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Evaporation Rate Calculator

Estimate liquid evaporation rate from surface area, vapor pressure, wind speed, and humidity.

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What Affects Water Evaporation?

Evaporation depends on how much moisture the air can still hold, how fast fresh dry air moves across the surface, and the area exposed. Pools, ponds, and open tanks lose water faster on warm, dry, windy days and slower when the air is cool and humid.

Evaporation Rate Formula

$$g_h = (25 + 19 \times v) \times A \times (X_s - X)$$

Here \(g_h\) is evaporation in kilograms per hour, \(v\) is wind speed above the surface in m/s, \(A\) is surface area in m², \(X_s\) is the saturation humidity ratio at the water temperature, and \(X\) is the current humidity ratio. The humidity ratio is the mass of water vapor per unit mass of dry air (kg/kg).

Humidity Ratio from Temperature

For air between about 0 °C and 30 °C, the saturation humidity ratio can be approximated as:

$$X_s = 3.733 \times 10^{-3} + 3.2 \times 10^{-4} T + 3 \times 10^{-6} T^2 + 4 \times 10^{-7} T^3$$

where \(T\) is temperature in degrees Celsius. Multiply \(X_s\) by relative humidity (as a fraction) to get the current humidity ratio \(X\). Multiply the hourly rate by 24 for a daily estimate using average conditions.

Related tools: Relative Humidity Calculator and Pool Calculator.

Frequently Asked Questions

Why does wind increase evaporation?

Wind replaces humid air sitting above the water with drier air from the surroundings. That keeps the humidity deficit high and sustains a faster evaporation rate.

How does relative humidity affect evaporation?

High relative humidity means the air already holds much of the moisture it can carry, so \(X_s - X\) is small and evaporation slows. At 100% RH, the deficit is zero and evaporation stops.

Can I use this for a swimming pool?

Yes. Enter the pool surface area, typical afternoon temperature, humidity, and wind speed. Switch to per-day output to estimate how much water to add over 24 hours.

What is the difference between kg/h and liters per hour?

For water, one kilogram is approximately one liter, so the mass rate in kg/h is a good estimate of volume loss in L/h.

Is this formula accurate at extreme temperatures?

The humidity ratio polynomial is tuned for roughly 0 °C to 30 °C. Outside that range, treat results as rough estimates and validate with local measurements.