Most efficient windscreen for remote canister stove

[DAN-Y]

Brad, my testing is consistant at 70 degree starting temps. Stovebench testing has to have set parameters.

[bradmacmt]

Brad, my testing is consistant at 70 degree starting temps. Stovebench testing has to have set parameters.

Dan, yes, I’ve been clear on that from the beginning.

[Jerry Adams]

I like that modification Chris.  I believe you mentioned that previously.

yeah, 2 g of fuel used isn’t useful without knowing temperature rise of the water.

Usually, when I heat water, it starts pretty cold.  Maybe 40 F or something.  If I had a target end temp it would be 195 F or so.  And if I’m comparing two conditions, like with and without windscreen, I’ll normalize with the actual temperature difference.  If my target was a 155 F temperature rise, but I only did 150 F on a particular test, I’ll multiply the measured fuel used by 155 / 150 and use that for comparison.

[Jon Fong / Flat Cat Gear]

A simple way of looking at it is that you need ~5 grams of fuel to raise 2 cups of water from 70F to 212F so that is about 5 grams to raise the temperature by ~150 F or 30 F per gram of fuel. So for every 30F lower than 70 F (water at 40 F) you need to use 1 gram more fuel.

Stove Efficiency is simply the Energy required to boil water/Energy available. Once you know the Efficiency, you can calculate fuel consumption based upon water volume, water temperature etc.

Energy required to boil water Mw*cp*ΔT
Mw=Mass (g)
Cp= Specific Heat Capacity (4.186 J/gC)
ΔT =difference in temperature (C)

Energy Available (Isobutane) Specific Energy*Mass
Energy Density (isobutane) – (45208.8 J/g)
Mg = Mass of gas (g)

Efficiency = (Mw*cp*ΔT)/(Specific Energy*Mg)

A stove boiling 2 cups (473 ml) of 70 F water to 209 F using 4.9 grams of fuel has an efficiency of 70%.
Now what happens if you are boiling 500ml of 45 F water at 10,000 feet, how much fuel is required?
The boiling point of water at 10,000 feet is 194 F (or 90 C) with a starting point of 45 F (7.2 C)

0.74=(500g*(4.186 J/gC)*(90C-7.2C))/(( 45208.8 J/g)*(Mg))
0.74=(473*4.186*82.8)/(45.2088*Mg)

Solving for Mg = 5.5 g of fuel

 

My 2 cents

[bradmacmt]

I was told there would be no math…

[DAN-Y]

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