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Re: Open Flames Burner Power

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  • Grayson Stewart
    ... the ... This sounds like it will give you a close approximation of the energy removed from the vapor and received by the condesor. There should be all
    Message 1 of 8 , Sep 2, 2004
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      > I have 2 digital thermometers mounted in compression Tees. One on
      the
      > water inlet to the condenser and one on the outlet. I then measure
      > water flowrate thru the condenser

      This sounds like it will give you a close approximation of the
      energy removed from the vapor and received by the condesor. There
      should be all kind of losses to the condensor jacket and any other
      part of the setup that you can place your hand on that feels hot.
    • 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 2 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 3 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 4 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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