blogfast25
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Spiral condenser requirements - fractionated distillation
I'm finally looking into upgrading my distillation set up, currently consisting of a 500 ml RBF, 500 mm Vigreux column and Liebig condenser.
I want proper control of reflux ratio with an overhead spiral condenser and timed flow splitter to control R and distillate take-off.
My question is about required dimensions of the spiral consenser. Knowing the vapour flow in mol/min, is there a design criterium that allows to
estimate spiral length (surface area), assuming cooling water flux is high enough so the water doesn't heat up significantly (isothermal spiral), to
ensure full condensation of vapour phase?
I'm bound to the spiral condenser type for practical reasons.
Would condensing capacity be affected by application of partial vaccuum?
[Edited on 14-2-2015 by blogfast25]
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Chemosynthesis
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Spiral like Graham condenser, or spiral in a different manner? Because I have some heat exchanger equations that should work for a Graham condenser.
Theoretically, the condensation should be effected by vacuum due to the differences in temperature, but I'm not sure if it'll be significant. http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/...
My completely amateur recommendation would be to get some thermal conductivities from Perry's Handbook, try to figure out if you need to adjust the
equations for goofy shapes, and then plug and chug.
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Crowfjord
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He said overhead - I think he means something similar to a Friedrich condenser.
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Magpie
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The critical parts for finicky distillations are:
1. Column design (type, packing, diameter, length, etc)
2. Reflux control (splitter, timer, etc)
Seems like most any condenser would do as long as it is compatible with the rest of the apparatus.
What am I missing here?
The single most important condition for a successful synthesis is good mixing - Nicodem
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blogfast25
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Graham, overhead.
Thanks Chemo, I remember that stuff from my engineering course and have the formulas in a book somewhere.
Constant wall temperature cross flow is probably what I need to be looking at as an approximation. Actual shape doesn't have that much influence.
It's a question of estimating minimum dimensions and then slightly over-sizing.
[Edited on 14-2-2015 by blogfast25]
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Zombie
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The simplest method is to figure the maximum BTU's the column will be subjected to, and match the thermal capacity of the cooling water to return to
room temp.
Without doing the math myself... if 1 liter of water can carry 1,500 btu's figure from there how large the column need to be. That number is just an
example, and most likely way off.
Precisely, water has to absorb 4.184 Joules of heat for the temperature of one gram of water to increase 1 degree celsius (°C). For comparison sake,
it only takes 0.385 Joules of heat to raise 1 gram of copper 1°C..
Using that as a basis there are plenty of on line conversion programs to get the correct answer.
Engineering toolbox is my favorite.
http://www.engineeringtoolbox.com/
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blogfast25
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Crikey, you yokels still use BTUs? 
It's not the column I'm concerned with but a suitably sized overhead Graham condenser to condense all vapour sent up in the column.
As a distiller you'll be familiar with that problem. You might even be able to tell me what size condenser you use on your own column(s) and for what
pot capacity.
Yes, engineeringtoolbox is good.
I'm an engineer but need to brush up on the relevant equations and material constants. It'll be a doddle! 
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Zombie
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"I want proper control of reflux ratio with an overhead spiral condenser and timed flow splitter to control R and distillate take-off."
This is the statement I am basing my answer on. I assumed (wrongly?) that by reflux, you were referring to liquid / vapor interaction inside the
column.
In distilling that is reflux. Condensing is done in the PC or product condenser. (again distilling).
Either way the math still applies. The maximum Yokel BTU's should be less than the stagnate capacity of the condenser.
I think the easiest way to sort it is fill your PC , and measure the amount of water / divide to find thermal capacity per mm or cm, and take it from
there.
For my own set up I have 5kw powering 18 gallons (pot / thumper), and a 30" x 1/2" product condenser inside a 1.5" water jacket.
Off the top of my head I forget the numbers but I do know I am approx 30% over the required "knock down" at full power.
You do want a margin for error in case the water stops or it gets heat soaked.
Key to all of it is keeping the water temp stable, ie: chiller / ice, whatever.
I know you knew 90% of this but I included it for others that might not.
Yokel huh... If I knew what that meant I might be flattered.
Edit:
I just saw the picture in my head. The PC or condenser will be standing on top of your column as the reflux agent.
I didn't understand that at first.
Either way the answer I posted is correct, and a few minutes on the conversions will sort it out for you.
[Edited on 15-2-2015 by Zombie]
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Fulmen
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I would start here: http://www.engineeringtoolbox.com/conductive-heat-loss-cylin...
| Quote: |
Conductive heat loss through a cylinder or pipe wall can be expressed as
q = 2 π k (ti - to) / ln(ro / ri) (1)
where
q = heat transferred per unit time per unit length of cylinder or pipe (W/m, Btu/hr ft)
k = thermal conductivity of the material (W/m.K or W/m oC, Btu/(hr oF ft2/ft))
to = temperature outside pipe or cylinder (K or oC, oF)
ti = temperature inside pipe or cylinder (K or oC, oF)
ln = the natural logarithm
ro = cylinder or pipe outside radius (m, ft)
ri = cylinder or pipe inside radius(m, ft)
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blogfast25
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Fulmen:
Something like that. I'm thinking of calculating the cooling capacity of one single torus, then work out how many tori are required. Approximating a
spiral by a series of tori, i.o.w.
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blogfast25
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Quote: Originally posted by Zombie  |
For my own set up I have 5kw powering 18 gallons (pot / thumper), and a 30" x 1/2" product condenser inside a 1.5" water jacket.
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A copper coil inside a water jacket? Very "Moonshiners" (I love that show!) What I'm looking at is very different: cooling water runs through a glass
spiral, vapours condense on it, drips down and a timer valve decides how much gets sent down (L) the column and how much gets tapped off (distillate
D). Reflux ratio R = L/D.
No offence meant but the rest of your post is non-scientific gobbledygook.
Out of sheer curiosity: is your thumper heated?
[Edited on 15-2-2015 by blogfast25]
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Fulmen
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That sounds like a reasonable approximation. Not entirely sure what else you should factor in, other than enthalpy of condensation and cooling to the
desired temperature.
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blogfast25
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| Quote: Originally posted by Fulmen | | That sounds like a reasonable approximation. Not entirely sure what else you should factor in, other than enthalpy of condensation and cooling to the
desired temperature. |
Well, heat transfer coefficients for one!
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Fulmen
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Of course, that's the "k" in the formula I linked to. I'm unsure if you need to factor the condensed liquid into the equation, it could alter the heat
transfer. Some margin of safety (this is never an exact science) would be prudent.
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blogfast25
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| Quote: Originally posted by Fulmen | | Of course, that's the "k" in the formula I linked to. I'm unsure if you need to factor the condensed liquid into the equation, it could alter the heat
transfer. Some margin of safety (this is never an exact science) would be prudent. |
Yes, I'll be assuming the cooling serpentine is wet.
The safety factor comes from choosing the right 'next size up', as these condensers typically come in 10 cm active length increments.
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Fulmen
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This might be useful: http://www.nzifst.org.nz/unitoperations/httrtheory8.htm
A vertical spiral should be close to a horizontal tube.
[Edited on 15-2-15 by Fulmen]
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blogfast25
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Yes, good find, thanks!
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