1
$\begingroup$

I understand Bolzmann definition of entropy well enough, but I keep struggling to understand entropy $S$ and it's relation to heat and efficiency in classical (non statistical) thermodynamics.

As Carnot said, heat engine efficiency depends only on temperature difference of it's hot and cold sinks. That is: $$η = 1 - T_c/T_h$$

"Well, what's a big deal" — I always asked my self. "*There are lots of differentials in other engines like:

  1. height differential in waterwheel
  2. pressure differential in .... some engines
  3. wind speed "differential" in windmills (Betz Law)

I mean, why $S$ wasn't invented for a waterwheel at first? Why thermodynamics lead the way for $S$ discovery?

*"But recently I understood that

  1. For many engines there is no known theoretical efficiency limit, so no one knows on what this possible limit depends: design or physical constrain.
  2. There must be engines efficiecy of which depend on design and not on some physical constrain. Examples?
  3. Even if for some engines there is a theoretical efficiency limit independent of it's design, still you can't derive entropy from this.

So this lead to conclusion that "there is something special about heat(energy transported through temperature differential)"... I know it's silly to figure this out when it's written in every book, but I'm quiet dumb in these matters.

Now the question: For Carnot case it is true that:

If $Q$ drops from $T_h$ to $T_c$ only $W=Q(1-T_c/T_h)$ amount can be extracted as work. Why it's not the case for any other engine (but without heat transfer) and their differential (pressure, velociry, voltage ect)?

Example: Let $X$ Joules of energy "drops" from high level $L_h$ to low level $L_l$ passing through a waterwheel. Then as far as I know, there is no such thing (and any other kind) as $W=X(1-L_l/L_h)$. Why? Is there a way for the blind (like me) to see what's SO SPECIAL about Q and nothing special about X(in waterwheel example).

$\endgroup$
3
  • 1
    $\begingroup$ Eliminate all the "engines" you described that are not operating in a cycle. The Carnot limiting efficiency applies only to engines operating in a cycle. $\endgroup$ Dec 5, 2018 at 12:37
  • $\begingroup$ What do you mean by “$Q$ drops from $T_h$ to $T_c$”? The $Q$ in your equation is the gross heat added in the cycle. For the Carnot cycle that is the heat added during the reversible isothermal expansion at $T_h$. $\endgroup$
    – Bob D
    Dec 5, 2018 at 14:06
  • $\begingroup$ Dops should be in quotes. Thats how Carnot thought about heat "dropping" just like "water drops from Level to lower level". I used the same wording to connect two examples (waterwheel and heat engine) $\endgroup$
    – coobit
    Dec 5, 2018 at 16:37

1 Answer 1

0
$\begingroup$

why S wasn't invented for a waterwheel at first?

Your question goes into the heart of thermodynamics and has to do with microscopic irreversibility. Classical mechanics is reversible, and large macroscopic mechanical systems whose performance is not immediately or appreciably affected by the molecular state, such as a waterwheel for example, are practically reversible, which is to say their entropy is zero: you do $X$ amount of work to rotate the wheel by some amount, it returns that work to you when you reverse the motion.

When the microscopic state is affected by the process, it is not possible to reverse it. When steam runs through a reversible Carnot cycle, starting from state $A$ and ending at the same state later, the macroscopic state (i.e., pressure and temperature) has been restored, but the internal state has not. In fact, it cannot be restored, because this would require us to bring every molecule to its original position and velocity. This microscopic irreversibility is the origin of entropy and heat.

To get more insights on entropy one needs to get deeper into statistical thermodynamics and away from classical thermodynamics and cycles.

$\endgroup$
2
  • $\begingroup$ Thanks, but Clausius didn't have any insight into statistical thermodinamics(well, exept for his desgregation term) but still delivered $S$. I know what Bolzmann entropy means (number of microstates given macrostate), I feel how it was derived by him, but Clausius $S$ just passes my mind without hitting any strings of my soul. :) I just don't get calusius meaning of $S$. $\endgroup$
    – coobit
    Dec 5, 2018 at 11:36
  • $\begingroup$ We can, as Clausius and the rest of the classical thermodynamicists did, infer the existence of entropy from macroscopic considerations. Giving it physical meaning is tougher. We may offer a non statistical interpretation: entropy is an arrow that points an isolated system towards equilibrium, never away from it. Still, one may argue that this only tells us what entropy is good for, not what entropy is. $\endgroup$
    – Themis
    Dec 5, 2018 at 12:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.