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Carnot efficiency, mathematical slight of hand

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  • Carnot efficiency, mathematical slight of hand

    How is the Carnot efficiency of a heat engine calculated, and is it valid?

    In seriously looking into and carefully studying the mathematics involved in calculating an engines "Carnot efficiency", (which is supposed to be a hard limit which no heat engine could ever surpass, a "Law" of nature), years ago, I made a remarkable discovery.

    Carnot believed that heat was literally an indestructible fluid. Like water, he imagined; (actually having no personal experience with steam engines and how they operate, as the steam engine had not yet been introduced in France); that a steam engine worked the same as something more familiar, a water wheel. The power that can be derived from a water wheel is proportional to the distance the water is able to fall. On that basis Carnot imagined that the power that could be derived from a heat engine is proportional to the difference in temperature between the steam as it went into the engine and the temperature of the steam as it was exhausted from the engine, or the "fall" from high temperature to low temperature.

    As far as "absolute zero", there was no such concept in Carnot's day. For all practical purposes, there was no point in discussing any temperature below zero centigrade, as far as steam engines were concerned.

    So, if water is elevated from zero centigrade up to boiling, then used to power a heat engine and is exhausted at a temperature of 80 centigrade then the engine can be calculated to be 20% efficient.

    Setting aside the fact that this is complete madness based on the fallacy that heat is a fluid that can in effect "run down hill" from a high temperature to a low temperature but itself remain unchanged, like water is unchanged after passing over a water wheel... I don't really have any issue with all this.

    If for example, a ball is raised from a height of zero feet to ten feet, the energy used to raise the ball can be recovered when the ball is dropped. How much energy? Only the 10 feet worth, naturally.

    Well suppose we lift the ball ten feet, then drop it into a hole another ten feet deep?

    Well, that could be done, but then to raise the ball back up it will need to be raised twenty feet. You can't ever really get out more than you put in.

    Well, sometime after Carnot, the idea of an absolute zero of temperature came along, so Carnot was reinterpreted.

    The percentage of useful work energy output hat can be derived from a heat engine, rather than using zero centigrade as the baseline, the idea was updated and improved. It was now recognized that the real baseline of heat was zero Kelvin, not zero centigrade., but, other than that minor correction, nothing else was changed as far as the original Carnot engine "efficiency" calculation based on the idea that the percentage of work output of an engine is proportional to the difference in temperature.

    Again, I don't personally have any real problem with this, it makes sense.

    If I take a cup of water at room temperature and heat it up to boiling, then put a model Stirling engine on the cup to run, then the most I can get out of the engine is exactly what I put in to bring the water to a boil in the first place. One the water cools back down to room temperature we are done.

    This is perfectly sensible, I think, for the most part anyway.

    If my Stirling engine can utilize all the heat energy, all the joules I put into the room temperature water to bring it up to a boil. without any loses to friction or vibration or heat otherwise escaping, without powering the engine, then the engine can be considered 100% efficient, right?

    Well no. Not according to the modern method for calculating "Carnot efficiency", which really has little if anything to do with Carnot, or his original hair brained idea of how a steam engine operated, based on complete nonsense.

    We have, today, several different temperature scales, which at times can be a bit confusing, and this is where we can see a bit of "slight of hand" when it comes to determining the efficiency of a heat engine.

    I will try to make this clear in another post.

  • #2
    Suppose I said that if I lift up a ball and drop it on some device for extracting energy, then I can only recover the potential gravitational force that I put into the ball by lifting it up off the ground. I can measure the height the ball is lifted with a ruler.

    I claim that my energy recovery device is 100% efficient, meaning that the device can extract all the gravitational energy put into any object that is lifted up and dropped onto it.

    No problem there.

    But suppose some stranger comes along and points out that in reality, gravitational force extends all the way down to the center of the earth as well as out into outer space above.

    Who is this wise guy?

    OK, I can agree with that. So what? What exactly is your point mister?

    The stranger replies, "well, your calculations are wrong. Your so-called gravitational force, energy recovery machine is nowhere near one hundred percent efficient, as you claim. If it were really 100% efficient it could extract all the gravitational force acting on the object all the way from the moon down to the center of the earth.

    Get out of here you bloody idiot, you're out of your mind. The object was never lifted up to the moon, and the machine is not located at the center of the earth, but if an object could infact be dropped from the moon down to the center of the earth, then my machine would be able to extract all that energy, it IS 100% efficient!

    No, it isn't.

    Yes it is!

    No, it is not!

    Is.

    Not.

    This is a foolish argument, however, it is exactly the situation we are in today when it comes to calculating the "Carnot efficiency" of a heat engine.

    For example, if I run a Stirling engine on a cup of boiled water, the temperature difference between ambient temperature and boiling is about 20% of the the temperature scale, all the way down to absolute zero Kelvin.

    So, if my engine utilizes all the joules put into the water to bring it to a boil and as a result the temperature of the water is reduced back down to room temperature, all the heat added having been completely converted to work, then my engine has a "Carnot efficiency" of only 20%..

    Wait a minute. But the engine converted every bit of heat that was added to the water into useful work output. How can you say that it is only 20% efficient?

    The stranger smirks and says, Sorry, but on the absolute temperature scale, your engine has only utilized 20% of the heat energy.

    Yes, sure, but the water, before we added heat to bring it to a boil was already eighty degrees Fahrenheit. Eighty degrees is the starting temperature. the baseline. We raised the temperature to boiling and the engine used all that heat bringing the water back down to eighty degrees. This engine IS one hundred percent efficient.

    No it isn't. To be 100% efficient your engine would have to bring the temperature of the water down to absolute zero.

    That's impossible!

    Exactly!



    Don't be fooled.
    Last edited by Tom Booth; 01-01-2022, 07:06 PM.

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    • #3
      What is the difference, and why does it matter?

      The difference is, in terms of temperature, in the case of a model Stirling engine running on a cup of coffee for example, is about 300 Kelvin.

      Talking about what constitutes a 100% efficient or "perfect" heat engine that is.

      It makes a difference in relation to Nicola Tesla's proposal of running a heat engine directly on the surrounding ambient air.

      Tesla proposed that a heat engine, running on cold, such as ice, dry ice or liquid nitrogen could run indefinitely because no energy is actually being drawn out of the ice, or added to the ice.

      The ambient heat entering into the engine is converted to a different form; mechanical action, or electricity, so if the engine were "100% efficient" then "no heat at all" would reach the cold sink.

      So, how much heat does the engine have to be able to convert, on a Stirling engine running on ice for example? How much would the engine have to cool the incoming ambient heat?

      Well, logically, if the ice is at "freezing", then the engine must use up or convert enough heat so as to reduce the temperature to 0 Centigrade, 32 Fahrenheit or 273 Kelvin, NOT zero Kelvin.

      Tesla believed that even a relatively inefficient heat engine, running on cold rather than heat, could derive SOME energy from atmospheric heat because the energy converted could be used to operate a refrigerating machine.

      This refrigerating machine would only need to make up the difference. That is, the engine, if only 75% efficient would put 25% of the heat into the ice. So if the 75% converted energy were used to remove the 25% "waste heat" from the ice, there would still be a 50% overall efficiency of conversion of ambient heat to some useful form.

      The advocates of "Carnot efficiency" however will rant and rave that to be 100% efficient, a heat engine would have to reduce the temperature to 0 Kelvin or absolute zero.

      While it may be true that at the temperature of ice, there is still a long long way to go to get to absolute zero, so there is still a lot of heat available, so an engine reducing ambient heat to freezing by conversion to another form is not utilizing "ALL" the heat, that is irrelevant.

      For Tesla's proposed "Self Acting Engine" to work, the ambient would only need to be reduced down to somewhere near the temperature of the cold sink itself. If ice is used, then that would be 0 Celsius not zero Kelvin.

      Tesla was trying to get to as low a temperature as practically possible, so his goal was apparently, to use liquid air as the sink. But in principle, any cold sink below ambient could be utilized.
      Last edited by Tom Booth; 01-04-2022, 05:21 PM.

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