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Re: Open Flames Burner Power

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  • Grayson Stewart
    It just occured to me that your original post was really a two part question. One to determine the boilup parameters and the other question being the
    Message 1 of 8 , Sep 2, 2004
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      It just occured to me that your original post was really a two part
      question. One to determine the boilup parameters and the other
      question being the magnitude of the burner itself.

      The boilup can be calculated with and without losses from my first
      post, but you can pretty accurately determine the magnitude of your
      burner on your next run. Weigh the propane cylinder as acurately as
      possible before and after the run and record the time spent with the
      burner on a particular setting.

      [Change in weight of cylinder in kg) times 12.9 KWh/kg divided by
      time in hours on a particular setting.

      This should give you the KW of that particular setting of your
      particular burner. Would be more acurate with good scales and
      averaged over a couple of runs.
    • BOKAKOB
      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 2 of 8 , Sep 2, 2004
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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 your own risk and finally...
        I can be wrong I must say
        Cheers, Alex...






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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 3 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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