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

[Herbert Schulz <herbs@wideopenwest.com>]

> On Mar 3, 2015, at 2:06 PM, John Denker <jsd@av8n.com> wrote:
>
> On 03/03/2015 11:07 AM, Carl Mungan wrote in part:
>
> > Define efficiency W/Qh = 0.5ln2/(ln2+0.75) = 0.24015.
>
> Algebra error.
>
> The denominator is Qh, which is just ln(2),
> not ln(2) + 0.75.
>
> I have no idea where that 0.75 came from.
>
> I get efficiency of 0.5, as expected.
>
> Hint:  I used a spreadsheet to run an example.
> That makes it particularly easy to check the work.

Howdy,

You can easily use the method you mentioned (breaking up an non-Carnot engine in differentially small Carnot engines) to prove that the efficiency of that engine MUST be lower than a Carnot engine acting between the same T(cold) and T(hot).

Good Luck,

Herb Schulz
(herbs at wideopenwest dot com)