Heat Engines & PV Diagrams - Video Tutorials & Practice Problems
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1
concept
Calculating Work in Heat Engines Using PV Diagrams
Video duration:
7m
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Video transcript
Hey everyone. So in earlier videos we saw how to solve PV diagram problems. And in more recent videos, we've looked at heat engine problems. Well, some problems like the one that we're going to work out below. We'll give you a PV diagram for a heat engine and then ask you to calculate something like the total work done by the engine. So we're gonna put those ideas together in this video and I'm gonna show you how to calculate the work in heat engines by using the PV diagram. And what we're gonna see is we're gonna see the relationship between the PV diagram steps and the quantities that we use in our energy flow diagrams. So let's go ahead and get started. We're actually just gonna jump straight into our problem because there's a lot of stuff that we already know how to do. So let's check it out in this problem. We have two moles of a gas and it's basically undergoing a cyclic process. We were told here, is that heat is removed from this gas at constant pressure from A to B. Remember that just means that it's an ice a barrick process. Then from B to C, we're adding heat at constant volume. So this is ice, a volumetric and then finally, the gas is going to expand ice a thermally. So that just means it's going to sort of ride along this curve right here, go back to a and then the whole thing is going to repeat over again. So that's actually the first point I want to make here, remember that heat engines are cyclic processes. They start and end at the same point at the same initial state. But what you need to know here is that on PV diagrams, they will actually always run in clockwise loops. Notice how we've gone through a cycle like this, but the overall pattern of this loop is in the clockwise direction. Alright, so that's always going to be true for heat engines. Alright, so let's take a look at this problem here, because the first thing we want to calculate in part A is the heat transfer of each process. So we have three processes. We want to calculate Q four. So we have Q from A to B, and then we have que from B to C and then we have Q from C. Back to A again. Alright, so we spend a lot of time developing our equations in this table here for special processes. That's the that's basically we're going to do here. We're just gonna look at each process and then figure out which equation of Q that we use from this row here. So let's get started. So from A to B. We know this is an ice a barrick process, constant pressure. So we looked through the table and this is the equation we're gonna use we're just gonna plug and chug, we're gonna fly through this real quick, this is gonna be N C, p times delta T. We have two moles. Cp just comes from this table here, it's a mono atomic gas. So we're gonna use five halves are 5/2 times 8.314. And now, what's the change in temperature? We're going between two ice. Affirms 200 back to 100. So this is basically just going to be delta T. Equals negative 100. So you go ahead and work this out. What you're gonna get is negative 41, jewels. Now, for the second one, we're gonna use a very similar process. Q. B. C. Is equal to uh this is an ice a volumetric process. So we're gonna use this equation over here very much the same process. We're just gonna use a slightly different moller specific heat. This is gonna be too. Now we're not gonna use five halves, we're gonna use three halves for mono atomic. Right? So there's gonna be three halves times 8.314. And now, what's the delta T. Well here we're going from 100 back to 200 again, so this is going to be delta T equals 100. So you can use 100 here. And what you're gonna get is 20 for 90 for jules. Alright, now, finally these last process here, cue from C to a is a nicer thermal one. So we look through equations and Q equals w whatever the work done in this process is, that's also the heat transfer. Now do we use this equation then. Well actually we don't need to because in this problem we're told that this process here or this gas does 2300 joules of work. So this is Q. Which equals W. Which equals 2300 jewels. And there's no calculation necessary. Alright, so that's all there is to it. Right. That's just all the heat transfer is using a bunch of equations we've seen before. So let's now go on to part B. We want to calculate the work done by the heat engine. So that's going to be the work done by the engine. Now, how do we do this? Well, you may remember from our video on cyclic processes that whenever you have a cycle, the work that's done is going to be the area that is enclosed within the loop. So, in other words, the work done is gonna be the area. So how do we do that? Well, unfortunately, there's a couple of problems here because one we have this sort of curved ice a thermal process. So we can't use something like a triangle or anything like that. And we also don't have any of the values for pressure or volume or anything that we need to calculate the area. So, while the work done by the engine is actually the area that's in this shape, we just can't calculate it. So we're gonna have to use a different equation now in more recent videos, remember we've talked a lot about heat engines and we've been using these energy flow diagrams. And remember for heat engines, the work done is just equal to Q H minus Q. C. Right in the heat engines, some heat comes in from the hot reservoir. The engine produces some usable energy work and then spits out the waste heat to the cold reservoir and the work done is just the Q N minus the queue out. We just have to find those. So how do we do that? Which numbers are going to be? Are they gonna be? It's gonna be this one? Is it gonna be this one or this one? The answer is we're gonna actually have to use all of them. So remember that we're gonna use Q H minus Q. C. And Q. H. Is really just the heat that's added into the system. Remember we have, we have heat added into the system from the hot reservoir but in this problem here, we noticed and that in part a we actually calculated to cues that were positive numbers 24 94 2300 in both of these processes. What happens is that Q is flowing into the system right as the heat is flowing into the system here. So we have multiple heats that are going into the system. What's happening here is that qh is actually just the sum of all of the positive cues over the cycle, that's all there is to it. So this process adds some heat and this process also adds some heat. Both of them, make up both of them combined to make up the heat flowing in from the hot reservoir. So this is gonna be 24 94 plus 2300. And this is just gonna be 47 90 for jules. Alright, that's the heat that's flowing in now. We do the same exact thing for the Q. C. In this problem, we calculated one value of Q. That was negative in this. In this process. What happens is heat is flowing out of the system and it's flowing to the cold reservoir. So Q. C. Is the heat removed out of the system To the cold reservoir and it's going to be the sum of all negative cues over the cycle. So this is just gonna be negative 41, Now, just be very careful here because whenever we use these energy flow diagrams, we always want these to be positive numbers. So you may see a bunch of absolute value signs. So, basically what's happening is that this negative sign? When you calculate it just means the heat is being removed. But when we use it in our energy flow diagrams, we just use it as a positive number. So you may just see these written with a bunch of absolute value signs. All right. So that just means that the work done by the engine is just 94 minus 41 57 because we're using the absolute value here. All right. Just be really careful with the signs here. You don't want to subtract this because this already has a negative sign, and if you mess this up, you're gonna get the wrong answer, right? So just be very careful, make sure you're subtracting them and you're gonna get 637 jewels. Alright, So that's it for this one. Guys, let me know if you have any questions.
2
example
Example 1
Video duration:
5m
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Video transcript
Hey guys. So let's take a look at this example problem. We have a heat engine that's shown as a cycle in a PV diagram. Remember heat engines are always going to be cycles. And we want to do with part A is calculate the work that's done over a complete cycle in this process, right? So let's get started with part A. We want to calculate W number W pops up in a bunch of different equations for heat engines. But probably the most fundamental one is gonna be this QH minus QC. The work that's is the qh it's the heat that flows in minus the heat that comes out. That's just what comes from our energy flow diagrams, right? So if you have a hot reservoir connected to an engine connected to a cold reservoir here, this is your hot and cold. What happens is you have heat flowing in, that's qh, you have heat flowing out, that's QC. And then the work that's done is always going to be the difference between those two numbers, right? So this work done is always QH minus QC. All right. So that's where that equation comes from. So let's see if we can try to use it. So what happens here is that we don't actually know from the energy flow diagram. We don't know actually what Qh and QC are all we have in this problem are just the cues for some of the steps that are going on in this process here. So we're probably not going to use this equation over here. The other thing is that qh the heat transferred from the hot reservoir is actually what we're going to calculate in part B, right. So this is where we're going to calculate in part B. So we're gonna have to use a different sort of method to calculate the work. And remember that for, we actually do know how to do that from our discussion on cyclic processes. Remember that the work that's done over a cycle is always equal to the area that's enclosed within the loop. So the area enclosed gives you the work. So in other words, the work that's done is really just the area that is enclosed within this rectangle. And the good news is because it's a rectangle, there's a pretty easy way to calculate the area. So the work that's done here is really just gonna be base times height. So in other words, this is the base over here and this is gonna be the height and I have all the numbers, you know, the pressures and volumes and things like that. So I actually can calculate this work this way. So this work that's done is gonna be base times height. So in other words, we're gonna use um the base, which is the difference between 0.45 and 0.60. So this base here equals 0.0 0.15. So this is gonna be 0.15. And then this over here is gonna be a height which is equal to 300. So this H equals 300 there's no need to convert or anything because all this stuff is in si units, we have pascals and we'll have killer pascals or anything like that. So you're gonna just multiply this by 300. If you work this out, what you're gonna get is you're gonna get 45 jewels. So essentially what happens is the area inside of this is gonna be 45 jewels. That's the work that's done. All right. So that's the first part. Now let's move on to the second part here. And the second part, we want to calculate the total heat transfer from the hot reservoir. So in other words, what happens here is we've actually calculated in our energy flow diagram that this W here equals 45. So then if we have, if we say if this is 45 jewels, then basically what is Qh, that's what we're trying to find here? All right. So can we use this equation again now that we actually have what? Qh is, well, let's see if we have W equals QH minus QC. Then what happens is we can rerate this equation here to solve for QH. So what happens is I'm just gonna move the QC over to the other side. So in other words, the work that's done and I'm gonna drop all the absolute values because they can get kind of annoying here. The work plus the heat that flows to the coal reservoir is equal to the heat that flows to the hot reservoir. All right. So we have what W is, what is QC? So in other words, what is this sort of heat? What is the total amount of heat that flows out of uh this, this sort of cycle here? And to do that, we're gonna actually take a look at these two arrows over here. If you notice what happens in this sort of cycle here is that there's two places where heat is flowing out of the uh of the heat engine here. And that's indicated by these science, this Q equals negative 90 Q equals negative 25. What happens at a heat engine in general? Is that any time you go up or to the right? You are adding heat. So in other words, these two processes here are gonna be plus QS, these gonna be plus QS and then the other two processes which already labeled here are gonna be where heat flows out of the cycle. So in other words, what happens here, this QC is actually just equal to the sum of Q that flows out of the cycle here. So this is gonna be W plus, this is gonna be W plus the heat that flows out and that's gonna equal your qh. So in other words, what happens here, we have our W which is just 45 that's our 45 joules plus. Now, the heat that flows out. Now, remember we're always gonna use positive numbers here. So this negative sign just tells you that the heat is flowing out, but we plug it into our equations. It's gonna be a positive sign. So this is gonna be 90 plus and this is gonna be 25 over here. All right. So everything is positive and this is gonna equal your qh. So in other words, when you plug this in, what you're gonna get here is you're gonna get 100 and 60 jewels. All right. So basically what happens here is that the QC, right? Is the total amount of heat that flows out of the cycle, which we, if you just multi, you know, if you just add the 90 the 25 is gonna equal 100 and 15. And the work that we just calculated to Q H is gonna be 100 and 60. And that should make some sense. When you look at this uh energy flow diagram, you have 100 and 60 that flows in 100 and 15 that flows out. And the difference is basically what the heat can use as we work, which is the 45. All right. So all that stuff makes sense in terms of our numbers here. So your final answer is 100 and 60 jewels. Let me know if you guys have any questions.
3
Problem
Problem
A heat engine follows the cycle shown below. What is the thermal efficiency (%) of this engine?
A
58.8%
B
99.9%
C
14.3%
D
12.5%
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