Vacuum distillation calculations ?

[Externet]

Hi.

Lost in space on how to calculate, or how to make a nomogram/chart; please some guidance.

 

1 litre of water at 20C in a regulable vacuum chamber, no heat added.

At what amount of vacuum will the litre evaporate in one hour ?

 

Or, if wanted to evaporate 1 litre in one hour, how much vacuum should be applied ?

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Now, the water can freeze if the vacuum is considerable ( I do not know the figure), so that should be avoided if evaporation is desired, as if frozen, the 'sublimation/evaporation' speed would considerably decrease -Am I wrong there ?

 

The optimal condition could then be to keep the boiling water from freezing (about not less than 5C) due to vacuum. Am I wrong here ? Is the water freezing at 0 C or at another temperature due to the vacuum ?

 

How to calculate/chart the amount of vacuum that will evaporate the litre of water faster ?

 

Will the litre of water evaporate faster under vacuum if its starting temperature is 80C ?

 

What would be the temperature of the water vapor flowing trough the vacuum pump suction hose ?

 

The X-Y chart could ideally fit initial water temperature, final water temperature, vacuum amount, time, volume evaporated...

 

Thanks for light on any or all of the above.

Miguel

Edited May 26, 2009 by Externet

[John Cuthber]

The rate at which the water evaporates is largely determined by how fast heat is supplied to it.

 

If you put some water in a Dewar vessel (which is very well insulated) and connected it to a vacuum pump the water would initially boil and get colder, soon it would freeze. The ice would then sublime and get colder still. Eventually the ice would reach a temperature so low that the vapour pressure would be the same as the limit of the pump's performance and the system would just sit there; very cold and not evaporating.

[Externet]

Thanks, John.

OK, understood that the only heat of vaporization is supplied by the amount of heat the litre of water has at the initial temperature. When it gets 'all used up', water freezes. So I have to supply heat if the initial water temperature does not provide the calories needed.

 

Is the heat of vaporization constant 540Kcal/litre also under vacuum ?

 

How do I calculate the amount of heat the 1 litre will yield at a given starting temperature?;

Or; At what initial water temperature should the process begin to have all the water vaporized with no freezing ?

 

Miguel

[John Cuthber]

"Is the heat of vaporization constant 540Kcal/litre also under vacuum ? "

I'm not sure but I don't think the latent heat changes much with pressure.

 

"How do I calculate the amount of heat the 1 litre will yield at a given starting temperature?;"

from the heat capacity of water and the temperature drop.

Once you get a little bit below 0C the water will freeze and that won't help.

If the water started at 100C it could drop its temperature by 100C before it froze.

The heat capacity of water is pretty close to 1 cal/g/c (it's exactly 1 at 15C by definition).

 

1 Kg of water cooling by 100C will liberate 100KJ of heat.

Clearly that's not enough to evaporate it because it's less than the 540KJ/litre letent heat.

 

You would need to superheat the water to something like 540C in order to have it all flash to steam when exposed to a vacuum.

(actually things get more complicated; you can't assume that the latent heat and heat capacity are constant over that big range also the stuff would turn into a supercritical fluid and the question of evaporation gets poorly defined, but the problem is still there.)

 

Even with a vacuum, you still need to supply extra heat to evaporate off the water.