[ 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

about refrigeration cycles. I'm going to talk about the

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

the second chapter in your book. They talk about something

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

heat transfer, minus your work,

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

the sign into the equation. And all of our heat transfer

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

denominator, your heat, your fuel that you have to buy. So, you always want to

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.

So, let's ask ourselves,

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

had your home air conditioner and you just reversed

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.

Whether we think about it

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.

So, let's talk about that. So, let's talk about

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

about the real world.

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.

And it talks about an

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

adiabatic work. And that's what's

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

6-9 of your textbook, it talks about what's called the

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

different refrigeration cycles. We'll start with

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 ].