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Efficiency of a Four-Step Engine

Patrick Ford
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Hey, guys, let's do an example. Ah, Proposed four step engine is produced when, in moles of a mono atomic ideal gas undergo the process shown in the following PV diagram. What would the efficiency of the engine be? Okay, The efficiency is given by the work output by the engine divided by the heat. Input it into the engine. Remember, there are two heats in a cyclic process and an engine. There's going to be amount of heat input it, which is gonna be positive heat and amount of heat output, which is gonna be negative heat. Okay, the work is easy to calculate. The work is just the area contained by the cycle with the appropriate sign. Now first, could this even be an engine? Yes, it can. It's a clockwise cycle, right? So it's absolutely an engine. Okay. The area is going to be the base times height, because it's just a rectangle. This is gonna be to you. Not this is gonna be one half peanut. So this is to V not one half peanut, right? That too, and that one have canceled. This is just peanut Vienna, and this is absolutely gonna be positive. It's gonna be positive. Right? Energy released. Bye, Engine. Okay, that's so that's just the work done by the engine. Now, in order to find how much heat is input, we have to analyze the steps individually. Okay, I'm gonna call this step one this step to this Step three and this Step four. Okay. Now, something important to know about P v diagrams. And you can show this for yourself if you want Is that when the process is going up into the right, the heat is always input into the system. So in process one and process to heat is gonna be added into the system. Okay. So cue was going to go in when the process is down or to the left. Heat is always leaving the system. All we need to know is how much he enters the system, right? This how much he enters the system. So we only have to look at steps one and two. You can use the same method, the same process that I'm using here to analyze steps three and four and show that the heats always gonna be leaving in those steps. Okay, for step one. First of all, the work is always going to be zero, because this is an isil coric process. This means that the first law of thermodynamics, which says the change in internal energy is the heat transferred. Plus, the work done just means sorry. Wrong color. It just means that the heat transferred is the change in internal energy four. Step one. Okay. Now, the internal energy off any ideal gas is f over to in our t an equation that we used a bunch of times because this is a model atomic ideal gas. The number of degrees of freedom is simply 31 for each translational directional. Now, the change in internal energy is what we're interested in. And this is gonna be free has in our delta t we're talking about in moles simply changing its state. There's no heat coming into the city. Sorry. There's no gas coming into the system. No gas leaving the system. So the only thing that changes when the internal energy changes is the temperature. So the temperature is the only thing that gets this delta. The question is, what is the change in temperature? Okay, Well, PV was nrt The ideal gas equation is gonna tell us how the temperature changes as the gas changes state. This is going to tell us because the volume is a constant. This is an I support process that V Delta P is in our delta T. The change in the left side is due entirely to a change in pressure. The change on the right side is due entirely to a change in temperature. So Delta T is v Delta p over in our now plugging that in two Delta U. Okay, okay, right. So I could say Delta t one is equal to this. This equals three halves in our times, V one Delta P one over in our okay. Notice that the n r in the numerator and then are in the denominator. Cancel. So this is three halves V one Delta P one. Okay, which is three halves. What's the volume for step one? It's just V not. What's the change in pressure and step one from peanut. It's from one half peanut too peanut. Okay, so this is just one half peanut. So this becomes 3/4 peanut Vienna. Okay, that's the change. Internal energy for step one and by the first law of thermodynamics. Therefore, the heat exchange in step one, which is Delta U one, is going to be positive. 3/4 Peanut Vienna. Okay, it's very important that the heat is positive because we're only looking for heat input into the gas. We're not looking for heat. Output it by the gas. Okay for step two, the pressure doesn't change for Step two Delta P zero. Okay, but Delta U is still equal to Q plus W. However, we do know that at constant pressure, W is equal to negative. P Delta V Okay, so this is for Step two. What's the pressure at Step two? It's just peanut, right? What's the change in volume? It's three v not minus two v. Not sorry, minus V not which is going to be positive to be not so this is negative. Two. Peanut Vienna, right? This is positive to Vienna. Okay, so that's how much work is done during step to now. What's the change in internal energy during step to? Well, it's still three halves in our delta t during step, too, right? What's Delta T during step to? Well, PV equals in our tea pressure does not change on Lee. The volume changes so we have peed Delta V equals in our delta t So p two Delta V two equals in our delta t two right in our delta T too, is this whole right portion. So Delta U two is three halves p two Delta V two, which is three halves peanut, right, The pressure for step two times once again the same change in pressure that we found here. So this is just three peanut V. Not so by the first law of thermodynamics. Hugh during Step two is just Delta U two minus w, which is three peanut V not minus negative to peanut V not which is five peanut Vienna. Okay, so for the cycle Sorry, The heat the total heat input during the cycle is just the heat input during steps one and steps to because steps three steps Step four release heat. Okay, so the amount of heat inputs just during step one Step two Step one had 3/4 peanut Vienna and Step two had five Peanut Vienna. If we find the least common denominator, this is 3/4 if you're not plus this common denominators. Four. This is 20/4 Peanut Vienna. So that's 23/4 Peanut Vienna. Okay, and what's the work done by the cycle we calculated it to be Peanut Vienna. This means that the efficiency, which is just the work over the amount of heat input, is peanut peanut over 23/4 Peanut view, not those peanut peanut's canceled. And finally, the efficiency is 4/23. Now 4 23 is an exact answer, and you can leave it like that if you want. Or you can approximate a number 0.174 which is 17.4%. So this engine actually has an incredibly low efficiency Onley 17.4%. Other engines can have much, much, much higher, um, efficiencies, right? I showed a case in an earlier problem where car, not engine, had an 80% efficiency when it was placed between a reservoir of 1000 Kelvin and a reservoir of 100 Calvin. This 17.5% efficiency sort of leaves a lot to be desired, but this is an application of the first law of thermodynamics on heat engines and how to find the efficiency using a PV diagram which could be very useful. Alright, guys, that wraps up this problem. Thanks for watching