[Stoves] How to convert thermal efficiency into fuel savings figures

[ajheggie at gmail.com]

[Default] On Thu, 11 Oct 2012 08:58:49 -0500,Paul Anderson
<psanders at ilstu.edu> wrote:

>Using Andrew's example below, 55% divided by 15% = 3.7 times the amount 
>of fuel for the 3-stone fire versus the better one at 55% efficiency.
>
>Or     15 divided by 55 =  .27 or only 27% of the fuel is used for the 
>55% TE stove compared to the referenced 15% TE 3-stove fire.
>
>These are reciprocals.      3.7 x .27 = 1.00       (precisely, it is    
>3.66666...   x    0.2727... = 1.000000 )
>
>So the better stove saves 73% of the fuel in this example.

Paul, I'd avoid using percentages in this sort of illustration, it's
too much like an advertising ploy and open to confusion.

The simple answer to your question is that there is a direct linear
relationship between fuel use and efficiency of converting the
calorific value released from the fuel into heat for the pot
contents..

As others have pointed out it is the difficulty in defining how to
measure or agree protocols that frustrates this.

Plus of course efficiency is not necessarily the most important aspect
of cooking.

AJH