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marky
climber
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Topic Author's Original Post - Aug 23, 2007 - 07:54pm PT
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I'm playing around with my stove (trying to estimate burn time for one of the bigger isobutane bottles), and I had the stove run for 20 minutes on medium-high.
When I killed the stove, I grabbed the bottle, which was cold and somewhat dewy (it's about 75 and sunny here). Is this typical? Seems kind of odd.
The stove is the Brunton (nee Optimus) Crux; the gas is Primus 4-season "Power Gas" (propane/isobutane/butane)
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Mighty Hiker
Social climber
Vancouver, B.C.
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Aug 23, 2007 - 07:56pm PT
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Yes. The fuel bottle is under pressure. As it loses pressure, AOTBE, it cools. Compression = heating. Decompression = cooling. Usually.
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paganmonkeyboy
Trad climber
the blighted lands of hatu
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Aug 23, 2007 - 07:56pm PT
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as the pressure leaves the bottle it will cool - pressure and temperature are directly related by the ideal gas law
PV=nRT
http://en.wikipedia.org/wiki/Ideal_gas_law
the wet is condensation forming on the bottle as it gets colder.
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marky
climber
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Topic Author's Reply - Aug 23, 2007 - 07:59pm PT
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oh, okay. thanks. I never noticed this with MSR bottles and white gas stoves.
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Burns
Trad climber
Nowhere special
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Aug 23, 2007 - 08:03pm PT
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I think it's related to the ideal gas law:
pV=nRT
p=pressure. Pressure is decreasing
V=volume. Volume remains the same
R=universal gas constant
n=moles of the gas. This decreases, but apparently at a rate such that:
T is also decreasing.
This is also why those gas canisters don't do as well in very cold temps.
'course, all the sciency guys on the forum will probably point how much of a tard I am.
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scuffy b
climber
The deck above the 5
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Aug 23, 2007 - 08:18pm PT
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Isn't the effect compounded by the evaporation of the liquid in
the tank? The fuel leaves the tank in vapor form, right?
So there is also the change-of-state energy cost.
EDIT: You should see what happens when a large propane tank is
used to fuel a heater at a cold job site. Often there is ice on
the outside of the tank, even if the air is above freezing.
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paganmonkeyboy
Trad climber
the blighted lands of hatu
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Aug 23, 2007 - 08:28pm PT
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i think scruffy nailed it - i didn't even look at the phase diagram...i think there is some enthalpy at work here too ?
http://en.wikipedia.org/wiki/Enthalpy
where's ed ? he'll tell us all we need to know i'm sure - this should be cake for him...
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climbrunride
Trad climber
Durango, CO
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Aug 23, 2007 - 08:36pm PT
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There's the Ideal Gas Law, but this is also eaasy to understand through the comparative formula:
P1 V1 / T1 = P2 V2 / T2
Or, "PV over T equals PV over T". So the initial pressure multiplied by the initial volume, divided by the initial temperature yields the same result as the final pressure multiplied by the final volume, divided by the final temperature. If you decrease both the volume and pressure, then the temperature too, decreases.
However, you don't feel the full "real" drop of temperature, since the (now colder) cannister is constantly absorbing heat energy from the surrounding environment.from the
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paganmonkeyboy
Trad climber
the blighted lands of hatu
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Aug 23, 2007 - 08:41pm PT
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does the fuel leave the tank in vapor form or does it pass through a heater tube like the white gas stoves ? i don't have any experience with one of these stoves, but i think the in msr the fuel leaves as liquid and has to heat up in the system to vaporize...
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| Messages 1 - 9 of total 9 in this topic |
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