Qfwfq Posted January 21, 2006 Report Posted January 21, 2006 See volume 1 of the Feynman lectures, in the 4th chapter he develops gravitational potential energy from a Carnot-like argument.Unfortunately, I've been unable to find my copy of vol 1 for years, I was quite convinced an old friend of mine still had it but she can't even remember me having lent it to her! :rainbow: If her memory ever gets refreshed, I'll surely remember to look up the point you cite. :rainbow: However, I disagree that the 1st of thermodynamics being the same as the principle of energy conservation, it simply puts heat as being yet another form of energy. It's right that the 2nd principle is a further limitation. Quote
Erasmus00 Posted January 22, 2006 Report Posted January 22, 2006 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. One mistake is identifying temperatures with energies. The carnot cycle tells you how much work you can harness from the energy you bring into the engine, usually designated Q or Qh. -Will Quote
Erasmus00 Posted January 22, 2006 Report Posted January 22, 2006 However, I disagree that the 1st of thermodynamics being the same as the principle of energy conservation, it simply puts heat as being yet another form of energy. It's right that the 2nd principle is a further limitation. Formulating thermodynamics from statistical mechanics, the first law carries over directly from classical mechanics. The second law limits the production of useful energy, but doesn't change the character of the first law at all. -Will Quote
Kayra Posted January 23, 2006 Author Report Posted January 23, 2006 One mistake is identifying temperatures with energies. The carnot cycle tells you how much work you can harness from the energy you bring into the engine, usually designated Q or Qh. -Will Hmm... the Carnot formula seems to deal exclusively with heat, and heat engines. I know that heat is just another form of energy. Perhaps yet another phrasing of the question then :lol: Does the Carnot cylce (and associated formulas) give us the efficiency with which we can use the energy put into the system, or the efficiency with which we cna harness the existing total energy in the system? Quote
Erasmus00 Posted January 23, 2006 Report Posted January 23, 2006 Hmm... the Carnot formula seems to deal exclusively with heat, and heat engines. I know that heat is just another form of energy. Yes, that is the case. But temperature and heat are not the same thing. You can't say that the energy of the system is Th or Th-Tc or anything of the sort. The energy that goes into the engine IS the heat, but that is a seperate quantity, usually called Qh. Does the Carnot cylce (and associated formulas) give us the efficiency with which we can use the energy put into the system, or the efficiency with which we cna harness the existing total energy in the system? The Carnot formula tells us how well we can utilize Qh, the heat input into the engine. -Will Quote
Kayra Posted January 23, 2006 Author Report Posted January 23, 2006 Yes, that is the case. But temperature and heat are not the same thing. You can't say that the energy of the system is Th or Th-Tc or anything of the sort. The energy that goes into the engine IS the heat, but that is a seperate quantity, usually called Qh. The Carnot formula tells us how well we can utilize Qh, the heat input into the engine. -Will I think I understand part of my misconception.Heat is energy, Temp is a measure of um... average molecular activity? A cup of water and an ocean can have the same temperature, but the ocean has a vastly larger energy content. With a known Volume, Density, and temp coefficient, the total heat energy of a system can be determined. I get that part. The part I do not get is that the Carnot efficiency is calculated based on an engine cycle operating between 2 temperatures (regardless of the amount of energy in the system), and how far above absolute 0 those temperatures are. (hence all values in the calculations use Kelvins). Quote
Qfwfq Posted January 24, 2006 Report Posted January 24, 2006 The second law limits the production of useful energy, but doesn't change the character of the first law at all.I don't think I claimed it does, I only said that the 2nd is necessary to answer the question. Quote
Erasmus00 Posted January 26, 2006 Report Posted January 26, 2006 The part I do not get is that the Carnot efficiency is calculated based on an engine cycle operating between 2 temperatures (regardless of the amount of energy in the system), and how far above absolute 0 those temperatures are. (hence all values in the calculations use Kelvins). Please forgive what I'm sure is too long an answer. There are two ways to think about this. The first way is to follow the cycle around, calculate for the isothermal heating/cooling and adiabatic expansion contraction assuming your engine material is an ideal gas. This allows you to establish the efficiency of a carnot engine without any prior knowledge of entropy. Using this, the carnot argument work as follows: 1. Establish using conservation of energy that any reversible engine is the most effecient (as I've tried to argue above)2. Design a reversible engine. This is our carnot cycle.3. Calculate the effeciency of our new engine using known laws (in this case ideal gas ideas), thus establishing the best efficiency possible. 4. Generalize by creating the concept of entropy, from which follows the second law. In this case we look at the efficiency limit as something that just happened as a result of our calculation,which isn't satisfying misses out on the "why does this happen" type questions, but has the advantage of being extremely powerful in the use of problem solving. In fact, before Boltzman and the like, this "how much can we learn from steam engines" approach was all there was to thermodynamics and it was thought a nearly complete discipline. However, the "thermodynamic" definition of entropy doesn't seem particularly fundamental. The second way of thinking about this is to start doing statistical mechanics. I can't derive it here, but one fundamental result of stat. mech is that we can attach the idea of entropy to "Boltzman entropy" (sometimes called statistical entropy, sometimes counting entropy). This is where the popular notion of entropy meaning disorder or chaos comes from, which is more or less the correct notion. The more disordered a state is, the higher its boltzman entropy. Now, using this new entropy deffinition we can establish the second law of thermodynamics as a statistical principle. In this case the Carnot argument simply allows us to connect the concept of statistical entropy to thermodynamic entropy (showing they are indeed the same thing). Finally, to get to the heart of your question, to understand why the carnot effeciency has its form, you must recast the second law in the form "heat only flows downhill."(i.e. between a temperature difference) We immediately see then that our effeciency should be proportional to Th-Tc. Why? The larger the difference, the longer we can keep the heat flowing untill it gets to the bottom. The longer the heat flows, the more work we can extract. Also, we realize from counting entropy considerations that at absolute 0, a system has 0 energy (and 0 entropy, which is called the third law). So if your bottom temperature is 0, we should recover an efficiency of 1 because we are extracting all our energy. How can we make this happen? We divide by Th. Thus, the form of the efficiency should be (Th-Tc)/Th. I hope this helps. -Will Quote
Kayra Posted January 26, 2006 Author Report Posted January 26, 2006 OK, Thanks guys. I especially appreciate your patience Erasmus00. Quote
arkain101 Posted January 29, 2006 Report Posted January 29, 2006 I was thinking of a design, that could be 100% efficient. The only problem here is if pressure on a material can cause a material to stay warm. For example, if we put a block of metal on a device that applied enough g-force to flatten the metal, could this pressure allow a material to remain warm?Most reasoning says, no because heat is always lost, but I wondered if enough pressure was around, could a material remain warm. Quote
jingalala Posted January 13, 2012 Report Posted January 13, 2012 Could anyone explain to me in non-math terms why a heat engine can not reach (or approach) 100%? I have a difficult time visualizing this limitation. we will get 100% efficiency (efficiency= wnet/qin = (qin-qout)/qin = 1-(qout/qin)) when qout=0, but when qout is zero there is no sink which means no gradient (we need gradient in order to flow something either heat or water or electricity). So we will never have 100% efficiency even though it is a reversible engine which is clearly evident from the kelvin Plank statement of 2nd law of thermodynamics. Quote
JOHNDJACOBY Posted October 9, 2014 Report Posted October 9, 2014 Kayra: To answer your question: "Could anyone explain to me in non-math terms why a heat engine cannot reach (or approach) 100%?". It is assumed that you mean why the efficiency of a heat engine cannot reach 100%. First of all, no engine can operate on the Carnot cycle. The Carnot cycle only defines the limit of efficiency for all engines operating within the temperature range between TH and TL. TH is the high temperature (of the heat source) and TL is the low temperature (of the heat sink). A Carnot cycle with a high temperature of 500 degrees Rankine and a low temperature of 0 degrees Rankine (Absolute Zero) would define the limit of efficiency to be 100%. The only time the Carnot cycle efficiency is 100% is when the low temperature is 0 degrees Rankine (Absolute Zero). The Carnot efficiency is defined by the equation: E = 100% (1-TL/TH). If TL were 0 degrees Rankine, then the limit of efficiency defined by the Carnot cycle would be 100%, no matter what the high temperature, TH, might be. Since there must be a temperature differential for heat to flow (Heat flows downhill), the Carnot efficiency of 100% could never be reached by any heat engine. Not only is there the Carnot cycle limitation, but there must be a heat sink with a temperature of TL. Nowhere on earth can you find a temperature, TL, of 0 degrees Rankine, necessary for 100% efficiency, defined by the Carnot cycle. The above, only refers to the thermal efficiency. There are other barriers to efficiency, such as friction. So, Kayra, do not waste your time trying to design a 100% efficient heat engine. If you were mathematically inclined, I expand on the Carnot cycle in my book "The Most Efficient Engine" (The New Carnot Cycle). You can read the description of the book and author on the web site: WWW.CREATESPACE.COM/3371409. Anyway, Kayra, I hope I answered your question. Sincerely, John D. Jacoby Quote
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