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Properties of Cyclic Thermodynamic Processes

Patrick Ford
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Hey guys, So we've solved lots of problems with PV diagrams, but one of the most important ones and common ones that you'll see is one where you start off with some position in the PV diagram, you'll do a sequence of steps and then you'll end right back where you started. And it basically just goes in a loop like this. So these are called cyclic thermodynamic processes and there's a couple of really important properties that you need to know about them. So in this video we're gonna dive right in, we're gonna solve this problem together and let's go ahead and check it out. So as a cyclic process is one where a system or a gas completes a sequence of steps, but it returns to its initial state. So that's what a cycle is. It returns back to the initial states. So there's two important properties here and rather than tell you want to show you. So let's just go ahead and take a look at our problem here. So we have this ideal gas and it goes through this process. We want to calculate the work that's done by the gas over the cycle. So the total amount of work done. So how do we do that? Well, let's see in part a if we want to calculate the total amount of work done, remember this is just a bunch of processes, sort of strung together. The way we calculate the network or the total work is you just have to add up all the work's done by all these individual processes, right, We've done that before. So basically if I have from A to B and B, two C and so on and so forth. The total work will just be the work done across all of these different steps. So the work done from A to B, B to C, C to D. And the work done from D. Back to A. Alright, so it might seem like there's a lot of terms we're gonna have to go calculate a lot of stuff, but let's just go ahead and refer to our table here just to make some of these things really easy. So, if you think about this, this sort of square here, a rectangle is really made up of two different types of steps. Aisa barrick and ice a volumetric. And what's easy about that is that we have the work equations for both of these. So it turns out what happens is remember in ice of violin metric processes, the work done is equal to zero. So that means that from A to B, there is no work done because there's no change in volume. Right? And then from C to D. This is also a vertical line. So there's no work done here. So, really, instead of four times, we really only have two. And these two are Sabbaruddin processes, which means we'll be able to use P delta v. Alright, so that just means that we're going to take this work equation and it simplifies to the work done from BBC, that's gonna be the pressure at B. It's the pressure here. So this is PB times the change in the volume. This is gonna be delta V. From B to C. Plus the work the pressure down the p. The pressure done here at D. Sorry, that's P. D. Here. And then times the change in the volume from D. Back to A. All right, so let's take a look at this. Basically, I'm just gonna start plugging in some numbers here at PBS 35. What's the change in the volume from B to C. We're going from 28 to 40 here. So what that means here is that this delta V. Is equal to 12 From here to here. Alright, so 28, you know the difference between 28 and 40? So this is gonna be 12. And then what about actually let's go ahead and plug it in, you know, 40 -28 just to do final minus initial. So then what about this PD here, this is gonna be 15 And then the change in the volume here. Now from D. back to a. Remember that this arrow here goes to the left, so my final minus initial is actually 28 -40. So it's gonna be 28 -40. Alright, so if you go ahead and work this out here, what you're gonna end up getting, is that the total amount of work done over this whole entire cycle is jewels. All right, So that's the answer. Let's move on to part B. Now, in part B. We want to calculate the area that is enclosed within the loop. The area that's enclosed is going to be this area that's sort of enclosed inside of this rectangle. So all I have to do is just find the area of this rectangle. How do I do that? Well, the area over here is just gonna be base times height. Right? That's the area of a rectangle. This is the area here. So base times height. What is my base? Well, my base here is 12 because that's the distance. B 2028 forties. This is 12. What about the heights? The height here is 15 minus 30 35 minus 15, which is 20. So the base here times the height is 20, sorry, 12 Times 20. And this is equal to 240 jewels as well. So, what happens here is you'll notice that these two numbers are the same. We calculated the work done. It's 2 40. The area is to 40 as well. And so that leads me to the first important property. You need to know which is that the total work done in a cyclic process is the area enclosed inside of the loop. The area enclosed inside here represents the total work done. And the reason for this is because if you think about what's happening for B. Two C, we have all of this area that's underneath the curve. But then from C to D. Right, well, sorry, from D. Back to a we're basically have this area underneath the curve here and this area here sort of just cancels itself out. Right. We have positive work done that, we have negative one work done. So this sort of effectively just cancels out and then what you end up with is just everything that's inside of the square or the rectangle or the loop in general, it actually works for any type of loop. So that's the sort of first process. The first property you need to know now what happens is in this process, this cycle here, we've gone in a clockwise loop, so that the top the work done at the top from BBC was positive and the work done at the bottom here was negative. So what that means here is that if the loop runs clockwise, the work done is positive. If we had run this loop in reverse, then what happens is that this work would have been negative and this one would have been positive. So your work done in the cycle would have been negative. So these are just some important properties you need to know if it's clockwise, it's going to be positive. If it's counterclockwise it's going to be negative. Alright. So now let's move on to part C. Now. So in parts that we want to calculate basically, we're giving all these heats from A to B, B to C, C to D and D. Today, we want to calculate the change in the internal energy for the cycle. So that just means we're going to calculate delta E. But now for the cycle, so how do we do that? Well, the only, the only equation we know for delta E is that the delta internal equals q minus W. But that was only for a single process. Well, it turns out what happens is instead of using delta equals q minus W. For each individual process, you're really over the cycle. Just gonna be adding a bunch of these, you know, q minus W. And then q minus W and q minus w. You're gonna be adding up all the delta ease for each one of these processes here. So you can actually sort of simplify this and you can actually just use delta E. For the cycle here is equal to Q. For the cycle minus the W. With the work done over the cycle, so you can do across multiple processes or cycles. It doesn't matter. Right? And so basically we can just use that delta E is equal to the total amount of heat transfer for the cycle, minus the total amount of work done over the cycle. And the reason this is helpful is because in a lot of problems you have to calculate the work done over the cycle first and then you'll have to use it in this equation here. So we already have with this delta, so that this w for the cycle is because we calculated that in part A. So in order to calculate the delta E. I just need to figure out what's the total amount of heat transfer over the cycle. Well, I'm actually told where all the heat transfers are for each of the different processes, the different legs or the steps here. So what I can do is just add up all of these things together here that will represent the total amount of heat transfer. So really this is just going to be 800 plus -600 -300 here, and then this is going to be minus. And then this is here. The work done is gonna be 240. Right? So this is this and this is this whole entire piece over here. When you go ahead and work this out, What you're gonna get is that the change in the internal energy over the cycle is equal to zero. When you plug in all these numbers you're going to get zero. And that's the second important property about cyclic processes that you need to understand because the internal energy depends only on basically where you are in the PV diagram and not how you got there and doesn't depend on the path that you take, the delta E equals zero over a cycle. There's no change in internal energy. As long as you start and end in the same place, it doesn't matter the path that you take, as long as you start and end in the same place, there's no change in internal energy. So what that means here is that there is no change in the internal energy. Then the total amount of heat transferred over the cycle is equal to the total amount of work done over the cycle. And that really just comes from this equation over here. If this is equal to zero, then these two things must equal each other. So that's another sort of important thing you need to know. Alright guys, that's it for this one. Let me know if you have any questions.