## Re: [Distillers] Re: Open Flames Burner Power

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• I am using a regular kitchen stove. I dont use propane cylinders. Whatever I wrote above is my subjective opinion There are no warranties of any kind Act on
Message 1 of 8 , Sep 2, 2004
I am using a regular kitchen stove. I dont use propane cylinders.

Whatever I wrote above is my subjective opinion
There are no warranties of any kind
Act on your own risk and finally...
I can be wrong I must say
Cheers, Alex...

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• ... Then you should be able to use the information from the first post alone. Only difference between yours and mine is I use internal elements. The losses
Message 2 of 8 , Sep 2, 2004
--- In Distillers@yahoogroups.com, BOKAKOB <bokakob@y...> wrote:
> I am using a regular kitchen stove. I dont use propane cylinders.

Then you should be able to use the information from the first post
alone. Only difference between yours and mine is I use internal
elements. The losses as you know are what is going to prevent you
from arriving at say 3800 W for a 3800 W element.

If you wanted to get really close, you could also well insulate
everything and then follow through with the earlier diescription,
but also calculate the energy to heat the amount of copper and
stainless in the setup.
*****************************************************************
specific heat of stainless steel=0.5 J/g C

specific heat of copper = 0.385 J/g C

mass of copper or stainless X temperature change X specific heat =
Joules

Joules divided by 3600 = Watthour

Watthour times the duration to boilup = Watts used
******************************************************************
Everything else is heat wasted to heat your kitchen, stovetop, etc.
Of course you wouldn't use the condensor during this boilup because
that would draw out additional heat. I'm sure you realize that,
just mentioning for others that may be reading the post.
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