This is a problem were we have a sealed tank.
So the volume is 60 cubic cm, and it does not change.
What it starts out with is water at 600 degrees and 8,000 kPa pressure, and then we cool down
the tank down from 600 degrees to 150 degree C, and what we want to know is the final pressure
in the tank and what phases we have present.
If there is more than 1 phase what is the fraction of water that is in each of the phases.
So this is a problem were we are going to use steam tables to get the properties of
water at the varies conditions, and at first we want to set up what we know.
So we have a fixed volume container.
It is 60 cubic cm, 600 degree C, and 8,000 kPa.
We are going to cool it.
The volume is not going to change, but my drawing of making the volume identical probably
changes a little.
So we still have 60 cubic cm, buy know we are at 150 degree C, and lets call this pressure
2.
This is one of our unknowns, and the other unknown is what phases we have.
So we can go to the steam tables, and we can look up 600 degree, and 8,000 kPa.
Now at this pressure the saturation temperature is 295 degree C. So we are well above saturation
temperature.
So this is superheated steam that we start out with.
All in one phase, and we can from the steam tables we can determine the specific volume
at these conditions.
So this is volume per gram, 48.5 cubic cm per gram.
So the fact that the actual volume is 60 cubic cm doesn't really matter for solving this
problem, because as you can imagine if the tank was half the size and everything was
identical we would get the same behavior.
Now for other steam tables if you look up the specific volume you might find in units
of cubic m per kg.
So it would be 0.0485 cubic m per kg.
Seems a little easier here to work in terms of cubic cm per g.
So our final specific volume per gram is the same, and so to determine our final condition
it helps to look at a pressure volume diagram to visualize what is happening.
This is not drawn to scale, but we know that we are above the saturation line.
So this line here is saturated vapor, and we are somewhere here.
This is point 1 were we start.
Now we are cooling this down.
We are keeping the volume constant.
So constant volume means our point 2.
Our final point is somewhere along this line as we lower the temperature.
We are going to lower the pressure, and so the question becomes are we above saturation.
So this is the saturation or are we bellow it, and so, for 150 degree C we can look at
saturation conditions in the stream table.
So what I have done here is listed values from the stream tables for 150 degree C, and
these are the volumes of saturated liquid and the saturated vapor at 150 degree C. Since
our volume is 48.5 it is in between these 2, and we know this volume, this volume 2
is 150 degree C. That means we are somewhere down here in the steam tables.
So remember this is a log scale.
So at 150 degree C the pressure.
The saturation pressure for water again from the steam tables.
This is 476 kPa.
So this is the line for 476.
This point corresponds to specific volume of 1.09 cubic cm per gram.
This point is 392.4.
We have answered part of the problem here.
Here is the final pressure, because we are in the 2 phase region, and now we need to
determine how much of each phase.
So what I am going to use.
I am going to use x as the fraction of vapor.
This of course is also known as quality for steam.
So the fraction of vapor times the volume of the vapor phase that is at saturation conditions
150 degree C plus 1 minus x.
The fraction of liquid times the volume of liquid should equal V2= 48.5 cubic cm per
gram.
So we are going to substitute these two values into this equation, and we can now solve this
algebraic equation for x; x=0.124.
So this says that 12.4% is vapor and the rest is liquid.
From our data we have 2 sig.
figs.
So 12% vapor 88% liquid. of course this is a percent by mass.
So this is the other part that we want to answer.
What percentages are in each phase.
So by using the steam tables, and understanding physically what the behavior is we can calculate
the final conditions.