# Thermodynamics: 2nd Law, Heat Engine & Refrigeration Cycles (16 of 25)

[ Music ] >> Well, good afternoon,
everybody. First of all, don't
forget to turn in the homework that's
due today. The solutions will be
posted right after class, in the case outside my office. Also, I have all the lectures, through last time,
on Blackboard now. You know, the video
recordings that are there. So, you should be able
to access all of those. Again, they're YouTube
videos, but they're private, so they're only accessible
through Blackboard. And only if you're
enrolled in this class. So, make sure you look at
those, if you need them. So, we're now talking about the
second law of thermodynamics. And hopefully we recognize that
the second law is very different than the first law, right? The first law is
conservation of energy. It tells us well, that energy
can't be created or destroyed. It can just be moved around. The second law is
totally different. The second law tells us that there are restrictions
on processes. Even processes that technically
satisfy the first law of dynamics, first
law of thermodynamics.

That still doesn't mean that
they're absolutely possible. We would have to see if they
also satisfy the second law. So, this is what we're
talking about now. Now, the second law
has a variety of forms or manifestations, if you will. There's no single equation
like the first law, that we're just going
to plug into and see if the second law is
then valid or not.

It doesn't work that way. What we're going to find is
that there's various forms of the second law, based upon
the different types of processes that we're going
to be analyzing. So, we've already looked at the
two forms of the second law, that are associated with
two different cycles. One is the heat engine cycle. And the other is the
refrigeration cycle, okay? What we saw there was there
are restrictions, right? What the restriction tells us, let's say for the
heat engine cycle, is that there's no process
that's possible — I'm sorry. Is that there's no
process that is possible, whose sole result is absorbed through the heat
for one reservoir. And production of all that
heat, into work, okay? That basically tells
us that there has to be some heat rejected from
the system, as well, okay? That was the Kelvin-Planck
statement of the second law.

The Clausius statement is
really the one that refers to the refrigeration cycle. Here, it basically, it tells
that no process is possible that results solely
in a transfer of heat from a cooler to a warmer body. In other words, if we want
to transfer heat from a cool to a warm thing, we have
to add some work to it.

So, these are restrictions. Now, what we're going to
have to do at this point, is try to figure out how to
quantify these restrictions. How do we solve a
problem for a heat engine? Or how do we solve a problem
for a refrigeration cycle? Such that at the
end of it we can say that this process is
or is not possible? If it is possible, then it
satisfies the second law of thermodynamics. If it's not possible, then
it violates the second law of thermodynamics. So, that's what we're
working on now. Now, you may recall that
we're looking, essentially, simultaneously at
the heat engine and at the refrigeration cycle.

And what I promised you last
time, is that we were going to take that box, if you will, that I just simply
call a heat engine. And we're going to
now actually look and see what's inside of it. So, from this then,
we'll move on and we'll develop
some relationships. And hopefully by the end of
today or first thing on Friday, we'll be able to actually
quantify whether one of these two different
cycles can work or not.

I mean, is it possible? Does it satisfy the second law? So, we've seen that
for a heat engine and this will be
heat engine first. [ Writing on Board ] So, one's too thin. And one's too thick. I brought my own today. Let's start over. Here's my box. Much better. This is a heat engine. And we know that
in a heat engine, we're going to be adding
a certain amount of heat. So, the amount of heat
that's being added, we're going to call Q sub H. And the H stands for
high temperature. And this is going to come
from some heat source. So, that's our high
temperature heat source. This could be combustion gases
from the combustion process.

It could be hot water coming
from a geothermal power plant or from a geothermal well or
a solar facility of some sort. Some heat source is
providing heat to our cycle. We know that we're trying
to produce some net work. So, we'll say net work
is coming off this cycle. And then we also know, from
our Kelvin-Planck statement, that there has to be some
heat rejected, as well. So, for the heat rejected, I'm
going to call this Q sub L. And L stands for low
or low temperature. So, this is our heat sink, okay? So, QH is the heat added. And QL is the heat rejected. And W net, well, obviously
is just the net work, okay? So, what's happening
inside the box here? Well, believe it or not, this
is something I showed you on the very first day of class
during our introduction lecture. We're going to have
— oh and by the way, this is just one
possibility for a heat engine for what's inside
the box, if you will. There's various types
of heat engines. And I'm not going to
get into most of them.

But this one is very,
very common. So, the first thing we're going
to do is add heat to a device. Well, let's add it to a boiler. And the boiler is going to
basically take liquid water and boil it into steam. Most likely super-heated steam. Of course, it doesn't
even have to be water. I mean, we can have
a heat engine cycle that operates on any fluid. It could be water. We could run a cycle
on a refrigerant. We can use some organic
compound like alcohol or — there's many possibilities. But none the less, usually
it's water, for our purposes. The steam coming out of
the boiler is now going to go into a turbine. So, here's my turbine and we understand now
how turbines work, right? They're going to spin under the
influence of the high pressure, high temperature, in this case
steam, coming through them.

So, some work is
going to be produced. Now, this says net work. That's not quite net work yet. We'll get to that in a minute. Anyway, what comes out of the
turbine, now we're going to have to send it into a device that
allows us to reject the heat. And that device is
called the condenser. So, we're going to take
heat out of our steam. By the way, we call
this a working fluid. That's the fluid
that does the work. It's moving through the cycle. Again, it doesn't
have to be steam. It's whatever our working
fluid happens to be. But none the less, we
pull heat out of it. And that condenses it
back into a liquid. And then we need to
pump that liquid back up to the pressure
of the boiler. So, here we have a pump. And we then feed the
boiler from that pump. Now, let's note that a pump
requires work, correct? The work from that
pump is actually going to come right off the turbine.

If you look at a real
steam power plant and granted, there's variations. But on many steam power
plants, the shaft that runs through the turbine
is not only going to be spinning an electric
generator on one end, it's going to be spinning
the pump on the other end. So, the turbine is going to produce a certain
amount of turbine work. Some of that turbine work is
going to operate the pump, which is why I've drawn
this line across here.

But what isn't used
to drive the pump, that represents the net
work output from the cycle. And that's the net work that then crosses a
boundary of the system. So, this is how a heat
engine would operate. And eventually, we're
going to be able to analyze all these
different devices. Granted, our major analysis
of this will be in ME302. You're not going to have to
deal with it in this course. But none the less,
these are the same kind of devices we've been talking
about for the last week, right? Aren't all of these
steady flow devices? We've got heat exchangers like
the boiler and the condenser. We've got a turbine. We've got a pump.

You know how to analyze each
one of these devices now, right? Or at least you're learning
how to analyze these devices. So, eventually we can
analyze all of them together. But again, we're
not quite there yet. So, this is the heat engine. Let's also look at the
refrigeration cycle. [ Writing on Board ] And in many ways, the refrigeration cycle is just
the opposite of a heat engine. So, let me show my box again. It's a little bit different. Now, for refrigeration
cycle, what we're trying to do is take heat from
a low temperature source.

So, here we're going
to have our source. And we're going to add
heat at low temperature. So, this is still the QL term. L for low temperature. But it is heat added. Okay. Now, we know that
in order to reject heat at a higher temperature,
we're going to have to provide some net work input. So, let's first show
the rejection. So, we're going to reject
heat at a higher temperature. So, this is Q sub H. And
this is heat rejected. And it's rejecting it into a high temperature
heat sink, all right? So, the source is where
the heat comes from. The sink is where the heat
that's rejected, goes to. So, you got QH and QL. And now let's look at
what's inside the box. So, we're adding heat
at low temperature. And we're going to do this
into typically a refrigerant.

Sometimes they're
just called Freon, just kind of as a generic
name for refrigerant. And you're going to turn a
liquid refrigerant into a vapor. So, we can either
call this a boiler or we can call it an evaporator. I'm not sure why it's customary to use the word evaporator
instead of boiler. But it's the same thing. We're turning the
liquid into a vapor. Now, we're going to take
this vapor and we're going to provide some work
input to it. This work input, because
we're talking about a vapor, is going to go into
a compressor. So, we use the word pump
for the heat engine cycle above because we moved a liquid.

Here we're moving a vapor,
we call it a compressor. But they are very
much the same thing. And it does require
some work input. So, let's show the net work. But let's note that the
arrow is pointing in. Unlike above, where the
arrow is pointing out. In fact, perhaps what
I should do is up above where I put net work, maybe I
really should have put net work output for the heat engine. And here, we're talking about
the refrigeration cycle, why don't I say net work input. So, we add work to the vapor.

And it compresses it. Its temperature rises. The temperature rises above the
temperature of the heat sink. Like our ambient
environment, right? And that allows us
to exchange heat. So, you have to have
a device to do it in. Well, we've got a vapor. And we're taking heat out of it. It's condensing that
vapor back into a liquid. So, we do call this a condenser. And then after we've rejected
the heat into the sink, we'll come out of the condenser. And typically, what's
done is we just run this through a throttle. Now, the purpose of the
throttle, is really just to reduce the pressure. So, we get back to the
pressure of the evaporator. So, here's a throttle. [ Writing on Board ] In the terminology of
the air conditioning or refrigeration business, this
is called an expansion valve. But here in thermal class,
we call it a throttle. Again, the same thing. But anyway, that leaves us
with the liquid that's going to go right back
into the evaporator. And the cycle is going
to start all over again. So, this is the refrigeration
cycle.

Now, let's just note here
that, unlike the textbook, where they first talk
entirely about heat engines. And then they talk entirely
two of them simultaneously. Just kind of going back
and forth between the two. Because quite frankly, they're
pretty much the same thing, right? Just they work in opposite
directions of one another. But they're more or less
the same type of device. And I've always found
it more convenient to discuss them both
simultaneously. So, this is the way it looks
like in the real world. These are the devices. We've got a source and a sink. We've got the various
mechanical components. We've got heat transfer. We've got work done. Now what we want to do,
is we want to figure out how do we determine
the performance of these two different devices? And then how do we
determine whether that performance satisfies or violates the second
law of thermodynamics? So, we move on. Now, let's just go back
to the heat engine. And what we're going to
ask ourselves is this. How do we express the
performance of a heat engine? [ Writing on Board ] So, how do we determine
the performance? And in fact, we're not just
looking at the performance of the heat engine, we also
want to look at the performance of the refrigeration cycle.

So, how do we determine the
performance of these two cycles? Well, let's start
with heat engine. You may recall from the — I think it's the first or maybe
that's called efficiency. And we did talk about
it very, very briefly. Basically, efficiency
was identified or defined as a desired output, divided
by the required input. Now, if we look at a heat
engine, then first of all, we still use the Greek letter
Eta, to describe efficiency. The efficiency we're talking about is called the
thermodynamic efficiency. Or it could be called
the thermal efficiency. It's the same thing, just
two words for the same thing. But the way that
we define this — by the way, I just
put the subscript TH. Think of it as thermal. Think about it as thermodynamic. But that is the efficiency
we're talking about. So, again, this is the desired
output, over the required input.

So, the question now is,
what is the desired output? And what is the required input? Well, it's a heat engine. What we're trying to
do is get as much work out of the system as possible. This is work to,
well, presumably, if it's an electric generator,
to spin the electric generator. But it doesn't have to be that. It could be the work required
to run an air compressor.

It could be anything. But the desired output is
that net work output, right? So, net work output
is what we desire. But there's going to be
a certain required input. And the required input, isn't that the heat
input from the source? I mean, aren't we paying for
fuel, presumably lots of fuel, to burn and then get the
heat input that we need, so that our powerplant
works properly? Or our heat engine, if
you will, works properly. So, the require input
then, is our heat input. So, therefore, the thermodynamic
efficiency is just the net work out, divided by the heat
input, which is Q sub H. Now, I'm going to modify
this a little bit here. But to do so, we need to look
at the first law for a cycle.

So, for a cycle, what we
would show, for the first law, is that if we integrate
over the entire cycle. And the way we show
that is by — with the integration sign
with a little circle in it. That means we're
integrating over the cycle. So, the integral
over the entire cycle of all the work that's
done in the cycle, has to equal the integral
over the entire cycle of all the heat transfer
associated with that cycle. Now, if you think about
it, that should make sense. I mean, it's the
first law, right? The first law in its most basic
form, you know, really tells us that the heat transfer,
minus the work, is going to equal the
change in the enthalpy, plus change in kinetic energy, plus change in potential
energy, right? But if it's a cycle,
we're beginning and ending at exactly the same
thermodynamic state. I mean, that's what
a cycle is, right? Which means the kinetic
energy is going to be the same at the beginning at the end.

The enthalpy is going
to be the same at the beginning and the end. The potential energy
is going to be the same at the beginning and the end. In other words, your
has to equal zero. In other words, for the
cycle, the net heat transfer and the net work have to
be equal to each other. I mean, otherwise
it's not a cycle. If those aren't equal
to each other, then you have some net change
in one of your properties and it's not a cycle
anymore, right? So, this statement
holds true for a cycle. But what we would note is that
for our heat engine cycle, there's only one
work term, right? That's the net work.

That's the only work that
crosses the boundary. So, the net work out is on the
left-hand side of the equation. And then as far as the heat
in — well, the heat transfer, there's actually
two heat transfers across the boundary, right? We've got heat added
at high temperature. So, QH. And then we have
heat that's rejected at low temperature. So, we have a QL. Now, we know that our basic form
of the first law should tell us that it's the heat transfer,
plus the other heat transfer. In other words, the heat
in, plus the heat out. And that the heat out would
simply have a negative sign associated with it.

One thing that has to
be made real clear, as we talk about cycles, we're going to actually put
and work terms are going to be absolute values or
magnitudes, if you prefer. So, from here on out, when we
talk about cycles, all the Q, all the W terms are
positive terms. And if there's a sign that has
to be considered, like in this, it has to be included
within the equation. So, the net work out,
equals QH, minus QL.

And again, let me just note. All Q and W terms
are absolute values. [ Writing on Board ] The sign will be
included in the equation. All right, so, it's
just something you want to keep track of. Well, all right, so, with all
this in mind, let's go back to our performance
characteristic for the heat engine, which was
our thermodynamic efficiency. So, back to the thermal
efficiency. So, what is a thermal
efficiency? It's a net work output,
divided by the heat input. Now, isn't the net work output, from the equation
above, just QH minus QL? So, we can write this as
QH, minus QL, divided by QH. Or we can certainly simplify.

And then this equals
one, minus QL over QH. So, this is the equation
that we're going to utilize, if we have a heat engine. We're going to have to find
its thermodynamic efficiency. Now, just to kind of give you
a hint as to where we're headed with this, eventually,
we're going to look at what's called an ideal
cycle or a Carnot cycle. And what we will find,
based on observation, is that nothing can have a
higher thermodynamic efficiency, than this ideal cycle,
this Carnot cycle. And that allows us to determine
whether the second law has been satisfied or not. Eventually what we're
going to do, is we're going to calculate the thermal
efficiency of our cycle. We're going to compare it
to the thermal efficiency of this ideal Carnot cycle.

And if our efficiency is greater than that Carnot cycle,
that's impossible. The ideal cycle is the
best possible cycle. It's going to be the one that
gives us the maximum amount of work output, for the
minimum amount of heat input. And that is kind of an
ideal situation, isn't it? You're always trying to
maximize your benefit, right? Maximize the work out. Maximize the electric
generations, so you can charge your
rate pairs for that, right? But you want to minimize what
you have to pay for, right? You minimize the
maximize what you're able to sell to your customers. And minimize your
associated costs. So, none the less, we
will eventually have a way to compare my actual cycle, using this thermal efficiency
equation, to this new cycle, this Carnot cycle, that we
have not yet talked about. We'll get to that in a moment. Let's go now to the
refrigeration cycle.

Remember, we're trying to look
at performance characteristics for two different
types of cycles. So, we've finished
with the heat engine. And now we'll look at
the refrigeration cycle. Now, for refrigeration cycle, we still have the
same basic definition. Note that I've left
the hand side of the equal sign blank
for the time being. We still have our same
performance characteristic, which is our desired output,
over our required input. However, we're not going to
call it the efficiency any more. We're going to use
a different term, which is coefficient
of performance. So, COP or coefficient
of performance. This is the performance
characteristic or the performance parameter
that we're going to use for the refrigeration cycle. And we're going to do much the
same thing here as we did above. We're going to find a
mathematical expression, in terms of just the heat
transfer and work terms. And then in the future,
we're going to have a way to compare this to the
ideal refrigeration cycle, to see whether the cycle of thermodynamics has
been satisfied or not.

what is the desired output? Well, the desired output
is the heat transfer of interest, right? [ Writing on Board ] So, heat transfer of interest. And the required input,
well, that's just going to be the net work input
that we have to pay for. So, this is our basic definition of the coefficient
of performance. Now, here we have to be careful because there's actually two
different refrigeration cycles that one must consider. And I'm not talking
about a refrigerator versus an air conditioner. Those are the same, okay? In both those situations,
the heat transfer of interest is the heat that we
can pull out of our cold space. In other words, QL. So, one type of refrigeration
cycle that we have to analyze is simply the
refrigerator or freezer or air conditioner or ice maker. And I'm sure there's other
devices that would all, you know, satisfy this type
of coefficient performance. But if we have this
device, then we use COP with a subscript R. Anomaly,
the R stands for refrigerator. But again, it could be any
device where the heat transfer of interest is the heat
that we're trying to take out of the food that we're
trying to keep cold or the air that we're trying to keep cold.

And put it into the substance,
into the working fluid of this refrigeration cycle. This is, we're putting
it into the refrigerant. So, it's QL. So, this coefficient performance
then, the heat transfer of interest is the heat
that's added into the cycle. It's taking out of what we're
trying to keep cool and put into the refrigerant
in the cycle. So, that's QL. And then we have to
divide this, of course, by the net work input. Okay. Now, once again,
I want to simplify. And in order to simplify,
I'm going to have to go back to my first law.

So, back to the first law. So, first law for a cycle. Again, the total
amount of work has to equal the total heat
transfer for the cycle. Again, we're using only
absolute values here. So, what is the work term? The only work term we have
is a single work input term. And we know work
input is negative. So, we have minus W net input. And then, as far as the heat
transfer term, we're adding heat at low temperature, so QL. And we're rejecting heat at
high temperature, so minus QH. We can rearrange this. We can just say that the net
work input is then just QH, minus QL, okay? Now, what I'd like
to do is I would like to do the same thing
here that I did above. Let's go back to our
coefficient performance for the refrigerator,
freezer, air conditioner.

And the QL is the term
that's in the numerator. We're going to leave that be. The net work input is the
term of the denominator. And you can see, I'm just
going to plug in from above. So, this is QH, minus QL. And then what I want
to do is I want to divide both the numerator
and denominator by QL. So, this becomes
[inaudible] over QL, which is one, over QH over QL. And then minus QH
over QL, which is one. And this then, is the
equation for the coefficient of performance for this first
type of refrigeration cycle, the refrigerator, freezer,
air conditioner type. Okay. Now, you know,
eventually we'll see how to solve these equations.

I have several example problems. We're not there yet. All right, so, I mentioned
that there was another type of refrigeration cycle. And unfortunately, this is a
type of refrigeration cycle that we just don't see very much
here in Southern California, but it's called the heat pump. And please do not confuse
heat pump with heat engine. They're totally different, okay? A heat pump is a
refrigeration cycle. But the purpose of a heat
pump is to heat a space that we're trying to keep warm. So, for instance, what if you
the processes? What if it's the middle of
winter instead of summer? Can't you just take heat out of
the cold outside air and send it into the refrigeration cycle? And then reject that
heat into the house that we're trying to keep warm? Sure, why not? I mean, it's just a
refrigeration cycle.

as a refrigerator, freezer, air conditioner, where
we're more interested in keeping the food cold. Or whether we're
thinking of a heat pump, where we're more interested
in keeping our house warm. They're both just
refrigeration cycles. The difference though, is the
desired heat transfer, right? The heat transfer of interest
is not QL anymore, it's QH. So, if we go back to our
equation for the coefficient of performance, well,
first of all, we're going to use
HP for heat pump. Not R anymore. And then the numerator is the
heat transfer of interest. So, this is QH. And then it would be over,
oops, I'm ahead of myself. This would be QH. Then over the network input. But the network input, just
like we saw previously, is still QH, minus QL. And if we just divide
numerator and denominator by QH, we get one over one,
minus QL over QH. So, this equation then
and I'll underline it, is our performance
characteristic. Our performance perimeter,
if you will, associated with the heat pump
type of refrigeration cycle.

Okay, now, anybody's whose
been in parts of the country where they don't have
underground natural gas pipes, you're only choice for heat in the house is either a big
giant propane tank outside or an electric device,
like the heat pump. And it is an electric
device, isn't it? The work input comes
through a compressor, which is being run
by an electric motor. So, it turns out it's much more
effective to use a heat pump, then just to use one of those
resistance heaters that some of you have in the wall of
your apartments in the bathroom or wherever they happen to be. Although, those resistance
heaters are incredibly cheap and heat pumps are
incredibly expensive. I mean, there's as expensive
as an air conditioning unit. So, you know, you don't find
heat pumps in most apartments. Because quite frankly, the landlord doesn't care
how much it costs you to pay or to run your heater
all winter long. He only cares about
what it costs him to build the apartment, so he
can rent it out to you, right? So, again, welcome
to the real world. And we move on.

So, these are our
performance characteristics. One note here, that
these equations above, the thermal efficiency,
the coefficient of performance equations, can
all be written as — I'm sorry. Can be written with
the rate form of the heat transfer
in work terms. In other words, if I just take
any of those above equations and do a timed derivative, then
my Q's would become Q dots. My W's become W dots.

There's no difference at
all, except they're in terms of the rates of heat transfer
or the rates work production, rather than the total amount of heat transfer,
total amount of work. So, the thermal efficiency,
the coefficient of performance equations can be
written with Q dots and W dots, in place of Q's and W's. Okay, so, just be aware of that. Look at the problems carefully. And know whether they're
talking about a rate equation or a total amount of heat
transfer or work equation. All right. Now, we move on and talk
about the ideal cycle. We have the equations we need
to determine the performance of the heat engine or
the refrigeration cycle.

Now, we need to look at what
is the best possible value of those terms, right? The efficiency or the
coefficient of performance, so we can see if the second
law has been satisfied or not. That's really all
there is to it. So, the way we do that is by
comparison to the ideal cycle. So, the ideal cycle is also
called the Carnot cycle. Now, as we talk about a Carnot
cycle, we need first to talk about the concept
of reversibility.

a reversible process. Okay, so, I will note, before I
even write down the definition, that the reversible process
is an ideal process, okay? It's a process that's done
in such a way that at the end of the process, we can reverse
it, we can go backwards. And both the system, as
well as the surroundings, can be restored to
their initial state. So, that's what we
mean by reversible. In the real world,
nothing is reversible. Nothing. Nothing is reversible. In the real world, you've got
all sorts of losses, friction, vibration, effects of
thick or viscous fluids.

And there's a whole long list. But in the ideal world,
remember, this as ideal cycle, the processes that are
involved are assumed to be these reversible
processes, okay? So, a reversible process is
one performed in such a way that at the end of the
process both system and surroundings can be
returned to their initial state. Now, again, we understand that that can't happen
in the real world. But we're not talking

We're trying to define a
theoretical limit here, right? If we want to use the second
law of thermodynamics, we need to compare our real
cycle to an ideal cycle. One that you couldn't
possibly do better than. So, that's the whole point of this concept of
the Carnot cycle. Now, again, why isn't
such a process possible? I'm not going to
write all the reasons. But in the real world, there
are real reasons, right? Friction represents a loss. Deformation represents a loss. Vibration, viscous
effects, expansion or throttling processes. Collapsing or snapping of a
wire or a bubble or a film. I mean, in those cases, you
hear that irreversibility. When you have a bubble
that collapses, you hear a sound, right? Or that's a pressure wave. That's energy that's lost
from your system, right? You can't get it back. Mixing is another process that
is not a reversible process. So, again, we understand
that it's not possible in the real world to have
irreversible process.

But again, it represents
a theoretical limit. I might note that if a
process is not reversible well, it is called irreversible, okay? So, that should make a
certain amount of sense to you. Now. Again, the reversible process
represents a theoretical limit that we're trying to analyze. So, there's two other things that are discussed
briefly in your textbook. And I'm not going to spend
a lot of time on them. The book talks about an
internally reversible process.

external reversible process. If something is both internally
and externally reversible, then both the system, in
other words, internally and the surroundings
externally reversible. Well, that means it's
completely reversible or we just call it a
reversible process. But we could theoretically
have processes where only the system itself
is capable of returning to the original state. The surroundings would not be. So, that's what we call an
internal reversible process. It's not completely
reversible, maybe think of it as partially reversible. We'll say that only the
system can be returned to its original state. And the other term is called
externally reversible. So, it's kind of a partially
reversible process, as well.

Here, only the surroundings
can be returned. Well, to their initial
state or original state. [ Writing on Board ] Okay, so, we're not
going to really get into internal reversible versus
external revisable right now. We just have to recognize
that those are possibilities. And we will get into
that a little bit more, when we start talking
about entropy. So, let us now look in
a little more detail at this Carnot cycle. [ Writing on Board ] So, first we're going to look
at the Carnot heat engine. And simultaneously,
refrigeration cycle. [ Writing on Board ] So, I'm basically going to put
a couple of boxes here again. For the heat engine,
well, we have our box. We're still going to be adding
heat at high temperature. We're still going to be rejecting heat
at low temperature. We're still going to have
some net work output.

The high temperature heat
source is going to be at a temperature we'll
just call T sub H. And the low temperature
heat sink, is going to be a temperature
T sub L. Again, L for low, H for high, all right? And then everything that's
inside the box is going to be just like what we
had in the box before. Now, you may say well, then
how is this any different than a real heat engine cycle? Well, it's really not. I mean, it's not any
different in the sense that it has all the same types
of mechanical components in it. It's still a boiler, still
a turbine, condenser, pump. That's all there. The difference is the way
that the processes take place. It turns out that in a Carnot
cycle, the process takes place in a very specific way that
is removed from reality. That beeps every day at
this time, doesn't, it? Okay. All right, so, there's
my Carnot heat engine. Let's also look at the
refrigeration cycle. And once again, that's not going
to look markedly different.

Actually, it's not going
to look any different. So, we have our box. We would have our
source and our sink. The sink is going to
be at high temperature. The source is going to
be at low temperature. We're going to add heat
at low temperature. We're going to reject
heat at high temperature. We're going to require
some net work input. Again, it doesn't
look any different than the refrigeration
cycle we saw before.

And what's inside the box
is the same as what we had, not entirely, but
for the most part, the same as what we had before. But again, the difference
is the way that the processes take place. So, what is so different
about the Carnot cycle that makes it reversible? That makes it ideal? Well, really, there's
a couple of things. So, what makes the
Carnot cycle unique? And for that matter, ideal? And by the way, reversible
is really just another name for ideal. It all means the same thing. So, I'm just going to put
reversible here, as well.

So, what makes the
Carnot cycle unique? Well, what is it
that makes it ideal? What is it that makes
it reversible? So, the first thing is this. Heat transfer does not cause
the temperature to change, okay? Now, if you think about that,
you add heat to something, unless you're changing phase, you would expect the
temperature to rise, right? You remove heat from something and you expect the
temperature to drop. It doesn't happen, okay? You can add heat all day long
from the source into the cycle and the temperature
will not change at all. So, basically, this
is an ideal process.

To have transfer that doesn't
affect the temperature, is a very idealized
process, right? So, this is basically called
simply isothermal heat transfer. So, this is one thing that
makes the cycle unique, okay? There's no change in the
temperature of the heat source. But that doesn't
make sense, right? The heat source should
lose temperature. The cycle should
gain temperature. But neither one is
changing, right? The heat transfer takes place
without any temperature change. That's part of the Carnot cycle. What's the other thing that
makes it rather unique? This has to do with the way
that work is done, okay? So, work is done with
no heat exchange. Now, if you think about
it, that's pretty rare. If you have a mechanical
device that's doing work, it's spinning, right? It's got some bearings in it. There's definitely going
to be some friction.

There's definitely going
to be some heat generated. In the real world, if
you're doing the work, it will have a heat
exchange associated with it. And if you don't believe me,
just feel the side of any pump. Feel the side of any compressor,
it's going to be warm. You're going to feel the
heat coming off of that. But in this idealized
Carnot cycle world of ours, work can be done with
no heat transfer. So, this is called
unique about the cycle. Now, one thing that I forgot
to do and I really need to change both these two
underlying things here. I should really put
reversible in there. Isothermal reversible
heat transfer, would be a more accurate
way to describe this. Adiabatic reversible work, would really be a more
accurate way to describe this.

But again, these are processes
that are ideal, reversible. And therefore, represent
the same theoretical limits that we're aiming for, right? Remember, we're trying
to find a cycle. Whether it's refrigeration
cycle or heat engine cycle that is going to be compared to the best possible
cycle, the Carnot cycle. So, we can see if the cycle of thermodynamics
is satisfied or not. None the less, studies have
been done over countless years, decades really, centuries
at this point. And this has led to what's
called Carnot's principle. [ Writing on Board ] And basically, what
Carnot's principle tell us is that no engine — I'm sorry. No heat engine or
refrigeration cycle, operating between
two reservoirs, can be more efficient or have a
higher coefficient performance than the Carnot cycle operating
between the same two reservoirs.

So, no heat engine or
refrigeration cycle. I'm just going to abbreviate
refrigeration cycle with refrig. Operating between two reservoirs
can have a higher thermal efficiency or coefficient
of performance than the Carnot cycle operating
between the same two reservoirs. And my writing is getting
a little bit sloppy. I apologize. But that's the Carnot principle. Now, isn't this the second law? Doesn't this place the kind
of restrictions on a process, exactly like what
we're looking for? If no actual heat engine or refrigeration cycle can
have a higher efficiency or coefficient of performance
than the Carnot cycle, then that's a restriction,
right? We have to find the value
of the thermal efficiency or coefficient of performance for our heat engine
or refrigerator.

Compare it to the thermal
efficiency or coefficient of performance for our Carnot
heat engine or refrigerator. Well, I should say refrigeration
cycle, then compare the two. If our cycle has a greater
efficiency then the Carnot cycle, it can't work. It violates this
Carnot principle. It violates the second law. So, this is just one
form of the second law. Now, one thing that we can do,
is we can actually now look at this statement and compare
it to our equations, okay? And this actually leads
to two corollaries.

[ Writing on Board ] I don't think I spelled
that right. Oh, well. And these are
the two corollaries. So, the first one is that no
heat engine can produce more work per unit of heat
input, than a Carnot cycle. [ Writing on Board ] And all we really have to do is
look back at the basic equation. The equation for thermal
efficiency is still the same as it was before. I mean, just because we're
talking about a Carnot cycle, doesn't mean that our
performance parameter changes. The thermal efficiency is still
the net work, divided by — well, the net work out,
divided by the heat input. So, if the maximum possible
thermal efficiency we can have is for the Carnot cycle,
then doesn't that mean that the Carnot cycle is one
that's producing more work per unit of heat input, than any
other possible cycle, right? Every other cycle has to
have a lower efficiency, such that per unit
of heat input, it must have a lower
amount of work.

So, no heat engine can
produce more work per unit of heat input, than
the Carnot cycle. And I just wrote this equation
over here for illustration. And in fact, there's
another way of saying that. Or requires less heat input
per unit of work output. Than a Carnot cycle. And again, that makes
sense, just by looking at the equation above, right? For a particular unit of work, since the Carnot cycle has the
highest possible efficiency, then it would also have to
have the lowest possible amount of heat input. So, this is one of
the two corollaries. The other one refers to
the refrigeration cycle. All right, so, the
other corollary is here. No refrigeration cycle
requires less work input. [ Writing on Board ] Well, yeah, I'm sorry,
that's right. Per unit of heat input,
then a Carnot cycle.

Okay. Or another way of
saying that is we can say or produces more heat transfer
per unit of work input. [ Writing on Board ] And I just realized I have
to make one change here. Up above where it
says heat input, it should really say heat
transfer and not heat input. Please keep in mind that in
the refrigerator, freezer or air conditioner, the desired
heat transfer is the heat input into the cycle out
of the food, right? In a heat pump, the desired
heat transfer is the heat out of the cycle and into the
space we're trying to keep warm. So, it wouldn't be appropriate to use the word heat
input or heat output. It should really just
say heat transfer.

So, these are just
two corollaries. And again, these allow you to utilize the second
law of thermodynamics. Now, the question becomes,
okay, we have the basic equation for calculating the
thermodynamic efficiency or the coefficient
of performance for both the heat engine and
the refrigeration cycle, right? But how do we do it for
the Carnot heat engine or the Carnot refrigeration
cycle? Now, this gets to be
a little bit tricky. And I'm not going to go through
the entire chapter of your book. If you look in section
thermodynamic temperature scale. And amongst other things, it talks about the
existence of absolute zero. And it talks about how one is
able to create the Kelvin scale and the ranking scale. But that's not of
interest to us. What's of interest to
us is the following. So, I'll just say from
section 6-9 in the book, it is shown that the
following applies. That the ratio of heat
transfer at the high source, to the heat transfer
at the low sink.

Or it could be the reverse. It could be a high
temperature sink and a low temperature source. It doesn't matter. Of course, one applies to
the heat engine and the other for the refrigeration cycle. But for a reversible
process, I'll put a little rev for reversible, is actually
going to equal the ratio of the high temperature
to the low temperature. Now, this is not something that you could say
for any cycle, okay? This is only for the reversible. In other words, it's
for the Carnot cycle. So, for the Carnot cycle,
the ratio of heat transfers, equals the ratio
of temperatures. Now, this is not
something that is going to be entirely obvious to you.

Go through section 6-9. If you read through it and
you still don't believe it, well, then read it again. But this is true, right? We have to accept
this as being true. It's been well, proven
for you in the book. And I just don't have
the time to spend on it. So, that gives us the ability
now to find new equations for the thermal efficiency
for the Carnot cycle, right? And coefficient of performance
for the Carnot cycle. All we have to do really, is go
back to those basic equations. So, therefore, we'll start
with a thermal efficiency. We know that the thermal
efficiency is one, minus QL over QH. But we note that for the
Carnot cycle, QH and QL — I mean, those are both
reversible processes, right? Heat transfer takes
place isothermally without any temperature change. That's the nature
of the Carnot cycle. So, therefore, the equation
directly above applies.

So, this has become one,
minus TL over TH, right? Now, what we can also do
is we could apply those two corollaries. We note that the thermal
efficiency is also defined as the net work output,
divided by the heat input. But the corollaries say that nothing can produce more
work output than the Carnot. So, we could write
this also as net work out maximum, over
the heat input.

Or the other way we look at it
is we say no heat engine can — well, no heat engine producing
a certain amount of work, will require less heat
input than the Carnot cycle. So, we can also write this
as the net work output, divided by the heat
input minimum. Okay, so, please note
that there's a max and there's a min here. And these statements
specifically refer to those corollaries of
the Carnot principle. All right. So, what about the
refrigeration cycle now? Okay, so, we have two
the refrigerator, freezer, air conditioner. And we know that the equation
that was developed was one over QH over QL, minus one. And therefore, this is
going to equal one over TH over TL, minus one, right? So, that is how we're going to
find the coefficient performance for the Carnot refrigerator,
freezer or air conditioner.

We're going to use temperatures
instead of heat transfer terms. And based on those corollaries,
we could then write this as QL over the net work input minimum. So, let's put min for minimum. Or we can write it
as QL maximum, maximum amount of heat input. >> Can you move it up? >> Yep, yeah. Yeah, this isn't working
really well today. This pen is very thick. So, I get very few
lines per page. Okay, thanks for reminding me. Keep doing so. Anyway, so, then this would be
over the net work input, okay? So, nothing will produce
more heat transfer per unit of work input, than
the Carnot cycle. And then lastly, we
would have our heat pump. So, the equation for
that was one over one, minus QL over QH, right? And therefore, for
the Carnot cycle, it's one over one,
minus TL over TH. So, that's how we're going to
find the coefficient performance for a Carnot heat pump. By the way, if you want to
make sure that you don't forget that these only apply to the
Carnot cycle, then just go up here and put a little
c next to the coefficient of performance or the
thermal efficiency terms.

You know, the c for Carnot will
make it obvious all the time. so, yeah, just add a
little c in there somewhere, just as a reminder to yourself. Anyway, as far as those
two corollaries go, we would then put QH over
the net input minimum. Or QH maximum over
the net input. So, again, no cycle can
have a greater coefficient of performance than
the Carnot cycle. But the corollaries to that
tell you that for any unit of heat transfer, the Carnot
cycle would have the minimum possible work input. Or for any particular
unit of work input, no cycle can have
more heat output. Those are the Carnot
cycle statements. So, now what we need to
do is utilize all of this in some example problems, so that we just simply
know what's going on here. So, first of all, let
me just pause this. This is a good time to
pause for any questions. Any question? Yes? >> So, in terms of like the
heat pump, if you have your pump in like a very cold setting,
it would be more efficient? >> Yeah, basically, right? >> And it's the same for the
other ones in their [inaudible].

Whenever would be
closest to one? >> Well, I mean, it turns
out that coefficient of performances can actually
become greater than one. Thermal efficiencies can't
get greater than one. They end up at a maximum at one. So, we're not trying to compare
our actual cycle to a value of efficiency or coefficient
of performance that equals one. We're comparing the actual
cycle's, again, efficiency or coefficient of performance to the Carnot cycles
efficiency or performance, okay? And the Carnot cycle,
it's not going to have a perfect
efficiency of one.

You're never going to be able
to turn all the work into heat. I mean, even the ideal case,
that's not going to happen. Well, I take that
back, it could happen if your high temperature
heat source is at infinity. And your low temperature heat
sink is at absolute zero. Well, then, everything
approaches 100%. But we're not going
to talk about that. So, these are the relationships
we're going to use. Any other questions? All right, so, let me just go
through some example problems. And I kind of wrote these out. So, here's our first
example problem. And then I don't have
to write it all again. These are not from the book. So, I couldn't actually
copy them from the book. So, the first example
that we're going to look at is a heat engine. And it says that we have a
heat engine that operates with 150-degree Celsius
heat source and a 20-degree Celsius
heat sink. What's the maximum
possible thermal efficiency? Well, that one's kind
of easy, isn't it? It doesn't say what is the
Carnot cycle efficiency.

But we know, from
the Carnot principle, that nothing can have a
higher thermal efficiency than the Carnot cycle. So, that when you're asked to find the maximum
possible thermal efficiency, it is the Carnot
cycle efficiency that we're looking for. So, the solution to this
is simply the Carnot cycle thermodynamic efficiency. Which is one minus TL over TH,
is the maximum possible value. So, we need to make sure that
we use the absolute temperature scale, right? When we can't use Celsius. So, the low temperature
sink is at 20, plus 273.15. The high temperature source
is at 150, plus 273.15. So, this converts
them both into Kelvin. And if we go through the
mathematics, we get 0.31. So, if you happen to have
a heat engine operating between these two limits.

And by the way, these are
pretty typical numbers for a geothermal power plant. If you look at the temperature
of a geothermal brine that comes out of the ground, yeah, 150
Celsius is pretty common. And if you look at
the temperature of the surrounding environment,
like the local river or an ocean or whatever, then 20-degree
Celsius is pretty typical. So, it turns out that for
a geothermal power plant, the maximum possible
efficiency is only 31%.

In other words, for every 100
units of heat that you put into that cycle, you're
only getting 31 units of work out of it, okay? And that's the maximum. Once you start throwing
in the real world, then this efficiency drops down
to closer to 22 to maybe 24%. Again, that's the
real world for you. So, this is the first
example problem. I see that I'm out of time. I have two more example
problems. So, I'm just going to
continue with those on Friday. And are there any last
questions here today? Yes. >> Yeah, if what the [inaudible]
specific heat [inaudible] temperature, if they
might have a different internal [inaudible]. >> It absolutely doesn't matter. Because for the Carnot
cycle, the heat source and the heat sink,
they stay the same. They're always at a
constant temperature. So, it really doesn't
matter what's happening. In the ideal sense,
nothing is happening. I mean, you're pulling as much
heat out of it as you want. And you're not getting
any change to that particular heat source. Or you're dumping as much heat
as you want into the heat sink and you're not getting any
change in that heat sink.

So, that's the nature
of these ideal cycles. Okay, so, good, that's
it for today. I will see you all on Friday. Please don't forget to turn in
your homework that's due today. I'll post the solutions
right after class. [ Music ].