Re: [Phys-l] entropy and electric motors

[kyle forinash <kforinas@ius.edu>]

> ----------------------------------------------------------------------
>
> > So would it be fair to think of the thermal noise as roughly parallel to
> > the Qout of a heat engine?
>
> I dunno.  There may be some connection there, but if so, I'm not
> seeing it.
>
> The disconnect comes from the following:
>  *) Heat energy, by definition, means your heat bath is at equilibrium.
>   That means it is in the maximum entropy macrostate, i.e. maximum
>   entropy per unit energy.
>  *) The electrical energy coming from (say) a battery is nowhere near
>   full thermal equilibrium.  The distribution of probability versus
>   energy level does not follow the Boltzmann distribution.  More
>   precisely, it follows the Boltzmann distribution *except* for
>   one *enormous* exception.  For details, including an explanatory
>   diagram, see
>     http://www.av8n.com/physics/thermo-laws.htm#sec-metastable-t
Very helpful, thanks.
> The electric motor is not a heat engine, and ideas (such as Qout) that
> apply to heat engines don't necessarily apply to electric motors.

Sorry to be dull witted- let me see if I can re-phrase my question.

1. OK, heat engines are not 100% efficient because entropy change in the 
hot reservoir must be compensated for by entropy change in the cold 
reservoir; final entropy can't be higher than what you started with so 
you don't get all the energy out as useful work (D. V. Schroeder in his 
'Thermal Physics' has a nice example with two Einstein solids that 
explains the necessary 'heat' flow to the cold reservoir entirely in 
terms of the  multiplicity of energy states; ie entropy. The word 'heat' 
need not be mentioned at all and temperature is defined as what is equal 
between two bodies when entropy stops changing).
2. Batteries (fuel cells, etc.) additionally have entropy changes 
related to chemical potential. So (I'm a little fuzzy on the exact 
calculation but) again tracking the entropy of the chemicals before and 
after they react in the battery we can say that we won't get all of the 
energy out in a useful form because entropy has to increase (or remain 
constant at best).
3. So in an electric motor, ??? the change in entropy of the electrons 
flowing through a perfect motor due to a potential difference is 
equivalent to thermal noise?? Or is this just not a useful way to think 
of motors? For a resistor the random thermal energy (temperature 
increase of the resistor) minus the original random thermal energy that 
was there in the wire before the current limits the efficiency- same for 
a motor if mechanic energy is the output instead of temperature increase?

I think I understand your reference to the Nyquist paper (which I now 
have read- great paper); the applied potential has to be larger than the 
equivalent thermal noise potential for current to flow and this (very 
small) limit limits the motor efficiency (not much for macroscopic motors).

Thanks for the help.

kyle
> This illustrates the point that the foundations of thermodynamics are
> energy and entropy.  Non-experts think it's about heat and temperature,
> but it's not.
>  ++ Energy is primary and fundamental.
>  ++ Entropy is primary and fundamental.
>  -- Energy and entropy are well behaved even in situations where
>   the temperature is zero, unknown, irrelevant, and/or undefinable.
>  -- Often it is unnecessary or impossible to quantify "heat".  It is
>   more practical to quantify energy and entropy instead.
>
> To repeat:  Anything you can do with "heat" you can do more easily
> and more precisely using energy and entropy instead.
>
>
> > Do you happen to know where the loss comes from in real electric engines?
>
> Mainly from "I^2 R" loss in the windings.
>
> Engineering is generally a multivariate optimization process, with
> lots of tradeoffs between the various variables.  If all you cared
> about was efficiency, you would use more wires and fatter wires in
> the windings ... but you also want to optimize the power-to-weight
> ratio and the power-to-capital-cost ratio, which argues for fewer
> and thinner wires.
>
> Keep in mind that this I^2 R loss is a nuisance loss, not required
> by thermodynamics.  It is like friction in your heat engine, which
> is not required by thermodynamics.  It causes an inefficiency in
> _addition_ to whatever inefficiency is required by Carnot.
--
------------------------------------------
 "When applied to material things,
the term "sustainable growth" is an oxymoron."
	Albert Bartlett

 kyle forinash   812-941-2039
  kforinas@ius.edu
  http://Physics.ius.edu/
-----------------------------------------