So in this problem we've got a mass of 5 kilograms of liquid and vapor, they're in equilibrium,
they're in a piston cylinder device and the piston has a weight, when the piston moves
it induces a pressure of 300 kilopascals and that's just because of the weight of the piston
pushing down on the air. Heat is added to the system and we've got two kilograms of
liquid water therefore we've got three kilograms of water vapor so heat's added, once the pressure
reaches three hundred kilopascal the piston moves upwards at that point and heat transfer
continues until the volume is 20 percent greater than the original volume, and the first thing
we want to do is draw this state transition on a P V diagram, try to draw your own isotherms
on this P V diagram, and when I do it I often screw it up but I'll draw a temperature like
this and I've drawn it wrong, I can tell at a glance because when I compress a liquid
if I squeeze on it and I decrease it's volume I would expect the pressure to shoot up rapidly
so this pressure is going down and I drew it wrong, so here are correctly drawn isotherms,
I've drawn it for three different temperatures. So we want to draw the state transition on
this P V diagram and there's two things that occur, the first is heating under constant
volume before the piston is able to rise, and the second one is heating at constant
pressure. So at constant volume the state transition can only move vertically but at
constant pressure it can only move horizontally, so in the initial state we've got some vertical
transition but what we don't know is at this fixed volume is it heated to beyond the two
phase region, so did we boil all the water into steam or not, and it turns out that we
will and I'll show you in a second how we can figure that out, once the pressure is
great enough the piston lifts up and we continue at constant pressure to a greater specific
volume so here's the state transition on a P V diagram. So we've drawn the state transition,
we want to determine the initial and the final temperature, and to solve these problems in
both states I compile a list of all the intensive variables that I know so for example at state
one I know that the pressure is 125 kilopascal, and because it's a two phase mixture in equilibrium
I can look up in the water table the initial temperature, so here's the pressure table
for saturated water, at this pressure I find the saturation temperature a 105.97 degrees
C and for state two all I know is that P2 is equal to 300 kilopascal and I can't tell
you anything about the temperature because I don't know what isotherm I'm encountering
here or we would expect that there is some other isotherm that this would intersect but
we don't know far to the right this arrow is going or what isotherm it lands on. I should
point out that within the two phase envelope you can specify either the pressure or the
temperature but you're not allowed to specify both. so state two the only intensive variable
that we know is that P2 is equal to 300 kilopascal, one thing I do know is that the total mass
in state one is equal to 5 kilograms and that means the total mass in state two, because
it's a closed system, is also equal to 5 kilograms. Another thing that we know is that the volume,
at state two, is equal to 1.2 times the volume at state one, so if I knew the volume at state
one I would know the volume at state two and then I could find the specific volume at state
two because it's simply the actual volume at two divided by the total mass in the system
which we know, and if I can get V2 that gives me another intensive variable to work with.
So in state one, because I do know the pressure associated with it I can find the total volume,
I simply take the mass of vapor multiply that by it's specific volume and add that to the
mass of fluid and multiply that by it's specific volume so in terms of an equation the actual
volume at state one looks like this, and I can look up the specific volume of the liquid
in the vapor up in the saturated water table and at a pressure of 125 kilopascal, specific
volume for the saturated liquid of .001 cubic meters per kilogram and the saturated volume
for the vapor of 1.3750 cubic meters per kilogram, and what I come up with when I plug in numbers
is 4.127 cubic meters and that's the volume at the initial state, so that's the total
volume of the liquid and the vapor beneath this piston before it starts rising, and V2
is 20 percent bigger so what I get for V2 is 4.953 cubic meters. So now I can find the
specific volume at the second state and I get a value of 0.9905 cubic meters per kilogram.
Lets try to determine the mass of liquid water that's remaining within here once the piston
first starts moving, so the way I've drawn this P V diagram, the temperature rose and
it rose until it was outside the two phase region so this diagram would suggest that
all of the water had boiled off before the piston started moving upward so we'll try
to confirm that. And one way we can do so is we know the pressure, let's figure out
what the specific volume of the vapor would be in this case and if the specific volume
of a saturated vapor at this pressure is less than the actual volume, or in other words
if the actual volume was bigger than the specific volume associated with the saturated vapor
that means its now a superheated vapor because we've moved to the right of the saturated
vapor line. When the piston first starts moving we know the volume, we know the mass we can
calculate a specific volume of 2.85 cubic meters per kilogram, so when the piston first
starts moving we have a specific volume .825 cubic meters per kilogram and a pressure of
300 kilopascal so we want to figure out is this specific volume, is it greater than the
specific volume of saturated vapor, so we come back to the saturated water pressure
table and we'll look for a pressure of 300 kilopascal, and this column here is the saturated
vapor column, so let's scroll down until we hit 300 kilopascal and the specific volume
of the saturated vapor in this case is about .6 cubic meters per kilogram, before the piston
even starts moving this specific volume is bigger than the specific volume of the saturated
vapor so that means before it even starts moving the system is in a superheated vapor
condition, so to answer the question how much water is remaining, in that case there is
no liquid water remaining once the piston first starts moving. So what we've got left
to find is the final temperature and the amount of work done. So lets think about the final
temperature, we know we're in the superheated vapor region and we know there's a pressure
and what else do we know, at state two, we've got P2 is 300 kilopascal and the specific
volume at state two is .9905, so we know two intensive variables and that is enough in
the super heated region to give us any other information that we're interested in, including
what we want in this case which is the temperature. So here's the superheated water table, and
we're at a pressure of 300 kilopascal or .3 megapascal and for each specific volume we
can back out a temperature and this first column is the specific volume, now we want
a specific volume of about point 99 so we're going to be somewhere between a specific volume
of .87 and 1.03, the first one corresponds to a temperature of 300 degrees C and the
second one corresponds to a temperature of 400 degrees C we expect to get a temperature
between 3 and 400 once we interpolate, so here's a general formula for linear interpolation,
plugging in numbers I come up with a temperature of about 374 degrees C so this is the final
temperature in the system, the only thing we have left to calculate is the amount of
work that was done lifting up this piston, and its important to recognize that no work
was done initially because the volume was fixed, there was no flow work, the only time
that work was done was when the volume increased 20 percent when the piston was lifted. So
the boundary work's the integral of P dV, in this case the pressure is constant so it
comes out, we're left with P delta V and the pressure is 300 kilopascal or 300,000 pascal
times delta V which is the volume at state two which we calculated minus V1 and what
we come up with is a value of 248 kilojoules.