23. The Second Law of Thermodynamics
The Otto Cycle
The Otto Cycle
Was this helpful?
Hey, guys. In this video, we're gonna talk about the auto cycle, which is the theoretic cycle that the gas undergoes in a four stroke internal combustion engine. All right, let's get to it. Now, remember, guys that the common gasoline engine cars is a four stroke internal combustion engine. Okay, now those four strokes are the intake stroke, where fuel air mixtures pulled into the cylinder, the compression stroke where the piston compresses. That fuel air mixture makes it very, very dense. Very high pressure in between these compression and expansion strokes is ignition. Ignition fits right up in here. Ignition is the injection of heat by a spark plug into that very, very dense, compressed fueler mixture that ignites it, causes it to change chemically into carbon dioxide and water, and released a bunch of free energy. That free energy causes the expansion stroke where the piston is pushed away from that gas, allowing the exhaust to expand okay and release all that energy into the piston. And finally, there's the exhaust stroke, where the piston pushes all of that exhaust all that burnt gasoline and air through the exhaust valve and out of the piston, allowing the intake stroke to start again. Okay, Now, the auto cycle, as I said, is the theoretic cycle. Okay? It's a very ideal ized cycle that the gas is supposed to undergo. Okay, in a four stroke internal combustion engine. In reality, it doesn't happen quite like the auto cycle, but the auto cycles a close theoretic explanation of it. Okay, now the auto cycle is given on a PV diagram above me, and it occurs in six steps. The first step is the intake stroke. Okay, where gas is pulled in at a constant pressure. The second stroke is the compression stroke. Okay, which is compressed very, very rapidly. Okay. The piston is moving very quickly in the cylinder during these strokes. Okay, Step three is the ignition stroke. And this actually happens at a constant volume because ignitions intended theoretically to take place instantaneously instantly. All of that era, a few layers converted into exhaust, and the pressure dramatically rises before the piston can move. Okay, step four. Is that expansion stroke? Just like the compression stroke. So both of these, they both occur very, very rapidly. Okay, Now, step five is actually kind of like the second half of the expansion stroke. The expansion stroke isn't technically finished until step five is done. Step five allows a depressed aeration of the depressurization of the exhaust by heat leaving the cylinder. When the heat leaves the exhaust, the exhaust drops in pressure. And this also occurs at a constant volume, like the ignition stroke. And finally, Step six is the exhaust stroke, which also occurs at a constant pressure. Okay, so in the auto cycle, these ideal ized theoretic steps are step one being at a constant pressure. So it's an isil. Berrick expansion. Okay, step to remember the compression and the expansion strokes occur very, very rapidly. The piss is moving very quickly, much too quick for heat to enter or leave the cylinder. So this is an idea, Batic Compression. Okay, Step three is is a coric. It happens at this constant volume. Okay, It's an ISO coric pressurization. The pressures increase at constant volume. Okay, step for the expansion stroke. Just like I said about the compression stroke, both of them occur very, very rapidly. So this is also a DEA Batic. Okay. And step five, which is sort of like that second half of the expansion Stroke allows heat toe leave at a constant volume. So this is also ice. Oh, coric. As the heat leaves, the pressure drops. So it's a depressurization. Okay. And finally Step six the exhaust stroke where exhaust is leaving against no resistance. This is ISO barrack. It occurs at just this initial pressure. Okay, so these air the idealize step I so barrick idiomatic s a cork. Idiomatic s a cork I so barrick. Okay, let's do an example Estimate how much work is done by the gas in the auto cycle shown in the following figure Is this worked on on or by the gas estimate the work done by finding the area enclosed by the cycle Now, normally we would find the area enclosed by the cycle but the shape is weird for that. The shape is kind of like this. Okay, where this height is larger than this height. I don't know what the area of that shape is, but we can break this down into the two steps that contribute tau work. Notice that this step the ice, of course, step in this step the other I support step Don't do any work, okay? Because they are at a constant volume, and I support processes. Never do any work. Now. These two is a barbaric processes do contribute to the work, but they contribute the same amount of work. They're both horizontal lines, right on top of each other in opposite directions. They contribute the same magnitude of the work. But since they're opposite directions, the signs of the worker opposite. So they canceled. So really, the only thing that contributes to the work is this step and this step so we can find each of them independently and then add them together. Okay, so the red step can be thought of like a triangle. It could be approximated as a triangle, sitting on top of a rectangle. Okay, so you can see it looks like a triangle sitting on top of a rectangle. That is a shape that we can absolutely find the area for. Okay. And it starts at 00005 and ends up 0005 Okay. And the triangle starts here at seven and goes up to 170 the base of the rectangles at zero. Just the bottom and we can approximate the green process going to the left as that same shape. Okay. A triangle sitting on top. A rectangle. Okay, now the volume numbers, they're going to be the same. But the pressure numbers are going to be different up here. At the top of the triangle is 25. Here at the bottom of the triangle is one. And obviously it also starts at zero. Okay, so all we have to do is add the area of the triangle in the rectangle for both of these figures to find the two works done by each of these processes. So the area is going to be that of a triangle, plus that of a rectangle. Now, they both have the same base, but they have different heights. So I'm gonna put a little prime on the rectangle height. Just toe. Indicate that it's different if you take the distance or sorry, the difference here on a calculator, you'll find that 00045 So this becomes one half 00045 The height of the triangle is 163 1 70 minus seven. But don't forget that this is times 10 to the five Pascal's. We need that the volume is just cubic meters, so there's no times 10 to the anything here right plus 0.45 The same base for the rectangle and the triangle. But the height is seven, right? Obviously times 10 to the five past scouts Plugging this into a calculator, we get 3982 5 jewels for the area of the red loop. Okay, that red process. Minimize myself for the second for the green process. It's going to be the same exact equation, right? The area of the triangle, plus the area of the rectangle. The only thing that's going to change our the pressure numbers because the volume numbers are identical. This is a change in pressure of 24. Okay, right, that's the height. The volume number is the same, and this is a change clearly of one. Plugging this into a calculator. We get 585 jewels. So now the work is going to be the some of each of these, but with their appropriate sign, that's very, very important. The sign is very important. Now let's look at the red process, the red processes to the right so that work is negative. Okay, so we get a negative sign in front of the first thing. The work that sorry, the green processes to the left, so that work is positive. Okay, so we get a positive sign here in between those works now plugging this into a calculator, we get minus 3397.5 jewels. So that is how much work is done. Bye. Sorry is done in this auto cycle. Now, since it's negative, this is by the gas, right? That's very important. This is by the gas because it's negative. And obviously the work done by the gas should be negative. Because this is an engine. The gas should under go a cycle that allows it to release work that allows the engine to release energy into the system. Alright, guys, that wraps up the idealize auto cycle as the cycle of the gas undergoes any four pissed Sorry for stroke piston engine. Thanks for watching guys
Finding the Compression Ratio
Was this helpful?
Hey, guys, let's do an example. Three moles of an ideal die atomic gas, which is treated with rigid molecular bonds, undergoes have fallen auto cycle. How much heats? Input? How much heat is output? What's the work done by the engine? Okay. And finally, what's the efficiency of this engine now? What's very important is do not estimate the work by finding the area enclosed by this cycle. We saw a problem before where you can estimate the area of the cycle by approximating two triangles. It turns out this is not a very accurate way to estimate the work done on when you're trying to calculate the efficiency, you need a very precise calculation of the work done. Okay, so we're gonna have to do this a different way. So let's break this up into four steps. I'm gonna call step one this step, which is the compression stroke step to this step, which is ignition Step three, which is that first part of the the expansion stroke and Step four, which is like the second part of the expansion stroke. We could ignore all contributions due to these steps because they don't do any work since its cyclic right, Since it starts and ends in the same position, there's no change in internal energy, which means by the first law of thermodynamics there's no heat exchange either. So nothing is going on in those two steps. They completely balanced each other out. Okay, so let's look at step one. Step one is a DEA Batic. Remember, that's very important to keep in mind for the auto cycle, which steps are idiomatic. Which steps or is a cork? Which steps are is a thermal. Which steps are is a Barrick, etcetera. Idiomatic by definition, means that there is no heat transfer for this step, so Q 10 Okay, now we know that the first law of thermodynamics says Delta U is Q plus W. And since Q four step one is zero, we can say that Delta you for step one is just the work done in step one. So if we confined the amount of work done, sorry if we can find the change in internal energy for step one, we confined the work done in step one. I don't actually want to encircle those. Let me highlight them. I want to serve all the final answers. So let's find the changing internal energy for the ideal gas during Step one. Rumor that the internal energy for any ideal gas is just f over to in our tea way were told that this was an ideal die atomic gas. Die. Atomic gas is either habit degrees of freedom of five or degrees of freedom of seven. Because we're treating the molecular bonds is rigid. It's five. Okay, this means that the change in internal energy for step one is on Lee going to be due to the change in temperature because the number of moles isn't changing, gas doesn't into relieve the cylinder, except for in the intake and the exhaust steps which were ignoring right those air. These steps which were ignoring so during the cycle gas doesn't enter relieve the piston. So the only thing that's changing is the temperature. So the question is, what's the change in temperature for step one? Well, if we want to relate pressure and volume like on a PV diagram to temperature, we need to use the ideal gas law during Step one. Both the pressure and the volume are changing, and that's responsible for producing the change in temperature. So the entirety of the left side is changing and that leads to a change in the temperature. So we can say that the change in the temperature is just the change of P times V during step one, divided by in our So how does P times v change in step one? Okay, so Delta t one is going to be Well, what's the final pressure times the volume, The final pressures 25 right times 10 to the five. Don't forget this times 10 to the five. The final volume is 0. during step one, minus the initial pressure, which was one times 10 to the five. And the initial volume, which was 50.5 Only three zeroes divided by the number of moles, which is three. And the Idol gas constant, which is 8.314 Plugging this into your calculator. This works out to be three Kelvin. Okay, so now that I know what the change in temperature is during step one, I can plug that into this equation, which tells me how the internal energy changes in step one for a given temperature change. The number of moles is still three, right? Died of gas. Constant still 8.314 And the temperature change is three. Kelvin, plugging this into your calculator, we get 187 jewels as the change in internal energy for the first step. But we don't care about how the internal energy cares. I mean changes. All we care about is the amount of heat input, the amount of heat output, the work done in the efficiency. Okay, so this tells us, remember, guys, that first law tells us that the work done during step one is the same as the change of internal energy for step one. So this is 187 jewels, okay? And we're gonna hold on to that for later. Now, let's look at step two. Oh, okay. This is Step two. This is an YSL coric process, right? I so cork processes mean that the change in volume for that step is zero. That also means that the work for that step is also zero. Okay, If the change in volume is zero, the work done. Zero now, looking at the first law of thermodynamics. If the work done is zero. Then the change in internal energy for that step is just the heat transferred for that step. That's what the ideal. Sorry. That's what the first law of thermodynamics tells us about this isil cork process. And finally, don't forget that any change in internal energy for this cycle since gas isn't coming in or leaving during the cycle, it's always going to be five halves in our times. Delta T. This is because the only thing that leads to internal energy change is a temperature change in order to find how the temperature changes with the pressure in the volume that we have on the PV diagram, we have to look at the ideal gas law Now for an YSL Coric process, right? The volume does not change on Lee. The pressure changes for an is a core process. We're sitting here at this constant volume so we can say that the left side Onley changes with pressure. There's only a change in pressure here. The volume is a constant and the right half changes with temperature. What? So that means Delta T two is V two Delta P two over and are right this volume that this is a court process occurs that is just this right, that minimum volume for the gas and the change in pressure is from 25 to 70. That's the increasing pressure that we're undergoing in this process. And this is divided by the number of moles, which is three times 8.314 Let me minimize myself here, which, if you plug into your calculator, is equivalent to 29 1 Kelvin. Okay, so now we know how the temperature changes in step two that leads us directly. Remember our equation for change in internal energy that leads us directly to how the internal energy changes in step to the number of moles is three died of gas, Constance 8.314 The change in temperature is 29.1, right? We just found that out. So plugging this into a calculator tells us the amount off and sorry, the amount of the change of internal energy for the second step is 1815 jewels. But we don't care about that. All we care about is how much heat is transferred. And don't forget our first law requirement for this step tells us that the heat transfer in this step is equivalent to the change of internal energy, which is 8 18 15 chores. Okay, so four, step one. And for step two, we know how much sheets transferred and how much work is transferred. All we have to do is match this for steps to and steps. Three. Sorry for steps, three steps, four. But remember, Step three is another 80 about it. Step. We already know all the requirements and all the equations for the idiomatic process, because we just use them in step one. We know that this means that for step three, the heat transfer zero okay. And we know what this means for Step three that the change in temperature can be given as the change in the pressure times the volume for step three over in our This is the same equation that we got from the ideal gas law in step one, which is another idiomatic process. All we have to do now is plug this in. Okay. The final pressure after step three is seven times 10 to the five. The final volume after step 30005 The initial pressure of Step three is 170 times 10 to the five Pascal and the initial volume is our very small volume. 00005 The number of moles is always the same. Three and the idea gas constant, obviously, always the same. Plugging this into a calculator. We're gonna get negative 20 Kelvin as the result of the temperature change. And remember our first law requirements for and a dramatic process. If you just check your notes, it's that the amount of that sorry if you change in internal energy, equals the amount of work done. Okay, So if we find the change in internal energy, which we know is just five halves in our times, a change in temperature For that step we'll find how much work is done. The number of moles is just three idol gas constant 8.314 right. We just found the temperature change was negative. 20. Plugging this into a calculator Negative jewels. Which tells us that the work done in this process, which is the same as a change of internal energy by the first Law, is negative. 1247 Jules. Okay, finally, we have stepped four. The last step that we have to look at Step four is also an ISO coric step. Just like step to waas. So we can use all the same equations in step two. We know immediately that this means that the work done in step four is going to be zero. So we have that. We know that the temperature change in step four is just going to be the volume in step four times to change in pressure of step four over in our this is the same equation that we used in step two, which we got from the ideal gas law for I support process. In this case, we're at the large volume Our final pressure is one times 10 to the five, right. That minimum pressure that we started on the ending point. Sorry. The starting point, which is the same as the ending point. The initial pressure okay, is seven times 10 to the five. Okay. The number of moles we know is three. And the ideal gas constant is always 8.314 plugging this into a calculator. We get negative. 12 Kelvin as our temperature change now, just like we did in every other step, we can plug in this temperature change into the change of internal energy. Right? The change in internal energy for any gas process is going to be due to the change in temperature because the number of bowls doesn't change in this cyclic process, the number of moles is three, right? As it's always been, 8.314 Is ideal gas constant? We just found the change in temperature is negative. 12. So this is going to come out to be negative. 748. Jules, when you plug it into a calculator, remember, guys, we just found this requirement by the first law Foreign is a court processes that says that the heat transferred during that step is equivalent to the change in internal energy during that step. So now we know the heat transfer during that step. Okay, so now we found how much work is done. How much heat is transferred in every step. We can answer the questions that we were given. How much heat is into the system how much heat is out of the system. What's the work done by the system, Right? And what is the efficiency? Those are our four questions. Well, we only had two steps. That heat was transferred Step two and Step three. The two is a cork steps because the step one Step three Sorry, Step two and Step four are the two is a cork steps. Step one and Step three were both idiomatic, so neither of them had a heat transfer. Step two had a positive heat. That's the only instance where heat is entering the system. So the heat input is just the heat for Step two, which is 1 815 jewels. Okay, now the Onley heat transfer that was out of the system that was leaving the system was the heat transfer and Step four. And this is a magnitude of 748 jewels. That's how much he left the system during this cycle. The work done was sorry. There was work done in both idiomatic steps. Step one and Step three, which is going to be, um, 187 jewels minus 1 247 jewels, which works out to be when you plug into your calculator. Negative. 10 60 jewels. Okay, but remember, what is this? This is actually the work done by the gas. This is not the work done by the engine. The work done by the engine is however much energy is released by the gas. If you concede e, since this is negative, 1060 jewels of work left the gas. That's how much work was produced by the engine. So the work output is positive. 1060 Jules. Okay. And finally, the efficiency is just the work out divided by the heat. In how much work do you get out for? How much heat do you put in? So this is 10. 60/18 15 right? Which was the amount of work output in the amount of heat input. And this works out to be 0.584 So the efficiency of the engine is 58. percent. Okay, this is a long process, but this is how you solve p V diagrams for these different properties of a heat engine. Alright, guys, that wraps up this problem. Thanks for watching
Additional resources for The Otto Cycle