to have to determine and I will leave that as an unknown
variable m2 that enters.
We do know somethings about this, it's a fully specified stream, Saturated steam at 5.0 bar.
So we can say for example that the per mass volume if we look it up in the steam tables
should be something like 0.375 m^3/kg.
Therefore the total volume of this stream if we determine the mass we can just multiply
by this 0.375 m^3/kg.
The same thing for the internal energy.
We can multiply by that mass if we determine what it is from the steam tables at 5.0 bar.
The internal energy of the vapor is 2561 kJ/kg.
m2 is still an unknown and hopefully we can figure it out by doing some kind of balance
as we go to the final condition.
We know that we know 1 state variable and that is that the final pressure is equal to
4.0 bar.
What we will assume and we will be able to of course in the problem solution determine
that this is right is that we are going to have a saturated vapor-liquid mixture here
in the final condition.
That is because we are adding a saturated stream to a system that contains already a
fluid that is mostly liquid but is in vapor-liquid equilibrium.
So we might expect that we are in vapor-liquid equilibrium here but as would see will if
we tried to solve this problem using the saturated steam tables and ended up not being able to
end up at a solution this would actually indicate to us that we needed to go use the super heated
steam tables or the sub-cooled liquid tables.
We can write out a couple of equations here.
One thing we know is the volume of the tank.
Again we do not know the mass of the tank but we can say that the final per mass volume
is going to be equal to the initial volume of 0.1 m^3 over the 1 kg that was in there
initially plus whatever mass we add throughout this process.
So that might be useful relationship.
The other thing we can do is write out our energy balance for this system.
The problem statement specified that it is a well insulated system and so we do not have
any heat transfer in this process but we do have a change in internal energy potentially.
We also have work that is done in this process.
So what is causing the system to contract and going from here where it occupies some
volume in the lines to the final state is an external pressure of 5.0 bar that is pushing
the fluid into the tank.
Because of that we will need to include that work and that work will therefore be equal
to the negative of the external pressure, 5.0 bar, times the change in volume and that
change is a negative change which is why I am cancelling out the negative signs.
It is actually just equal to the volume that this steam occupied initially.
That is the volume that you lose, that you have the volume of the tank in both the initial
and final states.
So that volume change is equal to the volume that occupied the steam originally in the
lines.
So that is equal to V2 but it is still something that we do not know but again we do know we
can substitute a little bit further.
We do know from this expression in the top left hand corner that that's equal
to 0.375 m^3/kg times m2.
Now we want to be able to solve for what the final internal energy is using the steam tables
so start solving for this delta U.
In the final state we know what the pressure is so we can look up the values of the internal
energy for the vapor and the liquid.
What we can basically say here first of all the quality of the stream if we were to know
the volume, the per mass volume of the system, we could again use the lever rule in the same
way that we did before to look up in the steam tables the volume, the per mass volume of
the liquid and of the vapor and apply the quality equation to solve for that.
If we had that then we could multiply, so if we are solving for delta U here, we can
multiply the quality times the total mass of the system which is m2 plus 1.0 kg times
the internal energy of the vapor at the final conditions at 4 bar.
Then add that to 1 minus the quality times the total mass again, m2 plus 1kg, times the
final per mass volume of the liquid which we can also look up in the steam tables.
Then subtract off from that the initial state which is m2 time 2561 as well as the initial
internal energy of the tank which has already been determined to be 540 kJ/kg.
We have as it turns out enough equations for our unknowns but maybe it is a little bit
hard to see that here that we use the steam table to actually be able to solve for this.
So let's outline the procedure that we can actually use to solve for basically these
two equations and two unknowns, the volume and the delta U.
So what we will do is here is one procedure.
We will guess the value of V hat f so the per mass volume.
We can use that to calculate the mass that enters the tank.
If we back up here we can see that if we know the volume, if we specify then we can solve
for m2.
Then we can solve for the quality using m2.
Then in the fourth step we will solve for what the expected internal energy of the final
state should be.
We can do that again looking up the known values for the liquid times 1 minus the quality
and the vapor times the quality.
That should give us the per mass final internal energy.
Then what we can do is, we can solve for q from the energy balance equation.
We will call this q in step 5.
We can ask the question now.
Does q from step 5 equal the q that we determined from step 3.
The first guess is probably not and so we need to go back and make a new guess.
We can keep iterating on this process until we do get acceptable convergence and then
that is our final quality.
If I do this process the final quality I get is approximately 0.18 and it turns out that
means we added about 0.18 kg from those lines that supplied the steam.
The easiest way to actually set this up is that you can basically set this up where you
guess Vf in excel spread sheet and then use solver or some other similar method in order
to keep varying Vf until you get very little difference between the q evaluated in step
5 and the q evaluated in step 3.