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Re: Open Flames Burner Power

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  • Grayson Stewart
    ... 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 1 of 8 , Sep 2, 2004
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      --- 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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