Carnot efficiency 3: Proving that it is the most efficient | Physics | Khan Academy

[@360wheelz5]

00:37 "Work's a good thing, so I'll make that in green."

[@amangoswami7919]

what is preventing us from applying same logic to some other engine than carnot ? with this logic, we can prove every engine to be most efficient !!!

[@陳柏廷-u5p]

guess there must be an effective reversible engine for that

[@joao_ssouza]

@@LuizCarazolli why? I need a proof.

[@LuizCarazolli]

@@joao_ssouza Actually I made a huge mistake. Carnot's theorem applies to all reversible engines, stating that regardless of the cycle, 2 reversible engines operating under the same conditions will present the same and highest possible efficiency. Irreversible engines are the ones that will always be less efficient. So, in reality, Carnot's cycle is simply a reversible cycle, which, just like any other one operating in a reversible engine, must present the highest efficiency allowed by the second law, and that's what the video is about

[@gabor6259]

"In this house we obey the laws of thermodynamics." Homer Simpson

[@MrTweetyhack]

Ha! I really polished the thing well and I really did achieve 41%!

[@davithonlynnington]

Why couldn't this argument be applied to any heat engine? Any engine could be made to take in Q1 and give out Q2 and W (which are arbitrary), and if it was connected in reverse to a "better" engine it would be shown to violate the second law.

[@skeetorkiftwon]

David Lynn The single difference being the Carnot model doesn't have any other detracting factors such as friction, turbulence, or vibration. Hence it is an "ideal" heat engine. We develop ideal models as the classic, see also the ideal diode, pulley, etc. Any other heat engine model would simply include more, and more complex, variables. A rose by any other name would still be the ideal heat engine model.

[@williamnathanael412]

@@skeetorkiftwon how does it relate with why does it needs two isotherms and two adiabats? Like why cant it be two isotherms and two ischorics or isobarics?

[@ameennaseem8936]

i know this was posted a very long time ago. But students are discovering it even now, so let me clarify. This argument showed by Sal not only proves that carnot engine is most efficient, it also shows that the all reversible engines (working within two tempratures) have the SAME efficiancy equal to that of carnot engine.(Sal should have mentioned that, he probably forgot) So your argument that there can be a "a worse and a better engine" is not applicable here. Also there can be less efficient engines, but these involve dissipative irreversible process and thereby lose energy and cannot be used for this perfect combination like a carnot engine.

[@joao_ssouza]

@@ameennaseem8936 ok, but all this just does not explain why carnot engine must have two adiabats and two isotherms, prove to me that a two isochoric and two transfers of heat with a constant volume is not more efficient. I am really struggling to swallow this down.

[@KyleDunnIt]

@@joao_ssouza Any isochoric steps would have losses (i.e. not *ideal*). The adiabatic and isothermal steps of the Carnot model are used because neither have a change in internal energy (zero losses).

[@Pr0x1mo]

Put these in order, Sal!!!

[@Skokiii]

@Annergize I try to answer the question. If you add heat Q+ at Th to a system an addition of entropy to your system comes along. The magnitude of this entropy you add equals Q+/Th = S. Entropy is a state variable. In a cycle you want to come back to your initial state eventually. Thus you have to reject the entropy S again. This is done by the rejection of heat Q- at Tc. Q-/Tc now has to equal S. With Tc = 299K and Th = 300K => Q- = 299/300 * Q+. Almost all the heat added has to be rejected.

[@MisterTutor2010]

There is simpler way to prove that a Greater than Carnot efficiency engine can't happen. This can be done by using the efficiency equation. W=Q1-Q2>Q1(1-T2/T1) -Q2>Q1*T2/T1 Q2/T2

[@thanasisconstantinou7442]

There is an error in your equation, cause you assume that efficiency is greater than it"s self.

[@srpenguinbr]

@@thanasisconstantinou7442 it is not an error. He used the equation 1-t2/t1=w/q1 and supposed there was some machine that could do some work W>w, i.e, W>q1(1-t2/t1). This lead to the contradiction.

[@nestorv7627]

@@srpenguinbr the op still made a mistake. -Q2/T2 is the change in entropy of the engine as it loses energy Q2. +Q1/T1 is the change in entropy of the engine as it gains energy Q1. Therefore Q2/T2 - Q1/T1 = -ΔS2 - ΔS1 = -ΔS(engine) 0 There is no contradiction here

[@sd0753]

41% efficiency is the greatest efficiency obtainable with the difference in temp between those two resiviors. If you look at the equation used to calculate efficiency you would see that a greater difference in temp will generally generate a greater efficiency. And if you could get the cold side down to absolute zero the efficiency would be 100% regardless of the temp of the hot side

[@MisterTutor2010]

All you have to do is make make the adiabatic expansion infinite to make the temperature inside the cylinder go to zero and have an absolute zero reservoir for the isothermal compression :)

[@LuisBarros-i4x]

This only proves that THERE IS indeed a maximum efficiency machine, in this case we are calling it the "Carnot's machine", BUT it doesn't prove that the "Carnot's machine" we are talking about in this "proof" is running under the Carnot's cycle.

[@absxn]

Exactly what I was thinking! How does this prove that an isothermal expansion, adiabatic expansion, isothermal contraction and then an adiabatic contraction is the most optimal way for an engine to do work, given a heat source and sink? Why not e.g. an isobaric expansion (heating), isovolumetric depressurization (cooling), isobaric contraction (cooling), and then isovolumetric pressurization (heating)? Still forms a cycle with net work done -- the process is just shaped like a rectangle instead!

[@ayandas8299]

Khan man you are the greatest teacher on the planet

[@MedusaMrigesh]

Your videos are great and help a lot! Thank you!

[@sheenshinestar]

woooww!!!.....couldnt get a simpler explanation than this!!...thnx a lot!

[@shobyanto6635]

Thanks for the presentation

[@exhalerwolf1272]

Watched at 2x speed. Very quickly explained but in enough detail 👌

[@wolterh6]

excellent explanation

[@Skokiii]

I think this is not a prove for the Carnot engine being the most effiecient engine. It just says, that two coupled engines have to have the same efficiencies in order not to violate the 2nd law of thermodynamics.

[@mtttony]

Great explanation, Surely the cold side can be made to absorb heat and the hot side to dissipate heat,like a heat pump..In this endless cycle you have shown.The unmentioned byproduct of your system is moving a huge amount of heat.Given you attached heat exchangers onto the hot and cold side.For one Kwh of electricity an efficient compressor will move up to 7Kw of heat by compressing the gas to a liquid causing it to give off all of its heat..The high pressure evaporator side drives a reverse compressor off of expansion gases and can be coupled to the compression side.This is scavenging like a turbo..Remember moving heat is a huge byproduct of compressing gas to a liquid and the evaporator side is not usually scavenged to turn the compressor..This would up the efficiency of the system..And to finish of.If you look at a heat pump scroll compressor type you will see the input energy to power in, is linear but the output is an exponential curve..How big do you have to make a heat pump before you can convey enough heat to power said heat pump..Food for thought..Thanks!

[@satyapratik]

Multiple Engineverse !

[@sciencenerd7639]

thanks

[@Tsardoz]

What happens if we have two bodies, one at 299K and the other at 300K and we use a Carnot engine between them. The max efficiency is 1 - 299/300 = 0.003. Why is this so inefficient when the temperatures are so close?

[@waltercatalfamo6028]

You can find the answer by yourself. Draw in a pV diagram a carnot cycle between these two temperatures. The area inside the figure is very small, then work->0, then the efficiency is ridicolous.

[@talgabay7230]

Sal you're the best!

[@smk_18]

Thanks a lot! Clearly understood 😄😇

[@azndude3600]

But isnt the work put in the system increasing the overall entropy, like Maxwell's demon?

[@ShawnSwander]

It's weird because this is just saying no one can make an engine more than 100% efficient. Which is just another way to state part of the 2nd law of thermodynamics. You could make a "better engine" than a 100% efficient engine by making one that does the work faster (more powerful). You also could make an engine that pulls energy from other systems because the second law of thermodynamics doesn't apply to open systems. So maybe it's a carnot engine with an AI that makes more carnot engines so it just keeps making more work per unit time. You could argue it would run out of materials or something but we're talking about an imaginary engine from the beginning.

[@vortyx090]

thanks man

[@JohanSol]

Which process would be more efficient then?

[@skeetorkiftwon]

JohanSol Magic.

[@bablabla9728]

you save my final exam bro

[@suhaylpatel9471]

legenddddd!!!

[@Skokiii]

@Annergize this leads to a very small efficiency

[@bubshab]

Awwww..... I wanna heat my house with ice. Damn universe, go back to where you belong!

[@shayanmoosavi9139]

Universe : fine. *[goes back to the singularity and destroys everything in the process]*

[@monalidutta3417]

Hahaha. This was funny!

[@vipindhaker1499]

yeah baby! you too

[@johngliha451]

You're trapped in your own little box, the Carnot cycle is NOT the most efficient.

[@jesuscastelan2218]

Jack Stowe what do you mean?

[@skeetorkiftwon]

Jesus Castelan He means he studies fantasy books.

[@swapnilloharkar9668]

Ha ha