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Evaporating water using heat and reducing vapour pressure

  1. May 19, 2016 #1

    jmarkwalker

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    I want to reduce the amount of water present in the digestate coming out of my Anaerobic Digester.
    The traditional method would be to boil it until I've removed as much off as I want. However, this is expensive from an energy consumption point of view.
    There's a great video on Youtube showing a glass of water boiling at room temperature, by putting the glass in a sealed container (transparent) and using a vacuum pump to remove the surrounding air. The water gets cold and although the video doesn't show this, it will turn into ice. If the vacuum pump is left on, the ice sublimes and eventually disappears. I believe this is how 'freeze drying' works.
    The rate at which evaporation of liquid water occurs is a function of the water's temperature and the vapour pressure above the liquid water. The vacuum pump boils the water by removing the vapour pressure.
    So here's the start of my questions:-
    1) If I have 1L of water at room temperature and I want to evaporate half of it off, I can easily calculate how much energy I would require if I was to boil it off. How much energy would the vacuum pump consume doing the same task? Assume that the ambient temperature would maintain the liquid at room temperature.
    In practical terms, I don't have a vacuum pump but I can source some kind of pump that will reduce the pressure within my container (not eliminate it). So...
    2) Going back to the 1L of liquid water (maintained) at room temperature model, what would a graph of pressure-reduction vs time to remove half the water look like? Linear? Does a small reduction in pressure increase evaporation rate? What pressure drop is required to achieve boiling?
    3) What's the most energy efficient way of removing half the water in my digestate? A small amount of heat in the liquid and a fan blowing fresh air over the surface would work, but would a sealed container with with a pump sucking the vapour out be more efficient? How much power would that consume?

    Apologies for the long post.

    Mark
     
  2. jcsd
  3. May 19, 2016 #2

    SteamKing

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    You've got some misconceptions going on here which should be cleared up first.

    Freeze-drying is exactly that: the stuff which is freeze-dried is first frozen, so that any moisture present turns to ice, then the ambient pressure is reduced so that the ice crystals formed by the freezing will sublime directly to water vapor and then be removed from the rest of the substance. To reconstitute the freeze-dried material, simply add the water back to it and re-heat if necessary. Works great for making a cup of coffee from freeze-dried coffee crystals.

    https://en.wikipedia.org/wiki/Freeze-drying

    Simply putting something under reduced pressure does not necessarily lower its temperature. The ambient pressure over a liquid like water does however correspond to the temperature at which the water boils and turns into vapor. At normal atmospheric pressure, the boiling point of water is 100° C. Water boils at 100° C because its vapor pressure at that temperature equals a pressure of exactly one atmosphere.

    At half atmospheric pressure, the boiling point of water is reduced to about 80° C. If the ambient pressure is reduced to about 1 psi, the boiling point of water drops to about 100° F (38° C), barely above body temperature. This is the operating principle behind flash distillation units or flash evaporators to turn seawater into freshwater: instead of heating a bunch of seawater at 100° C or greater to turn it into vapor, take the same water at a temperature of 50° C and inject it into a space where the pressure inside is lower than the vapor pressure of water at 50° C, which is about 1.8 psi absolute. The liquid will immediately convert to vapor, or flash into steam. The steam can then be easily removed and condensed in a separate chamber into fresh water. Much more economical than boiling the whole mess at atmospheric pressure.
     
  4. May 19, 2016 #3

    jmarkwalker

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    Hi SteamKing,

    Thank you for your reply. You've managed to answer most of my questions.
    Thank you also for explaining the freeze drying process - I didn't realise that freezing was done before pressure reduction.
    Having explained how desalination is done, it shows to me that it is more energy efficient to reduce pressure rather than heat to 100 degrees C.
    You've also given some data about the boiling point at three lower pressure values. This is very useful because digestate comes out of the digester at approximately 38 degrees, so a pressure of 1 psi absolute would give me a boiling liquid. That's something I could aim for.
    I would still like to quantify the energy consumed by the pump reducing the pressure. I imagine the equations behind this will be quite complicated, but an approximation would be extremely useful.
    There are many businesses around the world (even in developed countries like the UK) using kilns to dry wood. These businesses could be transformed if pressure reduction is significantly more efficient than just applying heat.
    I also want to 'get my head around' the concept of increased evaporation as opposed to boiling. If I can't use a pump that is capable of getting the pressure down to 1 psi, or the vessel that the digestate sits in can not maintain its structural integrity at 1 psi, can I still get a good rate of evaporation whilst having the liquid not boiling?
    You've given me three data points:- 80, 50 and 38 degrees C. Thank you for this. Are you able to point me at a graph to give me the entire spectrum of boiling points vs absolute pressure. I've not found anything on google.
    One last question; if reducing pressure increases evaporation, then surely it must decrease temperature. That being the case, energy still needs to be put into the liquid (my digestate) to achieve boiling or a good level of evaporation.
    So back to my original question:- how does the 'energy in to reduce the pressure' and the 'energy in to maintain the heat' compare with the energy consumed to boil the liquid at atmospheric pressure? How much of an energy saving do I get by reducing the pressure by how much?

    Thanks again SteamKing.

    Mark
     
  5. May 19, 2016 #4

    SteamKing

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    In addition to drying the wood, often it is heated under pressure in kilns to treat it with chemical preservatives which protect the wood against rotting and insect attack.
    The increased pressure helps to drive the sap out of the green wood and to increase the rate at which the preservative compounds diffuse into the wood fibers.

    There are a variety of method for treating wood, some of which do not rely on high pressure:

    https://en.wikipedia.org/wiki/Wood_preservation

    In the home, cooks use pressure cookers to raise the boiling point of water beyond 100° C so that food can cook at higher temperatures without drying out excessively. Cooking time is also reduced.
    You may not need to get pressure down to 1 psia to see significant savings.

    You can use this page to calculate the boiling point for water at various pressures:

    http://www.efunda.com/materials/water/steamtable_sat.cfm

    You can select different units for temperature and pressure for your convenience.
    Without knowing more details about your setup, I can't really say.

    It should take a lot less energy to run a small vacuum pump than to heat up a given quantity of material to boil off the moisture.
     
  6. May 20, 2016 #5

    jmarkwalker

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    Thanks for the link. I've bookmarked it now.
     
  7. May 20, 2016 #6

    CWatters

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    Not sure if this helps but...

    I recently saw a TV program about the making of Maple syrup. They normally boil it to remove water but the program mentioned some factories use reverse osmosis first to reduce the amount of boiling needed. If reverse osmosis is cheaper than heating/boiling for that application, perhaps also yours?
     
  8. May 20, 2016 #7
    Even though you are reducing the boiling point by using the vacuum pump, you are still going to have to supply heat of vaporization to the liquid. So the energy requirements are going to be greater than just running the vacuum pump. If you wait for the heat of vaporization to be supplied by the surrounding room, you will be waiting a very long time and run the vacuum pump very slowly. Ask yourself how long it takes water to evaporate at room temperature into the surrounding air. The heat of vaporization at room temperature is even a little higher than at 100 C. Unless you supply the heat of vaporization to the water, it's temperature is going to drop very rapidly.
     
  9. May 23, 2016 #8
    Energy consumption vs. pressure reduction in a given volume of gas is available from vacuum pump manufacturers. It will vary, depending on the efficiency of the pump and motor and gearing (if any). You would have to make several assumptions to calculate.

    There is a way to get a win-win based on Chestermiller's comment above. You could put the pump and motor outdoors, the vacuum vessel indoors in the summertime. Heat supplied to the vacuum vessel will be removed from the indoor space, a small contribution to air conditioning. In the winter you could reverse the placing of the system elements, a small contribution to heating.
     
  10. Jan 20, 2017 #9
    If you consider heat of vaporization of water is close to the same whether you boil the water under atmospheric pressure or under vacuum, your main energy savings will be reducing heat lost to the apparatus and the surroundings. If vaporizing under vacuum, your system temperature differential, and heat loss to surroundings, (un-insulated system) will be reduced and system more efficient.

    Consider a partially-closed-system vacuum solution. Exhaust of the vacuum pump (water vapor) passes over electric motor and pump to cool the machinery (pump fully encased in ducting) and then exhausts over your evaporation vessel. The latent heat in the water vapor, turning to liquid water and returning heat of vaporization back to your vessel, will maintain your evaporation vessel temperature, thereby reducing your overall system energy requirement.
     
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