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Water evaporation rates

  1. Jun 2, 2008 #1
    Not sure where this posts needs to go, as I am not familiar with engineering, so if it needs moved, please do so.

    I build decorative ponds, and one of the conversations I end up with is the evaporation rate of the water in a pond setting. I do know there are formulas out there on the rate of water that is not moving, but in my case there are several wild cards to add to the mix.

    Case in point.

    Pond has aprox 560 SF of surface area. It has a depth of 6 feet. There is a major water fall that is 6 feet wide, three feet long, with 22,000 gallons of water flowing over it per hour. That water flows to an upper pond, then into the lower pond via two more falls, aprox 3 feet wide and two feet long each.

    Then there is a skimmer return that roils the water surface and keeps it in motion, the flow of which is 3800 gallons per hour.

    I am aware of many of the situations that affect evaporation rates. Could there be a formula that I could use that would give me at least a rough idea of what to expect?

    While not uneducated, I am somewhat math challenged (wish like heck I had paid more attention in high school and college!)

    Thank you in advance!

  2. jcsd
  3. Jun 2, 2008 #2

    jim mcnamara

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    All of the splash activity that is sending water particles into the air is dramatically affecting evaporation rates. It is analogous to watering with a sprinkler during the day. Water loss from evaporation is very high.

    There is no simple formula for this problem, but it is amenable to other methods. Since you do know the square footage, fill the pool to an exact depth, say 72". Run the pool for four hours in afternoon conditions. Measure the depth. Add water back to 72". Measure the volume it takes to get it back to 72" (or use the calculations below) Run the pool late at night - 4 hours again. Measure the depth and water needed for recharge.

    Assuming the day temps and night temps are representive, and there was no rain, you now have a base line for how the pool behaves in terms of evaporation at the worst case and best case of operation.

    There are 0.0043 gallons of water in a cubic inch of water. The water lost to evaporation in your experiment is
    Code (Text):

    (inches lost in 4 hours) * 144 * area in sq ft  == cubic inches of water lost
    cubic inches of water lost * .0043 == gallons lost
    You also need to realize that the pool owner's "experienced" evaporative loss is less than what you calculated, because if you're in a temperate area with measurable rainfall during the pool oiperation season, then rainfall recharges the pool as well.
  4. Jun 2, 2008 #3

    Thank you so much for your reply!

    What we are experiencing is about 19 gallons an hour loss over the last 4 days.

    What I used for a baseline is a pond that we built three years ago. It is a polyurea lined pond with no seams, and no leakage. IT was finished the same way, with mortared stone all the way around.

    The pond that is three years old will lose right at 9 inches a week without a major event. That is on 269 SF of water surface with a flow rate of nearly 11,000GPH. The SF of the new system is 521SF, and slighly over 22,000 GPH flow rate over nearly 5 times the rock area as the first pond.

    So, at 9 inches a week, that is 201.75 CF, at 7.5 gallons/CF is 1513 gallons per week loss. Divided by 168 hours gives you 9 gallons per hour loss on that pond. It does run 24/7.

    So, using that benchmark for the newer pond, a water loss of 19 gallons per hour over the last four days would/should be expected. Or so I would think.

    One other variable that we have noticed was wind. Several times over the last two years, there have been days where the wind during the day caused the water to evaporate 7-12 inches in just 12 hours, depending on the wind speed etc.

    So with the info on the other pond, would it be safe to say that 19 gallons an hour would be the fair amount to expect lost to evaporation?

  5. Jun 6, 2008 #4

    jim mcnamara

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    That is how I would approach it. 19 gallons due to evaporative loss sounds fine.

    And I left out wind on purpose.

    Stand a few feet downwind from one your waterfall displays when the wind is on at 35mph+.
    It isn't evaporating, water droplets are blown out of the system completely. This has more to do with how far water falls before it hits the reservoir. And wind speed.
  6. Jun 6, 2008 #5
    Thanks Jim

    I just was hoping to get a formula that you could use, but with all the different variables I dont know if there really could be one?

    I appriciate the help!

  7. Jun 6, 2008 #6


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    Hi d,
    This isn't really anything I have experience with, but I happen to have a paper on it that might interest you. See attached.

    Attached Files:

  8. Jun 6, 2008 #7
    Interesting read. Thanks for posting it, it will become one of the things I will use on different jobs.

    One thing that struck me as relevant to my problem is the statement about the unoccupied pools vs. the occupied pools, and that the evaporation rate would be at about twice the rate for occupied vs. unoccupied.

    So getting back to the situation at hand, a water fall that pushes 22,000 gallons of water over a large water fall into a small shallow pond with a ton of turbulence, then over twin falls into the main pond, which has turbulence of its own, could be realistically be expected to have the evaporation rate that is much higher than just a few kids swimming in a pool. Or at least I would read that in that statement. Now my question is how much more. And I don’t know if there is a really good answer.

  9. Feb 25, 2010 #8
    please can one help me for calculat suitable dimention(depth,width,leignth) to water evaporation pond to handle 20000 BLL water per day
  10. Apr 26, 2010 #9
    Can one help me how to calculate the evaporation rate of water at 95 degree celsius (kg/hr.m2 ).
    Last edited: Apr 26, 2010
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