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Stop Evaporation
Calculation method

Aerodynamic Mass-Transfer (Dalton) Method

Estimate evaporation from the vapour-pressure deficit and a wind function — the robust workhorse for wind-exposed industrial ponds.

E=f(u)(esea)E = f(u)\,\big(e_s - e_a\big)

The aerodynamic mass-transfer method is a Dalton-type law: evaporation is proportional to the vapour-pressure deficit between the water surface and the air, multiplied by a wind function. It is the form behind the illustrative estimator on this site.

The equation

E=f(u)(esea)E = f(u)\,\big(e_s - e_a\big)

ese_s is the saturation vapour pressure at the water-surface temperature, eae_a is the actual vapour pressure of the air, and f(u)f(u) is an empirical wind function — larger wind, larger f(u)f(u), faster evaporation. The classic calibration comes from the Lake Hefner studies (Harbeck, 1962).

Inputs & data needed

Wind speed at a known height, air temperature and relative humidity (to get eae_a), and the water-surface temperature (to get ese_s). No radiation data is required, which is why it suits wind-driven sites where radiation instruments are absent.

Worked example (illustrative)

Using the simplified wind function f(u)=1+0.45uf(u) = 1 + 0.45\,u from this site’s estimator, with water/air near 25C25\,^\circ\text{C} so es3.17 kPae_s \approx 3.17\ \text{kPa}, relative humidity 40%40\% so ea=0.40×3.17=1.27 kPae_a = 0.40 \times 3.17 = 1.27\ \text{kPa}, and wind u=3 m/su = 3\ \text{m/s}:

f(u)=1+0.45×3=2.35f(u) = 1 + 0.45 \times 3 = 2.35 E=2.35×(3.171.27)=2.35×1.904.5 mm/dayE = 2.35 \times (3.17 - 1.27) = 2.35 \times 1.90 \approx 4.5\ \text{mm/day}

The coefficient here is nominal — real wind functions are site-calibrated against observed loss, so treat this as illustrative.

Accuracy & when to use

Mass-transfer is robust on windy, well-mixed ponds and the natural choice for industrial reservoirs. Its accuracy hinges on the wind function: a generic f(u)f(u) can be off by tens of percent until calibrated. Where radiation dominates and wind is light, prefer Priestley-Taylor; for a combined treatment, see Penman-Monteith.

Frequently asked questions

Why does wind speed appear in the mass-transfer equation?
Wind sweeps humid air away from the water surface and replaces it with drier air, keeping the vapour-pressure deficit high. The wind function f(u) captures how strongly that ventilation accelerates evaporation.

Sources

  1. Harbeck (1962), USGS PP 272-E — mass-transfer field technique