Need a formula to calculate BTU loss in a pond.

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SUMMARY

This discussion focuses on calculating BTU loss for a 15' x 3' ornamental pond with an average depth of 2', containing 673 gallons of water. To maintain a temperature of 72°F in worst-case ambient conditions of 0°F, it is estimated that approximately 7200 watts will be required for heating. Key factors influencing heat loss include evaporation, which accounts for about 90% of heat loss, and wind exposure, which significantly affects evaporation rates. The discussion also suggests using a circulator pump for efficient heating and references a free Heat Transfer Textbook for further information.

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Marshel
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I am looking at building an ortemental pond. The pond will be 15' x 3' with an average depth of 2'. I caculated it out to 673 gallons and 45 square feet of surface area. I plan to put tropical fish in it and will have to maintain a temperature of 72F year round.
Worst case scenario would be an ambient temp of 0F.
Based on the gallons, surface area and ambient temperature, is there a formula to calculate BTU loss and wattage of heaters to maintain the 72 degree temp?

The initial thought is to use a heater simalar to the ones we use at work to heat our outside hydraulic units. Maybe use a home hot water heater, but I don't believe that will work.

I plan to control pond circulation and temp with a PLC and have that end handled.

Thank you
 
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I think your number will be about 40 watts per meter-squared per degree C, or about 180 watts per deg C for your pond, or about 7200 watts when outside temp is zero F. For better info, download this free Heat Transfer Textbook available at (11 MB file)
http://web.mit.edu/lienhard/www/ahtt.html
and look at page 21. If you have a circulator pump, you could heat the water as it flows through.
 
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Having done some research on heating both pools and ponds, I've found some interesting and surprising facts.

Something that may play an important role in your calculations is this:

Heat loss from a body of water is only in small part due to temperature difference. By far the largest component (I think it's about 90%) of heat loss is due to evaporation. This is why, even a thin coating over a body of water such as a swimming pool - sometihn thick enough to prevent evaporation, but not thick enough to stop direct heat transfer - is sufficient to keep the water temperature up overnight.

I grant that it is counter-productive to put any kind of film over a pond with live creatures in it. However...

A very large factor in evaporation is wind. A body of water protected from the breeze will lose heat much slower than a body of water that is exposed to even a 5mph breeze.
 

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