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.. What is the thermal elliciency of an engine that operatesby taking $n$ moles of diatomic ideal gas through the cycle1$\rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 1$ shown in Fig. 16.19$?$

Calculate $W$ and $Q_{H}$ to get $e$ . Where $e=10.5 \%$

Physics 101 Mechanics

Chapter 16

The Second Law of Thermodynamics

Temperature and Heat

Thermal Properties of Matter

The First Law of Thermodynamics

University of Michigan - Ann Arbor

University of Washington

Simon Fraser University

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Let's analyze each process one at a time, so let's start with 122 in the diagram. So for 1 to 2, Delta V is equal to zero, and this immediately implies that the work is equal to zero. Since there's no volume changed, there's no work done. Now Delta V is equal to zero. Also implies that that he inputted in is equal to end CV Dr T. And this is also equal to in CVI t to blast you one. And now we're going to use thie. I don't guess a lot of simplify this a little bit. The ideal gas law says that PV equals and Artie. And when v is constant, this says that the one B is equal to in our team won and P two times being is equal to in our tea too. These both must be true during their respective times that once wanted points too. Now, taking the difference of this equation and this equation gives p two minus p one taffy is able to in our tea too, Unless do you want. And this career in a CZ Delta P B is equal to you in our delta t and this is true whenever we have, Delta V is equal to zero so constant volume processes. Now playing this into this for Delta T gives Q is equal to and CV the Delta p over in our that's just pulling in here says this is in C V Delta T playing in for adult ity based on this guy. And now I can some fights a little bit. It's turned into a C V over r. I'm just gonna put that in parentheses. Time is being dealt a p. All I did was canceled the answer. And then I actually want to take the Delta p for 122 here. And so this ends up being c v over r times V. Not since that's the V for 1 to 2 and to pin on my ski. Not is the Delta Peter and then this and flies to see V over r Vina peanut and we'll start there and we'll go on to process 223 for 23 Since Delta P is equal zero and not Delta V. We know work is equal to P V, which is equal to P for you three and last me too, which is equal to two Peanut. That's what the pressure is for 2 2 to 3 and then V three mice. He, too, is just equal to Vina. And so this is what the work is during this cycle, built A P is equal to zero implies that Q is equal to in CP Times Delta T. And then we're going to do the same sort of process that we just did accept. Now we know that we're in a constant pressure process, so things changed ever so slightly and Vee wants is equal to an arty one. And PV two is equal to in R T too, and then taking the difference of this minus this get his p Delta V is equal to in our delta t So this changed a little bit before was dealt a penal Delta bi. But now that this is a constant pressure process, we expect these changes happen. So now plaguing this in for Delta t here. So pretty much combining these two to cancel the Delta t give us that Q is equal to in CP Delta T, and we're going to do the same sort of simplification that we did last time, and so I'm not going to repeat all the steps because it's very similar. But when you get a C P over our times to be not, be not and we'll start there now, go really fast. That process 3 to 4 because it is a constant blind process. So it is exactly analogous to the first thing we did again. Work is going to be zero sense. The change of line zero and cue is going to be negative, too. In this case, it's negative because he's being drawn out of the system. He is being rejected by the guests similarly, and so Q is equal to negative two time CV over R times P, not Vietnam and then for processes. For one, it's a constant pressure process again, and you'll get that the work is negative. Be not, be not. And the reason that's negative is because Delta V is less than zero. Since the blind is decreasing for this process, this ends up being negative and that you is going to be equal to Negative CP over our time's peanut veena. And so now we can actually start to some of the works the total work work toll. Is this some of every single contribution from all four branches here? When you addled them up, you be not be. Not now. Our ultimate goal is to figure out what the efficiency is and the formula for the efficiency is the work over he going in. And so we don't even have to worry about this here. This here because this he is when he is leaving and won't be factored in into Q H. Similarly, for 3 to 4. We don't need the petition of this either, because this is not going to be a part of qh since he is leaving the system. Wait, What we do need to do is sum this up with this here. And when we do that, we get that Q H is equal to three c v plus two R over our times. P not you, not And after adding them up, I removed CPI by using CP is equal to C V plus our so that was used in bringing into this form. But now that we have Q h w, we can calculate the efficiency the efficiency ends up being be not be not. That's just the work over this thing here, which is three CV plus two R divided by Pena Veena. You can do a lot of simplification here by using the fact that CV is five are over too. And so what you do is you blowing this in and simplify everything and you simply everything you get that this is equal to two nineteen 19 and the decimal. This is eagle 0.105 and as a friend, or as a percentage, this is equal to 10.5% and so that's the efficiency.

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