## Condensers

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• 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 1 of 3 , Aug 2 12:14 PM
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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
• 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 2 of 3 , Aug 2 11:47 PM
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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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