Re: [Phys-L] irreversible quasistatic

["Daniel V. Schroeder" <dschroeder@weber.edu>]

Carl,

I think what you're missing is the reservoir(s).

Carnot's theorem is a statement about the maximum engine efficiency for any two given reservoir temperatures.

As you've explained, you can't tell whether a quasistatic heat transfer process is reversible just by looking at what's happening inside the "system".  So how do you tell?  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.

Your cycle B requires either a hotter hot reservoir or a colder cold reservoir than cycle A.  The maximum Carnot efficiency for the B reservoirs will therefore be higher than for the A reservoirs, and therefore your irreversible B cycle will not come as close to its Carnot limit as your reversible A cycle will come to its Carnot limit.  The two cycles have the same efficiencies, but the reservoir pairs have different Carnot limits.

Hope this helps.

Dan


On Jul 30, 2016, at 10:00 AM, phys-l-request@www.phys-l.org wrote:

> From: Carl Mungan <mungan@usna.edu>
> Subject: Re: [Phys-L] irreversible quasistatic
> Date: July 29, 2016 11:01:00 AM MDT
> To: PHYS-L <Phys-L@Phys-L.org>
>
>
> 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.