Re: [Phys-L] Carnot (?) efficiency of non-Carnot cycles

[Herbert Schulz <herbs@wideopenwest.com>]

> On Mar 3, 2015, at 11:38 PM, John Denker <jsd@av8n.com> wrote:
>
> On 03/03/2015 02:51 PM, Herbert Schulz wrote:
> > The interesting thing about the engines shown is that all the heat
> > in occurs at T(hot) and all the heat out is at T(cold). So, all
> > those cycles are Carnot cyclaes.
>
> That's not the accepted definition of what a
> Carnot cycle is.  A Stirling cycle is dramatically
> different from a Carnot cycle, yet it uses a single 
> heat source at T(hot) and a single heat sink at
> T(cold).
>
> A Carnot cycle is a rectangle in (T, S) space.
> A Stirling cycle is a rectangle in (T, V) space.

Howdy,

I think I dealt with this in a previous message.

Saying that my dealing with a more general situation (i.e., any reversible heat engine, which includes a Stirling Engine of course) is out of point.

Also, just because you can make a Carnot Engine and Stirling engine look like pretty rectangles with the right choice of coordinate axis doesn't negate the use of a P-V diagram where it is obvious during what parts of cycles (especially for a Stirling Engine) Work is being by/on the gas withing the engine.

> > Make a P-V diagram for an arbitrary, reversible cycle.
> > At the point of highest temperature, T(hot), draw a isotherm; 
> > similarly at T(cold), the lowest temperature of operation.
>
> You could draw such things.  However that's got 
> little to do with the question that was asked.
>
> The Carnot efficiency formula involves two temperatures,
> namely the temperature of /the/ heat source and /the/
> heat sink.  If you have N different sources and M 
> different sinks, you can model it with NxM different
> heat engines ... and yes, the efficiency of the most
> efficient one is an upper bound on the efficiency
> of the whole mess, and obviously so ... but again 
> that's got precious little to do with the question 
> that was asked.
>
> The point remains that a Stirling cycle is quite
> different from a Carnot cycle.  It is not however 
> an "arbitrary" cycle.  
> a) Like a Carnot cycle, it is reversible.  
> b) Also like a Carnot cycle, it has a single 
>  T(hot) and a single T(cold).

Unlike a Carnot Engine there is heat transfer during parts of the cycle where the temperature changes!

> All heat engines that meet these two conditions (a)
> and (b) have exactly the same efficiency.   Otherwise
> you could hook them in tandem and make a perpetual
> motion machine.

and the Sterling Engine does NOT meet those two conditions.

Remember that the definition of the efficiency of a heat engine is W(net out)/Q(in) and Q(in) isn't `part of the heat transferred to the engine I want to deal with' it's the total Q(in).

For a Sterling Engine using a monatomic ideal gas

W(net out) = nR(T(hot)-T(cold))ln(V_2/V_1)

where V_2 and V-1 are the upper Volume and lower Volume of the cycle. The

Q(in) = (3/2)nR(T(hot)-T(cold)) + nRT(hot)ln(V_2/V_1)

where the first term comes from the heat transfer which goes into increasing the Internal Energy of the gas in the engine during the isochoric process at V_1 and the second from the heat transfer that goes into Work(out) during the isothermal process at T(hot) since none of that goes into Internal Energy. NOTE: you CANNOT simply say that the heat transferred into the engine is eliminated by the heat transferred out of the engine during the isochroic process at V_2; one part of Q(in) and the other part of Q(out).

The efficiency of the Stirling engine is then given by

eff(Stirling) = eff(Carnot)*[ln(V_2/V_1)/((3/2)eff(Carnot) + ln(V_2/V_1))]

which can easily be obtained from the W(net out) and Q(in) above, cancelling out the nR in the numerator and denominator and dividing the numerator and denominator by T(hot) (knowing that eff(Carnot) = (1 - T(cold)/T(hot))).

It's clear that the term in square brackets is <1 so

eff(Stirling) < eff(Carnot)

as I stated. If you wish to neglect the heat transfer into the engine during the isochoric process at V_1 the (3/2)eff(Carnot) term in the denominator goes away and, indeed, you get your result. However that is NOT the definition of efficiency!
> This is the glory of classical thermodynamics.  It
> is one of the most elegant and most useful ideas
> in the history of the world.

It is the glory of thermodynamics that no heat engine (reversible or not) can have an efficiency greater than that of a Carnot Engine. AND, as a matter of fact, it must be lower than that.

Good Luck,

Herb Schulz
(herbs at wideopenwest dot com)