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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:

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:

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.

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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