Cooling water circulators

[Rhodium]

Those of you who have chosen not to supply your condensers with tap water, but instead are using another setup, what are you using?

* A professional refrigerated water circulator? What did it cost?
* A bucket of ice-water with an aquarium pump?
* A 12VDC portable (car?) refrigerator and a water circulator? (is this powerful enough?)

How much "cold" is needed to supply a typical 1000ml aqueous reflux per hour (using the ice-water bucket example, how many kg of ice is needed)?

Using 10 liters of cooling water circulating from a bucket without any ice added, how much does the temperature rise over time?

[goiterjoe]

I have a 48 quart (40L) cooler with an aquarium pump mounted in the bottom that works well for my cooling needs.  I just start off with 20L of cool water and have no trouble with up to a 3L nitromethane reduction.  The rise in temperature is minimal.  The only time I ever have to add ice is when I'm reclaiming DCM or ether, as they are such low boiling solvents.  I also added ice to the cooler once when I was doing a vacuum isomerization of safrole, but when the ice melted and the water warmed up to just over room temperature, I didn't bother adding any more ice.  For most of the purposes a water reservoir is used for in making MDMA and meth, ice really isn't necessary.
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[placebo]

Using running tap water can be pretty useless depending on your condensor's capabilities and the vapour temperature of whatever you are distilling/refluxing. SWIM has a 34/35 dual coil condensor, 24cm effective length, 44cm overall length with a surface area of 900cm². So it is not small or inefficient by any means, and running a large (7L) distillation with tap water running for 2 days can make a significant contribution to your water bill, if applicable. Plus it aint exactly environmentally friendly.

SWIM also tried a recirculating ice-water setup. Purchase the largest esky/ice-chest you can. (preferably with a tap) 100L is great. Fill with ice (say 4x5kg bags and a little water to start. Purchase the largest submersible fountain pump from the hardware. They can be a bit pricey, but well worth it for the "head" they produce. They also have good filters on them and won't suck up ice particles. (SWIM has tried many other pumps and they don't last long, the thought of using an aquarium pump sounds like a joke but I suppose it could work for smaller setups, as long as you keep the bucket up high. Submerse pump to bottom of ice and go for it. This setup can run for about 4-5hours, then tap off a 3/4 of water and put another 4 bags of ice in.

SWIM is currently sick of this setup and has purchased a 150L chest freezer, (sealed, stainless steel inside) place pump in the bottom. 3/4 fill with water, turn on, and you have recirculating ice-water indefinately. Turn off and on as required, or setup an electronic timer to do so, or setup a thermostat to do so, or set the factory thermostat to it's lowest setting and it should be fine.

No matter what glass setup you have, SWIM would still go with the latter setups for its capacity and reliability.
Intelligence is not the recall of knowledge, but the ability to use it. (Rainman was a retard)

[Rhodium]

Using running tap water can be pretty useless depending on your condensor's capabilities and the vapour temperature of whatever you are distilling/refluxing. SWIM has a 34/35 dual coil condensor, 24cm effective length, 44cm overall length with a surface area of 900cm2.

Is there any way to roughly calculate how much water would be needed to effectively cool a reflux or distillation, given condenser surface area, ambient temperature, boiling point of the solvent and reflux "speed" (as in grams of solvent condensed per minute)?

Extra point: The same as above, but with a fixed water volume of 5000ml, how strong cooling capacity would you need a submersible cooling coil to have to keep the water constantly at room temp?

[Chromic]

It depends on the latent heat of vaporization as to how fast the cooling water will heat. In my experience the only time you need to watch out is when distilling anything aqueous.

20L will heat up to an unreasonably hot temperature within distilling 1L of water. Keep in mind that you will ruin your pump if you exceed ~40-50C for your cooling water.

1L of water cools ~4J/K.g and the heat of vap is ~2kJ/g! So you distill 1L, that's ~2MJ, which would bring 20L of water up by 25C. (anyone want to check my math?)  

[Chromic]

There sure is... heat transfer equations (knowing the surface area, convective heat transfer coefficient, rates for condensation, conductive resistance of the borosilicate, etc) should let you estimate (+/- 20%) pretty much everything you need to know about the system and how it's operating. You'd probably be better off doing a simple set of experiments and determine UA for your condenser...

However, if your condenser is holding back 100% of the reflux, then you can simplify the equations down to the basic thermodynamic set of equations as I briefly mentioned above.

[Rhodium]

Rated as: F in Maths

Whoah... But this assumes a perfectly insulated water container, right?

What about a THF reflux? Assuming the reflux returns 1g of THF per second, refluxing for 1000 sec. (= 16.67 minutes) that would be equal to distilling 1 kg of THF. So the specific energy of THF times 3.6 (the number of the above 16.67-minute episodes in an hour) which would be the number of degrees 25 L of water would heat up? Noo... Help! The math is overloading my brain! I don't know shit about physical chemistry... What would the hourly temperature increase be in 25L of water in my example?

[goiterjoe]

The pumps I have aren't aquarium pumps, they are fountain pumps that pump to a height of 4ft.  It's been so long since I bought them I forgot what they were.  You need to pay attention to the height the pump can push water.  If it isn't capable of pushing the water up your condensor, then it's going to cause some headaches trying to make it work.

As far as your question about calculating temperature rise and other variables, it is very possible to calculate these, but is it really worth it?  If you were going to build a permanent large scale operation, then you would be required to do this, but not for the things that go on around here.  All you might need to guesstimate at is how much cooling water you will need to dissipate off the heat of condensation or refluxing, and that is pretty simple to do. 
All paths are the same: they lead nowhere

[Chromic]

Sure, but my assumption isn't that off. I know it assumes an adiabatic water container, but it's not like it'll be able to dissapiate that much heat anyways.

Your pump only needs to be able to reach 18" and that should be good, you can pull the water over "the hill" and a small pump will keep a small trickle going thru the line.

Diethyl ether has a heat of vap of 30 kJ/mol, so if THF is similar, that would be 0.42 kJ/g, 420 kJ/h, or an increase of 4C / hr in 25L of water. So that's almost nothing... pChem isn't that hard, I don't know why everyone complains.  

[Mountain_Girl]

Rhodium:

To put the thermo into a reasonable formula:

M = mass of water in resevoir, approximately equal to volume (density ~ 1 kg/l) = V
Cp = heat capacity of water
T = resevoir temp.
DT =  resevoir temp. change
DHv = heat of condensation of substance being condensed (refluxed)
m = rate at which substance is being condensed
t = time

[rate of increase in resevoir energy] = [rate of energy removed from condensor]

    d(M.Cp.DT)/dt = m.DHv

    So,   d(DT)/dt = (m.DHv)/(M.Cp)

    Simplifying, d(DT)/dt = dT/dt
 
Therefore rate of temp. increase of resevoir:
       
             dT/dt = (m.DHv)/(M.Cp)  

Your example:
THF condensing at 1g/s = m 
DHv for THF = 411 J/g (Chromic's guess was good)
V = 25 l, so M = 25 000 g (keeping units consistent)
Cp = 4.2 J/g.°C

          dT/dt = (1 g/s x 411 J/g)/(25 000 g x 4.2 J/g.°C)
         
                = 0.0039 °C/s

                = 14 °C/hr

This does assume the resevoir neither loses heat to surroundings (ambient temp. colder than resevoir), or gains heat from surroundings (ambient temp. warmer than resevoir).
In the former case your calculation will give a conservative answer. In the latter, you should probably add another 5 or 10% to correct for this effect. In either case you'd make the resevoir a little larger than calculated.
Mountain Boy

[Chromic]

Sorry, my calculation is off by a factor of ~3.6! I was thinking you meant 1 kg / h.

Good explanation M_G. You'd make a good T.A.  

[noj]

Instead of all that math, I like to buy a few bags of ice and just add some when needed. A 5 gal bucket and fountain pump is adequate up to 2x MM reductions. Larger reductions require some ice. Ice is cheap.
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[Rhodium]

Mountain_person: Thank you very much! Your post made a whole lot more sense than mine.

noj: I like your simplistic take on the whole matter, but it is not in every country you can buy ice in your local convenience store. Most europeans would go "say what?" about the notion of buying ice cubes and bring home... And if you are going to buy a cooling unit of some sort, you need to know a ballpark figure for the heat evolution of some common lab setups.

[goiterjoe]

You have to factor in Newton's law of cooling on the left hand side of your equation for the heat dissipating off of the body of water, T'=k(T_ROOM-T).  This will make a considerable difference it total temperature rise.  I loaned out all my old thermo notes to someone, but I'll pick them up today and write out some more equations on this later this week.

if the body of water wasn't cooling itself, then mCpÄT(coolant)=m*(latent heat of vaporization of refluxing liquid) would be correct with fixed temperature values.
All paths are the same: they lead nowhere

[Chromic]

To get an accurate answer, this is not a question for thermodynamics! This is a question for heat transfer.

For the cooling equations you'd have to take into account three heat transfer coefficients for convection, one for conduction and one for evaporation. It would be a moderately complicated set of equations to approximate from a physical setup. You could more easily measure the cooling by filling the container with hot water, then measuring the temperature as a function of time. Graph that, preform regression on it, then use that equation to factor in the cooling. It would be a very simple differential equation for Mathcad to handle for anyone who hates to do the math.

But what it comes down to, is that 25L of water near room temp isn't going to be cooled by much more than 5K / hour. The approximation of an adiabatic container isn't too far off for getting a feel for the system.

[Ritter]

Another important factor to consider when calculating how much heat the water will absorb is the type of pump employed.  Many common small submersible pumps produce so much heat that they must be submersed to prevent destruction from overheating.  I have seen one small particularly popular submersible heat up to over 50'C when rigged to draw its supply of water without immersing the actual pump.  Submersible pumps can heat a five gallon bucket of water up to 35'C in an hour with no trouble- not to mention the water will also be heated by the solvent vapor it is condensing. 

If you don't have ice readily available, the best thing to do is to have a couple 5 gallon (or 20 Liter)buckets of room temp water on standbye so when the temp reaches about 35-40'C the pump can be pulled from one bucket and placed into the next one  containing fresh cool water.

Wow, I can't believe how detailed this thread has become considering its most simple topic!

[Chromic]

Ritter, that's just another reason to go for a small pump that only gives ~18" of head. They're cheaper, put off less heat and are more easily replaced.

[goiterjoe]

You could modify your return system to allow for better cooling by attaching a dispersing nozzle to your return line and spraying the water through the air as small particles before they fall into the reservoir.  In a dry environment, this might work well.  If you live in a humid environment, your lab would quickly become too muggy to breath comfortably in. 
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[ClearLight]

Make sure you aquarium pump has enough "head" to lift it from the ground to your condenser, like goiterjoe said... mine didn't, so precious bench space consumed.... Also, plastic ones can shrink at really cold temperaturs and appear to come unglued...  they quit working!  Quick patch job w/ teflon tape and it was back working...

Remember pump head!


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[goiterjoe]

If I were going to go through that much trouble, I would rip a fridge compressor, cooling coils, and radiator off of an old junk refridgerator and build a water cooler that could actually cool the recirculating water to well below room temperature. 
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