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Confused about the Carnot engine?

  1. Jan 23, 2015 #1
    I am teaching myself thermodynamics (and really enjoying it!) but am slightly confused about Carnot's engine. From the equation efficiency=1-T(cold reservoir)/T(hot reservoir), I see that the most efficient engine is one where the difference in temperature between the cold and hot reservoirs is the greatest. However I also read this:
    "It was Carnot who realised that the most efficient heat engine of all was a ‘reversible’ heat engine. In other words – one that got the same amount of work out of the heat transfer as would be needed to operate a perfect fridge to undo its operation.

    In order to do this, it is necessary for all the heat transfers (between one object and another) to take place with as small a temperature difference as possible. If this is not done, heat will flow from hot objects to cold – a process which could have been used to do work, but wasn’t. Therefore not enough work will be done to enable the fridge to return the heat to the hot object. "

    And to me this Implies that the most efficient engine is where the temperature different is smallest because then less heat flows straight from the hot to cold reservoir without being used to do work.

    So my first question is about the above, but I also wanted to ask about how Carnot's engine links to the third law of thermodynamics. I know That the equations for Carnot's engine are supposed to somehow demonstrate that absolute zero can never be reached in a finite number of steps, but I don't understand how?

    Thank you in advance :)
     
  2. jcsd
  3. Jan 23, 2015 #2

    DrClaude

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    Staff: Mentor

    You have to distinguish between the difference in temperature between the hot and cold reservoir, and the temperature difference between the working substance and either reservoir when heat is transferred. The former gives the efficiency of the cycle, while the latter is required to make the cycle reversible. The temperature of the working substance must be brought to an infinitesimal ##\delta T## below that of the hot reservoir when receiving heat from it, and then have its temperature lowered to ##\delta T## above that of the cold reservoir to dump heat in it. Both serve to inihbit the creation of "extra" entropy, which reduces the efficiency.

    I'm not sure if this is the best explanation, but in a refrigerator, you need to lower the temperature of the working substance below that of the cold reservoir you are cooling down, and therefore would require a temperature below 0 K to lower the temperature of the cold reservoir to 0 K, which is of course impossible.
     
  4. Jan 23, 2015 #3
    Thank you so much for your reply! I like the refrigerator explanation!

    But I was wondering if you could explain the above further? Why would the temperature difference between two objects being infinitesimal for maximum work being done? Why would there be extra entropy created otherwise, and why would the extra entropy reduce the efficiency? I apologise the load of questions! I don't want to burden you with so many, but I haven't found any good websites explaining this...
     
  5. Jan 23, 2015 #4

    DrClaude

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    Staff: Mentor

    The change in entropy related to a transfer of heat into or out of a system is
    $$
    dS \geq \frac{Q}{T}
    $$
    where the equality holds only in the quasistatic or reversible case. Otherwise, the irreversibility means that something additional is going one with the system (it is not simply converting the heat into expansion work), and that means a greater production of entropy.

    Then, consider an amount ##Q## of heat leaving a system at ##T_1## to another at ##T_2 < T_1##. The entropy of system 1 is decreased by ##\Delta S_1 = - Q / T_1##, while the entropy of system 2 increases by ##\Delta S_2 = Q / T_2##. The total change in entropy of the two system combined is
    $$
    \Delta S = Q \left( \frac{1}{T_2} - \frac{1}{T_1} \right) = Q \frac{T_1 - T_2}{T_1 T_2}
    $$
    The only way to have no change in entropy is to set ##T_1 = T_2##. The problem is that there can then be no heat exchange between the two (it contradicts ##T_2 < T_1##), so we consider ##T_1 = T_2 + \delta T##, with ##\delta T## infinitesimally small. This also shows that while the Carnot engine is the most efficient, it wouldn't be practical as it would take an infinitely long time to transfer the heat during the cycle.

    That's what PF is for!
     
  6. Jan 24, 2015 #5

    Philip Wood

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    Gold Member

    Just a footnote. Doesn't add to excellent answers already given, except to point out a rather nice 'symmetry'…

    Conduction of heat down a rod: heat out = heat in. Investigated mathematically by Fourier (c 1820)

    Reversible heat engine: entropy out = entropy in. Investigated by Carnot (c 1820)

    Two French men, working at about the same time, investigating what might be called opposite extremes of thermal processes. It should be noted, though, that Carnot didn't have the concept of entropy and doesn't even seem to have appreciated energy conservation – no reason why he should have done - these ideas developed later.
     
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