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Re: Reflux Woes still no azetrope etho.

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  • Harry
    ... Hi Mason, If you re not getting higher proof, that means there s excess impurities/water still entrained in the rising vapor. Therefore either A) vapor
    Message 1 of 47 , Jan 4, 2009

      --- In Distillers@yahoogroups.com, "rye_junkie1" <rye_junkie@...> wrote:

      > Need some help here guys.  I have been messing around with Reflux
      > columns for near a year now but only recently got serious about trying
      > to make true neutral alcohol.<snip>


      Hi Mason,

      If you're not getting higher proof, that means there's excess impurities/water still entrained in the rising vapor.  Therefore either

      A) vapor speed is wrong, (too much power) or

      B) separation capability is poor (too little packing height).

      So we apply a process of elimination.

      Let's examine this further.  First vapor speed...

      To find vapor speed, we need to know how much vapor volume your 650watt input is generating, and apply that to the volume capacity of your 1.5" x 45" packed column.

      Vapor Volume:
       1 kW input produces 45.77 cu.ins vapor per second (Mike Nixon's equations, see Files section).

      Therefore your 650 watt input produces
      .65 of the above, or 45.77 x .65 = 29.75 cu.ins vapor per second.

      Now to find the vapor speed:

      Speed S = 45.77 x kW / [PiDD /4] (Nixon)

      S = 45.77 x .65 / (3.1416 x 1.5 x 1.5 / 4)

      S = 16.83 inches per second

      Mike Nixon advises that a vapor speed of between 12 and 20 ins per second is ok.  You are right in the middle so you're fine there.

      Therefore the problem is...not enough packing height to achieve good separation.

      This holds true if you compare your column capacity to a 2" x 36" packed column, which is the standard.

      Volume = Area x Height, or PiDD /4 x Ht

      Standard column vol = 3.1416 x 2 x 2 /4 x 36 = 113 cu.ins

      Your column vol = 3.1416 x 1.5 x 1.5 /4 x 45 = 79 cu.ins.

      So you need to boost the height of packed column by another 34 cu.ins. of volume.  To find the extra length required...

      For 1.5" tube, that's 34 / (3.1416 x 1.5 x 1.5 /4 )= 19" more packed tubing.  That makes a total of 64" of PACKED height to achieve the separation you are chasing, using the equipment you have.

      I'd bite the bullet and re-design in 2" tube, but that may not be an option for you.  At least you know what the cause is now (and the remedy).

      It may also help to revue some theory at Tony's site...

      regards Harry

    • duds2u
      I found that there was a very small amount of play in the T piece that allowed me to do it so I didn t have to bend anything. ... pooling ? ... fitting
      Message 47 of 47 , Jan 9, 2009
        I found that there was a very small amount of play in the "T" piece
        that allowed me to do it so I didn't have to bend anything.

        --- In Distillers@yahoogroups.com, "Zapata Vive" <zapatavive@...> wrote:
        > How did you manage to "raise the ends of the cross to prevent
        > I think that means you bent the two horizontal arms of the "T"
        fitting slightly up? I thought of that, but couldn't think of a good
        way, especially without throwing them out of round. I also didn't
        really feel like messing around with it too much, that fitting cost me
        more than my column!
        > Tips much appreciated on this, I've got a friend thinking of
        building, and if I could convince him his would be even better than
        mine, he'd take the bait for sure!
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