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0:00
[ 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.
0:44
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
1:16
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
1:46
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.
2:16
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
2:46
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?
3:13
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.
3:42
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.
4:19
I brought my own today.
4:23
Let's start over.
4:25
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.
4:52
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.
5:14
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.
5:28
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.
6:03
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.
6:32
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
6:50
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.
7:14
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.
7:50
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.
8:13
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.
8:41
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.
9:05
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.
9:23
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.
9:52
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.
10:28
So, we're adding heat at low temperature.
10:33
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.
11:07
This work input, because we're talking about a vapor, is going to go into a compressor.
11:14
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.
11:41
And here, we're talking about the refrigeration cycle, why don't I say net work input.
11:53
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.
12:25
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
12:53
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?
13:18
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?
13:45
And then how do we determine whether that performance satisfies or violates the second law of thermodynamics? So, we move on.
14:00
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?
14:34
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
15:08
about is called the thermodynamic efficiency.
15:13
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 --
15:27
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.
16:03
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?
16:36
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.
17:06
So, for a cycle, what we would show, for the first law,
17:17
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?
17:42
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.
18:11
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.
18:37
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.
19:09
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.
19:36
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.
19:55
And again, let me just note.
20:13
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.
20:39
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.
20:49
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,
21:04
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.
21:40
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.
22:08
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.
22:35
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.
23:00
So, we've finished with the heat engine. And now we'll look at the refrigeration cycle. Now, for refrigeration cycle,
23:13
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.
23:35
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.
23:54
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.
24:17
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
24:59
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.
25:10
So, one type of refrigeration cycle that we have to analyze is simply the refrigerator or freezer
25:25
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
25:58
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.
26:24
Okay. Now, once again, I want to simplify.
26:30
And in order to simplify, I'm going to have to go back to my first law.
26:38
So, back to the first law.
26:42
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.
27:07
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.
27:43
The net work input is the term of the denominator. And you can see, I'm just going to plug in from above.
27:49
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,
28:15
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?
28:44
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.
29:15
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,
29:42
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.
30:30
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.
30:56
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,
31:18
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.
31:33
And we move on.
31:36
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
32:07
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,
32:30
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.
32:48
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.
33:25
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.
33:43
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,
34:15
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
34:43
that at the end of the process both system
34:58
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?
35:42
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.
36:12
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?
36:42
So, that should make a certain amount of sense to you.
36:48
Now.
36:55
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.
37:14
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.
37:48
It's not completely reversible, maybe think of it as partially reversible. We'll say that only the system can be returned
38:03
to its original state.
38:07
And the other term is called externally reversible.
38:13
So, it's kind of a partially reversible process, as well. Here, only the surroundings can be returned.
38:31
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.
38:55
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.
39:15
And simultaneously, refrigeration cycle. [ Writing on Board ] So, I'm basically going to put a couple of boxes here again.
39:33
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
40:22
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
40:48
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.
41:06
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.
41:15
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.
41:48
But again, the difference is the way that the processes take place. So, what is so different about the Carnot cycle
41:56
that makes it reversible? That makes it ideal?
42:01
Well, really, there's a couple of things.
42:08
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,
42:56
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.
43:34
So, this is one thing that makes the cycle unique, okay?
43:39
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.
44:16
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.
44:42
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.
45:01
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.
45:10
Isothermal reversible heat transfer, would be a more accurate way to describe this.
45:18
Adiabatic reversible work,
45:21
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,
45:57
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,
46:20
operating between two reservoirs, can be more efficient or have a higher coefficient performance than the Carnot cycle operating between the same two reservoirs.
46:30
So, no heat engine or refrigeration cycle. I'm just going to abbreviate refrigeration cycle with refrig.
46:45
Operating between two reservoirs can have a higher thermal efficiency or coefficient of performance
47:06
than the Carnot cycle operating between the same two reservoirs.
47:20
And my writing is getting a little bit sloppy. I apologize. But that's the Carnot principle. Now, isn't this the second law?
47:38
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.
48:07
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.
48:42
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 --
49:43
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.
50:13
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.
50:45
Than a Carnot cycle.
50:48
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.
51:05
So, this is one of the two corollaries. The other one refers to the refrigeration cycle.
51:14
All right, so, the other corollary is here.
51:20
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.
51:59
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.
52:35
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.
53:08
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?
53:21
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.
53:49
What's of interest to us is the following.
53:55
So, I'll just say from section 6-9 in the book, it is shown that the following applies.
54:08
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
54:39
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.
55:17
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.
55:36
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?
56:08
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.
56:52
So, we can also write this as the net work output, divided by the heat input minimum.
57:01
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.
57:15
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?
57:39
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.
58:19
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.
58:45
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.
59:14
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,
59:26
we would then put QH over the net input minimum.
59:36
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.
1:00:10
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?
1:00:26
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?
1:00:47
>> 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
1:01:14
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?
1:01:39
All right, so, let me just go through some example problems. And I kind of wrote these out.
1:01:50
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.
1:02:22
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.
1:02:47
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.
1:03:00
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.
1:03:12
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,
1:03:38
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.
1:04:09
And are there any last questions here today? Yes. >> Yeah, if what the [inaudible] specific heat [inaudible] temperature, if they might have a different
1:04:21
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
1:04:44
and you're not getting any change in that heat sink. So, that's the nature of these ideal cycles.
1:04:51
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 ]
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