Why is the Carnot Cycle The most efficient heat engine?
#1
Posted 13 January 2006 - 11:09 AM
I have a difficult time visualizing this limitation.
#2
Posted 13 January 2006 - 08:49 PM
Kayra said:
I have a difficult time visualizing this limitation.
Not just Carnot cycles, but any reversible engine has the maximum efficiency. The reason has to do with conservation of energy.
First, what is a reversible engine? It takes X amount of energy as input and does y amount of work. Or in the reverse, you can do Y amount of work, and it moves X amount of energy. (I'm being vague on the type of energy. A carnot cycle is a heat engine, but theoretically engines could work by moving gravitational potential energy, electrical potential, etc)
So why must a reversible engine have the maximum efficiency? Well, lets say it didn't, and there was a more efficient engine out there. This engine takes X amount of energy, and does Y+Z work. The work is higher then Y, becuase our engine is more effecient then the reversible engine.
So if we start with X energy and run it through our efficient engine, we get Y+Z work, and then we take the Y work and run it backwards through our reversible engine and we move X energy. So now we have the same energy we started with, X, and we have still managed to do work. This means we have violated conservation of energy (and created energy). Since this can't be, we must assume that our efficient engine is impossible.
Hope this helped.
-Will
#3
Posted 15 January 2006 - 12:40 PM
Heat is always trying to get away. Aka, Put on some thermal goggles on your heat machine and you will see heat. Why is this bad? Because, that means millions of infrared electromagnetic radiation (light we cant see) is being expelled from the machine. In other words, you cant insulate it enough because the heat will just pass from thing to thing and if its not where it should be to do work, then its lost.
Heat is moving particles remember.. so think of it as energy.. they lose it all the time, and it is the state all matter strives for.. low energy/very cold, but the suns energy and gravity and all that keep earth warm.
#4
Posted 15 January 2006 - 06:07 PM
Anyways. About the answer that was provided. Could not this heat dispersion be significally reduced by creating a magnetic field? I do know however, that maximization is not achievable rarely aproached. Meaning perfection can not be attained.
#5
Posted 15 January 2006 - 09:37 PM
Atoms also arent purely elastic.. when they bump momentum is lost. It is lost because energy can neiter be created or destroyed just changed. So they bump and give off energy, and slow down.
So thermodynamics is just the way she goes.. and it would be really difficult as far as I undersatnd -and I am just a grade 12 graduate, who self educates .. lol... - to redirect energy back into itself.
#6
Posted 16 January 2006 - 06:31 AM
Erasmus00 said:
First, what is a reversible engine? It takes X amount of energy as input and does y amount of work. Or in the reverse, you can do Y amount of work, and it moves X amount of energy. (I'm being vague on the type of energy. A carnot cycle is a heat engine, but theoretically engines could work by moving gravitational potential energy, electrical potential, etc)
So why must a reversible engine have the maximum efficiency? Well, lets say it didn't, and there was a more efficient engine out there. This engine takes X amount of energy, and does Y+Z work. The work is higher then Y, becuase our engine is more effecient then the reversible engine.
So if we start with X energy and run it through our efficient engine, we get Y+Z work, and then we take the Y work and run it backwards through our reversible engine and we move X energy. So now we have the same energy we started with, X, and we have still managed to do work. This means we have violated conservation of energy (and created energy). Since this can't be, we must assume that our efficient engine is impossible.
Hope this helped.
-Will
Well Drat
Still not quite there
If we had a reversible engine that was 100% efficient, then with X energy we could still do Y work, and reversing the engine Y work should still produce X energy, No?
In reference to the other replies, I think I will need to keep this simple and not involve any losses (friction, radiation, convection, conduction, etc) until I have a handle on how a perfect engine might work, thanks
And welcome aboard biotech7, this is a pretty good bunch of folks here (so says the lurker)
#7
Posted 16 January 2006 - 11:27 AM
Kayra said:
Still not quite there
If we had a reversible engine that was 100% efficient, then with X energy we could still do Y work, and reversing the engine Y work should still produce X energy, No?
The question is, could you design a reversible engine that was 100% efficient? Restricting our attention to heat engines, it turns out all reversible engines that operate between the same two temperatures are equally efficient.
Why? If they weren't, you could still set it up in such a way to violate conservation of energy as per the argument above. If you have access to a university library try looking through Volume 1 of the Feynman lectures on physics. Chapter 4 develops gravitational potential energy using Carnot's argument.
-Will
#8
Posted 16 January 2006 - 11:43 AM
biotech7 said:
Perhaps some, but this would still not bring efficiency to 100%. Further, it would likely require energy to generate said magnetic field, thus causing additional losses with E0 (starting energy).
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#9
Posted 16 January 2006 - 12:19 PM
Every action has an equal and opposite reaction.. so the engine will generally just barely work if you set it up just right.
I have however been working a new concept of multi-free energy absorbtion techniques to allow a system to operate in self powering manners.
It may not be 100% efficient, but as a car design, which it is, it will allow a car to nearly drive forever, in our standards.
There is a topic here about it.
http://hypography.co...iting-news.html
#10
Posted 20 January 2006 - 06:39 AM
Erasmus00 said:
The 2nd of thermodynamics is a statistical principle, it really isn't the conservation of a quantity.
Hypography Forum PITA......... er, Administrator.
#11
Posted 20 January 2006 - 06:58 AM
I guess I have difficulty seeing why this limitation exists, although this link I found does help a bit.
http://hyperphysics..../seclaw.html#c1
Sorry for being so thick guys
#12
Posted 20 January 2006 - 09:22 AM
Qfwfq said:
The 2nd of thermodynamics is a statistical principle, it really isn't the conservation of a quantity.
The carnot cycle argument is entirely dependant on the first law of thermodynamics, which is conservation of energy. This isn't a statistical principle at all. See volume 1 of the Feynman lectures, in the 4th chapter he develops gravitational potential energy from a Carnot-like argument.
-Will
#13
Posted 20 January 2006 - 12:37 PM
I thought the second law was statistically based, and further, affected the scope of the first law. (IE, limited it even further). I guess that means it is based on it.
Statistically based meaning that while it is probable that heat will flow from a hot area to a cool one, it is not impossible for it to spontaneously flow in the other direction... just incredibly improbable. At least, that is how i understood it.
Could someone tell me.. Does the carnot efficiency tell me the maximum energy I can harness from the total energy in a closed system, or useable energy in a closed system?
Note: found this to be an excelent reference:
http://acnet.pratt.e...modynamics.html
The answer may resolve my visualization dimlema
#14
Posted 20 January 2006 - 01:06 PM
Kayra said:
Generally, a carnot cycle is set up with two large heat reservoirs, one at some hot temperature, one at some cold temperature. This implies your engine is not a closed system. If you swapped out the hot reservoir (as an example) and replaced it with a hot brick (just something of finite size) then to find the maximum work you could do as you cooled the brick you need to look at a quantity called the free energy.
-Will
#15
Posted 20 January 2006 - 01:46 PM
Oh my.. I did say closed system, didn't I
Let my try my question again.
Given Th (Temp Hot) and Tc (Temp cold) , does the carnot formula of (Th-Tc) / (Th) tell me how efficiently I can harness the available energy (Th-Tc) or the total Energy Th.

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