Re: [Phys-L] efficiency versus Carnot efficiency

[John Denker <jsd@av8n.com>]

On 03/19/2013 11:10 AM, Bruce Sherwood wrote:
> There seems to be a simple vocabulary issue.

I'm pretty sure there is a much deeper issue than that.  Of course,
there may be vocabulary issues /also/ ... but if so, I'm not seeing
them.

> Curzon and Ahlborn. To
> put it very concretely, they asked what should be the temperature
> differences between heat source and engine and between engine and heat sink
> in order to maximize the power output for given temperatures of the heat
> source and sink.

That is exactly the way I am using the "vocabulary" of max thermodynamic
power.

>  The
> value of the insight of Curzon and Ahlborn is to sharpen the theoretical
> limit on what can be achieved,

We agree that they highlighted an interesting and important question.
However, I still believe that they got the wrong answer.

AFAICT to make sense of the situation requires a three-parameter model,
including zeroth-order, first-order, and second-order contributions.

As explained in more detail (including diagrams) at
  http://www.av8n.com/physics/power-plant-efficiency.htm
if you keep only the first-order term, you get the classic Carnot formula
that gives an efficiency number that is independent of how hard you run
the engine.

> I remember being told as a student that "practical engines don't come close
> to achieving the Carnot efficiency" followed by some vague statement about
> friction as an explanation.

That's a two-parameter model:  waste heat and friction.

>  I was glad to see the role of thermal
> resistance highlighted by Curzon and Ahlborn. 

That's *still* a two-parameter model, because thermal resistance plays the
same role as friction, if we approximate the frictional losses as being
proportional to how hard we drive the engine.

As discussed at
  http://www.av8n.com/physics/power-plant-efficiency.htm
this family of two-parameter hypotheses -- i.e. where there are no zeroth-order
terms -- all lead to the suggestion that one should run the engine as slowly
as possible.  These hypotheses are dead on arrival.  The notion that engines
are expensive and fuel is cheap will not revive them.

The right answer is to go to a three-parameter model, including losses in
/parallel/ i.e. heat leaks -- in addition to losses in series i.e. thermal
resistances.

As soon as you cobble up the simplest three-parameter model, you get a nice
maximum-efficiency point ... Mama Bear is too slow, Papa Bear is too fast,
but Baby Bear is just right.

  http://www.av8n.com/physics/power-plant-efficiency.htm#fig-power-efficiency-cume
  http://www.av8n.com/physics/power-plant-efficiency.htm#fig-power-efficiency-out

It must be emphasized that the max-efficiency point is nowhere near the
thermodynamic max-power point.  The two have got practically nothing to
do with each other.

> There is of course a "maximum
> power" for an engine in the sense that running it at even higher power
> would melt the metal parts, or some such awful consequence.

That's what I call redline.  On real engines, this point is very commonly 
marked on the tachometer using ... (wait for it) ... a radial red line.  It 
is a safe bet that redline will be well above the max-efficiency point and 
well below the thermodynamic max-power point.

> I remember being told as a student that "practical engines don't come close
> to achieving the Carnot efficiency"

A) This is very simply explained by the three-parameter model.

B) In contrast, the claim that engines are expensive and fuel is cheap and
therefore one should operate at the max-power point is a completely wrong
argument leading to a completely wrong conclusion.

C) If other parts of their argument happen to be correct, it doesn't change
point (B) ... or point (A) either.