Evaporation rates are typically calculated as 0.5% to 1% of the GPH (gallons per hour)/LPH (liters per hour) per day. This is an average estimation, as evaporation varies daily based on local climatic conditions (e.g., relative humidity, wind direction and velocity, sun vs. shade, etc.). These formulas will help you understand approximately how much water may be lost in a pond per day in gallons/liters and also how many days to lose an inch/centimeter of water in that same pond.
Imperial (Feet/Gallons)
- Determine the approximate range of the amount of gallons lost in the pond due to evaporation
- Gallons per hour of the pump (after head loss) X 0.005 = Approximate minimum amount of gallons lost in pond per day
- Gallons per hour of the pump (after head loss) X 0.01 = Approximate maximum amount of gallons lost in pond per day
- Determine the amount of time it will take to lose one inch of water in the pond
- Length X Width = Total square feet of the pond
- For irregular shaped ponds, multiply total square feet X 80%
- Total square feet of pond X 0.62 = Total gallons in the top 1" of water of the pond
- 0.62 = gallons per square foot @ 1" depth
- Total gallons in the top 1" of water of the pond / Approximate minimum amount of gallons lost in pond per day = Number of days before the pond loses 1" of water
- Total gallons in the top 1" of water of the pond / Approximate maximum amount of gallons lost in pond per day = Number of days before the pond loses 1" of water
- Length X Width = Total square feet of the pond
Imperial Example
- A 3,000 GPH pump (after head loss) has approximately 2,500 GPH of actual output
- Low-end evaporation: 2,500 x 0.005 = 12.5 approximate minimum amount of gallons lost in a pond per day
- High-end evaporation: 2,500 x 0.01 = 25 approximate maximum amount of gallons lost in a pond per day
- 11’ x 16’ pond has 176 square feet. Adjusted for the irregular shape @ 80% = 140 square feet.
- 140 x 0.62 (gallons per square foot @ 1" depth) = 86.8 gallons in the top 1" of water of the pond
- 86.8 / 12.5 = 6.9 days before the pond loses 1" of water (approximate minimum amount)
- 86.8 / 25 = 3.5 days before the pond loses 1" of water (approximate maximum amount)
Metric (Meters/Liters)
- Determine the approximate range of the amount of liters lost in the pond due to evaporation
- Liters per hour of the pump (after head loss) X 0.005 = Approximate minimum amount of liters lost in pond per day
- Liters per hour of the pump (after head loss) X 0.01 = Approximate maximum amount of liters lost in pond per day
- Determine the amount of time it will take to lose one centimeter of water in the pond
- Length X Width = Total square meters of the pond
- For irregular shaped ponds, multiply total square meters X 80%
- Total square meters of pond X 10 = Total liters in the top 1 cm of water of the pond
- 10 = liters per square meter @ 1 cm depth
- Total liters in the top 1 cm of water of the pond / Approximate minimum amount of liters lost in pond per day = Number of days before the pond loses 1 cm of water
- Total liters in the top 1 cm of water of the pond / Approximate maximum amount of liters lost in pond per day = Number of days before the pond loses 1 cm of water
- Length X Width = Total square meters of the pond
Metric Example
- An 12,000 LPH pump (after head loss) has approximately 10,000 LPH of actual output
- Low-end evaporation: 10,000 x 0.005 = 50 approximate minimum amount of liters lost in a pond per day
- High-end evaporation: 10,000 x 0.01 = 100 approximate maximum amount of liters lost in a pond per day
- 3.5 m x 5 m pond has 17.5 square meters. Adjusted for the irregular shape @ 80% = 14 square meters.
- 14 x 10 (liters per square meter @ 1 cm depth) = 140 liters in the top 1 cm of water of the pond
- 140 / 50 = 2.8 days before the pond loses 1 cm of water (approximate minimum amount)
- 140 / 100 = 1.4 days before the pond loses 1 cm of water (approximate maximum amount)