[Stoves] Fan-blaster Combustors, Stove Camp 2006, and theDell-Point Pellet Stove

[Crispin at newdawn.sz]

Dear Tom'n'All

>...what is the minimum power requirement for
>a stove to be tested with the 5 Liter WBT?

Trying to be helpful here:

The heat required to boil 5 litres in 30 minutes at a thermal efficiency is 
30% is about 2.9 KW.  That is the continuous power required.  The total heat 
needed is about 5.25 MegaJoules (which is 5.25 megawatts but most people 
don't think that way).

To simmer for 45 minutes at an efficiency of 40% boiling off 150cc of water 
while doing so (maintaining the water temperature within 3 degrees of the 
local boiling point) would depend a lot on whether there was heat stored in 
the stove etc.

I suggest that it is likely that for most stoves, simmering needs about 1/3 
of the power required to boil.  That means another 1.3 MegaJoules (not 1.75, 
because of the higher efficiency).  That is a heat requirement of about 500 
watts for 45 minutes.  In a high mass stove this heat need not be generated 
continuously because it can be invested in the stove mass and collected as 
the water temperature moves in the 3 degree range. I tried this on Wednesday 
some considerable success, using a 40 Kg concrete and ceramic STOvs Cooker 
with 5 litres in the pot.

A small stove with a similar efficiency but lower power (like a 1.4 KW 
paraffin wick stove) would take 62 minutes to boil 5 litres. A 2.3 KW FSP 
stove with a skirt will boil it in 20 minutes because it has a higher 
thermal efficiency and higher power.

One problem with the boiling test is the demand to take off the lid. 
Obviously if the evaporation requirement + heat radiated and convected off 
the water surface + the losses from the pot sides, stove components + wasted 
heat in the exhaust stream comes to a total higher than the stove can 
generate, the pot will never boil, as has been demonstrated several times 
with stoves in common use during the UCB WBT.

The same applies to stoves with pots on which are too big for them, even 
with a lid on.  At some point the losses equal the input and the temperature 
stabilises.  As the temperature of the water goes up, so do the losses at 
the same Delta T^4 (Kelvin) that applies to collecting the heat in the first 
place.

Regards
Crispin



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