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Re: Open Flames Burner Power

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  • toddk63
    Here s how I do it for an actual run. I have 2 digital thermometers mounted in compression Tees. One on the water inlet to the condenser and one on the
    Message 1 of 8 , Sep 1, 2004
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      Here's how I do it for an actual run.

      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 ( at the outlet drain) with a 2
      liter pitcher and a stopwatch, measuring how much water I fill up in
      30 or 60 seconds. Then apply the following formula

      kW= (T2-T1)/1.8 * (V /t) * 4.18677

      where T1 and T2 are in degF
      V is liters
      t is seconds

      T2 and T1 need to calibrated against each other at the same temp, so a
      correction factor can be applied to one or the other thermometer to
      eliminate bias due to instrument innaccuracies.

      This is the actual power, disregarding all losses. I can then use
      this kW number to estimate the zero reflux rate, and then measure the
      actual distillate collection rate to calculate my reflux ratio. The
      theoretical numbers of this reflux ratio calculation match very
      closely to real world, I have found.

      Todd K.


      --- In Distillers@yahoogroups.com, BOKAKOB <bokakob@y...> wrote:
      > I would like to ask if someone could post a sequence of measurements
      and calculation formulas for computing power in a boiler placed on a
      burner. I am sure there is a formula to calculate the wattage required
      to heat the given volume of water (say water) from one temperature to
      another with and without losses. I would like somehow to calibrate or
      at least to know the magnitude of my burner. For example, fully open
      it produces 4000 watt and half closed it is 2300 watts. Are there any
      mechanical or chemical engineers? Thank you in advance, Alex...
      >
      >
      >
      > 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...
      > ®
      >
      >
      >
      >
      >
      > ---------------------------------
      > Do you Yahoo!?
      > Win 1 of 4,000 free domain names from Yahoo! Enter now.
      >
      > [Non-text portions of this message have been removed]
    • 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 2 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 3 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 4 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...






            ---------------------------------
            Do you Yahoo!?
            Win 1 of 4,000 free domain names from Yahoo! Enter now.

            [Non-text portions of this message have been removed]
          • 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 5 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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