Energy needed to distill...

[Externet]

Hi all.

A litre of water at 25℃ needs a certain amount of calories to become steam at 100℃ ∆t=75℃ (75Kcal + 540Kcal =615Kcal) if I remember the figures.

That is a 1 ATM.

 

Under partial vacuum, such energy requierement can be less. Let's say it boils at 45℃ under such partial vacuum ; ∆t=20℃ thus would need 20 Kcal + 540Kcal = 560Kcal

 

---> Are the calories needed at 1 ATM and under partial vacuum constants ?

 

The difference in energy needed to achieve the same are different under the different pressures. Some extra energy has to be used to run a vacuum pump.

 

---> Is the sum of energy to boil at 1ATM equal to boil under partial vacuum + the energy to run the vacuum pump ?

 

In other words, does vacuum distilling uses less energy ?

 

Sorry for my poor wording

[swansont]

Let's say I have a mole of water. The vapor pressure is small even at room temperature (~3.1% of atmosphere at 25 ºC, to use your number). I will have to evacuate a certain volume in order to allow the water to boil and fill that space, and it will stop when the pressure reaches the vapor pressure. An ideal gas at room at 1 ATM takes up 22.4L, but as the vapor pressure of water is smaller, we need about 32 times that volume. The work I have to do to create this volume is PV, working against the 1 ATM I have, or about 72,400 J of energy. (101J/L * 22.4 *32)

 

1 mole of water is 18g, and that will take 18(75+540) = 11070 cal = 46,000 J to boil

 

So drawing a vacuum takes more energy in this example.

 

Another (perhaps simpler) way of answering the question qualitatively is this: if drawing a vacuum did take less energy, could you build an engine to do net work with that system? If the answer is yes, you have perpetual motion, and that would disprove the notion by contradiction. (draw a vacuum, let the water boil, reintroduce air, let the vapor condense — does that conserve energy?)

[John Cuthber]

The heat needed to boil the water is the latent heat of vapourisation.

There's a graph of how it varies with temperature here.

http://en.wikipedia.org/wiki/File:Heat_of_Vaporization_(Benzene%2BAcetone%2BMethanol%2BWater).png

The heat needed falls as the temperature is raised.

If you raise the temperature to the critical point the heat needed to evaporate the water is zero because there's no difference between the "gas" and "liguid" phases.

[Externet]

Thanks, gentlemen.

The answers escape my level of comprehension; I will need to learn more about the basics to grab the concepts, and perhaps re-word the same question in the future.

 

From what I understand, the energy to run continuosly a vacuum pump to evacuate the vapors in partial vacuum distilling plus the heat energy supplied is more than doing the same at 1ATM, no pump.

 

Choosing partial vacuum distillation over atmospheric is not aimed to save some energy, but for other reasons as maintaining a lower temperature; of when the temperature of the heat resource is low.

Then, to distill seawater, the 'energy price to pay' for the task is not decreased by the lower temperature-partial vacuum technique.

 

Something left out is

Takes 1K cal to raise 1 litre of water 1℃ at 1 ATM

Takes 540K cal to evaporate 1 litre of water at 1 ATM

----> If the water is in partial vacuum, are the figures 1Kcal/Kg and 540Kcal/Kg still valid constants to use ?

 

Thanks for the patience

[Mr Skeptic]

Well, if you have certain foodstuffs you might want to go with vacuum drying so you don't cook it. You can also use a different option: allow it to evaporate. Rather than using a vacuum pump, you use a fan. This will be slower as the pressure will be higher than the vapor pressure (ie, it won't boil), but also takes less energy. Here what you are doing is replacing cool moist air with warmer dry air. A desert would be ideal for this.

 

On the other hand, for 1.6 kJ you could take your mole of water up Mount Everest, where the pressure is about 1/3 lower, and so would the work to boil it via vacuum. Then using a vacuum might be more effective, though up there the temperature will be lower as well.

[swansont]

Something left out is

Takes 1K cal to raise 1 litre of water 1℃ at 1 ATM

Takes 540K cal to evaporate 1 litre of water at 1 ATM

----> If the water is in partial vacuum, are the figures 1Kcal/Kg and 540Kcal/Kg still valid constants to use ?

 

The latent heat value depends on temperature. Not a constant.