Why does wind speed change the evaporation rate so much?

A thin saturated boundary layer always sits just above the water surface. Wind keeps stripping this layer away, so the bulk air composition controls the driving force. The result is that evaporation grows almost linearly with wind speed, captured in the Carrier formula by the (1 + 0.8 v) factor. Doubling the wind speed roughly doubles the evaporation when v is already a few m/s.

What assumptions does the ASHRAE/Carrier formula make?

It assumes a flat free water surface in continuous contact with bulk air. Water and air states are represented by a single characteristic temperature, humidity and wind speed. There is no strong solar radiation, no nearby high-humidity walls, and no atomization. It is best suited to indoor pools, cooling tower fill packs, and humidifier basins. Spray, droplets, or fast-moving mists require k_g or Spalding-number formulations instead.

How do I convert kg/(m²·h) into mm of water per day?

Take water density as 1000 kg/m³, so 1 kg/m² equals 1 mm of water depth. Multiply m_evap [kg/(m²·h)] by 24 [h/day] to get mm/day. For example m_evap = 0.6 kg/(m²·h) corresponds to 14.4 mm/day. Real pools usually evaporate 10–30 % more than this because of waves and swimmer activity; engineers use that margin when sizing make-up water systems.

Is the evaporative heat flux really useful for cooling?

Yes — strikingly so. The latent heat of water is roughly 2400–2500 kJ/kg, so even m_evap = 0.6 kg/(m²·h) corresponds to about 400 W/m² of heat removal. That is comparable to a large fraction of peak solar irradiance. This is exactly why cooling towers, mist fans, evaporative coolers, and human sweating are so effective, and why direct-evaporative HVAC can approach the wet-bulb temperature of outside air.

Free-Water-Surface Evaporation Rate Simulator

Parameters

Water temperature T_water

°C

Air dry-bulb T_air

°C

Relative humidity RH

Wind speed v above surface

m/s

All values are per 1 m² of water surface. Multiply m_evap [kg/(m²·h)] by the actual surface area for total flow.

Results

—

Evaporation rate m_evap

—

Vapor pressure diff. Δe = e_w − e_a

—

Latent heat L

—

Heat flux q = m_evap·L

Water surface and air state

Blue = water (at T_water) / grey = air (T_air, RH) / top arrow = wind v / cyan arrows = evaporating vapor.

Evaporation rate vs wind speed m_evap(v)

x = wind speed v [m/s], y = m_evap [kg/(m²·h)]; yellow marker = current v.

Theory & Key Formulas

The evaporation rate from a free water surface is proportional to the vapor pressure difference between the saturated surface layer and the surrounding air, multiplied by a wind-driven mass-transfer coefficient.

Saturation vapor pressure e_s(T) [hPa], with T in °C (Magnus form):

$$e_s(T) = 6.112\,\exp\!\left(\frac{17.62\,T}{243.12 + T}\right)$$

Surface and air vapor pressures e_w and e_a:

$$e_w = e_s(T_\text{water}),\qquad e_a = \frac{\text{RH}}{100}\,e_s(T_\text{air})$$

Evaporation rate m_evap [kg/(m²·h)] (simplified Carrier form, v in m/s, e in hPa):

$$m_\text{evap} = (1 + 0.8\,v)\,(e_w - e_a)\,\times 0.01$$

Latent heat L [kJ/kg] and per-area heat flux q [kW/m²]:

$$L = 2500 - 2.4\,T_\text{water},\qquad q = \frac{m_\text{evap}\,L}{3600}$$

Even at v = 0 a natural-convection term remains; the larger the Δe = e_w − e_a (hot water vs. dry air), the higher the evaporation rate.

About the Free-Water-Surface Evaporation Simulator

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Why does my outdoor pool keep losing water in summer even without anyone splashing in it?

🎓

Two things drive it: the vapor-pressure gap between the water surface and the air, and the wind that sweeps vapor away. Right above the water there is always a saturated thin film; if the bulk air is drier than that film, vapor diffuses outward. Try dragging the humidity slider down — you will see Δe grow and the evaporation rate jump.

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Oh, that is striking. And if I move the wind speed slider from 0 to 5 m/s, evaporation grows almost linearly. Is that on purpose?

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Exactly. That linear growth is the (1 + 0.8 v) factor in the Carrier formula. At v = 0 there is still some natural-convection diffusion, hence the 1, and wind adds an almost-linear contribution. The coefficient 0.8 is calibrated for indoor pools and cooling-tower fills; outdoor surfaces under strong sun typically evaporate faster than the bare formula predicts.

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The "heat flux" card shows 410 W/m² at the defaults. Is that a lot?

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It is a lot. Peak summer solar irradiance is around 1000 W/m², so this single pool is dumping nearly half of that figure into cooling itself, just through evaporation. Water has an enormous latent heat — around 2400 kJ/kg — which is why misting fans, cooling towers, and even human sweating are so effective. Raise the water temperature on the slider; both Δe and the heat flux climb steeply.

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Is this also the principle behind HVAC cooling towers?

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Yes. Cooling towers spray warm condenser water over fill packs to create huge amounts of free surface area in a small box, and force outside air across it. The colder the wet-bulb temperature of the entering air, the lower the achievable cold-water temperature — and dry climates therefore get the most benefit. The formula in this tool is the heart of that calculation.

Frequently Asked Questions

Indoor pools usually see wind speeds of 0.05–0.15 m/s and no solar gain, so the low-v region of this tool maps directly to design values. Outdoor pools see both wind and solar radiation; a simple practice is to add a solar load term on top of the (1 + 0.8 v) formula. Real pools also evaporate 10–30 % more than the formula because of waves and swimmer activity, and engineers add that safety factor when sizing make-up water.

Yes. Only the vapor-pressure difference matters, not the temperatures themselves. With cold water at 10 °C and dry warm air at 25 °C / 30 % RH, e_w is still larger than e_a and water keeps evaporating. When e_w drops below e_a, the process reverses: vapor condenses onto the water surface. This simulator clamps Δe to zero in that case; use a dedicated condensation tool to size that effect.

Using water density of 1000 kg/m³, 1 kg/m² equals 1 mm of water depth. Multiply by 24 to get mm/day. For example m_evap = 0.6 kg/(m²·h) → 14.4 mm/day. Outdoor pools with sun and wind in summer commonly show 5–8 mm/day; indoor pools in winter sit around 1–2 mm/day.

All three share the same Dalton-type structure: m_evap ∝ f(v)·Δe. Dalton wrote the earliest form; Carrier calibrated the wind function for indoor pools; Penman added a net-radiation term and is preferred in meteorology and agriculture. This simulator uses the simplified Carrier form, which is well suited to indoor pools, cooling tower fill packs, and direct evaporative-cooling units.

Real-world Applications

Indoor pools and bathhouses: evaporated water raises room humidity and threatens condensation, mold and corrosion. Multiplying m_evap by the surface area gives the daily evaporated mass, which then sizes the dehumidification load and the make-up water supply. A 25 m × 12 m indoor pool at m_evap = 0.4 kg/(m²·h) evaporates 120 kg per hour, almost 2.9 tons per day.

Cooling towers and evaporative condensers: these devices exist to maximize evaporative cooling. Fill packs break the water into thin sheets and films to multiply the free surface area; forced draft fans drive up v in (1 + 0.8 v). The drier the outside air, the larger Δe and the colder the achievable approach to wet-bulb temperature — which is why dry-climate cooling towers are so efficient.

Misting fans and direct evaporative HVAC: the same physics underlies "swamp coolers", outdoor mist fans, data-center indirect evaporative cooling, and traditional water sprinkling on streets. They all push the air state toward wet bulb by adding water vapor; the closer wet-bulb is to a comfortable temperature, the more this approach saves over compressor-based cooling.

Reservoir water budgets and agriculture: for agricultural ponds and reservoirs, evaporation losses must be subtracted from rainfall and inflow when modelling storage. The Penman and Penman-Monteith equations — close relatives of this Carrier form — are the standard worldwide for irrigation planning. Arid-region reservoirs can lose over 2 m of depth per year purely to evaporation.

Common Misconceptions and Caveats

The most common misconception is that "higher water temperature means proportionally more evaporation". The driving quantity is actually e_w = e_s(T_water), which grows exponentially with temperature. Going from 20 °C to 30 °C nearly doubles e_w (from about 23.4 to 42.4 hPa), and going on to 40 °C reaches about 73.8 hPa. Drag the water-temperature slider while watching Δe and you will see the curve steepen at the top. Evaporation is exponentially sensitive to temperature, not linearly sensitive.

The second misconception is that "spraying water always cools the air". The rate is proportional to Δe = e_w − e_a, so when the air is already humid (large e_a) the driving force shrinks rapidly. Take the simulator from RH = 30 % to RH = 90 % and m_evap drops to roughly a quarter. In humid summers like those of Japan, mist fans and water-sprinkling cool only about half as much as they do in dry climates. Evaluate the humidity first before counting on evaporative cooling.

Finally, remember that this formula assumes a free, flat water surface. Atomized sprays, drying garments, soil evapotranspiration and even hot polished surfaces use different mass-transfer correlations (Sherwood-number forms, Spalding numbers, drying curves). Under strong solar radiation a separate radiative-heat term must be added. Use this tool for the dominant "flat water + parallel air flow" case, and switch to dedicated models when those assumptions break down.

Free-Water-Surface Evaporation Rate — ASHRAE / Carrier

About the Free-Water-Surface Evaporation Simulator

Frequently Asked Questions

Real-world Applications

Common Misconceptions and Caveats

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