Thermodynamics - Why is 100% efficiency impossible?

[ironizer]

The fact (or is it?) that a heat engine will never attain 100% efficiency in transferring heat into work, even under ideal conditions, really annoys me. Imagine this:

 

A cylinder-piston arrangement is made out of a material that is non-thermal conducting, and there is no friction or air resistance between the piston and the air/walls. Now suppose that the cylinder contains a pocket of air that is at the same pressure as the air outside. The pocket of air is quickly heated by an external source so that the temperature in the pocket rises, thus increasing the pressure. This pressure difference will cause the piston to accelerate outwards, expanding the volume inside the cylinder until it cools and reaches outside pressure again; at that point the kinetic energy of the piston is promptly absorbed and turned into work.

 

Given the above scenario, wouldn't 100% of the heat energy put into the system be transfered into work?

[Reaper]

Yes it is true. The engine you are referring to is what is called a Carnot engine. And this is an idealized hypothetical scenario. The reason it can never happen is because of things like friction, second law of thermodynamics, wear and tear, etc.

[ironizer]

yeah but I read that even the carnot engine (which i think is different than what i described) can't be 100% efficient. Got this on wikipedia

 

A third formulation of the second law, the heat engine formulation, by Lord Kelvin, is:

 

It is impossible to convert heat completely into work.

 

I think that is because at the peak of expansion, the volume of the gas is greater than what it is when it starts, which means that there is heat trapped inside that cannot do work.

 

The thing I'm thinking about, is what if there is no outside pressure, and instead there is a vaccum. Then the piston can expand infinitely, creating an asymptote that nears 100%.

 

Carnot efficiency equation is

[math]\eta=\frac{\Delta W}{\Delta Q_H}=1-\frac{T_C}{T_H}

[/math]

 

I don't see how the temperatures affect the efficiency. Shouldn't it be the pressure outside/pressure inside ratio?

[BhavinB]

It's subtely connected to Entropy...basically saying entropy changes in a system restrict the engine from transferring all its energy into work.

[ironizer]

It's subtely connected to Entropy...basically saying entropy changes in a system restrict the engine from transferring all its energy into work.

 

OK, so explain exactly why the heat cannot be made into work.That's what i'm not getting.

[ydoaPs]

OK, so explain exactly why the heat cannot be made into work.That's what i'm not getting.

 

In a Carnot engine, the heat source is the same temperature as the system and the heat sink is at the same temperature as the system. That means you can't have heat transfer into or out of the system. So, the engine doesn't work.

[Bignose]

yeah but I read that even the carnot engine (which i think is different than what i described) can't be 100% efficient. Got this on wikipedia

 

A third formulation of the second law, the heat engine formulation, by Lord Kelvin, is:

 

It is impossible to convert heat completely into work.

 

I think that is because at the peak of expansion, the volume of the gas is greater than what it is when it starts, which means that there is heat trapped inside that cannot do work.

 

The thing I'm thinking about, is what if there is no outside pressure, and instead there is a vaccum. Then the piston can expand infinitely, creating an asymptote that nears 100%.

 

Carnot efficiency equation is

[math]\eta=\frac{\Delta W}{\Delta Q_H}=1-\frac{T_C}{T_H}

[/math]

 

I don't see how the temperatures affect the efficiency. Shouldn't it be the pressure outside/pressure inside ratio?

 

Have you read the wiki article? http://en.wikipedia.org/wiki/Carnot_cycle I thought it was pretty well done, really. It explains why the temperatures enter into the efficiency. Basically, the efficiency is the ratio of how much work can be done for a given quantity of heat that is moved from the hot reservoir to the cold reservoir. (Seems like a pretty logical choice for efficiency to me.) The work [math]\Delta W[/math]can be calculated and is equal to [math]\Delta W = (T_H - T_C)(S_B -S_A) [/math]. Same for the amount of heat put in [math]\Delta Q_H = T_H(S_B -S_A) [/math]. Putting those ratios over each other, canceling the entropy terms, and you get the efficiency just in terms of the temperatures.

 

If you're not understanding how those quantities are computed, check out a good thermodynamics book. I recommend Smith, Van Ness, and Abbott's Introduction to Chemical Engineering Thermodynamics, though I am sure that there are other very good books that cover the Carnot cycle in detail, too.

[ironizer]

That's helpful.

 

But I don't see why you need to move heat from a hot to a cold reservoir in order to do work. Imagine the cylinder/piston system I described above, except this time the system is located in a vacuum. The heat does not need to be moved anywhere, and work can still be done.

[Bignose]

That's helpful.

 

But I don't see why you need to move heat from a hot to a cold reservoir in order to do work. Imagine the cylinder/piston system I described above, except this time the system is located in a vacuum. The heat does not need to be moved anywhere, and work can still be done.

 

How is this different from the first step in the wiki article? The first step is: "Reversible isothermal expansion of the gas" And, you need a hot reservoir to heat up the gas.

 

Also, the outside pressure has nothing to do with the calculation. The calculation is based on the amount of work the gas inside the piston does. What is works against on the outside is irrelevant.

[DrDNA]

The fact (or is it?) that a heat engine will never attain 100% efficiency in transferring heat into work, even under ideal conditions, really annoys me. Imagine this:

 

A cylinder-piston arrangement is made out of a material that is non-thermal conducting, and there is no friction or air resistance between the piston and the air/walls. Now suppose that the cylinder contains a pocket of air that is at the same pressure as the air outside. The pocket of air is quickly heated by an external source so that the temperature in the pocket rises, thus increasing the pressure. This pressure difference will cause the piston to accelerate outwards, expanding the volume inside the cylinder until it cools and reaches outside pressure again; at that point the kinetic energy of the piston is promptly absorbed and turned into work.

 

Given the above scenario, wouldn't 100% of the heat energy put into the system be transfered into work?

 

Your description reminded me of a Stirling engine:

 

http://en.wikipedia.org/wiki/Stirling_engine

[Fred56]

Perhaps everyone is forgetting about something called equilibrium, quod vide(cough).

[ironizer]

A 100% efficient thermodynamic process would be non cyclable because the volume would have to be expanded infinitely (so 100% is an asymptote)

 

Stirling engine needs a cold reservoir to operate. I don't see why a cold reservoir is necessary (not talking about the stirling engine, just a thermodynamic process in general)

[Bignose]

That first step is called reversible for a reason. That is, it assumes that it is only infinitesimal amounts of heat are being transferred, which means that that same packet of heat can be taken back out. That is the basic definition of reversible, and its definition is exceptionally important to thermodynamics.

 

But, no real process is ever truly reversible. Because the ideal of reversibility is that an infinitesimal amount is transferred, it would take infinite time to accomplish anything finite. Some processes can be very close to reversible, but nothing is truly reversible in real life.

 

In fact, three of the four steps of the Carnot cycle are reversible. It's the fourth step that is irreversible, and represents the efficiency loss.

 

If your goal was just to look for thermodynamic transfers that don't lose energy, then you are looking for the concept of reversibility. A useful concept in theoretical analysis, but is a physical impossibility in the real world.

 

*************************

 

A cold reservoir is needed because that is what you are going to use to cool the gas. Compression of a gas heats it up. You want to compress the gas without it gaining in temperature, because if you let is gain in temperature, you haven't done anything -- you've just gone backwards a step. The whole goal of the Carnot cycle is to use a heat differential to generate work -- if you compress the gas without keeping it at a cool temperature, you're going to use up all your work that you just gained to re-compress it.

[Fred56]

The "answer" is that irreversible processes result in transfer out of the system (losses). This is an equilibrium-busting process.

[boma]

Let me add couple of words.

First, I would recommend never look at wikipedia. There are SO MANY errors in it, that one cannot distinguish what is true and what is not. I have checked randomly articles on Physics and Math. My estimate that about 30% have factual errors or simply wrong. I forbid my student to use it (or at least, if they use it I will not accept as an excuse if they make errors due to wikipedia).

Second, in your first example the piston is indeed can transfer 100% heat into work. Nothing forbids that. What physicists mean by "efficiency cannot be more than 1-T2/T1" is no engine can transfer more than 1-T2/T1 (even in IDEAL model, with no friction, etc.) into work if the engine (actually its working body, like gas, etc) returns into initial state. If you push (very slowly to keep the pressure constant) your pistol back, you need to do some work (so, you prefer to lower the temperature inside of cylinder to decrease the pressure and therefore the work you need to push the piston back). That work should be subtracted from the work you received in the first part of the cycle. The first part work depends upon the high temperature (temperature of expansion), while the second (actually the third) part work depends upon the low temperature. So, the efficiency depends upon both.

Third, it is not really matters if the process reversible or not. If the process is ideal and reversible, the engine may reach the 1-T2/T1 efficiency, while if the process is not reversible (like real-life processes), the efficiency cannot reach it in principle, it will be always lower.

Fourth, everything above applies to ideal processes (no friction, etc.). If there are un-idealities (friction, heat loss, etc.) the efficiency will drop even for Carnot cycle (which has FOUR reversible steps).

If you want to follow, just send me an e-mail, as I might not check the forum often.

[insane_alien]

boma, the general stance of this board(and many other places including the uni i attend) is that wikipedia is fine, as a starting off point. the references are usually of a higher quality.

[boma]

boma, the general stance of this board(and many other places including the uni i attend) is that wikipedia is fine, as a starting off point. the references are usually of a higher quality.

 

That has nothing to do with what I wrote about Physics.

 

That was my advise (I may give advises, may not I?). As I mentioned I found TOO many errors in it and I don't recommend using it (I forbid to use it in my classes). From my point of view it will make you more harm (in terms of understanding Physics and other sciences) than good. Wikipedia is NOT peer (on PhD/MD/doctoral level)-reviewed source and the information is not reliable (it is true in about 60-70% cases -- my estimate, I might be wrong on the number). My friends also discourage their students from a use of wikipedia. And we are working in some of the best Universities in the US. I have not seen even single professor who said that there is no errors/problems with wikipedia in his/her area of expertise.

 

However, the decision is yours.

I am not going to discuss the quality of wikipedia in following posts. We may discuss it somewhere else if you wish.

 

If you have questions about Physics I will be glad to answer them. Please send me e-mail.

[Mr Skeptic]

That has nothing to do with what I wrote about Physics.

 

That was my advise (I may give advises, may not I?). As I mentioned I found TOO many errors in it and I don't recommend using it (I forbid to use it in my classes). From my point of view it will make you more harm (in terms of understanding Physics and other sciences) than good. Wikipedia is NOT peer (on PhD/MD/doctoral level)-reviewed source and the information is not reliable (it is true in about 60-70% cases -- my estimate, I might be wrong on the number). My friends also discourage their students from a use of wikipedia. And we are working in some of the best Universities in the US. I have not seen even single professor who said that there is no errors/problems with wikipedia in his/her area of expertise.

 

However, the decision is yours.

I am not going to discuss the quality of wikipedia in following posts. We may discuss it somewhere else if you wish.

 

If you have questions about Physics I will be glad to answer them. Please send me e-mail.

 

Do they teach spelling and grammar in the "best Universities in the US"?

 

Also, could you give some examples of the errors you found in so many of the articles? Were they spelling/grammar errors?

[boma]

Do they teach spelling and grammar in the "best Universities in the US"?

They teach Physics, Chemistry, Finance, Medical Sciences, etc. Couple of them teach at MIT, one at UNC-CH, several at Duke, couple more at Caltech. Many others at other Universities. I think these are reputable Universities.

 

My native language is not English, so I may make English grammar mistakes, but I don't really care about grammar mistakes. If you can show me my error in physics it would be more appropriate in this topic. You know there people who start discussing personal features if they cannot tell anything on the subject. I hope you are not one of them.

 

Because I am professor of Physics I tried to help you better understand Physics. I like the idea that students discuss Physics out of classroom and I will try to do a lot to help them. However, it seems a little bit strange that you are so negative about such participation. If you don't like my advise you are free not to follow it (as I mentioned above). However, that was advise from best intentions as you will never understand Physics if you accept everything you read in wikipedia without THINKING. I hope couple of people will accept my advise and will not educate themselves using that unreliable source of information.

 

I promised I will not discuss wikipedia in that topic. Start another one. Send me an e-mail.