Re: [Phys-L] irreversible quasistatic

[Carl Mungan <mungan@usna.edu>]

Hmm... I think Dan is on to a good point and I think it supports what you
have been saying, John.

The issue is one of clearly defining the system. For example, consider a
Stirling cycle. The analysis of its efficiency *changes* depending on
whether the regenerator is included inside or outside of the system
boundary. If it is inside, the heat output during the isochoric compression
is stored inside the system and re-input during the isochoric expansion. It
thus does not count into Q_in or Q_out, whereas it does if the regenerator
is outside the boundary. (In like fashion, one has to decide whether your
brake counts as "part of" the engine or not.)

The formula for efficiency of an engine is not (Q_h - Q_c)/Q_h but instead
(Q_in - Q_out)/Q_in where Q_is the sum of *all* heat inputs and not just
the piece Q_h that comes from contact with the hot reservoir.

Of course the reservoirs have to be outside the system because otherwise
they don't count into Q_in or Q_out and one cannot compute efficiency.
However, when we switch to computing entropy instead of efficiency, the
system entropy is of no interest. It's obviously zero, since the system
will cycle back to its starting point periodically. What matters is what
happens to the rest of the universe.

So the reservoirs are excluded when it comes to computing efficiency, but
must be included when it comes to computing entropy.

At least that's how I'm thinking now. -Carl

On Sat, Jul 30, 2016 at 2:55 PM, John Denker <jsd@av8n.com> wrote:

> On 07/30/2016 10:05 AM, Daniel V. Schroeder wrote:
> > You compare the system temperature to the reservoir temperature.  If
> > they're infinitesimally close, you can call the process reversible.
> > If not, the process is irreversible.
>
> That's not true in general, and I'm pretty sure it's not relevant
> to the question that was asked.
>
> > Your cycle B requires either a hotter hot reservoir or a colder cold
> > reservoir than cycle A.
>
> That's not clear at all.  It is entirely possible to have
> an irreversible engine with only the usual two reservoirs.
> For example, imagine a system with two subsystems, namely
> a reversible engine driving a brake, where the brake is
> heat-sunk to one of the two usual reservoirs.
>
> Irreversibility is about entropy, not about temperature per se.
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Carl E. Mungan, Professor of Physics  410-293-6680 (O) -3729 (F)
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