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0:11
>> Well, good afternoon to all of you.
0:14
First of all, just a reminder that your homework is up here from last time. Please don't forget to pick it up. I might note that, in general, most students really did just fine.
0:26
No significant errors or problems that I was really able to identify. We had the usual unit errors and, you know, some math errors here and there. One thing I have noticed is that a lot of the solutions are looking remarkably similar to each other, and practically identical to the official solutions manual. So, it's pretty clear that quite a few of you have access to that. You know, especially that problem, number, what was it, 32? The one that I had reassigned and I even gave you a hint
0:53
as to how to solve it correctly. More than half the class gave me a solution that was identical to the one, or at least similar to the one that was in the solutions manual, which I don't really consider to be the proper solution. So, I just wanted to point that out, you know, don't rely so much on the solutions that, when you take an exam, like the one that's coming up a week from Monday, that you're not going to be able to know how to set up the problems and solve them appropriately.
1:20
Anyway, so again, no major problems on the homework. Please make sure you look at both the solutions if you're interested. So, we're now beginning to talk about the second law of thermodynamics. And what we hopefully understand is that the second law is very different than the first law. You know, the first law is very specific as far as giving us relationships that allow us to show that energy is conserved.
1:44
You know, heat transfer minus work equals a change in energy. That kind of thing. But here, it's not quite so, let's say, specific. Basically, the second law denies the possibility of using energy in a certain way or doing processes in a certain way. And with that in mind, there's a lot of different forms of the second law really based upon the type of problem that we're going to analyze. Now, we understand the general statement of the second law, at least with regards to cycles.
2:15
But eventually we're going to have to move beyond cycles and start looking at individual processes and that's something that perhaps we'll get to before the end of today. But for now, we're still looking at cycles. We understand that no heat engine cycle can have an efficiency greater than that of the Carnot cycle, right? We understand that no refrigeration cycle can have a coefficient performance greater than that of the Carnot cycle. So that when we're trying to find limits, you know,
2:41
if we're trying to find what is some maximum possible efficiency, or what's some maximum possible coefficient of performance, or what's the minimum amount of work input or some maximum amount of work output, something like that, then we have to utilize the relationships for a Carnot cycle. So, when you're reading a problem, and it uses those kind of words like minimum or maximum, or is this possible or is this impossible, you know that you have to somehow make a comparison to the Carnot cycle.
3:12
Now, we did one example problem last time where we found the maximum efficiency associated with a particular heat engine cycle operating between two temperature limits, we know that's just the Carnot cycle's thermodynamic efficiency. Now, let's look at a couple of additional problems. The first one is just written here on the top. Basically, what this says is that we have a house that's being heated using a heat pump. Remember, a heat pump is a refrigeration cycle
3:39
where the discharge from the heat pump is at high temperature, the QH term, and that heat is dumped into the house we're trying to keep warm. So, the house is being heated using a heat pump. Heat is lost from the house at a rate of 80,000 BTUs per hour. The house is maintained at 68 degrees Fahrenheit while the outside air is at 20 degrees Fahrenheit. And we want to find the minimum power required by the heat pump. By the way, ignore everything below this heavy line.
4:06
That's the second problem we will do shortly. So, since it's asking you for the minimum power, we know that somehow we're going to have to utilize relationships for the Carnot cycle, and specifically since we're asking about a heat pump, that means we need to coefficient of performance for a heat pump. So, let me start working on this problem.
4:26
It's not necessarily obvious what is even meant when it says
4:33
that heat is lost from the house at 80,000 BTUs per hour, and how that has any influence over our problem, so maybe I'll just start with a drawing of a house. By the way, if you guys can't read this blue pen, just let me know. I guess you can't even tell that it's blue on the screen, can you? But it's the only extra fine point pen that's up here. The rest are thicker and you see what happens when I try to write with a thick pen. Oh, what do you know? Well, blue is good.
5:01
We'll keep this here just in case. Is this your own personal pen? That's hers? Okay. Well, I'll give it back to you afterwards. Thank you. So, when it says that we're losing 80,000 BTUs of power from the house, BTUs per hour from the house, that is not the same as losing 80,000 BTUs per hour from the heat pump. The heat pump is connected to the house,
5:25
so here's my heat pump. I'll just put a little cycle-looking thing inside of it. And we need to reject heat at high temperature
5:33
into that house, which is being kept at the high temperature of 68 degrees Fahrenheit, and it doing so, we need to pull heat out of the cold outside air, which is being kept at the low temperature of 0 degrees Fahrenheit. And for this cycle to work, we need to provide some sort of network input. And it's the minimum amount of network input that we're trying to find in this particular problem. Now, still, what about this 80,000 BTUs per hour business?
6:03
Well, if the house is losing 80,000 BTUs per hour, and we're trying to maintain a constant temperature within the house, then we better also be adding 80,000 BTUs per hour into the house. In other words, rejecting it from our refrigeration cycle into the house at high temperature. So, really, it's QH that is 80,000 BTUs per hour. Furthermore, let me just note that this is actually a rate, right, 80,000 BTUs per hour. It's not incredibly, well, incredibly. It's not entirely accurate
6:38
to just use a Q. We really need dots, okay. Remember, when I developed all those equations last time, at the very end, what I said was that all these equations are just as applicable if we know rates. You know, rates of work, rates of heat transfer as they are, and if we know total amounts of heat or total amounts of work. So, this one clearly is written in a rate form, because we are given a rate of doing heat transfer, and we're trying to find a minimum power, which is the rate of doing work.
7:04
So, I replace everything with a dot on top of it. All right. So, we have all the information we need. I forgot to write down TH. TH is 68 degrees Fahrenheit. And now, we have to ask ourselves, well, how do we determine what is the minimum amount of power required by this particular heat pump? Well, we start with the coefficient of performance for the heat pump for the Carnot heat pump. And hopefully we recall that this is just going to be 1 over 1 minus TL over TH.
7:38
Please make sure you use this version, not, oh, sorry. Please make sure you use this version of the coefficient of performance, not the one for a refrigerator, freezer or air conditioner, right? So, this is how we're going to find the value that we're looking for. Further, let's note that, by definition, coefficient of performance for heat pump is the rate of heat rejection divided by the network input. But again, since this is a Carnot cycle, for any unit
8:08
of heat rejection, the cycle that has the minimum amount of work input is the Carnot cycle, so this shouldn't just say new input. It should really say the net minimum input. And you may recall that this equation is one of those equations that I just call the corollary to, well, to the coefficient of performance for the Carnot cycle, right? The basic equation, I'm sorry, the basic statement is that no refrigeration cycle could have a greater coefficient performance than the Carnot cycle operating
8:45
between the same two limits, right? One of those corollaries was this, that if we were trying to find the maximum possible coefficient performance, which is the Carnot cycle, then that tells me for any heat input we have the minimum required heat, minimum required work input. So, it's not really that hard to solve this problem, when you think about it. We know TH, we know TL, we know Q dot H. There's only one unknown, and that's the net minimum power input. So, now we just solve the problem.
9:17
All right, so, why don't we find the coefficient performance first, I'll just put a C for Carnot, so 1 over 1 minus. Keep in mind that we can't leave this in Fahrenheit. We have to add 459.67, actually 460 is good enough. And in the denominator, 68 plus 459.67. So, this converts Fahrenheits into rankine. And if we go through the math, we end up with 11.0. So, that has to equal above. So 11.0 equals Q dot H over the net minimum work input.
10:01
Q dot H is 80,000 BTUs per hour.
10:09
And, therefore, we can just solve and we find that the net minimum, which is what we're looking for, in this case it's power, not really work. Again, if we go through the math we end up with 72, 73 BTUs per hour. Now, frankly, BTUs per hour is an appropriate set of units but it's not the one that's customarily used in the British system. In the British system of units, what we would typically use is the horsepower. So, we basically divide by 2545, it's actually 2444 point something.
10:46
And that's going to be the number of BTUs per hour for horsepower, and then this gives me 2.86 horsepower.
10:55
So, the minimum power that this particular heat pump could require is 2.86 horsepower.
11:04
Now, when the heat pump salesman comes knocking on your door trying to sell you a heat pump and tells you that, oh yeah, my heat pump is cheaper to run than the neighbor's heat pump because mine only requires 2 horsepower and he's got a 3 horsepower, you can just run into your back room, get your thermo book, do a little calculation and say, well, of course yours is cheaper, it's using a pump or a motor that's just not going to work for my situation.
11:28
You know, why would I want to buy your heat pump. I need at least 2.86, and that's the net minimum, right? In reality, you'd probably need something closer to 4 horsepower to overcome all the various losses in the system so that the system will really give you the right amount of heat to keep the house warm on that particular day. All right, so, are there any questions on this problem? All right. Please don't get confused with a heat pump versus a refrigerator, freezer or air conditioner, right?
11:56
They're both refrigeration cycles, they're analyzed in essentially the same way. The only difference is that in a heat pump it's a QH term that is a heat transfer of interest to us, for the refrigerator, freezer, air conditioner. It's the QL term that's of interest to us. So, other than that, same analysis.
12:14
Okay. Let us move on and look at another problem, here, let me do this. Let me just leave the bottom half of this up there.
12:24
And we'll look at the next problem there. So, this one's actually a lot more straightforward. Here I've just shown you a diagram. You've got a heat source that has a temperature of 1000 Kelvin, 325 kilojoules of heat is being added into our cycle, so this little thing there is the cycle. That's my machine, if you will. There's a net 200 kilojoules of work output. There's 125 kilojoules of heat rejected into the heat sync at 400 degrees Kelvin. So, with this diagram, the question is simply,
12:57
is this machine reversible, that is ideal. Or is it irreversible, in other words is it a real or possible machine? Or is it simply impossible? Now, the best way to do that is to just simply compare the thermodynamic efficiency based on the actual heat transfer and work data that they've given me to the thermodynamic efficiency based on the Carnot cycle, which is only a function of the minimum and maximum temperatures. So, that's what we do. All right. So, why don't we just start by noting
13:34
that this machine is clearly a heat engine, right?
13:41
You can see that we are adding heat at high temperature, we're rejecting heat at low temperature, we're producing a network output. So, even though it doesn't say anywhere that it's a heat engine, the diagram shows clearly that it's a heat engine. Therefore, we have to use our thermodynamic efficiency and for the Carnot cycle, this is just 1 minus TL over TH,
14:00
and that's just 1 minus, and note that the temperatures that are given are given in Kelvin, so no conversion necessary. So, 1 minus 400 Kelvin over 1000 Kelvin, is just 1 minus 0.4. In other words 0.60. So, that's the maximum possible value
14:23
that the heat engine can have for thermodynamic efficiency. So, how do we tell whether it's possible or not? Well, again we compare it to the data based on the actual heat transfer and work for the problem. So, the thermodynamic efficiency, I'll just put actual, is not based on the temperatures but is based on the heat transfer and the work, right? We can either write it as the net work output divided by the heat input or we can write it as 1 minus QL over QH, right?
14:53
Or we can write it as 1 minus QL over QH. Why don't we just leave it in this form for the time being? So, the net work output as shown in the diagram is 200 kilojoules, the heat input is 325 Kilojoules and therefore our mathematics tells us that this is 0.615.
15:15
So, now we just have to draw a conclusion based upon these results. It turns out that the actual thermodynamic efficiency is greater than the Carnot cycle efficiency. That's impossible right? It's greater than the Carnot, and therefore impossible.
15:36
Now, if we found that it was exactly equal to the Carnot cycle efficiency, you know, then we would say that it's reversible or ideal, right? On the other hand, if this efficiency was less than that of the Carnot cycle, well then we would say that it's irreversible or real or a possible system. We could also use the second law of thermodynamics to make this same conclusion. It satisfies the second law of thermodynamics as long as the actual thermodynamic efficiency is not greater than
16:08
or equal, I'm sorry, the actual one has to be less than or equal to the Carnot cycle efficiency.
16:17
That would say that we've satisfied the second law of thermodynamics. But this one doesn't satisfy the second law, right? It can't satisfy the second law if the calculated efficiency is greater than the Carnot cycle efficiency. So, we can say that this violates the second law of thermodynamics. So, I could have written this problem in a bunch of different ways. I could have said, does this machine satisfy or violate the second law of thermodynamics? I could have said, is this possible,
16:40
is this machine possible or impossible? I could have said, is this machine reversible or irreversible or impossible? I mean, there's a lot of ways to say exactly the same thing. So, make sure you recognize that. The second law tells us that the machine is either reversible or irreversible. If ir doesn't satisfy the second law, it's impossible. Okay.
17:05
[ Inaudible question ] Is the Carnot cycle impossible? [ inaudible ] In the real world, sure. We're not able to design one. That would mean that you would have to get rid of all inefficiencies, all losses if you will. All irreversibilities, as the book would say, and we, as humanb beings, just don't have the ability to do that. But that doesn't mean that the Carnot cycle isn't useful, right? This represents the theoretical limit. Now, by the way, just because I said that this is real
17:34
or even possible, that still doesn't mean we can make the thing. If we did conclude that this was possible, we would still have to design the right machines and test it out and make sure that we can actually build the thing, so, the, just because it satisfied the second law doesn't mean we can actually make it. It just means that it's possible that we can make this machine. All right.
17:56
So, again, any questions? Right. Now, one problem with this whole business, at least up until now, is that we don't have the ability to determine whether the cycle satisfies the second law or not, unless we find the efficiency twice, right? Once we have to find it for the actual cycle, and once we have to find it for the Carnot cycle. Wouldn't it be nice if there was some methodology whereby we could determine whether this particular cycle, and this would apply both to a heat engine as well
18:31
as a refrigeration cycle, wouldn't it be nice if we had one equation that we could just solve that shows us definitively whether the second law is satisfied or not. So, that's the next thing that we're going to do. Now, I might note that in your homework problems from this particular chapter, chapter 6, which we had actually just finished and are about to go into chapter 7 now, chapter 6 just wants you to use a method like I just described.
18:55
You know, using the Carnot cycle efficiency or coefficient of performance and comparing it to the actual efficiency or coefficient of performance. So, that's what chapter 6 is all about, so when you do those problems, you don't really want to use the method that I'm about to describe. You can, if you want to. it will be perfectly accurate, but not necessary. So, with this in mind, we'll begin chapter 7,
19:19
and chapter 7 deals with entropy.
19:27
Now, the first thing we're going to do has nothing to do with entropy. But that's what chapter 7 is all about, so, perhaps I should just say entropy.
19:38
Before we really get into entropy, I'm going to derive this equation that I implied and the equation is called the Clausius Inequality. So, the derivation of Clausius Inequality.
19:55
So, here what we have to do is we have to have a little bit of understanding of cycles. So, let me just start with a diagram, and let's first assume that we have a reversible cycle. Remember reversible cycle Remember reversible is another name for ideal.
20:17
But for a reversible cycle, okay, here's my cycle I've got a high temperature heat source, I'm transferring a certain amount of heat, QH, I am rejecting heat QL, into a low temperature sync, TL. And at the same time, I am producing some work. and because this is a reversible cycle, I'm just going to put REV instead of new work output. This is just the word associated with the reversible cycle, itself. So, I'm just going to put REV for that. All right, so this is my cycle.
20:56
I put a little circle thing in there. Now, for this cycle, what I'm interested in doing is looking at the integral mover the entire cycle or this term, delta Q over T. Now, I will note here that we saw this notation before with the interval sign with the circle on it. Again, that just means we're taking the interval over the entire cycle. So, we integrate over the entire cycle delta Q over T, but we would recognize that there's only two heat transfer terms,
21:27
and it's heat transfer from two reservoirs of known temperature, so we could write this if we wanted to, just as the integral of delta QH over TH, then minus the integral of delta QL over TL
21:48
and this can also then just be re-written as QH over TH,
21:57
and then minus QL over TL, okay. Now, I will note, put a little but here, but this is a reversible cycle, right? And because it's a reversible cycle, the ratio temperatures is equal to the ratio of heat transfers, right? That's supposed to be a Q. So, TH over TL equals QH over QL for a reversible cycle, all right, so that's something we know. And we can actually rearrange this, and as we rearrange this, we would not then, that QH over TH is going to equal QL over TL,
22:46
and then you can see that if we just simply plug this in above.
22:57
What we get is therefore that delta Q over T equals 0, right? QH over TH and QL over TL are the same, so, this equals 0. Okay. So, this applies for the reversible cycle only.
23:17
Okay. So, let's just remember this. And we'll come back to it in just a moment.
23:34
All right. Now what we want to do is we want to look at an irreversible cycle.
23:42
So, for an irreversible cycle, we would have the following.
23:54
Now, first of all, let's assume that we've got the same two reservoirs.
24:06
You know, the ones that are at TL and TH, and we've got the same heat input, QH.
24:14
So, the cycles are certainly similar, but since one's reversible and the other isn't, the reversible one is going to have more work output and less heat rejection at QL and the irreversible one is going to have to have less work output and therefore more heat rejection. That's just the nature of the irreversible cycle. So, let us just note that the thermodynamic efficiency for the reversible cycle, we know has to be greater than the thermodynamic efficiency
24:45
for the irreversible cycle, and again, I'm just going to use some abbreviations here. Let's also remember that the efficiency is 1 minus QL over QH, right?
24:59
So, I'll say but, efficiency is 1 minus QL over QH, so therefore we can just write 1 minus QL over QH for the reversible process. Therefore has to be greater than 1 minus QL over QH for the irreversible process. So, that should make sense, right? Now, again, the cycles are similar in that they both have the amount, the same amount of heat input QH, so if they got the same QH, the only way for 1 minus Q over QH reversible to be greater than 1 minus QL
25:43
over QH irreversible is for the irreversible heat transfer to be greater, I'm sorry, I should put here irreversible heat rejection, QL, has to be greater than the reversible QL. I mean, there's no other way, right? If they have the same QHs, then QL, irreversible has to be greater than QL reversible. Such that the QL over QH term is greater. Such that the 1 minus the QL over QH term is less than on the irreversible side than it is for the reversible side, right?
26:20
So, I mean, this absolutely has to be. All right? So, therefore what could we say? Well, let's go back to our, well, not go back to it. Let's note this. Sorry. Okay.
26:41
So, hence we would note the following. If I were to look at QH over TH minus QL, thank you, over TL for the irreversible process and I compare it to QH over TH minus QL over TL for the reversible process, well, is it less than or is it greater than? Well, it has to be less than, right? If QL irreversible is greater and all the other terms are the same, then we're subtracting a bigger number from the same QH over TH term and therefore the irreversible has to be less than the reversible.
27:28
So, I mean, we're just using logic based on what we know about the cycle. But, didn't we already show that QH over TH minus QL over TL for reversible cycle is 0?
27:43
So, we would note then that this equals 0, so therefore, QH over TH minus QL over TL has to be less than 0. Now, why is that important to us? Well, let's go back to this delta Q over T for the cycle. And we're specifically looking at the irreversible cycle, here. Okay. So, this is supposed to say now, all right.
28:21
So, we're going to look at this particular term and just like we saw with the reversible, we'll just show this as delta QH over TH minus the integral of delta QL over TL.
28:37
And again, this is for irreversible.
28:42
Or we can, you know, just show this as QH over TH minus QL over TL. But didn't we just show above that QH over TH minus TL over TL for the irreversible cycle is less than 0? I need to put a little irreversible in here, and another irreversible in here. Just to make sure all this makes sense. So, anyway, we showed that this is less than 0 from above. Right from there. Okay, so that tells me then that the integral of delta Q over T
29:29
for an irreversible cycle has to be less than 0.
29:34
And this, then, leads us to the Clausius Inequality. Quite frankly, we can only have a reversible or an irreversible cycle, right? So, can't we just say, therefore, for any cycle, can't we just note that the integral of delta Q over T has to be either less than or equal to 0? And we would note here that the less than 0 refers to the irreversible and the equal to 0 refers to the reversible cycle, right? Again, we can use the word real, we can use the word ideal,
30:19
that's for irreversible or reversible, respectively. I might also note here that, it can't be greater than 0, right? I mean, any process either has to be real or, at the very extreme case, ideal, right? It can't be impossible. So, I might also just note that if these are greater than 0,
30:39
then it's simply impossible.
30:44
Okay? So, this is the Clausius Inequality.
30:54
And it's this that we're going to be able to use to solve certain types of problems. Now, let me just note, there we are again. let me just note that even though I derive this for a heat engine cycle, it's just as applicable for a refrigeration cycle. Okay. I don't want to have to derive it all over again for refrigeration cycle, but it does apply to both type of cycles, if you're interested, okay?
31:35
All right, so, I don't want to have to do through a derivation all over again, but the derivation would be essentially identical except we'd use coefficient of performance instead of efficiency.
31:45
So, this is what I was looking for. Remember, the whole point of this was to find a single equation that would allow me to determine whether the cycle that we're analyzing could work or not. Is it possible or is it not? Does it satisfy the second law or does it not? So, this is what we need to utilize and what I want to do is just show by example how to use this. So, why don't we just use the same one?
32:15
It's the last one that we just did where you had the little diagram and you had a certain amount of heat and work and temperatures and all that, okay. So, it was the one where, what did I say? I said that is this machine reversible, irreversible or impossible and I just gave you that information. So, if we wanted to solve this particular example problem,
32:38
we don't have to separately find the thermodynamic efficiency of the Carnot cycle and the actual cycle, why don't we just solve for the integral of delta Q over T and see what happens? Well, we know that there's only two heat transfers, right? One is coming in at high temperature from the source at TH, the other is coming in, I'm sorry, is being rejected at a low temperature to the heat sync at the temperature TL, and we would just plug in our numbers and see if this less than or equal to 0.
33:17
If it is, it's possible. If not, well, it's impossible. So, well just plug in numbers. So, this is equal to, and again, we need the numbers from before. 325 kilojoules came in from 1000 degree Kelvin source and we're rejecting 125 kilojoules to a 400 degree Kelvin sync, and if we go through the mathematics, guess what, we get 0.0125. I'm just going to show the sign.
33:51
Just to make sure this is really obvious, it's positive.
33:58
That's impossible, right? Clausius Inequality tells us that this has to be less than or equal to 0, but we showed that it's greater than 0, so therefore, impossible.
34:11
So, this is just a slightly simpler way to solve this particular type of problem. And again, you can do the same thing for refrigeration cycles,
34:20
using exactly the same relationship. So, are there any questions then on this particular example problem? Okay. You could use it all the time, yeah. Yeah. Now, you know, I might note that there's quite a few problems where, you know, I might not only ask you if it's ideal, real or impossible. I mean, there's other problems where you specifically have to calculate heat transfer rates, power, work, that kind of stuff, in which case you're going to have to go
34:50
through and use your thermodynamic efficiencies and it really doesn't save you any time by using this method versus, you know, the other equation, but you know, just read the problems carefully and what you may want to do is just solve these problems both ways, just to see that it really does work both ways. You know, at least until you get comfortable using the Clausius Inequality. All right, so, questions? Yeah.
35:22
[ Inaudible question ] Yeah, you would have to have an issue with the sign. Heat is being added at high temperatures, so you would actually have a positive QL term and then you'd have a negative QH term. So, maybe what I said before isn't completely accurate, the equations are generally the same, but since the heat transfer is in the opposite direction, it's QL heat that's added, and it's QH that's the heat that's rejected. And as such, you would have to essentially swap those out.
35:56
So, any other questions? Good. Finally, we get to entropy.
36:08
Now, I could, if I wanted to, just tell you that entropy is a property and just force you to use it in the appropriate way. And I don't like to do that. I would much rather just derive the existence of the property entropy so at least you know that it's a real property, and then we can start figuring out how to use it. So, derivation of the property entropy. [ Writing ] So, where do we start?
36:56
What I'd like to do is consider two cycles that operate between state points 1 and 2. So, let's just do this. Let's say we have state point 1 and 2, and in one of those cycles we're going to go from 1 to 2 along a path that we'll just call Path B or let's call it Process B. And we'll return to make it a cycle along this other path that we'll just call Process A. Okay. So, let's look at two cycles, okay. So, the first cycle is where we use A and B. Okay.
37:45
But there's another cycle that we also want to look at. The second cycle is this one, where we're going to look at a different process going from 1 to 2, and instead of going from 1 to 2 along a path that we call Process B, let's look at a different one called Process C. So, this other cycle we're going to look at uses A and C to complete our cycle. So, remember, it's a cycle. We're starting at 1, we're going through one process until we get to 2 and then we're reversing
38:16
and we're going back on the other process. Now, in this case, let's note that both of these cycles are internally reversible. Okay. So, both cycles are internally reversible. You know, for some reason, this blue one is just working better. Just a different designed tip, isn't it? Oh, well. Thank you, anyway.
38:54
So, let's look at the cycles. Okay, so let's start with the first cycle. So, let's put a little 1 over here. And what I want to do is I want to look at the integral of delta Q over T for this entire cycle. All right. Now, there's two processes, right? There's the process we call A, and then there's the process that we call B. So, we go from 1 to 2 using process B, so integral from 1 to 2 of Delta Q over T and I'll just put a little B there to remind us that we're going through that process.
39:33
And again, with the understanding that this is internally reversible. Right. Both cycles are internally reversible. If they're internally reversible, that means both processes within the cycle have to be internally reversible, as well. All right, so, that's how we go from 1 to 2 and then we complete the cycle by going from 2 back to 1, and we're interested in delta Q over T as we return on Path A. Now, we know that this has to equal 0, right?
40:03
Because we know that the integral over a cycle of delta Q over T for an internally reversible cycle is equal to 0, right? So, again, it's because it's internally reversible that this equals 0. Now, let's go to the cycle 2. And again, we're interested in finding delta Q over t and again, we're going to travel from 1 to 2 and back to 1, but this time it's a different cycle, right? We're going to follow a different path from 1 to 2. Yeah, it's the same path from 2 back to 1 again, but still,
40:34
it's a different cycle. So, we go from integral 1 to 2 of delta Q over T using process C, and then we return from 2 to 1,
40:46
integral delta Q over T on process A again, and again, this equals 0, right? Both these cycles were considered to be internally reversible, so they're both equal to 0. And again, it's because they're internally reversible that this integral of delta Q over T is equal to 0, right? We just saw that when we were looking at the Clausius Inequality. Okay. Now, I might note that both equations are equal to 0, right? So, why don't I just equate them?
41:13
So, equate the two equations and you might note that if we equate the two equations, the integral 2 to 1 delta Q over T, A term cancels and therefore we get the following that the integral from 1 to 2 of delta Q over T following the path that we call process B has to equal the integral from 1 to 2 of delta Q over T following the path that we call Process C. Now, again, these are both internally reversible. I'll just put a little internal reversible here to remind us.
42:01
Okay. Now, believe it or not, I'm done.
42:10
What's left is for us to interpret this. Now, keep in mind that those two processes, B and C, were completely arbitrary. I mean, they're both internally reversible, yet they're still completely arbitrary processes. I could do this again. I can put on there, Process D, and then Process E and Process F and it doesn't matter. As long as they're internally reversible processes, if I take the integral from 1 to 2 of Delta Q over T for any of those irreversible processes, the values are all going
42:42
to be equal to each other. In other words, the integral of delta Q over T doesn't depend upon the process. I've used arbitrary processes, it only depends upon the state points. It all depends upon where I start at point 1 and where I end at state point 2. Doesn't that mean that it's a property? I mean, how could it not be a property? A property would be anything that is only a function of the thermodynamic states and not a function of the processes that occur.
43:15
So, it turns out that the integral delta Q over T for an internal reversible process is the same for any internally reversible process between any two state points and therefore it's a property. So, let's see if I can state that so it's in your notes. So, my choice of process from 1 to 2 was arbitrary. I could choose any internally reversible process, like D or E or F, et cetera, et cetera, et cetera, et cetera.
44:00
And the values of the integral delta Q over T for the internally reversible process would be the same. In other words, it only depends on the state, that is, 1 or 2, and not the process. you know, such as B, C, D, et cetera, right? And therefore, it is a property.
44:50
Guess what we call it? Well, you know. Entropy. So, it is called entropy,
44:59
and given the letter S. So, S is a thermodynamic property that we simply call entropy. Okay. Now, I can re-write the entropy in a variety of ways, I mean, quite frankly, this is really in its incremental form. We can show this as DS and that's delta Q over T for the internally irreversible process. Okay, so this is just one way that we can write it, we can also integrate it.
45:45
Well, maybe we'll do this. We write entropy in these ways. So, written in its differential form, and note that this is a D now, right? We don't use a delta. We only use delta if we're talking about something that's not a property, like heat transfer. Heat transfer is not a property, it's very dependent upon the path taken. So, we use a delta. But here, this is a D. This is an exact differential. This is a point function, as one might say in mathematics. So, it's a differential quantity.
46:18
So, this is DS. Or we can write it in this fashion. We can say that the entropy change is just going to be S2 minus S1 and this will be just the integral from 1 to 2 of the S, which is therefore just going to be the integral from 1 to 2 of delta Q over T for the internally reversible process. Okay? So, these are just a couple of different ways that we can actually write entropy.
46:50
Now, if you were thinking that entropy was going to be the result of some long, complicated derivation, well, clearly you're wrong. It's not anything that's that difficult to understand. That, well, it's not easy to conceptually understand. This tells us nothing about what entropy really means. Most people consider entropy to just be a measure of randomness or a measure of disorder in a particular process. Something that has a large change in entropy would be done
47:17
in a very disorderly fashion, in other words, with a lot of losses associated with it.
47:24
Something that has a very small change in entropy would be done in a relatively orderly fashion, one that has very few losses associated with it. In fact, what one can do is one can compare different processes that perform the same function. Let's say you're looking at two pumps and you wanted to know which pump would be the better one to purchase. You can do a calculation and figure out what the entropy change would be and it's very possible that one will have a much different entropy change
47:47
than the other. The one with the small entropy change is a well-designed pump with minimal losses. Where a large entropy change would be a poorly designed pump with significant losses, although it would probably be much more inexpensive and would be made overseas somewhere. But that's whole different story and we're not going to talk about that. But one could think about those things. Which is better? A cheap pump that has, you know, more losses
48:13
and is probably going to break down more quickly, but is going to be cheaper or a more expensive pump that's going to last for a lot longer. Again, these are questions you're going to have to ask yourselves as an engineer. And it really comes down to entropy.
48:24
Well, are you going to have to go through this integral every time you're solving a problem associated with entropy, and the answer is no. Once we're concluded that entropy is a property, we can quantify it using empirical relationships from laboratories and present it in the property tables. So, you may have noticed that the very last column in the water tables, the very last column in the refrigerant tables is entropy data. Now, actually it's specific entropy data,
48:57
and note that this entropy would have the units of heat transfer over temperature, so, these units would be kilojoules per Kelvin, or BTUs per Rankine, but we can also have specific entropy. So, the entropy data that's in our tables is always going to be intensive data. What I've derived is the extensive property entropy. So, maybe one thing to note is that entropy will have an intensive form, specific entropy.
49:41
And specific entropy is just lower case s. In fact, now you can see why I put these little wings on the s term above. That's actually capital letter S. And the specific entropy, like all of our other specific properties, is just the total entropy divided by the mass, and this data is in our tables.
50:05
So, turns out, no, you don't have to go through integral calculus every time you're trying to find an entropy change. All you need to do is go into the property tables and look up the values of entropy to see how the entropy changes for any one particular process. Okay. And then, one other thing that I suppose I should do here is write down the units. So, this is going to be kilojoules per Kelvin per kilogram, usually will say kilograms, I'm sorry, kilojoules per kilogram per Kelvin.
50:34
Or, it could be BTUs per pound mass per Rankine. So, those are the units. This is just or, I mean, you know, these are the units for the SI or the British system, the way they want to use. And then one last thing I'll note, even though the derivation of entropy was done, specifically by looking at a reversible cycle like a couple of different reversible cycles, the reality is that once we have concluded that it's a property, it's applicable for all processes.
51:07
It doesn't only apply to internally reversible processes, it doesn't only apply to cycles. This applies to all processes. Again, it's a property, right?
51:21
S is a property. And as such, it could be used in analysis of any process.
51:40
Right? Not just cycles, not just internally reversible processes.
51:53
Okay, so this is really important. Big letters, lots of exclamation marks. It's a property, we're going to have to figure out how to use it. Okay. I might also note that this applies to both open and closed systems as well.
52:21
Okay, so that's also important, okay? It'll apply to any process. Now, typically, the, well, we use it differently for different types of processes, and that's the thing that really we're going to be covering for the next couple of weeks. So, let me not talk about that in any more detail as for now. But, nonetheless, let's continue. There's one special case that I want to mention and just derive an expression for.
52:58
And this will be the special case of if we have an internally reversible process that's also isothermal. So, the internally reversible isothermal process.
53:15
Now, let's assume that the temperature is T0 for this process. And what I would like to do is I would like to find the entropy change.
53:29
So, my entropy change S2 minus S1 is you prefer, is just going to be the integral of delta Q over T. And the integral of delta Q over T, and again, this is for an internally reversible process.
53:51
Well, if we have a process that's isothermal, that means that the temperature's constant, and it's constant at the temperature of T sub 0, so can't we just pull the temperature out? And that gives us just the integral of delta Q. Oh, by the way, I forgot my 1s and 2s here as the limits of integration. So, if T0 is constant, after all, isothermal means constant temperature, it comes out of the integral, and what I am able to do then is integrate delta Q, well, that's just Q, right?
54:22
So, this is just going to be Q1, 2 over T0. So, if it turns out that you have an internally reversible process that is also isothermal, you don't actually have to look into the property tables and find entropy values at state 1 and state 2 and take the difference between them. All you have to do is know how much heat transfer is taking place and at what temperature that heat transfer is taking place, and you can just calculate the entropy change directly.
54:53
Now, again, this is just one special case. It's by no means the best case. Most of the problems we're going to look at are not going to be isothermal processes, quite frankly. They're just not. So, we're going to have to use the actual property data in the tables for the majority of our problems. Well, moving ahead.
55:20
We now have to get to what really is the important part of all this. I mean, quite frankly, all I've done so far is just tell you that entropy is a property, right? I've derived it and I find it's a property we're going to have to learn how to use it now. And that's what is going to begin, or that's what I am going to begin discussing now and this gets to what's called the increase in entropy principle.
55:54
Now, as it turns out, and this is something that will be derived. Not sure I'm going to get to the whole thing today. I may, we'll see. But what we're going to find is that for a real process, entropy always increases. Okay. That's why it's called the increase in entropy principle. Some of you have probably heard of this before. And some of you have probably been told, oh yeah, entropy is always increasing. And in fact, this is what social scientists use to predict the end of the world.
56:23
Because, if entropy is always increasing, eventually we are going to get to a point where it's going to increase, and remember, entropy is a measure of randomness or disorder as we keep moving through this world of ours, things become more and more disordered, until, you know, we're just not functional anymore. So, that's the end of the world for you. Please don't worry about that. if the end of the world comes, well, we're on the same boat and it's not a big boat and it's not a good boat.
56:53
But that's not something we should worry about. And plus, this is thermodynamics, right? We can't interpret thermodynamic relationships in terms of how our society works. So, yeah, people in social sciences can talk about the increase in entropy principle all day long and it really doesn't affect us at all. From our standpoint, we're trying to determine whether a process will work. I mean, that's what this is all about, right? We're trying to figure out whether the process
57:20
that we're trying to analyze is a possible process? Is it something we can actually design and build? We're also trying to compare two different processes that perform the same function, like two pumps, two turbines, whatever, so that we can make the best choice as engineers. So, this is really where the increase in entropy principle comes in. So, what we want to do is something that's actually similar to what was done earlier in deriving the existence of a property entropy.
57:45
Again, I'm going to look at two cycles.
57:48
So, let's look at two cycles.
57:53
But this time, the cycles are not going to be quite the same way they were. We're still going to start at points 1 and 2. So, I start at point 1 and go to point 2. We're still going to go from 1 to 2 using a process that we'll call B and then come back using a process that we call A. We're going to look at another cycle where we're going to go from 1 to 2, but this time, it's going to use a different process we're going to call C, and we're still going to come back along the path we call
58:28
process A. But the difference is this. That unlike previously where we were only looking at internally reversible processes, here we're not.
58:44
Here, we'll note that Processes A and B are internally reversible.
58:55
But Process C is not internally reversible. It's irreversible.
59:03
Or if you will, it's real or it's possible. How about that? Okay. So, it's slightly different than we had before, right? So, what I want to do is I want to look again at the two cycles. The first cycle is going to use Processes B and A, and we're still going to look at the integral of delta Q over T for this entire process. But, if we realize that Processes A and B are both internally reversible, then we know that the integral Delta Q over T for internally reversible process is equal to 0.
59:51
Right? Again, I can write that this is internally reversible, because all processes are internally reversible. Okay. So, this is equal to 0. So, we'll look at this, then. We'll say the integral from 1 to 2 of delta Q over TA, and then plus the integral from 2 to 1 of delta Q over TB and that would be equal to 0, right? However, let's look at the other process. So, let's look at the one that uses Processes C and A. Now, we're still going to look at the integral of delta Q over T,
1:00:41
but it is not a internally reversible cycle anymore, right? One of the processes in the cycle can be considered internally reversible, but the other one is specifically not. It's a real process and therefore is the interval delta Q over T equal 0? No, remember, we just looked at the Clausius Inequality. In fact, we had to derive the Clausius Inequality first, otherwise I couldn't have developed this increase in entropy principle, right? We know that this has to be less than 0,
1:01:11
because this is not internally reversible. This is the irreversible process. So, it's less than 0. So, therefore, we would note that the integral from 1 to 2 of delta Q over T, oh, I have my subscripts backward up here, don't I? If you would, just correct this, we went from 1 to 2 on Path B, and then we returned on Path A. So, if you could just switch those around up there.
1:01:43
Anyway, we note that the integral over the cycle for the irreversible process is less than 0. So, we added the integral of delta Q over T
1:01:52
as we went along Process C. and plus the integral from 2 to 1 of delta Q over T as we came back along process A. And this is not equal to 0, it's got to be less than 0, right? Above this is equal to 0, below here it's less than 0. And this is just a Clausius Inequality. And I have sloppy writing, but we can deal with that. All right, so now what I would like to do is I would actually like to subtract the latter equation from the former equation.
1:02:29
So, subtract latter from the former equation.
1:02:47
So, if we do so, we would note again that the delta Q over TA terms are going to cancel.
1:03:12
And we basically get the following. That the integral from 1 to 2 of delta Q over T as we follow path B or process B bas got to be greater than the integral from 1 to 2 as we look at delta Q over T following process C. Okay. So, this equation, I'm just going to call equation A. Now,
1:03:43
I'm not done with this. I've only got a minute left, less than a minute left, and quite frankly, it's going to take quite a bit more than a minute to finish this off. So, why don't we do this? Let's leave that as a, let's call it a thermodynamic cliffhanger, how about that [laughter]. And we will deal with this on Monday and eventually develop the increasing entropy principle. Are there any questions then, now, on today's lecture? You know, read this at home and let me just note lastly that,
1:04:11
if you just can't follow the logic, then don't worry about it, okay? What really counts is that you can use what will be the increasing entropy principle that I'm about to develop. Okay, maybe it will take you some time to get a better grasp on the whole idea of entropy and how it was derived, but that's not my emphasis. I'm not going to ask you to go through a derivation like this on one of your exams. You know, maybe if you guys were at Cal Tech or MIT or Stanford or something like that, well,
1:04:37
that's a totally different story, right? But, you're being taught there to be the next generation of academics. Here, we're trying to teach the next generation of engineers. So, I want you to know how to use entropy. If you can derive entropy, great, all the power to you. But that's not my emphasis. All right, so now, I'm really done for the day. Have a good weekend, everybody. [ Music ]
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