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Re: Rate of collection equation

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  • Harry
    ... so ... Borrowed from Mike Nixon s excellent paper in the Distillers/files section...
    Message 1 of 6 , Nov 1, 2006
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      --- In new_distillers@yahoogroups.com, "tyler_97355" <kd7enm@...>
      wrote:
      >
      > I need the equation to find out the rate that i can theoretically
      > collect distillate from a pot still running no reflux at a certain
      > power input. I understand that no still will run without heat loss,
      so
      > the numbers might be off a little. I was just hoping to be able to
      > calculat the collection rate for a particular heat input, without
      > having to actually make it. Might also help when trying to determine
      > the rate of collection for a given reflux ratio.
      >
      > -Tyler
      >



      Borrowed from Mike Nixon's excellent paper in the Distillers/files
      section...
      http://groups.yahoo.com/group/Distillers/files/Vapor%20Speed%20101%20-%
      20Mike%20Nixon.htm
      ...or TinyURL http://tinyurl.com/y797ow

      <quote>
      The latent heat of vaporisation of both ethanol and water are about
      the same... 40 kJ/gram mole
      (this is just a happy coincidence ... other substances have different
      LHVs, so please don't generalize)
      So 1 kW vaporises 60/40 = 1.5 gram moles of either ethanol or water,
      or any mix of them, every minute.


      1 gram mole ethanol = 46gram
      1 gram mole water = 18 gram
      So 1 kW vaporises 1.5 x 46 gram ethanol or 1.5 x 18 gram water every
      minute.
      1 gram mole of any vapor occupies 22.4 liters at STP (standard temp
      and pressure)


      Vapor expands when heated, and at around 90 deg C 1 gram mole of any
      vapor occupies 30 liters.
      So 1 kW, which vaporises 1.5 gram mole of either ethanol or water each
      minute gives 1.5 x 30 = 45 liters vapor/minute ie. 45/60 liters/second


      But 1 liter = 61.024 cubic inches, so 1 kW will give 61.024 x 45/60
      cubic inches of ethanol or water vapor each second, = 45.77 cubic
      inches/second

      </quote>

      Does the brain still hurt, Tyler? :-)


      Slainte!
      regards Harry
    • sonum norbu
      I knew that!!!!! :) blanikdog ... From: Harry To: new_distillers@yahoogroups.com Subject: [new_distillers] Re: Rate of collection equation Date: Wed, 01
      Message 2 of 6 , Nov 1, 2006
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        I knew that!!!!! :) blanikdog

        ----- Original Message -----
        From: Harry
        To: new_distillers@yahoogroups.com
        Subject: [new_distillers] Re: Rate of collection equation
        Date: Wed, 01 Nov 2006 09:13:12 -0000

        --- In new_distillers@yahoogroups.com, "tyler_97355" <kd7enm@...>
        wrote:
        >
        > I need the equation to find out the rate that i can theoretically
        > collect distillate from a pot still running no reflux at a certain
        > power input. I understand that no still will run without heat loss,
        so
        > the numbers might be off a little. I was just hoping to be able to
        > calculat the collection rate for a particular heat input, without
        > having to actually make it. Might also help when trying to determine
        > the rate of collection for a given reflux ratio.
        >
        > -Tyler
        >

        Borrowed from Mike Nixon's excellent paper in the Distillers/files
        section...
        http://groups.yahoo.com/group/Distillers/files/Vapor%20Speed%20101%20-%
        20Mike%20Nixon.htm
        ...or TinyURL http://tinyurl.com/y797ow

        <quote>
        The latent heat of vaporisation of both ethanol and water are about
        the same... 40 kJ/gram mole
        (this is just a happy coincidence ... other substances have different
        LHVs, so please don't generalize)
        So 1 kW vaporises 60/40 = 1.5 gram moles of either ethanol or water,
        or any mix of them, every minute.

        1 gram mole ethanol = 46gram
        1 gram mole water = 18 gram
        So 1 kW vaporises 1.5 x 46 gram ethanol or 1.5 x 18 gram water every
        minute.
        1 gram mole of any vapor occupies 22.4 liters at STP (standard temp
        and pressure)

        Vapor expands when heated, and at around 90 deg C 1 gram mole of any
        vapor occupies 30 liters.
        So 1 kW, which vaporises 1.5 gram mole of either ethanol or water each
        minute gives 1.5 x 30 = 45 liters vapor/minute ie. 45/60 liters/second

        But 1 liter = 61.024 cubic inches, so 1 kW will give 61.024 x 45/60
        cubic inches of ethanol or water vapor each second, = 45.77 cubic
        inches/second

        </quote>

        Does the brain still hurt, Tyler? :-)

        Slainte!
        regards Harry



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      • tyler_97355
        Damn you Harry! I had to break out the headache pills on that one! lol. Ok, well i think you helped explain to me about why the collection rate slows down on
        Message 3 of 6 , Nov 1, 2006
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          Damn you Harry! I had to break out the headache pills on that one!
          lol. Ok, well i think you helped explain to me about why the
          collection rate slows down on my pot still when the tail start to come
          through. If 1 kW vaporises 1.5 x 46 gram ethanol or 1.5 x 18 gram
          water every minute, that is quite a noticeable difference, and a good
          way to tell when your tails are coming. However, i'm still trying to
          wrap my head around those equations. You spit all that out just to see
          how fast my head would spin, didn't you? As far as calculations go,
          1.5 x 46 = 69grams = 2.43oz, now this is by weight. Does 16 fluid oz
          weigh 1 pound? Depends on what the fluid is. Anyway, it would come
          out to 71.97 ml. Does that sound about right? If so, does that mean
          that 2kw would be 3 x 46 = 138?

          -Tyler


          --- In new_distillers@yahoogroups.com, "Harry" <gnikomson2000@...> wrote:
          >
          > --- In new_distillers@yahoogroups.com, "tyler_97355" <kd7enm@>
          > wrote:
          > >
          > > I need the equation to find out the rate that i can theoretically
          > > collect distillate from a pot still running no reflux at a certain
          > > power input. I understand that no still will run without heat loss,
          > so
          > > the numbers might be off a little. I was just hoping to be able to
          > > calculat the collection rate for a particular heat input, without
          > > having to actually make it. Might also help when trying to determine
          > > the rate of collection for a given reflux ratio.
          > >
          > > -Tyler
          > >
          >
          >
          >
          > Borrowed from Mike Nixon's excellent paper in the Distillers/files
          > section...
          > http://groups.yahoo.com/group/Distillers/files/Vapor%20Speed%20101%20-%
          > 20Mike%20Nixon.htm
          > ...or TinyURL http://tinyurl.com/y797ow
          >
          > <quote>
          > The latent heat of vaporisation of both ethanol and water are about
          > the same... 40 kJ/gram mole
          > (this is just a happy coincidence ... other substances have different
          > LHVs, so please don't generalize)
          > So 1 kW vaporises 60/40 = 1.5 gram moles of either ethanol or water,
          > or any mix of them, every minute.
          >
          >
          > 1 gram mole ethanol = 46gram
          > 1 gram mole water = 18 gram
          > So 1 kW vaporises 1.5 x 46 gram ethanol or 1.5 x 18 gram water every
          > minute.
          > 1 gram mole of any vapor occupies 22.4 liters at STP (standard temp
          > and pressure)
          >
          >
          > Vapor expands when heated, and at around 90 deg C 1 gram mole of any
          > vapor occupies 30 liters.
          > So 1 kW, which vaporises 1.5 gram mole of either ethanol or water each
          > minute gives 1.5 x 30 = 45 liters vapor/minute ie. 45/60 liters/second
          >
          >
          > But 1 liter = 61.024 cubic inches, so 1 kW will give 61.024 x 45/60
          > cubic inches of ethanol or water vapor each second, = 45.77 cubic
          > inches/second
          >
          > </quote>
          >
          > Does the brain still hurt, Tyler? :-)
          >
          >
          > Slainte!
          > regards Harry
          >
        • Harry
          ... come ... good ... to ... see ... oz ... Sore head or no, methinks you re beginning to understand it. :-) Slainte! regards Harry
          Message 4 of 6 , Nov 2, 2006
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            --- In new_distillers@yahoogroups.com, "tyler_97355" <kd7enm@...>
            wrote:
            >
            > Damn you Harry! I had to break out the headache pills on that one!
            > lol. Ok, well i think you helped explain to me about why the
            > collection rate slows down on my pot still when the tail start to
            come
            > through. If 1 kW vaporises 1.5 x 46 gram ethanol or 1.5 x 18 gram
            > water every minute, that is quite a noticeable difference, and a
            good
            > way to tell when your tails are coming. However, i'm still trying
            to
            > wrap my head around those equations. You spit all that out just to
            see
            > how fast my head would spin, didn't you? As far as calculations go,
            > 1.5 x 46 = 69grams = 2.43oz, now this is by weight. Does 16 fluid
            oz
            > weigh 1 pound? Depends on what the fluid is. Anyway, it would come
            > out to 71.97 ml. Does that sound about right? If so, does that mean
            > that 2kw would be 3 x 46 = 138?
            >
            > -Tyler



            Sore head or no, methinks you're beginning to understand it. :-)


            Slainte!
            regards Harry
          • tyler_97355
            Using the equation 1.5 x 46 = 69grams = 2.43oz = 71.97ml to estemate the amount of distillate produced per hour in a still that theoretically has no heat
            Message 5 of 6 , Nov 10, 2006
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              Using the equation "1.5 x 46 = 69grams = 2.43oz = 71.97ml" to estemate
              the amount of distillate produced per hour in a still that
              theoretically has no heat loss, I am going to attemp to do some
              thinking out loud.

              If i made a 50 gallon (189.26ltr) wash at 14% ethenol, and ran it
              through a stripping run to collect 62.9ltr of 40% ethenol, i need to
              find out how long it would take.

              If i am using a 3kw electric heated boiler with a non-packed stripping
              column, suing the equation above, (1kw = 71.97ml per min.) x 3 = (3kw
              = 215.91ml per min.), i assume that i sould use the equation: "62.9 /
              .21591", which would give me 291.3 minutes = 4.855 hours.

              So if my messed up calculations are correct, i can strip a 50 gallon
              wash from 14% to 40% in 4.855 hours at 3kw heat input.

              Did i do that right?

              If that is right, then a 4.8kw heater should distill it at 345.456ml
              per min, and get it done in 3 hours.

              I guess at this point i need to run my still at that heat input, and
              measure the collection rate. When i compare that to my calculations,
              it sould let me know how much heat loss my still has.

              I was thinking, If i am using a non packed, non refluxing column, i
              shouldn't really have to worry about the column "choking", right? It
              seems to me that i just need to make sure that i can effectively slow
              down and cool the vapors coming into the condenser so i don't have
              steam coming out.

              Well now that i gave myself a headache, could someone tell me if i am
              right on any of this, or do i need to go back to middle school?

              -Tyler
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