RogueRose
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Evaporating water under vacuum - less energy intensive?
I'm trying to figure out how to determine how much energy is required to evaporate a gallon of water by boiling or by putting it under vacuum. I have
no problem determining the energy required to boil the water off, but don't know where to start when looking at the performance under vacuum. I would
assume the amount of energy will be determined by the efficiency of the pump.
I'm also wondering if somehow the heat of the motor could be used to heat the water and that alone would be enough or close to enough to evaporate the
water under pressure. I know my pump gets upwards of 150 if it has no air running through it and it can probably get even hotter.
How do I figure this out?
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JJay
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There are a lot of variables to consider if you're trying to calculate energy consumption without doing any experiments... it would probably be more
accurate to measure evaporation and energy use and do your calculations with those data points.
If you are simply looking for a low-energy method to evaporate water, you could use millennia-old techniques involving sun and wind.
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phlogiston
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The total amount of energy needed to go from one state (liquid water at temperature T and pressure P) to another (steam at temperature T2
and pressure P2) is always the same, regardless of what route (through pressure/temperature space) you take.
However, how much of that energy is in the form of heat, work (pump) etc will different for different methods.
[Edited on 30-1-2018 by phlogiston]
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"If a rocket goes up, who cares where it comes down, that's not my concern said Wernher von Braun" - Tom Lehrer |
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RogueRose
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Quote: Originally posted by phlogiston  | The total amount of energy needed to go from one state (liquid water at temperature T and pressure P) to another (steam at temperature T2
and pressure P2) is always the same, regardless of what route (through pressure/temperature space) you take.
However, how much of that energy is in the form of heat, work (pump) etc will different for different methods.
[Edited on 30-1-2018 by phlogiston] |
Thank you for the explination, that is what I was looking for and wasn't sure if liquids under vacuum had a different requirement for heat compared to
those under standard pressure.
But I would guess that if the water will boil at 140F under vacuum, it will still require ~540 calories per gram to convert from liquid to gas? It
seems that if there is any energy saved it would be from the lowering of the BP by 72 deg F, but IDK if the vacuum requires more or less energy to
create that pressure differential.
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JJay
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If any energy is saved, it would be due to less being wasted when heating the water. After all, heat would flow more slowly from a 140 F surface into
room temperature air than it would from a 250 F surface.
If your vacuum pump and heat source are perfectly efficient and no energy penetrates the closed system, it would take the same amount of energy to
evaporate the same amount of water with the vacuum pump as with the heat source. Of course, at some point the water would freeze and would no longer
evaporate if no heat were applied, and without a vacuum, some heat would be required to raise the water to its boiling point.
The heat of vaporization actually isn't quite constant at different pressures and temperatures, but it's pretty darn close to constant, and
imprecision due to changes in the heat of vaporization is not a significant source of error in many applications.
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SWIM
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The only really major savings in energy I can think of that you can get with vacuum is by using it to recycle the heat energy through multiple batches
of water.
The higher the vacuum, the lower the boiling point, so condensing water vapor at STP is hot enough to boil water at low pressure.
This can be done a few of times before the whole diminishing returns problem kicks in.
This actually reclaims the heat of vaporization as the water condenses and uses it to evaporate another portion of water.
I believe sugar boilers came up with this trick.
EDIT
Modern engineers, who build these things for industrial use call them multiple effect stills.
They can cut energy requirements for distillation from 50% to 60%.
[Edited on 2-2-2018 by SWIM]
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Chemvironment
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The goal is to raise the vapor pressure to that of the surrounding environment. As said before that will be the same whether you raise the water temp
and therefore vapor pressure or lower the surrounding pressure via a vacuum or a combination of the two.
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Chemetix
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Despite the lower temperature you can't escape the need for the amount of energy required to change a liquid to a vapour. To convince myself of this
and to check how valid the published values might be in a practical sense, I made a small apparatus that was two spheres connected to each other on a
sort of U tube.
I put a known mass of methanol in it and then sealed it under vaccuum. With a couple of styrofoam cups and a data logging thermometer the amount of
energy lost to evaporating the methanol was pretty much spot on the given value of the enthalpy of vapourisation of methanol.
I'm often doubting the given values of certain physical constants, trying to replicate them is almost impossible at times.
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AJKOER
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| Quote: Originally posted by JJay | If any energy is saved, it would be due to less being wasted when heating the water. After all, heat would flow more slowly from a 140 F surface into
room temperature air than it would from a 250 F surface.
......
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I suspect reducing wasted heat is the key to faster boiling.
I once observed very surprisingly what happens when water in an open vessel being strongly heated is completely covered with a thin layer of olive
oil. Normally, absence an oil layer, steam is increasing visible indicating some loss of heat. However, not so with the oil layer in place. Energy is
not dissipated, more is stored and it increases with heating. Eventually, the entire mass of water appears to erupt (shoot up) at almost once!
Caution, not a safe experiment to attempt, especially in a shallow pan!
So the water likely boils faster, but I do not recommend this technique for boiling water for normal food preparation.
[Edited on 4-2-2018 by AJKOER]
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