Re: [Phys-L] irreversible quasistatic

[John Denker <jsd@av8n.com>]

On 07/29/2016 12:17 PM, Carl Mungan wrote:
> I want to know what happens if one or more of the four processes are
> irreversible; however, all four remain quasistatic so that we can
> draw well-defined PV or TS diagrams. For specificity, assume one of
> the isochoric processes is as I described in my previous email.
>
> Now what do you say?

As I said in the previous message, there are additional variables
in the equations.

In particular, there is a key assumption that goes in to the
analysis of reversible engines, namely that the entropy taken
from the hot reservoir is equal to the entropy ejected into
the cold reservoir.

If you operate a brake (or the equivalent) that creates new
entropy, this equality is broken.  You need an additional
variable to keep track of the newly-created entropy.

This changes everything, including the compressibility.  The
compressibility has to do with the pressure, which is by
definition -∂E/∂V *at constant S* ... and if you mess with
the S you have to re-evaluate everything you know about
pressure.

Let's be clear:  To move along some leg of the cycle takes
energy.  This can be expanded as

	dE = ∂E/∂V|S dV + ∂E/∂S|V dS		[1]

assuming the state of the system (or subsystem) in question
can be described as a function of S and V.  Equation [1] is
a mathematical identity, basically the chain rule, if we
treat dE, dV, and dS as vectors.  (They are one-forms, not
pointy vectors.)

The classic Carnot cycle is exceptionally simple because on
the RHS of equation [1], the first term drops out on two of
the legs, and the second term drops out on the other two legs.
In contrast, if you are doing an irreversible expansion, the
second term does not drop out.  If you ignore it, you violate
conservation of energy.