Evaporation and heat loss of your pond

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This is an interesting subject that I think a lot of people have questions about, so I hope we can use this thread to answer some of those questions.
Please refer to the link in my signature below "Pond Heat Loss Formulas" for what we can use for a resource for this discussion.
The more people that can participate in this discussion, the better.:)

Evaporation is the single largest source of heat and water loss for a pond.
From the article:
Evaporation is generally the largest component of the total heat loss from the pond. Considering evaporation, the loss of volume generally comes to mind rather than the loss of heat. However, in order to boil water (and hence cause evaporation) heat must be added. The quantity of heat required to evaporate one pound of water varies with temperature and pressure, but under normal atmospheric conditions the value is approximately 1,000 British thermal units (Btu). When water is evaporated from the surface of the pond, the heat is taken from the remaining water. As a result, as each pound of water evaporates from the surface, approxiately 1,000 Btu are lost with escaping vapour. Losses can occur by evaporation even when the water temperature is at or below the surrounding air temperature.

The formula for calculation is Wp = (0.097 + 0.038v) x (Pw - Pa) x A

Where:
Wp = rate of evaporation (lbm/h)
A = pond surface area (ft2)
v = air velocity, (mph)
Pw = saturation vapor pressure of the pond water
Pa = saturation pressure at the air dew point

For Pw:
@ 60oF water, Pw = 0.256 psia
@ 70oF water, Pw = 0.363 psia
@ 80oF water, Pw = 0.507 psia
@ 90oF water, Pw = 0.698 psia

For Pa:
Use this calculator in the KPA lbs per square value:
https://www.weather.gov/epz/wxcalc_vaporpressure


We'll start with my pond values as an example:
Water temperature = 60F
Pond area = 899 square feet
Air temperature = 63F
Average wind velocity = 7.5 mph (from my Davis weather station)
Pw = 0.256 ( taken from the above given value amounts)
Pa = 0.29 ( I arrived at the value of 0.29 by using the Pa calculator, entering the air temperature in Fahrenheit and taking the Kpa pounds per square value)

Using the above formula, Wp=(.097+.038v)x(Pw-Pa)xA
Wp=(.097+(.038x7.5))x(0.256-0.29)x899
Wp=.097+(.285x-.034)x899
Wp=(.097-.00969)x899
Wp=.08731x899
Wp=78.49

Still with me?:)

So, Wp = rate of evaporation in lbm/h (pound water mass per hour)
1 pound/hour = .002 gallons/minute
78.49 x .002 = .157

So - my pond loses 0.157 us gallons/minute with the above weather conditions. If the above weather conditions remains constant for 24 hours, my pond will lose 226 gallons in a 24 hour period (about 1/2 inch)
For a 6000 gallon pond in my location, that seems about right given our latest weather conditions. Of course, weather conditions vary during the day and night but hopefully that gives people an idea of how to calculate water loss due to evaporation.
Fun.:)

Even if my calculations are off by 30%, I'm still within a reasonable water loss rate before I start to consider that I may have a leak.
Calculating heat loss value isn't far from calculating evaporative water loss.
Maybe we can revisit heat loss and whether it's worth putting in a water heater this fall.
If I've made a mistake in the above calculations, please point it out.

Thanks.
 
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addy1

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Nice................I feel like I am back in college!
 
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This information was the basis for me covering my pond last winter, with a solar cover. I was amazed when I'd peek under the cover. The fish were more active, as opposed to just resting on the bottom and the water celery, continued to grow.
 
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If anyone else is interested, and would care to post the required information for calculating their pond water loss through evaporation, feel free to post.
What we need:
-Water temperature
-Air temperature
-Pond area (sq ft)
-Wind speed
-An approximate idea of what your pond loses in daily evaporation (in inches)

I wouldn't mind going through the numbers of a few other people's ponds, to make sure I haven't missed anything.

.
 
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Ill play!

If anyone else is interested, and would care to post the required information for calculating their pond water loss through evaporation, feel free to post.
What we need:
-Water temperature about 70- 75 degrees
-Air temperature Consistently average of 90 F lately (some days a little higher some a little lower).
-Pond area (sq ft) 326
-Wind speed 5 MPH
-An approximate idea of what your pond loses in daily evaporation (in inches) 1/2" maybe

I wouldn't mind going through the numbers of a few other people's ponds, to make sure I haven't missed anything.

.
 
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Thanks pecan.
Off the top of my head, it looks like you have about 3 times the evaporation rate of my pond here.
That sounds about right, I'll run through the numbers later this afternoon.
 
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Thanks pecan.
Off the top of my head, it looks like you have about 3 times the evaporation rate of my pond here.
That sounds about right, I'll run through the numbers later this afternoon.
I often wondered if humidity played a role as well. We are the second driest state in the US, next to Nevada. We are dryer than Arizona (which is number 3). I considered that less water in the air may account for more evaporation, but I could be wrong. Just a theory.
 
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No, you're correct.
The amount of water vapour in the air space immediately about the pond water surface determines your water evaporation rate.
If you have any wind, that water vapour is carried away and your evaporation rate will continue to be high.

For me, I equate it to ice skating.
When you skate on ice, what you are actually moving on is a thin layer of water that is created between the skate blade and the rink ice.
It's a small distinction, but a significant one.
 
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Great topic. Knowing normal evaporation rates would ease a lot of worries .
how does a stream and waterfall affect the calculation? do you add the streams surface area and the stock tanks surface area (for the waterfall) to the ponds SA ?
 

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If anyone else is interested, and would care to post the required information for calculating their pond water loss through evaporation, feel free to post.
What we need:
-Water temperature
-Air temperature
-Pond area (sq ft)
-Wind speed
-An approximate idea of what your pond loses in daily evaporation (in inches)

I wouldn't mind going through the numbers of a few other people's ponds, to make sure I haven't missed anything.

.
How many readings would you need? As for wind speed.....there is a neighbor who has a weather station, but I'll to get that web address from my hubby -- he keeps up with that stuff more than I do. Is that OK? And pond square area, do you want surface area?
 
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Great topic. Knowing normal evaporation rates would ease a lot of worries .
how does a stream and waterfall affect the calculation? do you add the streams surface area and the stock tanks surface area (for the waterfall) to the ponds SA ?
I would include the surface areas of "bogs", streams and waterfalls.
At this point, I don't know how plant transpiration rates would figure in.
 
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How many readings would you need? As for wind speed.....there is a neighbor who has a weather station, but I'll to get that web address from my hubby -- he keeps up with that stuff more than I do. Is that OK? And pond square area, do you want surface area?
All we are looking at is a snapshot in time. An average number for wind and temperature would be fine for now.
Total surface area of the pond and associated streams, waterfalls.
Thanks
 

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That is why my pond looses water it is always windy here .I live up on a hill and have not seen many non windy days .Even with higher humidity it looses water .I can get a fog over the pond with the cool nights and hot days .
 
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Ill play!
Ok pecan, here's what I come up with:
Your pond would lose about 78 gallons per day through evaporation, which compared to my pond is actually about the same evaporation rate.
Do your air temperatures get much cooler at night than 90F?
My air and water temperatures remain fairly constant during the day/night and I lose about 3.97 gal/sq. ft/day
If your temperatures remain fairly constant, you would lose about 4.1 gal/sq. ft./day
I would be interested in your night water temperature and night air temperature readings.
I think if your air temperatures drop significantly during the night, your evaporation rate will also drop significantly.
With my pond, I'm looking at a difference of about 3F in water/air temperature difference.
With your pond, we see a difference during the day of 18F.
Your pond also is subject to a lower wind speed, which could also explain some of the discrepancy.
Thanks
 
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I just tried pecan's numbers with a 7.5 mph wind speed and calculated a loss of 105 gallons per day. A 26% increase in water loss.
Wind speed seems to make a big difference.
Also, pecan's pond is protected by a fence, so the 5 mph wind may not be actually reaching the pond surface.
Interesting.
 
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@MitchM Thanks! Our temps do drop at night we can have a 40 degree swing between highs and lows. I think lately it's been about a 30 degree swing, getting down into the 60's at night.
 
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I just tried pecan's numbers with a 7.5 mph wind speed and calculated a loss of 105 gallons per day. A 26% increase in water loss.
Wind speed seems to make a big difference.
Also, pecan's pond is protected by a fence, so the 5 mph wind may not be actually reaching the pond surface.
Interesting.
I have noticed when we have hot dry strong winds, my water levels drop very fast.
 
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Ok Mitch..... I tend to agree with your formula, let's throw in a few more variables.... 18' long, 16" wide...... 3000 gallons, almost zero wind due to fences and trees, direct sunlight from 10:00 am to about 2:00 pm. Now throw in a waterfall that is flowing at 90 gallons a minute, dropping approximately 18" to the water.... A 40 cfm aerator....Median temperature during the day about 82 deg. 70 at night .... In these conditions I am losing about an inch a day
 
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Ok Mitch..... I tend to agree with your formula, let's throw in a few more variables.... 18' long, 16" wide...... 3000 gallons, almost zero wind due to fences and trees, direct sunlight from 10:00 am to about 2:00 pm. Now throw in a waterfall that is flowing at 90 gallons a minute, dropping approximately 18" to the water.... A 40 cfm aerator....Median temperature during the day about 82 deg. 70 at night .... In these conditions I am losing about an inch a day
The formula relies mostly on the vapour pressure immediately about the water surface.
Wind will constantly reduce the vapour pressure, and I would imagine an aerator will also reduce vapour pressure, which is why aeration is so effective at cooling off a pond.
Can you specify the temperatures of your pond water and ambient air?

Here's an explanation of vapour pressure vs. relative humidity.
https://laulima.hawaii.edu/access/content/group/2c084cc1-8f08-442b-80e8-ed89faa22c33/book/chapter_5/humidity.htm
 

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All we are looking at is a snapshot in time
This is an extremely important point to remember. Given that the prevailing weather is always in constant flux any computations that are done are already out-dated as the factors involved in these computations have changed. As such, any result obtained may lead to wrong assumptions concerning the overall evaporation rate over day, weeks or months.
The use of this and other formula have falling out of practice in agricultural circles in favor of Pan evaporation rates used in conjunction with the pan coefficient for that particular geographical area.
Maps and charts should be available for each state.
As pondkeepers, it is more important to know what the expected weekly rate may be for each of our locations..
In addition, water loss due to transpiration of any aquatic plants must be taken into account over and above the evaporation rate.
In this information age, it is quicker to consult an existing chart.
uspanevap.gif
 
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