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Re: [Phys-L] irreversible quasistatic



Okay great. I thought it is an example, but wanted to hear what others think.

At the risk of flames over the term “heat," I am going to call this an example of an “irreversible quasistatic heat transfer process” (IQHTP). I assume if there is one such example, there are others. For example, although I cannot immediately think of a good example of how to realize an IQHTP for an isothermal expansion of an ideal gas, I presume there are such examples. In addition, we know there exist examples of "irreversible quasistatic work transfer processes.”

But here comes a surprise (to me). Once we have quasistatic processes, we can draw them on PV or TS graphs as continuous curves. But once that is done, and assuming the process acts on a monatomic ideal gas, we can then calculate all heat and work inputs and outputs to the gas from such curves. I am now going to connect a set of such processes together to form a cyclic process, let’s say it’s an engine for definiteness.

I will draw two identical cycles (in the sense of having the same vertices and connecting curves). Cycle A will consist purely of reversible processes. Cycle B will consist of at least one irreversible quasistatic process; the other steps in it could be either reversible or they could also be irreversible quasistatic.

Thus cycle A is reversible and cycle B is irreversible. Yet both have the same heat and work inputs and outputs. So both cycles have the same efficiency. That seems like a violation of Carnot’s theorem to me. How can the irreversible engine be just as efficient as the reversible one? Normally I think of irreversibility as being connected to dissipation or wasted opportunity in some form.

What am I missing? -Carl

ps: I am not worried about entropy. The entropy of the universe for cycle A is zero, whereas that for cycle B is positive, so all is well, when you include the entropy changes of the thermal reservoirs and not just of the gas. But only the gas need be involved in calculating the heats and works. For example, the area under a PV curve tells me the work done on the gas, regardless of what external agent (my arm, perhaps) might happen to be effecting that work transfer.

On Jul 29, 2016, at 12:20 PM, Daniel V. Schroeder <dschroeder@weber.edu> wrote:

Carl, I can't speak for all authors, but your example is exactly the type of situation that *I* would call irreversible and quasistatic. More generally, the two objects are weakly coupled in the sense that their internal relaxation times are short compared to the relaxation time for them to come to equilibrium with each other--and therefore we can consider each object to be internally in equilbrium at all times.

Cheers,

Dan


From: Carl Mungan <mungan@usna.edu>
Subject: [Phys-L] irreversible quasistatic
Date: July 28, 2016 2:48:25 PM MDT
To: PHYS-L <phys-l@phys-l.org>


I'm often uncertain about the idea of irreversible quasistatic processes,
mostly because thermo textbooks don't give many examples.

Would the following process be an example of such a process? Explain why
not if you don't think so.

One hot plate at temperature T_H is parallel to a cold plate at T_C. The
plates are oriented vertically with a small pendulum ball suspended between
them. The ball is set swinging, so that it hits the plates at the end of
its motion back and forth. The ball carries small increments of thermal
energy dE from the hot plate to the cold plate. The time t when the ball is
swinging through free space between the two plates allows the two plates to
re-equilibrate internally. In this way each plate is ever only
infinitesimally out of thermal equilibrium, yet I am accomplishing a net
energy transfer from the hot to the cold plate.

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