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Open Flames Burner Power

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  • BOKAKOB
    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
    Message 1 of 8 , Aug 31, 2004
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      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...






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

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    • Grayson Stewart
      ... measurements and calculation formulas for computing power in a boiler placed on a burner. The following is an example of how to calculate the theoretical
      Message 2 of 8 , Aug 31, 2004
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        --- 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.

        The following is an example of how to calculate the theoretical boil
        up time for 10 gallons of water. The example uses two internal
        elements, simply insert your assumed burner W in place of the
        elements:
        *********************
        10 gallons = 36,291 g
        1 cal = 1.163 x ee-6 kilowatt hour
        1 cal is needed to raise 1 g water 1 degree C.
        Using 1500W and a 3800W element = 5,300 W or 5.3 kilowatts

        Raise 36,291 g of wash from 21 degree C to 78 degree C
        36,291 g x 57 degree C = 2,071,836 cal needed.

        2,071,836 cal x 1.163xee-6 kilowatt hour / 1 cal = 2.409 kilowatt
        hour

        2.409 kilowatt hour / 5.3 kilowatt = .45 hour or 27 min.

        ************************************
        This is strictly theory and it will take longer to reach boil temp
        due to losses to environment, heating the boiler itself, etc.

        It would be difficult to directly calculate the magnitude of your
        burner because of the losses and no way of knowing the extent of
        those losses.

        You can however record the actual power and time it takes to reach
        boil temp in your setup then back through the above calculations to
        determine the amount of energy wasted to losses, ie. the energy to
        heat a 15 gallon stainless keg should be the same each time.

        After determining the energy losses plus the energy to heat the
        water you should be able to determine the magnitude of the burner.

        If I didn't explain this well just let me know, I was in a hurry.
        Hope it helps.
      • dean
        The specific enthalpy of water is 4.19 KJ/Kg. To heat 1Kg of water from 0-100 Q = 4.19 (kJ/kg.K) 1.0 (kg) (100 - 0)(K) = 419 (kJ) 1 watt = 1 KJ/second I think
        Message 3 of 8 , Aug 31, 2004
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          The specific enthalpy of water is 4.19 KJ/Kg.
          To heat 1Kg of water from 0-100
          Q = 4.19 (kJ/kg.K) 1.0 (kg) (100 - 0)(K) = 419 (kJ)
          1 watt = 1 KJ/second
          I think thats right Tech college was along time ago

          Dean.

          Grayson Stewart wrote:

          >--- 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.
          >
          >The following is an example of how to calculate the theoretical boil
          >up time for 10 gallons of water. The example uses two internal
          >elements, simply insert your assumed burner W in place of the
          >elements:
          >*********************
          >10 gallons = 36,291 g
          >1 cal = 1.163 x ee-6 kilowatt hour
          >1 cal is needed to raise 1 g water 1 degree C.
          >Using 1500W and a 3800W element = 5,300 W or 5.3 kilowatts
          >
          >Raise 36,291 g of wash from 21 degree C to 78 degree C
          >36,291 g x 57 degree C = 2,071,836 cal needed.
          >
          >2,071,836 cal x 1.163xee-6 kilowatt hour / 1 cal = 2.409 kilowatt
          >hour
          >
          >2.409 kilowatt hour / 5.3 kilowatt = .45 hour or 27 min.
          >
          >************************************
          >This is strictly theory and it will take longer to reach boil temp
          >due to losses to environment, heating the boiler itself, etc.
          >
          >It would be difficult to directly calculate the magnitude of your
          >burner because of the losses and no way of knowing the extent of
          >those losses.
          >
          >You can however record the actual power and time it takes to reach
          >boil temp in your setup then back through the above calculations to
          >determine the amount of energy wasted to losses, ie. the energy to
          >heat a 15 gallon stainless keg should be the same each time.
          >
          >After determining the energy losses plus the energy to heat the
          >water you should be able to determine the magnitude of the burner.
          >
          >If I didn't explain this well just let me know, I was in a hurry.
          >Hope it helps.
          >
          >
          >
          >
          >
          >
          > Distillers list archives : http://archive.nnytech.net/
          > FAQ and other information at http://homedistiller.org
          >Yahoo! Groups Links
          >
          >
          >
          >
          >
          >
          >
        • 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 4 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 5 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 6 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 7 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 8 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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