Hey guys, in this video we're going to talk about the Otto cycle, which is the theoretic cycle that the gas undergoes in a 4-stroke internal combustion engine. Alright, let's get to it. Now remember guys that the common gasoline engine in cars is a 4-stroke internal combustion engine. Okay. Now those 4 strokes are the intake stroke where fuel-air mixture is pulled into the cylinder, the compression stroke where the piston compresses that fuel-air mixture makes it very very dense, very high pressure. Okay. In between these compression and expansion strokes is ignition. Okay so ignition fits right up in here. Ignition is the injection of heat by a spark plug into that very very dense compressed fuel-air mixture that ignites it, causes it to change chemically into carbon dioxide and water, and release a bunch of free energy. Okay. 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 Otto cycle, as I said, is the theoretic cycle. Okay. It's a very idealized cycle that the gas is supposed to undergo, okay, in a 4-stroke internal combustion engine. In reality, it doesn't happen quite like the Otto cycle but the Otto cycle is a close theoretic explanation of it. Okay? Now the Otto cycle is given on a PV diagram above me and it occurs in 6 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 3 is the ignition stroke, and this actually happens at a constant volume because ignition is intended theoretically to take place instantaneously. Instantly all of that air at fuel-air is converted into exhaust, and the pressure dramatically rises before the piston can move. Okay. Step 4 is that expansion stroke. Just like the compression strokes, so both of these, they both occur very very rapidly. Okay? Now step 5 is actually kind of like the second half of the expansion stroke. The expansion stroke isn't technically finished until step 5 is done. Step 5 allows the depressurization of the exhaust by heat leaving the cylinder. When the heat leaves the exhaust, the exhaust drops in pressure. Okay. And this also occurs at a constant volume like the ignition stroke. And finally step 6 is the exhaust stroke, which also occurs at a constant pressure. Okay? So in the Otto cycle, these idealized theoretic steps are step 1 being at a constant pressure, so it's an isobaric expansion. Okay? Step 2, remember the compression and the expansion strokes occur very very rapidly. The pistons are moving very quickly, much too quick for heat to enter or leave the cylinder. So this is an adiabatic compression. Okay step 3 is isochoric. It happens at a constant volume. Okay? It's an isochoric pressurization. The pressures increase at a constant volume. Okay. Step 4, the expansion stroke. Just like I said about the compression stroke, both of them occur very very rapidly so this is also adiabatic. Okay? And step 5, which is sort of like that second half of the expansion stroke, allows heat to leave at isochoric. As the heat leaves, the pressure drops. So it's a depressurization. Okay? And finally, step 6, the exhaust stroke, where exhaust is leaving against no resistance. This is isobaric. It occurs at just this initial pressure. Okay? So these are the idealized steps. Isobaric, adiabatic, isochoric, adiabatic, isochoric, isobaric. Okay let's do an example. Estimate how much work is done by the gas in the Otto cycle shown in the following figure. Is this work done 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 2 steps that contribute to work. Notice that this step, the isochoric step, and this step, the other isochoric step, don't do any work okay because they are at a constant volume and isochoric processes never do any work. Now these 2 isobaric processes do contribute to the work, but they contribute the same amount of work. Right? 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 in opposite directions, the signs of the work are opposite so they cancel each other out. So really, the only work done is by the adiabatic changes; the compression and expansion strokes. Okay so the red step can be thought of like a triangle. It can 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, and it starts at 0.0005 and ends at 0.0005. Okay. And the triangle starts here at 7 and goes up to 170, and the base of the rectangle is at 0, just at the bottom. Alright? And we can approximate the green process going to the left as that same shape. Okay? A triangle sitting on top of a rectangle. Okay. Now the volume numbers are going to be the same, but the pressure numbers are going to be different. Okay? Up here at the top of the triangle is 25. Here at the bottom of the triangle is 1, and obviously, it also starts at 0. Okay, so all we have to do is add the area of the triangle and 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 going to put a little prime on the rectangle height just to indicate that it's different. If you take the distance or sorry the difference here on a calculator, you'll find that it's 0.00045, so this becomes one half of 0.00045. The height of the triangle is 163, right? 170 minus 7. But don't forget that this is times 10 to the 5 pascals. We need that. The volume is just cubic meters, so there's no times 10 to the anything here, right? Plus 0.00045, the same base for the rectangle and the triangle, but the height is 7. Right? Obviously times 10 to the 5 pascals. Plugging this into a calculator, we get 3,982.5 joules for the area of the red loop. Okay? That red process. Let me minimize myself for the second one. 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 are the pressure numbers because the volume numbers are identical. This is a change in pressure of 24. Right? That's the height. The volume numbers are the same, and this is a change clearly of 1. Plugging this into a calculator, we get 585 joules. So now the work is going to be the sum of each of these but with their appropriate signs. That's very very important. The sign is very important. Now let's look at the red process. The red process is to the right so that work is negative. Okay. So we get a negative sign in front of the first thing, the work, the sorry, the green process is to the left so that work is positive. Okay so we get a positive calculator we get minus 3,397.5 joules. So that is how much work is done in this Otto cycle by the gas. 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 undergo a cycle that allows it to release work.
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The Otto Cycle
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