Thermodynamics: 1st Law for Open Systems, Uniform Flow; 2nd Law, Heat Engine & Refrig (15 of 25)

Thermodynamics: 1st Law for Open Systems, Uniform Flow; 2nd Law, Heat Engine & Refrig (15 of 25)


[ Music ]>>Well, good afternoon. [inaudible response]
So as we start today. First let me just note that I have your new
homework set for the week. So these problems are — this week’s material
I’ll do next Wednesday. And remember that
you have — what? About seven or so? Eight problems from last week
that are due on Wednesday. So don’t forget to
bring that with you. And I have some more homework
to be picked up here as well. Just a note or two on the
homework that you’re doing now; one student had asked the
question just a minute ago, and I thought I’d
answer for all of you, if you have a mixing chamber, and in a typical mixing
chamber the pressure’s going to be the same everywhere
in the mixing chamber. Think of a mixing chamber just
like a big, open container with two pipes coming in
and one pipe going out. There’s really no possible
way for the pressures to be different in any of
those particular fluid streams. If let’s say the
discharge pressure were less than the inlet pressure then
the fluid would be pushing back up stream and up those
inlet pipes, right? You have to have
the same pressure at each section of
a mixing chamber. So when you’re looking at — there’s one more problem that
dealt with mixing chambers. It doesn’t specifically tell you that the outlet pressure is the
same as the two inlet pressures, but that’s really
what they’re implying. So please make sure
that you recognize that, you know the two inlet and
exit pressures are always going to be the same for a
mixing chamber unless you’re told otherwise. But for most problems they will
always be exactly the same. Let me also remind you that I
had reassigned problem one — I’m sorry, problem 32 from
the previous homework set to the homework set
that’s due on Wednesday. And remember that to solve that one you need
two Sevelking’s [phonetic] Equations. You have your mass fluid
equation, which you’ll combine with your ideal gas
equation of state and that’ll give you
one equation in terms of the unknown temperature
and velocity. And then you’ll have
a second equation, which is just your first
law of thermodynamics, which will also have an unknown
temperature and velocity. So two equations, solve for two
unknowns, so that’s the hint that I gave you a week ago, but
I’ll just give it to you again because it was, you
know arguably a somewhat confusing problem. Anyway, let’s just move
on to new material now. Oh, one other thing
I wanted to mention. I’ve been updating
Blackboard so I think we’re up to lecture number
13 now that’s available as a recording on Blackboard. Also, I’ve included all the
example problems that I’m going to be working on — the
ones we’ve already done, but also the ones that you’ll
see over the next month or so. I really never thought to
make these available to you until some students were
saying that it’s kind of hard to read a problem that’s on
one board and my lecture that’s on the other board up here, and this way we should
all have access. So I put those up just
right before class today. They’re all in image
format so you should be able to access the image on
your Smartphone, your IPad, or anything you bring
with you to class. As long as you can access
Blackboard then just go to “course information” and there’s a new topic
there that’s called “example problems.” So they’re all shown there. Again, they’re all images. So anyway, moving on. [laughter] So we’re actually
done, believe it or not, with the steady flow problems
associated with the first law of thermodynamics
for an open system. And we’re going to move on now
and talk about another type of problem, which you call
uniform flow problems. By the way, another thing that
I learned last time, afterwards, was that some students are
having trouble just reading the writing that I have
because the thickness of the pen was not
really that thick. Again, I didn’t realize
that was a problem. I guess it’s more of an issue
once we have one board going to the computer and then
the other board that’s just a lecture. Those of you sitting over here
probably have some difficulty reading over here. So I’m using a thicker
pen starting today and hopefully that’ll
make it easier for you to read the material that’s
going up on the board. All right, so uniform flow. First of all, we’re
still talking about process involving first
law or a control volume, or an open system if you will,
but we’re no longer talking about steady flow problems. And there’s a couple
of main differences between the uniform flow and
the steady flow problems. So first one difference is that the state will actually
change inside the control volume — and again, I’m just going
to use CV as an abbreviation for control volume, okay? — although it remains uniform
throughout the control volume. So the state will change
inside the control volume but is uniform throughout. And then the other difference is
that the state of the entering or exiting substance is
going to remain constant, but the flow rate will vary. So we’re saying that the inlet
and exit of them is constant. The flow rate — I’ll just
rate this as [inaudible] — but the mass flowrate varies. Okay, so this is a
little bit different than our steady flow problem
in that [inaudible] problems that we use this method for
are also going to be different. These types of problems are
typically used for filling or for emptying a vessel, like if we’re discharging a
fire extinguisher or for filling up our propane tank down
at the U-Haul Store. Those would be kind of typical of these particular
types of problems. So again, these are
primarily used for filling or emptying vessels. We often call these charging
or discharging problems. This is about charging
and discharging problems. And what I’m going to
do is we looked both at the continuity equation
as well as our first law and modify them from the general
form to be more accommodating of these types of problems. So let’s just move on with that. All right, so first
thing’s first. Let’s just remember the
continuity equation. And the continuity equation
basically tells us that a change in control volume — mass
with respect to time — is going to equal the difference between the rate
that mass flows in. So that’s the sum of all inlet
streams and that’s the rate that the mass flows out. So that’s the sum
of all exit streams. I left out DET. Now this could be
written in different ways. Sometimes we put the two
terms on the right hand side over onto the left-hand
side of the equation, but this should make
sense to you, right? The difference between
what enters and exits is what
remains, right? So if the entering flow
streams have a greater mass than the exiting, like
let’s say you’re filling up the propane tank, then
the change of the mass of the control volume
will be positive. If we’re emptying the
propane tank then the mass that leaves is greater
than the mass that comes in and we have a reduction in
the mass control volume. So this is one term
we’re going to note, or one equation we’re going to
note, but let’s also again keep in mind what it says above;
the mass flow rates will vary. You are definitely going
to have a different rate that mass is going to flow
in verses what it flows out and in fact in many
problems we’re going to have nothing flowing
in or nothing flowing out. Again, if we’re talking about
like filling up a propane tank, you don’t empty the propane
tank while you’re filling it. If we’re talking about a fire
extinguisher being discharged, you’re not filling it while
you’re discharging it. So it’s very easy for one
or more of these terms to go to zero and we’ll talk
about that here shortly. Anyway, if we take this
continuity equation and now just integrate it with time then basically
the DMDT term for the control volume
— well that — you differentiate that
and you just get the mass. So the mass evaluated
between the initial and final state is
going to be m2 minus m1. So that’s what you get on the
left hand side of the equation. And on the right hand
side of the equation — now remember, the dot
represents the time derivative. So if we’re integrating over
time then we’re getting rid of the time in the denominator
and this ends up then with a sum of our exit streams
of me minus sum of overall inlet streams of mi. And note that i and e and 1
and 2 are not the same things. Okay? Unlike the single
stream steady flow problems where I’ve replaced i
with a 1 and e with a 2, here they mean different
things, right? Here 1 and 2 represent, you
know the final and initial state in the control volume, yet the i and the e represents what’s
entering or exiting, right? So e represents what’s
flowing out of control volume and i represents what’s
flowing into the control volume. So no more equating 1 and i, 2
and e. They’re different now. Anyway, here is a sensory — a version of the
continuity equation — that we’re going to use for
these types of problems. Let’s now look at
the first problem. If we look at the first law —
and I’ll just write this in, well, a very general form — you would get the rate of e transfer plus the sum
overall inlet flow path. So m dot i, then hi plus ei
squared over 2 plus 2i — and these are all i’s here. And this is an equal rate
in change in the energy of the control volume. And remember, this term
would have gone away if we were talking about
a steady flow process because with a steady flow
process there is no change within the control volume,
but now there is a change in the control volume. I mean we’re always uniform
throughout the control volume, but it’s still changing
over time. So it is a change in energy
overtime with respect to — well, for the control
volume itself. And then we also
have the exit terms. So sum overall exit flow
paths, dot e and he. That’s ve squared plus 2 plus
g2e and then that’s [inaudible]. So this is the general
form of the first problem. Again, it’s not the
single stream steady flow, not the steady flow, it’s
just the general flow. Okay? So now what I’d like
to do is just make a note. So if we’re looking at
the control volume — The energy associated with
the control volume is — well, we see the
definition of energy. Energy is going to be the
internal energy plus the kinetic energy. So we’ll take me squared 2
plus the potential energy. So I’ll just write that as tz. That’s right, mtz. So this can be rewritten
as the mass — m’s the internal energy — I’m sorry, specific
internal energy — then plus e squared over 2
plus t. And then what I would like to do is I would like
to differentiate this. I mean, after all this is
going to have to go back in to the first [inaudible]
equation above. So if we take the change
in energy plus the time for that control volume, now
this is [inaudible] of mass. [inaudible] energy as
kinetic [inaudible] Well, at this point
what I’m going to do is [inaudible]
couple of steps all at once. I’m going to go back to
my first law equation, but now I’m going
to integrate — The first law — me over time. So we just go back up to the
equation above here and looking at this [inaudible]
equation here — And let’s take that and
[inaudible] over time. Note that as I do this,
at the same time I’m going to be plugging in this
equation that’s directly above. So this basically
goes in up here, but then it too gets integrated. So I just wanted to make sure
that you’re aware of that. So let’s integrate the first law
over time and let’s just be sure to include the equation
above in these equations. So if we do so, what
are we going to get? So let’s just take these
one term at a time. First of all the integration
of q dot over time, well that’s just q right? I mean q dot is qt. If we integrate the qt
we just get q again. So that’s just transfer
the control volume. If we integrate the
next term, keep in mind that these uniform flow problems
there is no change in the state at the inlet or at the exit. Whatever it is at the beginning
of the problem, it’s the same as it is at the end
of the problem. So there’s no change in
any of these terms, right? There’s no change in the
empathy, velocity, the height. The state is the same. The only thing that that
integrates is just the mass flow rate term. And whatever the mass flow
rate is really is dmpt, so if we integrate that then
we just get the mass again. So this summation
then [inaudible] mi, note that the dot
is gone here again. And then time is hi plus
[inaudible] q [inaudible]. And then on the right hand side,
well here we’re going to have to be a little bit careful. First of all [inaudible]
but here we’re looking at the [inaudible] term right? So we’re integrating that —
well, if we integrate that, again this is an e. The integral
just becomes the mass time internal energy plus the
kinetic input times the energy and we have to evaluate that
between points one and two. So what we get here then is
m2 and then e2 plus e2 squared over 2 plus 2d2 and
then minus m1. Oops, that’s not
supposed to be i and 1 plus e1 squared
over 2 [inaudible] 1. So that’s the change
in the internal energy or the control volume
integrated over time. Okay? And then we have to add to it the integral
of the exit term. So again this is
one of my equations, but I have to keep
spreading it down here. So next we have some overall
exit speeds and the mass of the exit and the [inaudible] of the exit plus t
squared over g plus gt. Okay? And then last but not
lease we have [inaudible] term or w dot term. [inaudible] w plus
w [inaudible]. So this long, long, long equation right here is
the first law for uniform flow. Now, as you’re looking at this
equation you realize there’s only like 18 separate terms. Now how can that be to
solve the problem, right? Obviously I’m being facetious. We don’t have problems that
use this entire equation. What we typically do is
simplify this into problems where we have some — well, let’s say useful
problems to solve. So I mentioned already
some of them right? I talked about fire
extinguishers and I talked about filling and
emptying a propane tank. What’s unique about those is that they have only a single
inlet or a single exit and what we can do
is we can take a look at the equation above,
then just modify it based on this special case
of a single inlet and exit plus some
other simplification. And you’ll see that
by the time I’m done with all this we’re going to
get rid of this lengthy equation and it will be replaced by
something that’s a lot more — let’s call it user friendly. So what do we want to do then? Let’s look at this special case. And the special case is going
to have either a single inlet — Or exit, not both
at the same time. Either one or the other. And let’s note that some further
simplifications we’re going to make is that we’re going to
elect all kinetic energy change and potential energy changes. Okay? For these types of
problems you’re filling or emptying a vessel, its
speeds are going to be low. Not only is their going
to be low speed associated with what we’re filling or
emptying, but then there’s going to be essentially no velocity
associated with what’s inside that particular container. And there’s not going to be
any height change associated with either of these processes. I mean the exit is
where the exit is. The entrance is where
the entrance is. The tank sits where
the tank does. So there’s no potential
energy changes either and if you look only
at a single in or out at a time then we have
a great simplification. So let’s look at the case
of first the single input. So let’s say we’re filling
up a fire extinguisher. Well, let’s relate this to
the general equation above. Right, so we’ve got the
general equation above. I mean, you can’t really put
everything there all at once, but note that; wow, eight of
the terms cancel out right away. Then you’ve got four kinetic
energy terms, all the d squared and the 2 terms, right? We also have four
[inaudible] e terms. So those all go away. And if we’re dealing with only
a single input then the entire exit stream goes away. There’s hardly anything
left for our equation right? What we get is the problem. We get the heat transfer
minus the work — I’m just using work term for the
positive on the right hand side over the negative on
the left hand side. And then we have
our inlet flow path, but only the [inaudible] right? There’s no [inaudible] So — and
there’s only one of them right? We can even drop the
summation and end up with mihi and then plus m2 minus m1e1. Absolutely everything in place. Absolutely everything
else goes away, right? So this is the first
[inaudible] or uniform problem where there’s only
a single input. I will note that when we solve
problems involving uniform flow, we still have to
solve a continuity equation simultaneously. And if we go back to the
continuity equation and get rid of the term that
deals with the exit — this obviously is
[inaudible], but if we look at the continuity equation
then the continuity equation is really also pretty
simple, right? It basically tells me
that the mass that flows in is just the difference
between what we end with and what we start with. I mean really you have
to know that’s called the continuity equation. It just has to do with
common sense, right? What flows in, this
is [inaudible] to the final and the initial. I mean, that should make sense. You can go back to the
continuity equations and get rid of the summation signs and
note that whatever mass is of t [inaudible] only
[inaudible] based on what flows in and out. So that’s going to have
to be used simultaneously. Yeah? [ Inaudible Question ] Good question. Yeah, it should be negative
on that side, but [inaudible]. You’re right because
I had it as a positive on the left hand
side, so as I move to the right hand side it’s
going to have to be negative. So that’s a good observation. That’s a typo. So, yeah put a minus
sign there to make sure that we’ve done it correctly. So thank you for that. Anyway, so this is an equation
that we’d use for single input and then we have
equally simple equations if we have a single exit. Only a single exit and no
inlet at the same time. So we go back up
to our equation. [inaudible] first law, and we still have our heat
transfer minus the work. Again, work moves over to
the left hand side and then on the right hand side,
well we have our mtvt and then minus m1 u1. All of [inaudible]
terms go away. All the inlet terms go away. It’s using only a single
exit and then the last term on the right hand side will
be exit term [inaudible] plus [inaudible] and the sign
I think is correct here. This should be fine. So this is the first
line equation. [inaudible] And that is — again, we’ll
have to look at the continuity. But again, we can go to
the continuity equation or we can just, you
know use our logic. If flow means leaving
[inaudible] have some mass flowing out, the mass that flows
out had better be the difference between what we start
with and what we end with. I mean, where else is it going? Right? So clearly
it’s just m1 minus m2. Now, again you go back to
the continuity equation and write this or you can
just use logic and say; well, obviously what’s flowing out has
to have started on the inside. The difference between what
you start with and end with — m1 minus m2 — had better
equal what’s flowing out. So these are the equations,
the two that are underlined, that you can use for either the
charging or discharging problem. You can see that the one that’s
[inaudible] isn’t that charging? So we’re charging
the fire extinguisher or we’re charging our propane
tank when we’re filling it up. So single input; these
are the charging problems. And single exit, this is
when we’re discharging. [inaudible] Thinking of just filling or
emptying a vessel, charging or discharging, single in,
single out, I mean think of them however you want,
but these are the types of problems that [inaudible]. Now, there’s no more
theory to talk about. Really, I think all
we need to do is go through an example problem and help illustrate how one
solves these kinds of problems. So, are there any questions? [inaudible] the theory? All right, so the example
problem that we’re going to do is problem 125
in chapter eight. So it’s on the computer and
let’s get that [inaudible]. Okay, so this one is from
the [inaudible] again, all this is on Blackboard now. So [inaudible] provide
students with [inaudible]. Let me read it to you anyway. [inaudible] So, it says
that we have a concrete tube and a rigid tank that contains
r134a at 14 degrees Celsius. At this state 55 percent
is in the vapor phase. In other words, [inaudible]
quality. And the rest in the
liquid phase. Obviously, what else
could it be, right? This says the tank is connected
by a valve to a supply line where refrigerant
at 1.4 mega pascals and 100 degrees Celsius
flows steadily. It says the valve is now open
slightly and the refrigerant is about to enter the tank. When the pressure reaches 1
mega pascal all refrigerant in the tank is in the
liquid phase only. In other words, you finally have
turned that two phase mixture into just all vapor, which
is saturated vapor and that’s when the valve is closed
and the problem ends. So it wants us to find the
temperature of the tank, the mass of the refrigerant
that has entered the tank and q to be transferred between the
tank and its surroundings. So let us start this one
with the usual slide. All right, so we have a tank
that looks bigger than this tank and it is connected
to a supply line and they give you all
sorts of information. So they do tell me the
volume and it’s rigid. So this is common. Most of these problems
will be rigid tanks, but you know I’ve seen problems
where they’re not rigid. You know perhaps you have a
[inaudible] cylinder device and you’d have to include
some sort of boundary work. But most of the time
they’re rigid. So it tells me that the
bottom is 0.3 cubic meters. Its refrigerant will
be a. It gives me the initial temperature. [inaudible] plus 14 Celsius. And the [inaudible]
is 55 percent. It says that the supply line
[inaudible] the supply line has an inlet pressure
of 1.4 mega pascal and [inaudible] temperature
of 100. And it says that the
refrigerant flows steadily in the supply line and that’s
really just the author’s way of saying that there’s no
change in inlet [inaudible]. But frankly, all these
uniform problems will show you that the inlet conditions
are uniform or the exit conditions
are uniform. There’s no change in them. The author didn’t really
have to tell us that, but it doesn’t hurt
I guess to do it. It says the valve is open, the
refrigerant’s entering the tank and it finally gets to a final
pressure of 1 mega pascal. All the refrigerant’s
in the vapor phase only. So 0.2 [inaudible]. And they would like us
to find the temperature and mass [inaudible]. And then transfer [inaudible]. Now [inaudible] call it c.
But this is the heat transfer to the control volume
during this particular task. All right, so I suppose
[inaudible] now how do we even solve this problem? Well, this is really
a charging problem and there’s just
the single inlet. So it’s pretty clear
what the version is of the first [inaudible] about basically [inaudible]
equal [inaudible]. So this is the version of the
first problem that [inaudible] and of course you’re also going
to have use some continuity. So [inaudible] Now do we
have the ability to solve for anything up here yet? Yes. First of all you might
notice that we know the initial and final mass as long as we
know the volume and the initial and final state — and
we do know the initial and final state, right? We have the temperature
and the quality in SOU [inaudible]
graph would allow you to find the initial
specific volume. And we know that the pressure
[inaudible] at the final state. So we should be able to
look at our graphing tables and find [inaudible]
volume and know that the initial mass
is just the volume of h as the specific volume. And we know that the
final mass [inaudible] is that same volume [inaudible]. So that’s something very common
to these types of problems, just to find this
relationship between the volume and the specific volume
in order to get the mass, but we do this not to
get the [inaudible] mass. Right? We do it to get the
initial mass or the final mass. The difference between the
two in the continuity equation to solve for the inlet mass. And nonetheless, at least we
see that we can find all this. Let me start putting
little checkmarks here. I know the volume and I know
the initial and final state of the [inaudible] volume. So yes [inaudible] m1 and m2. And finally these are not dots that are going [inaudible]
Anyway, these are my checkmarks. So we know the inlet mass. Okay, so let’s just
check off inlet mass. By the way it’s a rigid vessel. There’s no work associated
with it. There’s no mixing devices. There’s nothing. So we can always put zero
there on the work page. And again, getting back
to what we were talking about a moment ago, we
know our initial state. So we observe it by [inaudible]. We know our final state. We’ll observe it by [inaudible]. We know our inlet state, both
pressure and temperature. So we observe it by hi. And, as we just mentioned
a moment ago, we know m1, we know m2, we know mi. There’s nothing left
to solve for but the heat transfer, right? So if we’re trying to solve
for the heat transfer, really it’s a matter of
finding the specific volume and inputting [inaudible] data
— I’m sorry, [inaudible]. We can now plug it in to
all three of the equations and show them near the bottom
and we can get [inaudible]. By the way, as far as p2, I don’t know that’s
really chapter two or three probably isn’t it? If it was saturated
vapor it could pressure, the temperature is obviously
just the saturation temperature, right? So I don’t know why they
ask us that but [inaudible]. So, any questions guys? Let’s just keep in mind
that as we’re talking about refrigerant all of our
data comes from day 11, 12, or 13, just depending
on what the state is. Let’s just solve the problem. We’ll start [inaudible]
property data. So — So at p1 and [inaudible] 1
it’s a two phase mixture. We know that we have
these [inaudible] 11. It’s a two phase mixture
[inaudible] temperature. So I’m just going to go
down to the 14 degree line. I don’t know that I need
to take the property tables and put them up here
[inaudible]. Oh, we know how to do this. [inaudible] So, you’re
looking for two things, right? You need the internal
energy of one and you also need the specific
volume so we can find the mass. So p1, there’s vf. [inaudible] at 14
degrees Celsius it’s [ Inaudible ] Find mu one, you
get the same thing. [ Inaudible ] So, again it’s just a matter
of filling in this other table. 70.56 is [inaudible] 167.3 [ Inaudible ] As long as we have
it we might as well as [inaudible] So
m1 is the volume. 0.3 cubic meters is
the specific volume. [ Inaudible ] By the way, again
this is kind of sloppy with a big pen, but [inaudible]. We get mass of 12.36. Now let’s move on to state 0.2. We’re doing much the same thing, although state 0.2
is a saturated vapor. So here [inaudible] vapor. Now it can go to [inaudible]
the saturated pressure table [inaudible] at this pressure, 1 mega pascal, which
is [inaudible]. [ Inaudible ] So eg is 250 and 71. [inaudible] and as long as we have the specific
volume data stated we may as well find target mass. So 0.3 divided by
[inaudible] 0.76 And, last but not least
let’s find our [inaudible]. So [inaudible] you’ll see that
this is [inaudible] vapor. So we take away 13 we go to
the 1.4 mega pascal site table. We read down to the 100
degree Celsius line. I mean again we’ve done
this many times before, and we just pull out the value. We get [inaudible] 232. Next we may as well find m2 — I’m sorry, [inaudible] mi which is the difference
between m3 and m1. So we do that we get 2.40. Okay? [inaudible]
We’ve also been asked to find the temperature, 0.2. I mean honestly I should
have done that above right? As long as I drug in
the saturation table to get 0.2 I should
have found it there. So I’m just going to
go back up and read through [inaudible]
1 mega pascal. So we can read that
right off the table. It’s 30.94 degrees Celsius. And then the last thing we need
to do is solve the first law and find the unknown
heat transfer. So fill in the first law
[inaudible] And now — you know, this is written
into your notes above. We don’t have a big
blackboard here. I drew a big arrow
across half the board, but [inaudible] the first law. [inaudible] I’m not
going to rewrite it. I’ll just go with
the [inaudible] So we have the negative
[inaudible] it’s really pretty straight forward. Maybe to save time I’ll just
let you plug away yourselves. Etcetera, etcetera. If you go do the mathematics
you’ll end up with 899 joules. And you will find that
this is a positive number. If you go through the
mathematics it’s positive. So heat is actually being
added to this system. It is [inaudible]. So are there any questions
as to [inaudible] problem? Hopefully you don’t mind that
I didn’t actually plug away at all the number here. [inaudible] notes. [inaudible]. I will be doing this a lot more
as we progress with the class. You know I like to show
you, you know about units and conversion factors
and all that, but you’ve seen most
of that already. So I’m generally just trying to save time now
and not [inaudible]. But I went through all
the important steps today. Is there a question?>>Yeah [inaudible]
between what you’re finding at the subscript i for the
inlet and — verses 1 or 2?>>Well, let me just
pull this down. You know we have a
tank that’s connected by a valve to a supply line. So the supply line is your inlet
conditions and those are going to always be the same. And that’s your i. But what’s
in the tank is different, right? I mean we’re starting
with something at a markedly lower temperature and [inaudible] vapor
to the mixture. Those are your initial
[inaudible] and then as you open that valve and fill up
whatever’s in that tank, but then see you
get the [inaudible] So those are [inaudible]
condition. So there shouldn’t
be anything confusing between what represents
the inlet and what represents the
initial filent [phonetic] state within the flow volume itself. The inlet is not
[inaudible] flow volume. The inlet is just a pipe
that’s connected to it to provide [inaudible]
substance [inaudible] but it’s not the inlet,
it’s not the exit. It’s the same substance. But [inaudible] So yeah, make sure you read whatever
problems [inaudible] clearly and then so you know
the difference between the [inaudible]
conditions and the filent conditions. One thing I also wanted to
comment on; if we have a problem that involves an exit sometimes
those problems are a little bit confusing. For instance, let’s say you
started with a two phase mixture and we had an exit pipe in
the bottom of that tank. Well, if you are a
two phase mixture, because of the higher
density of the liquid portion of the two phase mixture the
liquid is going to always stay at the bottom of that tank
and if you have a valve at the bottom of the tank then
you’re actually only allowing the saturated liquid portion of
the two phase mixture to exit. On the other hand, if you
have exactly the same problem, the two phase mixture, but
your valve is on the top, well again the density of the
two phase saturated liquid — the saturated portion
of the mixture is such that the vapor’s going
to always rise to the top. So if your valve’s on the top
then it’s only the saturated vapor that’s going
to be leaving. So when you look at all of the
involved discharging for many of those problems you’ll
have to pay close attention to where it’s discharging. You’re actually pulling the
saturated liquid portion off the bottom or are you pulling the
saturated vapor portion off the top? So again that’s just
a [inaudible]. Anyway, any other problems? Any other questions
about this problem? Or [inaudible] after five?>>So is it always going to be
the same substance either way?>>Oh, absolutely. Right, these are still only
pure substance problems. It’s going to be water or –>>[inaudible]>>Yes, it’s never going
to be anything else. [inaudible] Any other questions? Right. Well, then let us
move on to chapter six. This is everybody’s favorite, which is the second
law of thermodynamics. I guess the first thing
that I want to say is that this material
can be quite confusing and it gets even more
confusing as we start talking about entropy over
the next few weeks. If you’re not showing
up to class and you’re not hearing
the lectures and you’re not hearing
my take on the whole idea of the second law and entropy
then you’re really doing yourself a disservice. Most students who just rely on reading the textbook
are generally just not able to get it entirely. The second law doesn’t
have to be complicated and hopefully you’ll see as
I present it that we’re going to use the second law
in a very specific way for a very specific
type of problem. But still, you do need to show
up to class on a regular basis. So if you’re not here today —
I’m probably kind of talking to the wrong people — but if
you’re not going to be showing up to certain classes in
the next couple of weeks, make sure you get the
notes from somebody. Make sure you understand
the notes. If you need to come into
my office and ask me to interpret those notes for you
so that nobody’s left behind. All right. So, the second law
of thermodynamics. Well, first of all
the second law’s only around because of the first law. The first law tells us that
for any process energy must be conserved. But it really doesn’t
give us any restrictions. It just tells us that as
long as we can keep track of all the energy [inaudible]
paths then entropy process is possible, but that’s
not really true. I mean, for instance let me
just give you an example or two. Let’s say we have a
— I don’t know — some substance, a gas maybe,
and let’s say we put some sort of a mixing device and
we add a lot of work — mechanical work — we
add energy into it. As we’re adding energy
input it’s pretty likely that that gas is
going to heat up. Now, that can happen, but we’re
going to reverse that process and we take a gas that has a
mixer in it can we cool that gas and will that start the
mixer into rotation? I don’t think so. But the first law doesn’t tell
us that that can’t happen. The first law just
tells us that as long as the energies all work out, as
long as you have the same amount of work energy coming in
and that equals the amount of heat that’s being
transferred, then the first law
works out just fine. But we know that it can happen. Right? We base it
on our observations. We know that just because the
first law says something could happen doesn’t mean
it can actually happen in the real world. That’s really what the
second law’s all about. The second law provides us
restrictions, if you will, as to the nature of
certain processes. It basically tells us which direction processes
can occur in. Right? So we need
the second law. You know just because
you can solve something with the first law doesn’t
mean it’s going to work. So that’s what the nature
of the second law is. That doesn’t tell you how
to use the second law. It’s just telling you what
the second law is all about. Okay? So again, the second
law basically tells us — What direction certain processes
are capable of occurring in. So, you know this law gives us
— this is called restrictions. As to what processes
that we [inaudible]. Some people like
to think of these as what direction
they can occur, but we know that not
everything can happen. [inaudible] Another good
example is heat transfer from something cold
to something hot; we know that heat is only
transferred from the hot thing to the cold thing, but there’s
nothing in the first law that says that that
has to happen. Right? The first law just
tells us that as long as the heat transfers and we’re
[inaudible] energy change, all that, why can’t
you transfer the heat from something that’s
cold to something hot? But we know that that
can’t happen, right? We know that if we take a
hot thing and we touch it up to a cold thing the
heat is always going to go into the cold thing, right? The cold thing slowly
gets warmer and the hot thing
slowly gets cooler. If you touch a hot
thing to a cold thing and watch the hot
thing getting hotter and the cold thing
getting colder? Well, why not? The first law doesn’t
say you can’t do that, but we can’t do that. We know that that’s not
the way it really works. So that’s what the
second law’s all about. It’s nothing more
complicated than that. Well, it is because
you have to use it and mathematics is
more complicated, but in essence that’s
the second law. Now, what’s typically done — what I’m going to do here is
give you two examples of systems which pretty clearly illustrate
the need for second law. So two processes [inaudible]. So one of those processes is
what they call heat engine and the other of those
processes they call a refrigeration [inaudible]. Now, quite frankly the
heat engine is a cycle — sometimes it’s also just
called heat engine cycle. But these are two different
types of processes that I want to illustrate the
new [inaudible]. And we’ll just go through
them one at a time. And the first one that
we’ll talk about is going to be the heat engine. So let me make a sketch
of the heat engine. So there’s my heat engine. And the way heat
engine works is you have to have some heat source. And if you provide a heat
source then we should be able to do a certain amount
of network output. Now, there’s many
devices that do this. I mean, like your
automobile engine is one. We do burn fuel to create
heat and that does work. Steam powered train
is another one. You burn fuel to create heat. You turn water into steam,
you let that steam run through a steam turbine, and
that rotates and [inaudible]. So we know that this
is not exactly wrong, but it’s also not complete. Anybody who’s ever worked with
any of these heat engines knows that the only way for
you to turn heat engine into work energy is to simultaneously reject
some heat out of the system. Okay? Why are power plants
always situated on big bodies of water like an ocean or a like
or why do they have that big, tall tower called
a cooling tower? They need to reject
heat somehow. What about your engine? There’s a radiator, right? Even if you don’t have
a water cooled engine, like you have your motorcycle
or maybe, you know VW Bug or something like that. You still have heat rejected, you just have water
are called fins, basically metal protrusions
on the side of the engine that facilitate heat transfer. But you have to have
a way to reject heat. If you don’t believe me
then just go to your car, drain the radiator, and then
[laughter] drive your car for a while and you’ll
see what happens. It’ll work for a while
until the engine seizes up and you pretty much have
to throw away your car. You don’t want that,
but you’re engineers, you know that you need
to reject heat, right? This is just common sense that
this is based on observation that you can’t just
turn heat into work, you also have to reject heat. That is what our
experience shows us, right? To produce work we must be able
to extract heat from one source and reject heat to a
heated [inaudible]. Right? That’s what our
experience is telling us. Nobody’s ever going to be able to observe anything
different, ever, ever, ever. Okay? So that’s an
illustration of the heat engine and what our experience
tells us. Our experience says — when I
say our experience I’m talking about collective
experience from years based on observation and
experimentation. But our experience says
that to produce work — We must extract heat from
a source and reject heat in what we call a sink. We can call it a heat
source and a heat sink, that would also be appropriate. So this is just an
example of a heat engine. And again, this illustrates
the absolute need for a second law
of thermodynamics. The first one says
that we can turn heat into work all day long. It doesn’t say anything
about rejecting heat, but we know in the real
world we have to reject heat. So we got to use the second law. The second law will allow us to determine whether
the heat engine — and we’re analyzing [inaudible]. Let’s look at the other
process [inaudible]. And this is the refrigeration
cycler. And in the refrigeration cycler, again in the illustration what
a refrigeration cycle does is we extract heat from a low
temperature heat source. A heat source at low
temperature, and we reject heat into a high temperature
[inaudible]. So here’s our heat source and
this is that high temperature. Now, if you think about
the refrigeration cycle, I’m talking about things like
an air-conditioning system, a refrigerator, a freezer, and this is exactly
what they do, right? Let’s look at the refrigerator. The refrigerator, you want
to run it about, what? 35 degrees Fahrenheit. The kitchen’s going
to be at maybe 65 or 70 degrees Fahrenheit. You are actually taking heat
out of the food in order to keep it cold, but
you have to return that heat somewhere, right? You reject that heat out into
the kitchen and that’s okay. Right? I mean you’ve all stood
there with your bare toes in front of the refrigerator
and you feel that warm air kind of blowing on your toes or maybe
you feel the top or the back or the sides of the refrigerator
and it’s warm, right? It’s warm because the
refrigerator is taking heat out of the cold food
and rejecting it into the warmer outside air. But we know, again,
based on observation that the new rule
doesn’t work that way. If you could simply pull
heat out of something cold and send it into
something hot well, you know you just invented
a perpetual motion machine of the type and we won’t’
get into that right now, but you know it just
doesn’t work, right? Again, if you take
the cold thing next to the hot thing
you’re never going to make the hot thing hotter
and the cold thing colder, which is exactly what would
happen if you’re taking heat out of the cold thing and
putting it into the hot thing. It’s not how a refrigerator
works. Well, because we have to
provide more input, right? Within that refrigerator
is the refrigeration cycle. We’re actually taking the
heat out of something cold and then doing work to it
so there’s always going to be some work input. And it’s that work input then
allows us to transfer heat from the cold source
to the cold sink. So again, our experience
tells us that the only way you’re
going to transfer heat from the cold thing to a hot
thing is to apply more heat. Now, again if you don’t
believe me just go unplug your refrigerator at home and see
how good your cold food is going to stay cold without
the work input from the electric motor
spinning a compressor. This will not work. So that’s what our
experience says, right? So our experience — Basically says that to
extract heat from a cold source and reject it to a warm sink
we need to provide work. Reject it into a warm space. Anyway, you could use
the word [inaudible]. You must provide work. Again, we need the
second law, right? The first law doesn’t
help us out here. The first law doesn’t care if we
add work as long as the amount of heat coming in from the
source is somehow equivalent to the heat that’s
rejected to the sink, plus any energy change
[inaudible]. Then the first law tells us
that that should work just fine. But we really know
that it doesn’t. So it basically is these
observations and our inability to show otherwise that
lead to the second law of thermodynamics
whereas the first law of thermodynamics denies
the possibility of creating and destroying energy, the second law denies
the possibility of using energy in
a particular way. So that’s the relationship between the first law
and the second law. Let me repeat that; it’s these
observations and our inability to show otherwise that
lead to the second law. Whereas the first law denies
the possibility of creating and destroying energy, right? Conservation of energy. The second law denies
the possibility of using energy in
a particular way. Now, this is all
observations, right? Hopefully you all accept
these observations. If you don’t accept
the observations, well you know you’re young,
you’re learning to be engineers, you’ll have to accept
them eventually. [laughter] It’s also
good being an engineer because you don’t always have to completely understand
the theory. That’s what physicists
are for, right? We really just have to know
how to use the theory in order to create the machines
to advance technology and better human
life and all of that. I mean that’s really what
engineering is all about. Some of you may not always
have the mathematical skills or conceptual skills to
fully understand the theory or the mathematical
relationships, but don’t let that get you down. I mean, you’re an engineer. You have to accept
things as being true. If you don’t understand any of
the theory that we’re talking about from here on out,
well join the club. Most people don’t understand the
concept of entropy, but you have to know how to use the equations
that’ll come with that. So that’s what I’m
emphasizing in this class. So two statements
of the second law. There’s actually two common
statements of the second law. The first one is called the
Kelvin-Planck statement. And what the Kelvin-Planck
statement says is this; it says no process
is possible — Whose sole result — Is the absorption of heat — From one reservoir — And [inaudible] So the sole result
is the [inaudible]. Okay? That’s the statement. What does that mean? Well, to us that means that in
order for us to convert work to heat we have to have
a heat sink, right? So this basically tells
us that we need — this is a statement that the
role is use a heat source. What’s the other statement? The other statement of the second law is called
the Clausius statement. And what the Clausius
statement says is that no process is possible — That results solely — In the transfer — Of heat. From the cold — To a hot reservoir. In other words, we
need work input. Okay? We need work input. So the “we need work input,”
the “we need heat sink,” as I stated above, those are
not part of the statement. That’s just what we
observe from this, right? We know from the
Clausius statement that you can’t just
transfer heat from the cold to a hot thing. So isn’t this Kelvin-Planck
statement basically just giving us some restrictions
on the heat engine? And isn’t the Clausius statement
really just a statement that gives us restrictions
on refrigeration cycle? So these two statements of the second law are exactly
what we’ve been talking about before, okay? This upper one refers to heat
engine and this lower one refers to refrigeration cycle. And I messed it up. My writing is messy. I can’t hardly read it. This says heat and this
says refrigeration. All right? So what we really have to do
now then is look inside the box. If we go back here to the basic
box that is refrigeration cycle and the basic box that is heater
cycle and say, “Well, okay. How do we do this then?” We know that we need
to reject heat somehow. What device do we need? We know that we need
to find work input for the refrigeration
cycle somehow. What devices do we
have inside the box? I’ll give you a hint. These are cycles, right? These are both called
cycles for a reason. We’re going to start
with a substance at a particular thermodynamic
state and then we’re going to allow it to go through
different processes until we get right back with the
same state that we started with. So that’s the next step
and I’m out of time. So we’ll work on this
more on Wednesday. So if this seems like we’re kind
of in the middle of something, then we are right in
the middle of something and you probably don’t want to start your homework
yet on this material. We just haven’t gotten
to that point. So anyway — I mean you can
start working on the homework that deals with uniform
flow problems, but not the chapter
six material for sure. [ Music ]

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