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RE: Condensers

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  • Tony & Elle Ackland
    Posting on behalf of PSagnelli ... ... I would like to try to sell 2 condensers. I feel they would be perfect for home distillation if someone was building a
    Message 1 of 3 , Mar 28, 2000
      Posting on behalf of PSagnelli ...

      > From: PSagnelli@...[SMTP:PSAGNELLI@...]

      I would like to try to sell 2 condensers. I feel they would be perfect for
      home distillation if
      someone was building a still. They are all copper, tube in shell
      construction and pure Tin coated on the steam side. Made by the Barnstead
      Co. New never used. About 24" long and 3" dia. and can condense 10 GPH .
      My phone is 203-846-4720
    • tnoftsger@hotmail.com
      Hi all, I am considering switching to a condenser much as described in new book, Making pure corn whiskey - a professional guide for amateur and micro
      Message 2 of 3 , Aug 2, 2001
        Hi all,
        I am considering switching to a condenser much as described in new
        book, "Making pure corn whiskey - a professional guide for
        amateur and micro distillers" that was discussed on this board a
        couple of weeks ago. I though i would try to determined the surface
        area needed in m^2 in my coiled 3/8" tubing. I have come up with 3
        meters of tubing with a flow rate of 400ml/min will give a heat
        transfer rate of 11,139.45 watts /m^2 of surface. Does this sound
        right? Anybody familiar with these calculations?, Thermodynamics is
        not my strong point.
        I am going through this trouble in an attempt to model each component
        of my still in an Excel Spreadsheet. It should also help give the
        user some fairly close flow rates and expected performances. Thanks


        Tom
      • Tony & Elle Ackland
        Tom, Thermodynamics .... I ve got a wee calculator at http://www.geocities.com/kiwi_distiller/cond_calc.htm to help with sizing condensors ... but here s the
        Message 3 of 3 , Aug 2, 2001
          Tom,

          Thermodynamics .... I've got a wee calculator at
          http://www.geocities.com/kiwi_distiller/cond_calc.htm to help with sizing
          condensors ... but here's the theory behind it ...

          Q = UA dt

          where

          Q = heat transfered (W)
          U = overall heat transfer coefficent
          A = surface area
          dT = temperature difference

          Q is the amount of heat to be removed. Provided you've got no other heat
          sinks in the system (eg losses to atmosphere via the walls of the column or
          pot), then this should be pretty close to the amount of heat you are
          putting into the system via the element.

          now... first problem (of which I'm a little unsure).. dT sometimes this is
          taken as the difference between the saturation temperature of the vapour
          (eg its dew point) and the wall temperature of the cold condensor (eg dT =
          Th - Tw). Other times, when you have the cooling liquid heating up (as it
          does), and the condensing vapour cooling down, you'd use the "log mean
          temperature difference", eg
          dT = (dT2 - dT1) / ln (dT2/dT1) , where dT2 is the temperature change in
          liquid 2, and dT1 is the temperature change in liquid 1. (its for this
          reason why you run the cold water in from the bottom of the condensor, so
          that it will have the most effect). In my calculator page I've used the
          "log mean temperature difference" - but am open to suggestions.

          A = surface area. Easy - length of the tubes x their circumference. Their
          circumference is pi (3.1412) times their diameter. Thus a 30cm length of
          6mm diameter tubing has a surface area of 0.3 x 0.006 x 3.1412 = 0.0056 m2

          U = overall heat transfer coefficient. This is a number which describes
          how easily the cold water gives its heat/cooling to the wall of the
          condensor (eg fast flowing cooling water better than slow), how easily the
          heat is transfered through the wall of the condensor (eg skinny walls of
          metal are far better than thick plastic walls), and how easily the
          heat/cooling is transfered to the vapour (eg vertical tubes vs horizontal,
          vapour flow velocity, etc). Each of these terms need to be calculated
          seperately, and then combined together. Sometimes, if one is really large,
          and the others are small, then you'd only bother with the large one. My
          calculator assumes a fixed value of 850 W/m2C. I took this from a table of
          typical values for "organics" being condensed in a shell-and-tube heat
          exchanger. They tend to range in that situation from 700 to 1000 W/m2C

          one expresion used to estimate it for vertical tubes is ..

          0.93 (( kf^3 x rho-f^2 x g x lambda) / (dTo x L x mu-f)) ^ 0.25

          where
          kf = thermal conductivity of the film
          rho-f = density of the fluid
          g = 9.81 m/s2
          lambda = heat of vapourisation
          dTo = change in temperature of fluid from outside surface to wall
          mu-f = absolute viscosity of the fluid

          I spent an evening once trying to work all this through for the case of our
          simple condensors, had a wee whisky, and decided that I'd simply use the
          approximation of 850 W/m2.C

          If say our "log mean temperature difference" is for cooling water entering
          at 10C, and exiting at 40C, and the dew point is 82C, then dT = (40-10)/
          ln((82-10)/(82-40)) = 55.6, then the heat transfer U.dT = 55.6 x 850 =
          47,310 W/m2

          This is quite a bit larger than your value of 11000 W/m2

          Yours is more like that obtained for a coil through a pool of water - eg
          like a coiled condensor in a bucket of cold water.

          So... the short answer is - try using more like 47000 W/m2 than 11000 W/m2

          But I'd suggest that you use the 850 x the LMTD so that you can juggle
          different inlet/outlet temperature combinations to see their affect.

          The other thing to calculate while you're in there is the amount of cooling
          water you'll need. This lets you judge whether or not you want a short
          condensor with heaps of cooling water, or a longer condensor with less
          water flowrate.

          here Q = m cp dT

          where

          Q = heat transfered (which should be the same Q as before)
          m = mass flowrate of water in mL/second (eg 400 mL/min = 6.66 mL/sec)
          cp = heat capacity of the water = 4.184 J/mL.C
          dT = change in water temperature ...

          or to rearrange for m; m = Q / (cp x dT)

          so... for a 1500W heat, and the water going from 10C to 40C,

          m = 1500 / (4.184 x (40-10)) = 11.95 mL/s = 717 mL/min

          remember for Q of the final condensor - if you have already had some
          refluxing coils take away some of the heat, then the Q for the final
          condensor will be the heat input - heat removed at reflux

          Tony
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