The Otto Cycle - Online Tutor, Practice Problems & Exam Prep

The Otto cycle describes the idealized process in a 4-stroke internal combustion engine, consisting of six steps: intake, compression, ignition, expansion, depressurization, and exhaust. Key processes include isobaric, adiabatic, and isochoric transformations. Work done by the gas can be calculated by finding the area enclosed in a pressure-volume diagram, revealing that the gas performs work during expansion. Understanding these concepts is crucial for grasping thermodynamics and energy conversion in engines.

The Otto Cycle

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.

Finding the Compression Ratio

Hey guys. Let's do an example. 3 moles of an ideal diatomic gas which is treated with rigid molecular bonds undergoes a fallen auto cycle. How much heat's 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 2 triangles. It turns out this is not a very accurate way to estimate the work done, and when you're trying to calculate the efficiency, you need a very precise calculation of the work done. Okay. So we're going to have to do this a different way. So let's break this up into 4 steps. I'm going to call step 1 this step, which is the compression stroke. Step 2 this step, which is ignition, step 3, which is that first part of the expansion stroke, and step 4, which is like the second part of the expansion stroke. We can ignore all contributions due to these steps because they don't do any work since it's cyclic. Right? Since it starts and ends at 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 2 steps. They completely balance each other out. Okay. So let's look at step 1. Step 1 is adiabatic. Remember that's very important to keep in mind for the auto cycle which steps are adiabatic, which steps are isochoric, which steps are isothermal, which steps are isobaric. The law of thermodynamics says delta u is q we know that the first law of thermodynamics says Δu=q+w and since q for step 1 is 0, we can say that Δu for step 1 is just the work done in step 1. So if we can find the amount of work done sorry, if we can find the change in internal energy for step 1, we can find the work done in step 1. I don't actually want to encircle those. Let me highlight them. I want to circle the final answers. Okay. So let's find the change in internal energy for the ideal gas during step 1. Remember that the internal energy for any ideal gas is just f2nRT. We were told that this was an ideal diatomic gas. Diatomic gases either have degrees of freedom of 5 or degrees of freedom of 7. Because we're treating the molecular bonds as rigid, it's 5. Okay? This means that the change in internal energy for step 1 is only going to be due to the change in temperature because the number of moles isn't changing. Gas doesn't enter or leave the cylinder except for in the intake and the exhaust steps which we're ignoring, right? Those are these steps which we're ignoring. So during the cycle, gas doesn't enter or leave the cylinder. So the only thing that's changing is the temperature. So the question is what's the change in temperature for step 1? Well, if we want to relate pressure and volume like on a p v diagram to temperature, we need to use the ideal gas law. During step 1, 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 temperature is just the change of p times v during step 1 divided by nr. So how does p times v change in step 1? Okay. So Δt1= well what's the final pressure times the volume? The final pressure is 25, right, times 10 to the 5. Don't forget this times 10 to the 5. The final volume is 0.0005, right, during step 1, minus the initial pressure which was 1 times 10 to the 5, and the initial volume which was 0.000 5, only 3 zeros, divided by the number of moles which is 3, and the ideal gas constant which is 8.314. Plugging this into your calculator this works out to be 3 kelvin. Okay? So now that I know what the change in temperature is during step 1 I can plug that in to this equation which tells me how the internal energy changes in step 1 for a given temperature change. The number of moles is still 3. Right? The ideal gas constant still 8.314 and the temperature change is 3 kelvin. Plugging this into your calculator we get 187 joules as the change in internal energy for the first step. But we don't care about how the internal energy changes. All we care about is the amount of heat input, the amount of heat output, the work done, and the efficiency. Okay? So this tells us remember guys that first law tells us that the work done during step 1 is the same as the change in internal energy for step 1. So this is 187 joules. Okay, and we're going to hold on to that for later. Now let's look at step 2. Okay. This is step 2. This is an isochoric process, right? Isochoric processes mean that the change in volume for that step is 0. That also means that the work for that step is also 0. Okay? If the change in volume is 0, the work done is 0. Now looking at the first law of thermodynamics, if the work done is 0 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 isochoric 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 5 halves nr times delta t. This is because the only thing that leads to an internal energy change is a temperature change. In order to find how the temperature changes with the pressure and the volume that we have on the p v diagram, we have to look at the ideal gas law. Now for an isochoric process, right, the volume does not change. Only the pressure changes for an isochoric process, right? We're sitting here at this constant volume. So we can say that the left side only changes with pressure. There's only a change in pressure here. The volume is a constant. Δt2= v2 Δp_2∗ v_2∗Δp2+/r right? This volume that this isochoric process occurs at is just this. Right? That minimum volume for the gas. And the change in pressure is from 25 to 170. That's the increase in pressure that we're undergoing in this process. And this is divided by the number of moles which is 3 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 2. That leads us directly to how the internal energy changes in step 2. The number of moles is 3. Diatomic gas constant is 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 of in, sorry, the amount of the change of internal energy for the second step is 1815 joules. 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 in internal energy which is 81815 joules. Okay? So for step 1 and for step 2, we know how much sheets transferred and how much work is transferred. All we have to do is match this for steps 2 and steps 3 sorry, for steps 3 and steps 4. But remember step 3 is another adiabatic step. We already know all the requirements and all the equations for the adiabatic process because we just used them in step 1. We know that this means that for step 3 the heat transfer is 0. Okay. And we know that this means for step 3 that the change in temperature can be given as the change in the pressure times the volume for step 3 over nr. This is the same equation that we got from the ideal gas law in step 1 which is another adiabatic process. All we have to do now is plug this in. Okay. The final pressure after step 3 is 7 times 10 to the 5. The final volume after step 3 is 0.0005. The initial pressure of step 3 is 170 times 10 to the 5 pascals and the initial volume is our very small volume 0.00005. The number of_mo}}">

Here’s what students ask on this topic:

What are the four strokes in an Otto cycle engine?

The four strokes in an Otto cycle engine are:

1. Intake Stroke: The fuel-air mixture is drawn into the cylinder at constant pressure.

2. Compression Stroke: The piston compresses the fuel-air mixture, increasing its pressure and temperature. This process is adiabatic, meaning no heat is exchanged.

3. Power (Expansion) Stroke: Ignition occurs at constant volume, causing a rapid increase in pressure. The high-pressure gases push the piston down, performing work. This is also an adiabatic process.

4. Exhaust Stroke: The piston expels the burnt gases at constant pressure, completing the cycle.

How is work calculated in the Otto cycle?

Work in the Otto cycle is calculated by finding the area enclosed by the cycle on a pressure-volume (PV) diagram. The work done by the gas during the cycle can be determined by summing the areas of the different processes:

$W=W{1}_{\to }+W{2}_{\to }+W{3}_{\to }+W{4}_{\to }$

Since isochoric processes (constant volume) do no work, only the isobaric (constant pressure) and adiabatic (no heat exchange) processes contribute to the work done. The net work is the difference between the work done during expansion and compression.

What is the significance of the Otto cycle in thermodynamics?

The Otto cycle is significant in thermodynamics as it models the idealized process of a 4-stroke internal combustion engine, commonly found in gasoline engines. It helps in understanding the principles of energy conversion, efficiency, and work done by the engine. By analyzing the Otto cycle, engineers can optimize engine performance, improve fuel efficiency, and reduce emissions. The cycle also illustrates key thermodynamic processes such as isobaric, adiabatic, and isochoric transformations, making it a fundamental concept in the study of thermodynamics and heat engines.

What are the key processes in the Otto cycle?

The key processes in the Otto cycle are:

1. Isobaric Process: Occurs during the intake and exhaust strokes, where the pressure remains constant.

2. Adiabatic Process: Occurs during the compression and expansion strokes, where no heat is exchanged with the surroundings.

3. Isochoric Process: Occurs during ignition and depressurization, where the volume remains constant while the pressure changes.

These processes are idealized to simplify the analysis of the engine's performance and efficiency.

How does the Otto cycle differ from the actual operation of a gasoline engine?

The Otto cycle is an idealized model that simplifies the complex processes occurring in a real gasoline engine. In reality, the processes are not perfectly isobaric, adiabatic, or isochoric. For example, ignition does not occur instantaneously, and there are heat losses and frictional effects that are not accounted for in the idealized cycle. Additionally, the actual engine operates with varying efficiencies and under different load conditions, which are not considered in the Otto cycle. Despite these differences, the Otto cycle provides a useful framework for understanding the fundamental principles of internal combustion engines.