Clutch Prep is now a part of Pearson
Ch 24: The Second Law of ThermodynamicsWorksheetSee all chapters

The Otto Cycle

See all sections
Sections
Heat Engines and the Second Law of Thermodynamics
Heat Engines & PV Diagrams
The Otto Cycle
The Carnot Cycle
Refrigerators
Entropy and the Second Law of Thermodynamics
Entropy Equations for Special Processes
Statistical Interpretation of Entropy

Concept #1: The Otto Cycle

Transcript

Hey guys, in this video we're going to talk about the auto cycle which is the theoretic cycle that the gas undergoes in a four stroke internal combustion engine alright let's get to it now remember guys that the common gasoline engine in cars is a four stroke internal combustion engine. Now those four 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 in between these compression and expansion strokes is ignition 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 that free energy causes the expansion stroke where the piston is pushed away from that gas allowing the exhaust to expand 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 now the auto cycle as I said is the theoretic cycle it's a very idealized cycle that the gas is supposed to undergo in a four stroke internal combustion engine in reality it doesn't happen quite like the auto cycle but the auto cycle is a closed theoretic explanation of it. Now the auto cycle is given on a P.V diagram above me and it occurs in six steps the first step is the intake stroke where gas is pulled in at a constant pressure the second stroke is the compression stroke. Which is compressed very very rapidly the piston is moving very quickly in the cylinder during these strokes 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 air fuel is converted into exhaust and the pressure dramatically rises before the piston can move step four is that expansion stroke just like the compression stroke so both of these they both occur very very rapidly. Now step five is actually kind of like the second half of the expansion stroke the expansion stroke isn't technically finish until step five is done step five allows a depressurization 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 so in the auto cycle these idealized theoretic steps are step one being at a constant pressure so it's an isobaric expansion step two remember the compression and the expansion strokes occur very very rapidly the piston is moving very quickly much too quick for heat to enter or leave the cylinder so this is an adiabatic

compression step three is isochoric it happens at this constant volume it's an isochoric pressurization the pressures increase at a constant volume, step four the expansion stroke just like I said about the compression stroke both of them occur very very rapidly so this is also adiabatic and step five which is sort of like that second half of the expanssion stroke allows heat to leave at a constant volume so this is also isochoric as the heat leaves the pressure drops so it's a depressurization and finally step six the exhaust stroke where exhaust is leaving against no resistance this is isobaric it occurs at just this initial pressure so these are the idealized steps isobaric, adiabatic, isochoric, adiabatic, isochoric, isobaric, Let's do an example. Estimate how much work is done by the gas in the auto cycle shown the falling 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, 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 to two steps that contribute to work notice that this step the isochoric and this step the other isochoric don't do any work because they are at a constant volume and isochoric processes never do any work. Now these two 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 their opposite directions the signs of the work are opposite so they cancel 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 so the red step can be thought of like a triangle can be approximated as a triangle sitting on top of a rectangle 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 0.00005 and ends at 0.0005 and the triangle starts here at 7 and goes up to 170 and the base of the rectangles at 0 just the bottom alright and we can approximate the green process going to the left as that same shape.

A triangle sitting on top a rectangle, now the volume numbers are 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 1 and obviously it also starts at 0 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're different heights some I'm going to put a little prime on the rectangle height just to indicate that it's a difference if you take the distance or sorry the difference here on a calculator you'll find that's 0.00045 so this becomes one half 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 obviously times 10 to the 5 Pascals plugging this into a calculator we get 3982.5 joules for the area of the red look that red process let me minimize myself for this segment 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 that's the height of the volume numbers is 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 sign that's very very important the sign is very important now lets look at the red process the red process is to the right so that work is negative so we get a negative sign in front of the first thing the work sorry the green processes to the left, so that work is positive so we get a positive sign here in between those works now plugging this into a calculater we get -3397.5 joules so that is how much work is done by 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 undergo a cycle that allows it to release work that allows the engine to release energy into the system. Alright guys that wraps up the idealized auto cycle as the cycle of the gas undergoes any four sorry four stroke piston engine. Thanks for watching guys.

Example #1: Finding the Compression Ratio

Transcript

Hey guys let's do an example three moles of an ideal diatomic gas which is treated with rigid molecular bonds undergoes the following Otto cycle. How much heat is input? How much heat is output? What's the work done by the engine 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 the 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 and when we are trying to calculate the efficiency you need a very precise calculation of the work done, so we're going to have to do this a different way so let's break this up into four steps. I'm going to call step one this step which is the compression stroke, step two this step which is ignition, step three which is that first part of the expansion stroke and step four 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 its cyclic, since it starts and ends at the same position there is no change in internal energy which means by the first law of thermodynamics there is no heat exchange either so nothing is going on in those two steps they completely balance each other out. So lets look at step one, step one 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 icothermal which steps are icobaric etc. Adiabatic by definition means that there is no heat transfer for this step so Q1 is 0. Now we know the first law of thermodynamics is delta U is Q plus W and since Q for step one is zero we can say that delta U for step one is just the work done in step one so if we can find the amount of work done sorry if we can find the change in internal energy for step one we can find the work done in step one I don't actually want to encircle those let me highlight them I want to circle the final answers, so let's find the change internal energy for the ideal gas during step one remember that the internal energy for any ideal gas is just F over 2 N, R, T. We were told that this was an ideal diatomic gas, diatomic gas is either have a degrees of freedom of 5 or degrees of freedom of 7 because we treat the moleculer bonds as rigid it's 5.

This means that the change in internal energy for step one 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 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 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 P.V. 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 N, R so how does P times V change in step one so delta T1 is going to be 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 0.00005 right during step one minus the initial pressure which was 1 times 10 to 5 and the initial volume which was 0.0005 only three 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. So now that I know what the change in temperature is during step one I can plug that in to this equation which tells me how the internal energy changes in step one for a given temperature change.