[Stoves] Crispin´s kiln-was Re: Traditional Charcoal Making

[Crispin Pemberton-Pigott]

Dear Friends

Tom wrote
>After studying comments on this list and other sources it is clear that a
>well designed and well built 24 ft3 (0.7 m3) wood fired kiln of the kind
>that Manny Hernandez has been building can do all that Crispin needs.

Actually I need a far bigger kiln and will probably go to 3 cubic metres on
the next unit.

>It is also clear that the problems with the electric kiln may be due to
many
>other reasons besides the fuel source. These same reasons are probably why
>we hear that most kilns in developing areas do not reach temperatures
higher
>than about 750C or 950 C.  

The main reasons seem to be poor combustion and then high excess air.  Damp
fuel (sawdust) I have noted in some places is a significant limiting factor
reducing the flame temperature (CIMOC in Boane, Mozambique being one
example).  Poor insulation is a factor but only because the total power
rating is too low.  Increasing the power level more than compensates for
poor insulation.  Unless there is a real concentrated effort to limit the
excess air, and like a good stove, preheat the secondary air, there is
little one can do about the low temps in rural kilns.

>From what I have read, in the past people compensated for high excess air by
greatly increasing the power input.  OK...it works but is very wasteful.

How does the math look if the exiting gases (chilled to the internal temp of
the kiln) are say, 900 C, and are used to preheat the secondary air to 700
C.  Initially the secondary air flow would remain the same for an equal gas
input.  Being hotter, the amount of gas can be reduced by an amount equal to
the quantity of heat recovered from the chimney.

I'll bet Tom Miles has done this type of calculation before.  What's the
practical limit?  Is it determined by the heat exchanger materials on the
hot end?

T1= flue temp
T2= heat exchanger wall temp
T3= secondary air preheated temperature
T4= old (initial) secondary air inlet temperature

T3-T4 = the temperature increase obtained for secondary air.

(T3+273)/(T4+273)= relative energy content ratio.

Gas heat value at ?? MJ/Kg

Reduce gas volume produced by an amount equal to the increased heat content.

Reduce the secondary air needed to burn the reduced gas volume.

This would probably increase the secondary air temperature slightly.  Run
the programme again a couple of times.

Note the final gas requirement.  Check the system efficiency gain.

Smile.

I think I can handle this, certainly on a spreadsheet.

Thanks
Crispin