Thermodynamics: 1st Law for Closed Systems, Specific Heats (9 of 25)

Thermodynamics: 1st Law for Closed Systems, Specific Heats (9 of 25)


[ Music ]>>What I want to do
today then is just begin with another example problem. In fact I’ll end with an
example problem as well. Basically that’s the
same thing as this. This is just my copy
in my notebook. So I’ll move this now. So we’ll do two example
problems today. And hopefully after you see
these two examples you have a much better feel for the
nature of the kind of problems that you’re going to be
dealing with in this class. By the way the midterm
exam is going to cover through section 42 which is,
you know, problem solving for a first law for
closed systems. And again it’s through
today’s material, through the homework
that’s due next week. All right so we’ve looked a
couple example problems already. Both of those were sealed
rigid vessel types of problems. And now we’re going to look at something a little
bit different. I actually put this on the board
last time but today we’re going to actually solve
through this problem. This does say problem
132 but it’s actually from the 6th edition, we’re
up to the 8th edition now. So you’re not going to
find this in your textbook. So I put it up here and
I’ll just leave it up there for a little while longer as I’m
starting to set up this problem. So as usual let’s
read through it. So we have a mass .2
kilograms of saturated R-134a. And it’s contained in a
piston cylinder device which is at kilopascal. It says initially 75% of the
mass is in a liquid phase which of course means 25% vapor. It says now heat is
transferred to the refrigerant at constant pressure until the
cylinder contains vapor only, all right. Now I will note that the author
could have just simply said that this was a frictionless
vertical piston cylinder device which automatically means that it’s a constant
pressure process. Instead the author gave the
initial pressure and then said that heat transfer occurs
at constant pressure. So either way it’s a
constant pressure process. Anyway show this process
on the P-v diagram, determine the volume
occupied by the refrigerant at the initial state,
the work done and the total heat transfer. All right, so that
is this problem. So 4-132 edition 6. Okay. All right, so the
first thing I always like to do is draw a
sketch and make sure that I show all the information
that’s known and unknown. So this one is a
piston cylinder device. So basically I would
show a piston cylinder. We’re given the mass,
so 0.2 kilograms. It’s R-134a, so I
note the substance. It says it’s saturated and then
it tells me the percentage mass in the liquid phase,
oh I’m sorry. First it tells me
the initial pressure so I’ll just put P1 is
equal to 200 kilopascal. All right now it tells
me that initially 75% of the mass is in
a liquid phase. And we all know that the
author did this to us on the last problem, right. We need the quality which is the
percentage vapor phase not the percentage liquid phase. So 75% liquid obviously
means 25% vapor. So the quality is given us .25. It says heat is transferred. It doesn’t tell us which
way heat is transferred just that heat is transferred. So I’ll show Q with kind
of ambiguous arrows. And then it says transferred
at constant pressure until the cylinder
contains vapor only. All right, so constant pressure. So I’ll just say P1
is also equal to P2. And now it starts as a two
phase mixture at the point where it contains, where
it contains vapor only is where we hit the
saturated vapor line, right. So we definitely know that
we’re at saturated vapor at .2. So 2 is saturated vapor. And we want to show
this on a PV diagram. And we want to find the
volume occupied initially. And we want to find
the work and we want to find the heat transfer. All right, so this pretty
much sets up the problem. Now let’s just start
one by showing it on the PV diagram even though
I don’t necessarily have to do this immediately. It does make a certain
amount of sense to do so. We note that we start as a reasonably high
quality two phase mixture. And since we’re moving
at constant pressure and this is a PV diagram then
it means that we have to moving to the right, right, moving along the
constant pressure line. And at the point where it
finally becomes all vapor that’s right there on the
saturated vapor line. So there’s my state point two. If this problem said
that heat was added until it becomes super-heated
well maybe we could perhaps show an entire constant pressure
line as we look like that and that way we show .2
being somewhere up here above saturated vapor line. But not for this problem, right. I’m sorry, I didn’t
do this right. This is a PV diagram,
I’m thinking TV diagram. Constant temperature lines
kind of work like that. And nonetheless it
doesn’t matter. If it’s super-heated vapor then
clearly we’ve added heat beyond the point it’s a saturated vapor and that’s not what
we want, right. This problem ends when the
cylinder contains only vapor and there you have it. Yes? [ Inaudible Student Question ] You know what? I didn’t really say
that properly. Yeah, 75% would be high quality but that’s the percentage
of liquid. Twenty five percent is a
relatively low quality. Honestly it’s still
probably going to be near the right hand
side because after all even through the vapor’s only 25%
by mass, it’s specific volume at this pretty low
pressure is going to be about 1,000 times greater than the specific
volume of the liquid. So even though it’s only 25% by
mass it might be more like 80 or 90% by volume and this
is a volume axis here. So it is very possible at .1
is far to the right like this. I don’t really know
for sure until I go through the numbers but, you
know, at least this is accurate. You know, horizontal
[inaudible]. All right, so what
next do we need to do? Well let’s make sure
we know what to assume, what not to assume. First of all there’s no kinetic
energy change as is the case with essentially all
these problems, right. There’s no velocity associated with the substances
in that cylinder. What about the potential energy? Clearly there’s going to
be a height change, right. I mean the volume is increasing so the piston does
have to move upwards. However, what you find is that
potential energy changes have to be what, I’m sorry
let me rephrase that. Height changes have to be very
significant before there’s any meaningful and measureable
change in potential energy. Usually you’d have to have a
height change of, you know, something on the order
of several meters before that potential energy
is not negligible. So here we’re talking about a
really small mass of material. I don’t know what
the total volume is but it’s not that great. And, you know, even if the cylinder rose
significantly there’s still not going to be several
meters of height change. So it makes sense to assume
that there’s no change in the potential energy, okay. So that’s typical of
these kind of problems. I might also note I
have no way to quantify that even if I wanted to. I don’t know if this is a short
squat cylinder which is going to increase a fraction or if it’s a tall skinny
cylinder that’s going to increase significantly. They don’t tell me anything about the geometry
of that cylinder. So even if I wanted to find
the height change I have no way to do so. So I just have to assume
that there’s no change in potential energy, okay. Now there definitely will
be some heat transfer. There’ll definitely
be some work. We know that work in this
problem is all boundary work, right, they’re not
electric resistance work, there’s nothing like that. And for a constant pressure
process this is mass times pressure times change
of specific volume. Remember this is boundary work but for constant
pressure only, all right. So that’s how we’re
going to find the work. And then we’re always going
to need the first law. The heat transfer is the change
in internal energy plus changes in kinetic and potential
energy plus the work. And by the way if I was
able to quantify the kinetic and potential energy, I
would definitely write out the equations but
here we know they’re zero so I mean why bother. We’ll just show delta
KE and delta PE and just show them
equal to zero. So this is the basic equation
that we need to solve for. So now the question is
can we find everything that we’re looking for? Not quite yet. What about the volume
at point 1. The only way to find the
volume is to know the mass and the specific volume. And fortunately I
can find that, right. So I’ll also note that to find
the intial volume is just mass times specific volume that
is initial specific volume. I’m given the mass
and I need V1. So with these equations we’ll
ask ourselves do we have enough information now to
solve the problem? So before I even solve anything
we just say well yeah, right. We know that, as long as we
know my two state points, state points 1 and 2, I
can into my property tables and I can find any
property I want. Now quality counts
as a property, right. So we know the pressure and
that the quality is 25%. So yes we know state
.1, we can find V1 as just Vf plus quality
times Vfg. We can find U1, just the
quality times Uf, I’m sorry, just Uf plus the
quality times Ufg. So we know state point 1. State .2 is given, I’m sorry, pressure 2 is given,
it’s the same as 1. What about state 2? Well again we need two
properties and we have them. We know that the pressure is
still 200 but we also know that it’s a saturated vapor
meaning that the quality’s 100%. We know that at state .2 V
is just Vg, U is just Ug, right, saturated vapors. So we have both [inaudible]
states. We can find everything
shown down here on the page. So now it’s just
a matter of going through a little
bit of mathematics. I will note that because
we’re talking R-134a and we’re in the SI system then
we’re going to have to use tables A11 and A12. A13 is a super-heat table but we don’t have super-heated
vapor anywhere in the problem. So we’ll just use A11 and A12. And in fact we really don’t
have the temperature anywhere in the problem either. We’re given pressure
at points 1 and 2. So really it’s just A12
that we’re going to use or our saturation
pressure table. So all the data’s going
to come out of there. Now we can work on this problem. So let’s just find our
thermodynamic property data first. And we’ll start at state .1. So at P1 and at the quality
at 1 we go into table A12. And now let me get rid
of the problem statement and put the property tables up. [ Pause ] Okay, so A12, all right. So here’s table A12, the
saturated R-134a pressure table. And all the data we need is
going to be at 200, right. So we go to the right table. We know the pressure. We read across. We need both Vf and Vfg. Of course we don’t have Vfg,
it’s not printed in the book. I don’t know why the author
didn’t just add one more column and give you the Vfg
data but he didn’t. So V1 is Vf plus the
quality times Vfg but we know Vfg is
just Vg minus Vf. [ Pause ] All right, so this is the
equation we have to solve for. So let’s just pull the
numbers out of our table. You can see them
right over here. So we go down to
the 200 psi line, I’m sorry, kilopascal line. We read across and Vf is
.0007532, the quality is 25%. By the way I should
put X1 over here just to make sure we know this
is the initial quality. Anyway, quality’s .25. And then Vg is .099951 minus Vf. So point triple zero 7532. And if we just go through the
mathematics we end up with V1 of .02555 cubic meters
per kilogram. So there’s V1. And we do U1 exactly
the same way. This is Uf plus the
quality times Ufg and there is Ufg
data in this book. So again just go to the 300,
I’m sorry, 200 kilopascal line and read across to the
5th and 6th columns. Uf is shown as 38.26 and then
plus the quality times Ufg which is 186.25, oh yeah. And then again the units are
all kilojoules per kilogram. We go through a little math. And we get 84.82
kilojoules per kilogram. By the way one thing I want to note sometimes I may have
some incorrect numbers in some of my problems that I’m
solving in class or even some of the problems I’ve
posted on the Blackboard or on the Board S in my office. You know, conventions are held
like every five years or so by the international community
and new research is presented. And the tables change, you know. New data becomes available for refrigerants or
water or whatever. So periodically the various
governing bodies usually say Sme. We’ll then print
new property data. So if you look at the property
tables from the 6th edition of this textbook which
is I think about 7 or 8 years old new compared to
the property tables that are in the 8th edition, the numbers
will be slightly different. Not all of them. The refrigerant tables didn’t
change, the water tables changed or maybe it was the
other way around. It doesn’t matter. I fixed everything on this
problem and I’ve tried real hard to fix everything on all
the problems that I’ve used in the past but I might
miss something a little bit. I have noticed that some of you are indeed using
[inaudible] property tables presumably from some
old book that you have. I know that because some of your
numbers are subtly different. You know, maybe just off down
in the 3rd or 4th decimal place. But some of you are using
numbers that aren’t correct. So that’s okay, I’ll
still accept that. But I just want you to
be aware the numbers do change periodically. All right, so back
to the problem. Let’s also find the
volume at this point. So that’s the mass times
the specific volume. So .2 kilograms times .02555
cubic meters per kilogram. And that’s going to give me
.00511, oops, cubic meters. So kilograms cancel and there’s
the first thing we’re looking for, the volume. All right, next let’s find the
thermodynamic property data at state .2. So at P2 saturated vapor. We’re still in table A12. So V1 is equal to Vg and
again we just go to the table. In fact Vg’s already here. It’s the same .09995
that was up at the top. So .0999.>>It should be M2
right, not P2.>>Oh yeah, yeah,
yeah, I’m so sorry. It is .2, I mean, I
put P2 saturated vapor but then I wrote and said V1. This is .2 saturated vapor. So V2 is Vg so it’s .099951
cubic meters per kilogram. And similarly U2 is
Ug which is then going to be 224.51 kilojoules
per kilogram. So we have all the data that
we need for the problem. Just two things left, right. We can find the work, we
plug it into the first law. We find the heat, we’re done. So work first. So the work and I’ll just
rewrite the equation. So we have .2 kilograms, our
pressure is 200 kilopascals, then .099951 minus plus .02555
meters cubed per kilogram. You know, as usual you
look at those units and the units clearly
don’t look correct. I mean, where’s the
kilojoules and all this. But as usual let’s just remember that a kilopascal is the
same thing a kilonewton per square meter. And a kilojoule is the same
thing as a kilonewton meter. So we can just cancel
out all the units here and everything cancels. Mass cancels, cubic meters
cancel, kilopascals cancels, kilonewtons cancel,
so we’re just left with kilojoules, right. So as long as you’re in
the SI system solving for work there’s no unit issues. Everything’s an identify
with regard to these conversion factors
so it works out easily. Caution, in the British
system it’s not going to work out this easily. It will definitely not
work out this easily. You know, you’ll have Psia
for your pressure units and then cubic feet per pound. A Psia cubic foot is not
the same thing as a BTU. You’re going to have to go into
the conversion factors which is in the inside back
cover of your book or in my case the last
page of the handout. And you’ll find the
conversion factor you need. In fact it’s right here where
the cursor is, why cursor turned to an x, I don’t know. But it’s right here. It says 1 BTU equals
5.40395 Psia foot cubed. So all the conversions you
need are in the book somewhere. But follow through
with the units. If you don’t follow through with
the units there will be errors. All right, no unit issues. We go through the mathematics and we find the work
is 2.98 kilojoules. And it’s a positive number
which is supposed to be, right. The volume is increasing. The piston is moving upwards. Work is being done by
the system as it pushes into the piston and
moves it upward. So work done by the
system is positive. And that’s a positive number. So that gives us
some confidence. Lastly heat transfer. All right, so heat transfers
found with the 1st law. Again I’ll rewrite the equation. So it’s really just
a matter of plugging in all the numbers
we already have. So .2 kilograms, U2 minus
U1, so 224.51 minus 84.82, those are both kilojoules per
kilogram plus 2.98 kilojoules. The units work out just fine and
we go through the mathematics. And we end up with
30.9 kilojoules. So that is this problem. Any questions? There is an alternate way to find the heat
transfer isn’t there? Remember that is we have a
constant pressure problem we can use [inaudible] instead
of internal energy as long we do it properly. I would note that for a
constant pressure process. [ Pause ] We still have the
same equation, right. I mean this is our
basic equation. However, the boundary work
is mass times pressures times change in specific volume. We remember that U2 and P2V2
is H2 and U1 plus P1V1 is H1. So it turns out that if
we wanted to we could just as easily have found
the total heat transfer for a constant pressure
process only and this is a constant
pressure process, right. But we could have found it just
as easily by finding H2 and H1. I could’ve gone back to state .1 and I say H1 equals Hf
plus quality times Hfg, H2 is just Hg, right. And I could have found
the heat transfer in a slightly more
direct fashion. Now you may wonder
well why didn’t I do it if it’s a little easier
to do it that way? Well the reason I didn’t do it
is because the author asked me to calculate the work. In fact to calculate the
work anyway what does it really matter? I mean I’m going to
need to find most of this thermodynamic
property data. It doesn’t really
save me any time. As long as I have to
find the work I might as well just use
the internal energy. But you can do it both ways. In fact I would suggest
do it both ways. For constant pressure problems if you don’t get the same answer
you’ve done something wrong. So it’s a good learning tool. Anyway, questions? Good. So this brings us
to a difficult problem. As if this last one
wasn’t hard enough, right. These are not really that complicated once
you’ve tried them. There’s a lot of them and they’re all a
little bit different. But this next one is a little
more complicated than most. So it’s this problem here. [ Pause ] Well, there we go. So this is out of the current
edition of this textbook. And this is problem 4-139. So let me read it. You have a piston
cylinder device which initially contains
.35 kilograms of steam at 3.5 megapascals, superheated
by 7.4 degrees Celsius. By the way that’s
not the temperature. That’s the number of degrees above the saturation
temperature. When it tells you
superheated by a certain number of degrees that’s
the temperature above the saturation. So we know we can go into the
steam tables at any pressure and find the corresponding
saturation temperature. The temperature initially is
7.4 degrees higher than that. So make sure you’re
aware of that. It says now the steam loses
heat to the surroundings and the piston moves down
hitting a set of stops at which point the cylinder
contains saturated liquid water. So the first part of the problem and what makes this hard
is it’s a two part problem. But the first part of the problem honestly is
not a whole lot different than the last problem. You know, you start
with a piston. Unlike the last problem
where you add heat and the piston moves up,
so far this problem starts with something superheated
and you remove heat and the piston moves down. So it’s the same problem as
the last problem but only up to that intermediate point. And this is where
things get difficult. It says the cooling
now continues until the cylinder contains
water at 200 degrees Celsius. By the way when the author uses
the word water he doesn’t mean saturated water, he
doesn’t mean liquid water, it doesn’t mean vapor
water, it could be anything, it could be any phase. So all we know is that it’s at
200 degrees at that final point. So the [inaudible]
will continue. We finally get to 200
degrees Celsius and that’s when the problem ends. Now isn’t the second part of the
problem really the same problem as the one we did at
the end last time? It’s just a constant
volume problem isn’t it? Once that piston comes down and hits the stops it
can’t move anymore. The volume is constant
inside that cylinder. The mass is clearly constant. So this is a constant
volume process. So this particular problem
is a two-part problem. You do the first part, let’s
call it from one to two which is a constant
pressure process. And then you do the second part
which is from two to three, let’s just call it last .3 which
is constant volume problem. All right, so 4-139 and
we know that from one to two is constant
pressure and from two to three is constant
volume, okay. So in other words P1
equals P2, V2 equals V3. I’ve seen problems like
this that are reversed where it starts sitting
on the stops as a constant volume problem. And then when enough heat is
added then the liquid starts to vaporize and the piston
starts moving upwards. So, you know, I can
certainly look at this problems
in both directions. All right, so we now have some
information about the problem. First of all any
questions just conceptually? Everybody should understand how
this is just a constant pressure problem followed immediately
by constant volume problem. At the end of the constant
pressure problem is the beginning of the constant
volume problem, right. All right. So now let us create
the right sketches. [ Pause ] All right, so here’s my
piston cylinder device. And then at some point
down here are the stops. This represents state .1. As the piston moves down and
sits on top of the stops here, I’ll just show this as a
dash line, then that’s going to represent state .2
and it’s also going to represent state .3, right. All right, so this is
just the illustration. So now let’s write down what
we know and what we don’t know. Oh by the way there’s
one thing I should point out about this problem
that I forgot to mention. I’m not really happy with
the way the author worded the problem. He really should have used
the word frictionless vertical piston cylinder device. Now granted there’s
a diagram in the book so it should be very obvious. But it may not be obvious to you as you’re first learning
this material. The fact that it’s a
frictionless vertical piston cylinder device means that it’s
a constant pressure process, all right. We know that. We’ve seen problems
like this before. But they don’t tell me that. The previous problem said
that the piston moves under constant pressure. So it was obvious. In this problem the
author has not done well. So if I give you a
problem like this or any piston cylinder
problem on the exam, I’ll make sure you know whether
it’s a vertical frictionless piston cylinder device or just
some random piston cylinder device that is a
moving horizontally at some odd angle set to
the pressure’s not constant. So I’ll make sure
it’s clear to you. In this problem anyway it’s
a constant pressure process from one to two. All right, back to the problem. So we know the mass
as .35 kilograms. We know that it’s steam,
in other words H2O. By the way when I write my
own problems I just use H2O, I don’t even use the word
water because I don’t want to confuse anybody, so I’ll
just say H2O and let you figure out what the phase is. All right, so it’s
H2O, it’s superheated by 7.4 degrees therefore T1 is
equal to T set plus 7.4 degrees. And we’ll figure out what
that value is shortly. But at least we know that T1 is
not the saturation temperature. It’s a certain number of degrees above the saturation
temperature, okay. Steam loses heat to
the surroundings. So there’s heat transfer
out of the system. And the piston moves down so
there is some heat transfer from 1 to 2. Eventually hits the stops at which point the cylinder
contains saturated liquid water. So 2 is sat liquid, all right. Again we know now that as
the cooling continues the temperature’s going
to drop some more. The volume’s going
to remain constant. But finally we get
to .3 and it tells me that T3 is 200 degrees
Celsius, all right. And let’s see, oh yeah I forgot
to write down the steam pressure at .1 so up here at 1 I should
have said that P1 is equal to 3.5 megapascals that was
given in the problem statement. All right, okay. So what are we trying to find? We’re trying to find
the final pressure and the final quality
if it’s a mixture. And let’s keep in mind that
the final is .3 not .2, right. And then we also would note that
we’re trying to find the quality if indeed it’s a
2 phase mixture. It also asks us to
find the boundary work. It also asks us to find
the amount of heat transfer when the piston first
hits the stops. So that’s really Q12. But it also asks me to find
the total heat transfer. So that’s the sum of the heat
transfer from 1 to 2 and 2 to 3. So Q total is Q1-3 which by
the way is equal to the sum of the heat transfer for
the first part from 1 to 2 and then plus the heat
transfer from 2 to 3. Okay so this is something
that we’re trying to find for this particular problem. So before we actually go
through let me show this also on a appropriate, how
to use a TV diagram. I’ll just put this
over here on the side. There’s my saturation line and there’s a constant
pressure line. All right, so we know that we
start as a superheated vapor. It’s not that superheated. I mean only 7 degrees above
saturation so it’s going to be pretty close to
the saturated vapor line. It says that .2 is
when it hits the stops and it’s all saturated liquid. So it’s a constant
pressure process, we’re going to follow
pressure line down to the saturated
vapor line all the way over until we finally get to
the saturated liquid line. So there’s state .2. And then we have a constant
volume process, right. Now notice a closed system. The volume’s constant. The mass is constant. The specific volume
is there for constant. And because we’re cooling the
temperature’s going to drop and we’ll end up with state
.3 that’s right there directly below state .2. So here’s our initial setup. Not really quite
done with the setup. Mostly done with the setup. What about some assumptions
to make and some equations and all that? Well like the previous
problem we’ll note that there’s no velocity
associated with the water that’s
inside this container. So there’s no kinetic
energy change. We’ll also note the height
change is probably minimal but we don’t know for sure, just because we don’t
have any geometry data. So we’ll just have to assume that this is pretty
small system. I mean it’s 1/3rd of a
kilogram, that’s not much, so. We can note that we’ll
assume there’s no change in potential energy, right. What about the work? That’s something we
should talk about. Well the work from 1 to 2 is
constant pressure work, right. So that’s just going to be
the mass, well I’m sorry. Let me first stay it’s
just boundary work, right, there’s no electric resistance,
there’s no other forms of work. So the work from 1 to
2 is all boundary work. And this for constant pressure
process is mass times pressure times change in specific
volume, okay. Again constant P only. So that’s the work
we have from 1 to 2. What about the work from 2 to 3. Yeah it’s zero, right. It’s a constant volume process. Again I should first write
that this is the boundary work and no other forms of work. And then that’s just the
integral of PdV from 2 to 3, but there’s no change
in volume so it’s zero. I mean we talked
about this before. You don’t have to write all
this out if you recognize that constant volume problems
have no work then just say work is equal to zero. So this gives us our work terms. And then what about
the heat transfer? Well the heat transfer,
first we’ll look at 1 to 2 and then we’ll move
onto the next part. So the heat transfer
is the change in internal energy plus
the change in kinetic and potential energies
and then plus the work. We know there’s no kinetic
or potential energy changes. So the heat transfer
from 1 to 2 is just this. And of course the work we
found earlier in the equation. So this is going to be how we’re
going to find the heat transfer. By the way since the
author specifically asked me to find the amount
of work I don’t have to use the [inaudible] version
of the 1st law that applies to these closed systems for constant pressure
processes, right. I could write Q equals
simply mass times H2 minus H1 but again there’s no need. I already have the
work so I’m just going to leave it in this form. So there’s the heat
transfer from 1 to 2. Now throughout the
heat transfer from 2 to 3 maybe I’ll just
do it this way. We’ll note that the
heat transfer from 1 to 3 is just the sum of
the heat transfer from 1 to 2 plus the heat transfer
from 2 to 3, all right. We know the heat transfer
from 1 to 2 is above. So M U2 minus U1 and
plus mP V2 minus V1. So there’s the heat
transfer from 1 to 2. What about the heat
transfer from 2 to 3? Well again there’s no
kinetic energy change. There’s no potential
energy change. There’s no work. So the only heat
transfer we have from 2 to 3 is just a change in
the internal energy, right. So this is just M
U3 minus U2, okay. Maybe what I should do to
make it more clear is note that this is equal
to the following. And all three of these
terms are now zero. So we got this. And then we’ll note
that the M2 U2’s cancel and we simply get M U3 minus U1. And then plus the
boundary work from 1 to 2. And this is how we’re going
to find the heat transfer. Thank you. So the equations are
the same equations. If this isn’t clear then
you just need to study more. There’s not a whole lot of
difference between this problem, the last problem, the
two we did last time. They’re 1st law problems
for closed systems. They use the same general
version of all the equations. They’re slightly different. The work sometimes
slightly different, the heat transfer might be
found slightly different. But they’re all the same. All these problems here in Chapter 4 are the same
problem but with variations. Yes?>>So when it says
that it hits the stops when it’s saturated
vapor [inaudible] that V2 is going Vm, right?>>Yes. That liquid, yeah.>>Okay, all right, so
what we want to do next? Not much more to do, right,
except get our property data. Do we have two independent
intensive properties at each of the three state points? And if the answer is
yes then we’re done. So let’s think about it. We know the pressure and we
know the temperature at .1, although I haven’t
calculated the temperature yet but I know them both. So yep state .1 is
known, superheated, pressure temperature
independent intensive. We have state .1. We can find V1, we
can find U1, okay. What about state .2? Yes, right, we know
that the pressure at 1 and 2 are the same. We have the pressure and we
know it’s a saturated liquid. So state .2 is known. We have Vf, we have
Uf is we need it, although we don’t need it. And then what about a .3? Well we know that it’s a
constant volume process. So we know the specific
volume is state .2, that has to equal the
specific volume at state .3. So we know that information. And we’re also told
that the temperature of .3 is 200 degrees Celsius. So again we have two properties. We have volume and
temperature, specific volume and temperature I should say. Again independent
intensive properties, so state .3 is known. So as long as we recognize
that we’re using the 1st law in its variations and we’re
using the data that we all have, the rest of this problem
is really just a matter of pulling data out of
tables and chugging away. So we’ll do that next. Let me pull the steam tables up. By the say SI units
we’ve got superheat, we’ve got two phase mixture. You know, we’re going to use A4, 5 and 6 as we solve
this problem. So that’s worth noting
here as well. All right, so hopefully you
have all this written down. [ Pause ] I’ve never owned
a PC in my life. I’m a Mac person. Although I do one of those
Surface Pro computers which is kind of a,
I won’t say anything. They work nice but
my iPad’s better. Okay, so I had to say something. Yeah, I don’t know, I guess PC’s
are just cheaper aren’t they? Okay, so [inaudible]
there we go. All right, so 4, 5, and 6. So let’s just start
at A5, so here’s A5. We really would like to
know what the temperature is at state .1. So let’s just start there. So the temperature at 1 is
the saturation temperature at the pressure of 1 plus 7.4. So the saturation
temperature at, was it 200? No, 3.5 megapascals,
quite a bit higher. So onto the next page,
there we go, 3,500. So that’s 242.56 plus 7.4 and why don’t we just call
it 250 degrees Celsius for simplicity, all right. So now I have the temperature,
I have the pressure. So now I need to go into
table A6 at P1 and T1. I’m going to table A6. So onto A6. Remember that we have these
sub tables for superheat. So here is the 3.5
megapascal sub table. We read down to the 250 line
which is the third line down. And it’s this data right
here where the cursor is. So we need two things, right. We need specific volume and we
need specific internal energy. So let’s just write
them both down. V1 is .05875 cubic
meters per kilogram. And U1 is 2623.9
kilojoules per kilogram. So there’s my data at state .1. Again I want to remind you that if I’ve made a
math error somewhere or pulled some data incorrectly
out of the tables I would like somebody to let me know
that so that I can fix it after all it’s all
getting recorded. So yes?>>On the tables so if we
needed like a 225 degrees for example, how
would we do that?>>That is interesting question. First of all you wouldn’t
because it turns out that at, well I mean maybe I
should take that back. What you could do is
you could interpolate between the saturated
vapor conditions and your superheat conditions. So you would have to go
in here at saturation which now corresponds
to 242, right. So these are saturation
conditions. These are conditions at 250. So you’re interpolating
between let’s say 242 and 250. Now that’s why I said
225 wouldn’t happen because 225 already put you
into the two phase region. But let’s say the
temperature was 245, right. So 250 obviously isn’t right. The data at 242.56
is not right either, you’re in between the two. So you would just
do the interpolation between those two entries, okay. All right, so next. We have our data at .1. Let’s just find all of our
thermodynamic property data. It’s going to be
easiest that way. So at .2 we have
a saturated liquid at 3.5 megapascals, right. It’s a constant pressure problem
because it only hits the stops when we finally get
to saturated liquid. So at P2 which equals
P1 and saturated liquid. All right, so this
is the pressure data. So we have to use table A5. We don’t need the
internal energy at .2 so, I think we do don’t we
because they have asked me to find the heat
transfer from 1 to 2. So I guess I better
write it down. All right, so we need V2 which
equals Vf and that is .001235. And U2 is Uf which is 1045.4
kilojoules per kilogram. All right, so that’s all the
data we need at state .2. Let’s do the same at .3. So at V3 which equals
V2 correct and at T3 and T3 was 200 degrees Celsius. Well now we’re given
temperature data. So that means we’re going
to have to go to table A4. I don’t really need to the test
because I simply know based on the problem statement
we’re a saturated liquid and then we’re cooling, you
know, you have to move downwards and that’s going to
put you definitely into the two phase region. I could do the test if I
wanted to just to prove that but not really necessary. You may want to do
it just to make sure. Nonetheless what
we would find is that at 200 degrees Celsius
my specific volume is greater than Vf and less than Vg. Okay, I mean we can see that
if we go back up to the table. I mean I have to go up to the
table anyway to get my numbers. But, all right so here’s A4. My temperature is, oops,
I need the other page. Yeah, here we go. So, you know, Vf is .0011
and my specific volume at .2 which equals my specific
volume at .3 is .01235. So .012 is greater than
.011 so clearly I’m above Vf and it’s a two phase mixture. So it’s a two phase mixture. How do we find the data we need? Well first keep in mind
that we really only need, well actually we need
two things don’t we? Really three things. They’ve asked me
for the pressure. They asked me for the quality. But then they also want to know
what the internal energy is. Actually I want to know
what the internal energy is so that I can solve my 1st law. So three things I need. The first one’s easy. It’s two phase. So P3 is just P sat at
200 degrees Celsius. So that gives me 150, I’m sorry,
1,554.9 kilopascals, okay. To find the quality
I’m going to have to use the specific volume
data for the two phase mixture, V3 is Vf plus the
quality at 3 times Vfg. So this is how we
find the quality. V3 is V2, it’s known to me. So this is just .001235
at 200 degrees Celsius. Vf can be looked up. Vf is .001157. The quality is what
we’re trying to find. Vfg is just the difference
between Vg and Vf. So Vg is .12721, so that’s Vg. And Vf is the same
number we have, .001157. Anyway, go through some math. We find that the
quality at .3 is .00062. So it’s extremely low quality. But still it’s a
two phase mixture. So now we have the quality and now let’s find
the internal energy. So U3 is Uf plus the
quality times Ufg. So again just look
up the numbers. Everything should be in
the table over there. Uf is 850.46, actually we had
that from before didn’t we? No we didn’t, we’re
at a lower temperature so we didn’t have
that from before. The quality is .00062. Ufg is 1743.7. And if we just go through
our math here we get 851.55 kilojoules per kilogram. All right, so not much
left to do but to go through our various equations. Next let’s find the work. So the work for this
particular problem is let’s see, we wrote it down before. It’s just our boundary
work from 1 to 2. I mean technically it’s the work
from 1 to 2 plus the work 2 to 3 but theirs is no
work from 2 to 3 because it’s constant
volume, right. So maybe I’ll just write that. Boundary work is what,
work 1, 2 plus work 2, 3 but that’s equal to zero. So that’s work 1, 2 which is
mass times pressure times V2 minus V1. So let’s just plug in the
numbers that we have here. Mass is .35 kilograms, pressure,
I’m going to put the pressure into kilopascals
instead of megapascals so that all my conversion
factors are just identities, just 1’s. In other words I don’t have
to worry about conversions. And then the specific
volume change. So we have .001235
and then minus .05875 and those are meters
cubed per kilogram. Again the units here are
the same units we had in the previous problem. Kilopascals basically convert into kilojoules per cubic
meter is what it converts into. And we end up with
minus 70.45 kilojoules. Now before we move on does
the minus sign make sense? Well it should, right. I mean negative work means work
is being done to the system. In other words the piston
is pushing into the system. The work is clearly
being done to the system. The volume is shrinking. So yes, the negative
sign is proper. And next we want to find
our two heat transfer terms. So again these equations
are already in your notes. So we have change in internal
energy plus the work from 1, 2. So again it’s just a matter
of plugging in the numbers, .35 times what are our numbers? I’ve lost my numbers. Okay, here we go, yeah,
thanks, 145.4 and U1 was 2623.9. I forgot my units. I don’t want to forget my units. And then the work
is minus 70.45. So if we go through
our mathematics we end up with minus 622.9 kilojoules. [ Pause ] And then lastly we find the
heat transfer from 2 to 3. And the heat transfer
from 2 to 3, well again this is just mass
times change in internal energy. Oh I’m sorry, I’m doing 2 to
3, I’m doing 1 to 3, right. I’m trying to find the
total heat transfer. I can find Q23 if I wanted
to but that’s not asked of me so I’m not going
to waste the time. So Q13 is mass times
U3 minus U1. And then plus the work from 1 to
3 which equals the work from 1 to 2 which still equals
the boundary work. So it’s everything
we had before. So .35 kilograms. Yes, and then U3 is 851.55. U1 is 2629, 2623.9 and
then plus the 70.45 again. And this gives me
minus 690.8 kilojoules. So there’s a nice
long fun problem. Yes?>>You could also find the
total [inaudible] 1 to 3 within Q12 plus [inaudible].>>You know, I’m not going to,
I’m not going to say for sure. I might have to go back and look
at the equations and see but, you know, we went through a
derivation of these equations and it would seem that there
are certainly ways to simplify if we needed to or to just
give us a different version. So yeah, I’m not
going to answer that. I would have to write
the equation down and see if it works out properly. But yes, there are certainly
ways you can get your answers that are a little bit different
then the way I’ve done here. For instance sometimes
the students were trying to find the work. They calculate the heat transfer
based on the [inaudible] for a constant pressure process. And then they calculate the
heat transfer using the internal energy plus the work. And they equate them and then
they get the work that way. So they never actually have to
calculate the boundary work. They just solve for
heat two different ways and the boundary work
is one of the unknowns. So yeah, there’s different ways
to solve some of these problems. Questions? All right. Well one of other thing
then that I want to do today and this is just to be
consistent with your syllabus. By the way we’re done
now through section 4-2. And we’re not going to talk
anymore about this material. Everything that we covered up until right now is fair
game on the midterm exam. And again the midterm
is a week from Monday. So you got plenty of
time to study for it. Again it’s a closed book, closed notes exam except
you’ll have property tables that I’ll hand out to you. Anyway so starting now we move on to the next section
of Chapter 4. And this deals with something
that’s called specific heat. So we’re not completely
leaving behind these closed system problems. But we have to ask ourself
what if I have something other than water or the
refrigerant, 134-a? For instance what if
I have, I don’t know, how about a gas like
an ideal gas. How do I solve 1st law
problems for ideal gases? They’re just as applicable
to those types of problems as they are to any
other type of problem. Well the answer is
specific heat. It’s not going to be clear yet what specific heats
are, why we use them. But I will tell you that
we are going to have to solve 1st law problems. Some of these 1st law
problems are going to be involving ideal gases. And because there’s so many
gases there isn’t nearly as much property data available
for this wide variety of gases as we have for say the
refrigerants or for water. So often times you just don’t
have property tables like these, like the ones we’ve been
looking at here on the board. We’re going to have
to use another method. And that involves what’s
called specific heat. [ Pause ] Okay, so specific heat. Now specific heats
are properties. [ Pause ] And there’s actually
two of them, okay. One is called specific
heat at constant pressure. And one is called specific
heat at constant volume. [ Pause ] The one we call specific heat
as constant pressure we’re going to use CP and the other one
that we call specific heat at constant volume
we’re going to call CV. Now specific heats
are properties. But these are different
than most of the other properties
we’ve seen. Remember when we
talked about [inaudible] and I defined [inaudible]
as U plus PV. And I simply said
that since U and P and V are all properties then
any mathematical combination of those properties
is also going to be a thermodynamic property. That’s really what
specific heats are. They’re different combinations
of other properties. We define the specific
heat at constant pressure as the partial derivative
of the specific [inaudible] with respect to temperature
holding pressure constant. And we define the specific
heat at constant volume as a partial derivative of
the specific internal energy with respect to temperature
holding volume constant. The reason we call these
specific heat at constant volume or constant pressure is not
because it has anything to do with constant volume or
constant pressure processes, that’s not at all important. They’re called constant
pressure and constant volume because of the way we take the
partial derivatives, right. When you write a partial
derivative like this that means we’re taking the
derivative of one variable with respect to another holding
this third variable constant, right. Specific heat at constant
volume, we’re taking the partial of U with respect to T but we’re
holding the volume constant. So that’s why we call
these specific heat at constant volume
or constant pressure. So please make sure
you recognize that. That we’re not talking
about constant pressure or volume processes,
we’re talking about new thermodynamic
properties that are combinations
of other properties. And we’re going to have to
figure out how to use them to solve some realistic
problems. Now this will have to
wait until next week. So we’ll talk about it later. Anyway at least you’ve
been introduced to this and we’ll deal with
it in more detail. So I’ve got homework
up here from last time. Please don’t forget
to pick up your work. If anybody has problems with anything please come
see me in my office hours. So far I mostly just have
to sit there by myself in the office hours because
nobody has questions. So that’s great. I guess you know the
material really well. But if anybody has
questions please come see me. Your exam’s in just over a week. And well it’s just one midterm. It’s still worth a quarter of
your grade so it’s significant. So if you have questions
or problems with. [ Music ]

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