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Water evaporation using solar reflector dishes

  1. Mar 17, 2009 #1
    Hi, we are running experiments with solar power; any inputs on below problem are highly appreciated.
    TNX Robby


    Project:
    We are experimenting how to maximize evaporation, using Solar Power. We have set up parabolic dishes with a fixed focus (Scheffler Reflectors). Each reflector can produce 3,4 KWh Thermal.

    A closed vessel having 300 liter capacity is filled with 200 liter of water. The vessel has an opening on top to which a pipe is connected. The pipe is connected to a vacuum pump which is producing a constant vacuum of 0,5 bar.

    The surface of the water inside the vessel is 1 square meter.

    The vessel is heated constantly with 3,5 kWh thermal (one reflector).

    The water evaporates and the vapor is "sucked out" via the vacuum pump.

    We are running a test series and will get the actual results. How can we do the theory? (we are more the hands on type of guys)

    Question 1.) How to calculate the evaporation rate?

    Question 2.) How can one maximize the evaporation, keeping the heating rate constant, and the vacuum constant at the same levels?
     
  2. jcsd
  3. Mar 17, 2009 #2
    No takers ? Ok some more input:

    We have done loads of experiments under pressure with up to 3 stages of condensation. Using superheated steam for the evaporation of the next stage. Unfortunately we did not get the desired results. We want to distill as much water as possible using solar energy.

    So now we will try doing it under low pressure.

    Basics for achieving maximum evaporation:


    1.) Pressure: Reducing the pressure means less exertion on the surface (right now we have a pump which can produce approx 0,5 bar low-pressure)
    2.) Boiling Point: At low-pressure the boiling point will come down and the available heating energy will be utilized better. (at 0,5 bar the boiling point should be reached at 81,35 Celsius)
    3.) Surface: The larger the surface area the more surface molecules will be able to escape. (in the vessel we currently use the water surface is 1 square meter – this we could change easily by adapting the design)
    4.) Temperature: The higher, the hotter the substance the faster evaporation will happen. Here we are limited to the input of our solar reflector dishes. We use 16 square meter Scheffler reflectors. So we can get only 3,5 KWh Thermal – assuming our normal conditions. (we are experimenting in a very sunny and arid area)
    5.) Flow rate of air: Fresh dry air moving over the surface will increase the evaporation, and keep the concentration of vapor in the vessel down. We have openings on the side of the vessel, closed with valves. They will open when the negative-pressure drops below 0,5 bar allowing a constant flow of “fresh” air from outside. This flow will draw the vapor from the vessel to the condensing unit.
    6.) Concentration: The lower the concentration of vapor in the air the higher the evaporation rate. – Covered by point 5.


    So I think we basically covered the theoretical issues. The question remains. How to calculate the amount of water which is evaporating?

    Option a)

    The formula below is for evaporation levels of Pools:


    • W = Rate of evaporation at the surface of the water level (kg/h m2)
    • Pw = Vapor pressure at saturation taken at the temperature of surface of water, in kPa
    • Pa = Vapor pressure at the dew point according to the temperature of the ambient air of the room, in kPa
    • V = Air velocity above at the surface of water, in m/s
    • Y = Latent heat necessary according to the change of state of the water vapor at the temperature of surface of water, in kJ/kg

    Option b) Shah equation: (here A is the water surface)
    M/A + (95+0.425V) x (Pw – Pa) / Y


    Any feedback? TNX Robby
     
  4. Mar 17, 2009 #3

    Danger

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    Gold Member

    Patience, lad. Although this is a global forum, a large percentage of us are in North America. The only reason that I saw this thread was because I woke up for a bladder relief. You can't expect a deluge of answers when most of your audience is asleep.
     
  5. Mar 17, 2009 #4

    Astronuc

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    1. Well - using conservation of mass - the evaporation rate would have to equal the rate at which one collects the distilled water, unless there is condensation back into the feed/effluent stream.

    2. As one mentioned, lowering the pressure reduces the energy required to vaporize the water, and adding a stream of air would improve the vapor removal rate.

    Does one use the feedwater as a cold fluid in the condenser?

    Basically one uses min = mout (mass conservation) and

    min hin + Q = mout hout, where h is the specific enthalpy and Q is the net heat input (the m and Q terms should actually be dotted to reflect mass and energy flowrates). Does one account for heat loss?
     
  6. Mar 17, 2009 #5
    Astronuc, TNX for your reply

    1.) There is no feed back to the system. The distilled water will be used. This project is aimed to reuse grey water.

    2.) Yes, the feedwater will be used in the condenser, so basically the feed water will be preheated befor it is fed into the vessel.

    And for the formula; I will have to digest this. I am a mechanical engineer and my physics lessions in school are long time back

    Cheers Robby
     
  7. Mar 17, 2009 #6

    Astronuc

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    Staff: Mentor

    What one is describing is a mass flow-heat transfer problem in mechanical engineering. So the feedwater to the evaporation chamber is preheated, and that heat comes from the effluent discharge. In effect, there are two heat exchangers then, the condenser and the evaporation chamber.

    This thread may do better if it is moved to one of the Engineering forums, e.g. Mechanical & Aerospace, General Engineering or Engineering System & Design.
     
  8. Mar 17, 2009 #7

    russ_watters

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    1. Divide the power by the latent heat of vaporization of water. That's the theoretical maximum.

    2. A. Get rid of the vacuum pump. It uses energy and lowers your efficiency. If you are boiling water, it will leave the vessel on its own and won't be hindered by the vapor pressure around it.
    B. Vacuum insulate the vessel with glass, like a thermos. That way, you won't lose heat to convection on the outside of the vessel.
    C. Depending on the configuration of the mirrors (can't really tell by the description), you may want to build a radiation shield around the vessel to block radiation away from it. That would basically just be a reflective tube pointed at the mirrors.

    By doing these things, you really should be able to achieve an evaporation rate very close to the theoretical maximum. I bet 95%+.
     
  9. Mar 20, 2009 #8
    Quote:
    2. A. Get rid of the vacuum pump. It uses energy and lowers your efficiency. If you are boiling water, it will leave the vessel on its own and won't be hindered by the vapor pressure around it.
    Unquote

    Yes I could, but we want to do it as fast as possible. With a vacuum pump the process would be faster - right ?
     
  10. Mar 21, 2009 #9

    russ_watters

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    Sure, but by a very small amount. For the added cost/complexity of the pump, you could probably get another mirror instead and that would be a bigger help.
     
  11. Mar 23, 2009 #10
    An interesting endeavor. I, not being a math type either, can't help you out on the formula your looking for. However I know something about boilers, and the suggested improvements that have been made are good ones. Especially about insulating anything not impinged by direct radiation from the mirrors.
    One aspect about the gray water distillation that hasn't been mentioned is, how are the non distilled leftovers being removed?
    I'll tell you right up front, the gray water should be filtered "very well" before being fed into your evaporator. Not only to remove any solids, but as much of the disolved contaminates as well. Something on the order of a good pool filter system. Otherwise any contamination will plate out as scale in your evaporator, piping and condenser. As it has to go someplace if the water carrier is removed.
    If this plating out, or sludge is not removed, it will form a highly efficient insulation coating on inside of everything. If this occurs it will be extremely hard to remove. Normally to do so requires and acid treatment, tuned to the type of scale to be removed. I mention this because you seem disappointed in your evaporation rate already.
    And I might also mention.....as the concentration of the fluid being evaporated changes in your evaporator, (becomes more concentrated) your rate of vaporization will change due to changes in the created fluids latent heat of evaporation. In short, you soon end up boiling something else other than just water. Therefore you need a blow down system to drain off the concentrates at various times.
    I think your idea is good, and you have put some thought into it. However if your not careful on the concentrates, you will ended up with a worthless pile of plumbing. This is one of the major hangups with large scale distillation projects.

    boab
     
  12. Jun 17, 2011 #11
    hi robby,
    i have been looking at trying something almost identical to your idea. how did you go with evaporation rate? i was looking to get a minimum 5l/min and was wondering if i could achieve this using a vacuum to lower the boiling point of the water. is there any ratio of pressure unit vs boiling point of water? i can't seem to find it on the web. i have energy required to bring water to boiling and i assume you can use the same equation to calculate energy input to bring water to boil as it is a temperature differential. you also have a requirement of 2261kj of energy to vaporize 1l of water at 100°C. this is actually the biggest energy requirement, not the bringing of the water to 100°C (tbc i am an amateur and have only come to this conclusion by my research on the net). does this change in a vacuum? if so how? you may also know; is the energy input for the vacuum pump and condensation unit less than what it would be to direct boil the water under normal condidtions?
    sorry more questions than answers...
     
  13. Jun 24, 2011 #12
    conn77, This may have what you want.
    http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/watvap.html
    Also increasing the surface area will increase evaporation rates.
    Look up Steam injector Air conditioners, they did a lot of work on cooling water
    by lowering it's vapor pressure. By spraying the return water in a fine mist, they increased the surface area by many times. This made it easier for the excited(hot) water molecules to leave.
     
  14. Jun 24, 2011 #13
    thanks john,
    yes i have been investigating spraying the solution. if you are using a fine mist with impurities in a vacuum will you not end up removing the impurities with the water? i am looking at trying to separate the solution leaving the water one side and solutes on the other. probably will be a case of experimenting with droplet size.
     
  15. Jun 25, 2011 #14
    Hi.
    I think you can further enhance your efficiency by pumping out water instead of vapour to maintain vacuum.
    Coz work = P*dV
    Lesser the V, lessser will be the Work..
     
  16. Jun 26, 2011 #15
    but wouldnt the same volume of water as the volume of steam need to be pumped out in order to achieve the same vacuum level? so, due to the density of water, you will need more energy to pump per unit volume of water, and due to the viscosity of water, there will also be alot more drag , hence more energy losses from pumping water than pumping steam
     
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