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The maximum theoretical efficiency of a heat engine

  1. Dec 14, 2012 #1

    This is one of those threads where the OP proposed a device which will not work and hopefully by the end understands why. So here's the device...

    Air enters through a long intake tube. At the end of the tube it enters a combustion chamber where fuel is added and burned, increasing the temperature of the air. It then exits along an exhaust tube which runs parallel to the intake tube. A series of thermocouples is positioned to harness the temperature difference between the intake and exhaust tubes. The first one would be in contact with both tubes close to the combustion chamber, the second a little further away, and so forth. In this arrangement the first thermocouple the intake air would pass would be the last one the exhaust would pass.

    Heat energy would flow from the exhaust, through the thermocouples into the cool intake air. It would then pass through the combustion chamber (where more heat energy would be added) and into the exhaust, right back where it started. The only ways energy could exit the apparatus would be either as electricity generated by the thermocouples, or as heat left in the exhaust after the last thermocouple.

    Making the intake and exhaust tubes arbitrarily long would bring the exhaust temperature arbitrarily close to the intake temperature, and the efficiency arbitrarily close to 100%.

    Why does this machine not work? I'm obviously no thermodynamics buff or I would immediately see why my machine won't work. Please try to dumb it down for me if you could. My math skills are limited to very basic calculus.
  2. jcsd
  3. Dec 14, 2012 #2


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    I'm not tracking what your claim is. How does your arrangement maximize T(hot)-T(cold)?
  4. Dec 14, 2012 #3


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    Thermocouples aren't very efficient. Their efficiency is also limited by the Carnot efficiency (though they don't get very close), so if you made the exhaust and intake lines very long, the temperature difference across the thermocouples would be very low and the thermocouple efficiency would be very poor.
  5. Dec 15, 2012 #4
    In broad terms efficiency is output over input.

    Mechanical efficiency is mechanical advantage/velocity ratio and can be theoretically 100% in the absence of friction.

    Thermodynamic efficiency is a comparative term that compares the output achieved to the theoretical output of a perfect cyclic machine (Carnot cycle). Thermodynamic efficiency can never be 100% - that is one version of the second law.

    Your rather complicated heat recovery scheme will increase actual efficiency (either definition) but it will never approach 100%.

    This is because it takes some of the energy input to move the input and exhaust gases through your delivery tubes.

    So what is your output and what is your input?
    Last edited: Dec 15, 2012
  6. Dec 15, 2012 #5
    It does not, therefore it does not work. I'm trying to understand why it does not work.

    It shouldn't matter what the efficiency of the thermocouples is. The waste heat gets sunk into the intake air which eventually becomes the exhaust air. The idea is that the heat gets trapped, going in circles. A little bit gets turned into electricity each time around but it shouldn't matter how many times around it goes.

    I had forgotten about kinetic energy of the air, thanks for bringing that up.
    My input is the chemical energy of the fuel, although any heat source would work.
    My outputs are the kinetic energy of the exhaust. Since kinetic energy is proportional to V2 this can be reduced by running the engine at a slower speed.
    A second output is the heat energy left in the exhaust after the last thermocouple. This can be brought arbitrarily close to zero by making the intake and exhaust tubes very long, using a very large number of thermocouples, and running the engine very slowly.
    The third output is the electricity generated by the thermocouples.
    That's all I can think of. Minimizing the first two seems to me to be an engineering issue. I don't see any principal that prevents them from getting arbitrarily close to zero.

    If my input air is 300 kelvin and the combination of conduction through the thermocouples and energy from my fuel raises it to 900 kelvin then, according to the wikipedia quote, my maximum efficiency should be 600/900, or about 67%. I don't understand where the other 33% is going?
  7. Dec 15, 2012 #6
    Have you considered the expansion work on the gases from 300°K to 900°K?
    Last edited: Dec 15, 2012
  8. Dec 15, 2012 #7


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    It sounds like you want to use the thermocouples as thermoelectric coolers. These do opposite things. Thermocouple takes a temperature difference and generates electricity. Thermoelectric coolers take electricity and generate a temperature difference.

    If you connect the exhaust to the intake, eventually the exhaust and intake temperature will be the same, and efficiency is zero. You want to generate a temperature difference between the exhaust and intake, so you need to supply work (electricity). A thermocouple isn't capable of generating the temperature difference between the exhaust and input. You need a thermoelectric cooler for that.
  9. Dec 15, 2012 #8
    You should look at the Brayton cycle and the Ericsson cycle engines.

    These types of thermodynamic engines, also known as the constant pressure engines, are the type you are proposing.
  10. Dec 15, 2012 #9
    Most of the thermocouples aren't at the exhaust end. Most of the thermocouples are on the hot end of the tube. The temperature on the hot end of the tube is much higher than the ambient temperature seen at the exhaust.

    The thermocouples at the beginning of the series aren't 100% efficient. This is because there is a finite, maybe large, difference between the temperatures at both ends of the thermocouple. So they would make a large amount of entropy.

    Thermocouples are irreversible engines. They make entropy. The bigger the difference in temperature at the two reservoirs, the faster they make entropy. So the thermocouples at the beginning of the series (the hot end) would be extremely inefficient.

    The efficiency of a thermocouple is still limited by the Carnot formula. Therefore, the thermocouples at the exhaust end will be highly inefficient for a different reason. They don't make entropy, but they still have to move the entropy from hot to cold. The temperature difference is small at the exhaust end. So they will still be inefficient.

    Thermocouples are not 100% efficient. Generally, a good thermocouple runs at about 30% of the Carnot efficiency. Since the laws of thermodynamics are "local", the theromocouples can't respond directly to the exhaust temperature. The relevant temperatures for each thermocouple is inside the tube in the immediate area and the temperature outside the tube.

    Even if you replaced each thermocouple by a Carnot engine, the efficiency of the total series will be less than 100%. The efficiency of the entire series would have to be less than the efficiency of the first Carnot cycle in the series.

    So you are making too errors here.
    1) The efficiency of a thermocouple does not exceed the efficiency of a Carnot engine.
    2) Thermodynamics is local, not global.

    Here is a link to an article on thermocouples.

    “Thermoelectric junctions are generally only around 5–10% as efficient as the ideal refrigerator (Carnot cycle), compared with 40–60% achieved by conventional compression cycle systems (reverse Rankine systems using compression/expansion). Due to the relatively low efficiency, thermoelectric cooling is generally only used in environments where the solid state nature (no moving parts, maintenance-free, compact size) outweighs pure efficiency.”
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