How this instrument works
Evaporation happens because the water surface and the air above it disagree about how much moisture that air can hold. The Tetens formula converts a temperature into the saturation vapor pressure at that temperature — the maximum partial pressure of water vapor the air could carry before it starts condensing back out. Compute it once at the water's own temperature and once at the air temperature, scale the second figure by relative humidity, and what remains is the actual pressure gradient pulling molecules off the surface.
The rate equation is a mass-transfer cousin of Newton's law of cooling: flux is proportional to a driving difference, multiplied by the area it acts over and a coefficient that absorbs everything else — the diffusivity of water vapor through air, the thickness of the still boundary layer clinging to the surface, and above all the wind or ventilation sweeping that layer away. Double the airflow across a pool and the coefficient roughly doubles too, even though neither temperature nor humidity moved at all.
Tetens published his approximation in 1930 as a fit to the Clausius–Clapeyron relation, trading thermodynamic exactness for an expression simple enough to evaluate by hand; it stays within about 0.1 kPa of the exact curve across ordinary outdoor and pool temperatures. The model also assumes a calm, uniform surface — direct sun, splashing, and fountains all add evaporation the equation cannot see, which is one reason measured water-loss bills often run higher than a first calculation predicts.
- Set Water surface temperature, °C to the surface itself, which can run warmer than the bulk water after sun exposure or heating.
- Set Air temperature, °C and Relative humidity, % for the space or open air just above the water.
- Enter Water surface area for the exposed patch that is actually losing moisture, not the tank's total wetted area.
- Choose a Mass transfer coefficient — higher for wind or forced ventilation over the surface, lower for still, sheltered water.
- Read Evaporation rate, alongside the two saturation vapor pressures the instrument worked out along the way.
Worked example — a 50 m² pool at 27 °C
Take an outdoor pool with 50 m² of exposed surface, water sitting at 27 °C, air at 25 °C and 60% relative humidity, and a mass transfer coefficient of 0.09 kg/(h·m²·kPa) — a reasonable figure for a lightly breezy site. The Tetens formula first gives the saturation vapor pressure at the water surface: Pws,water = 0.6108 · e^(17.27 × 27 / (27 + 237.3)) = 3.565 kPa. The same formula at the air temperature gives Pws,air = 0.6108 · e^(17.27 × 25 / (25 + 237.3)) = 3.168 kPa.
Scale Pws,air by the 60% relative humidity to find the vapor pressure already present in the air: 3.168 × 0.60 = 1.901 kPa. Subtract that from Pws,water and the driving pressure difference is 1.665 kPa. Multiply by the coefficient and the area — 0.09 × 50 × 1.665 — and the pool loses about 7.491 kg of water to the air every hour, close to 180 kg over a full day if conditions hold steady.
Questions
Why compute two separate saturation vapor pressures?
Because evaporation is driven by the gap between them, not by either temperature alone. Pws,water represents the vapor pressure right at the wet surface, where the film of air touching the water is effectively saturated at the water's own temperature. Pws,air, scaled by relative humidity, represents how much vapor the surrounding air already carries. Warm water next to warm, humid air can evaporate slower than cooler water next to cold, dry air, because it is the difference that moves molecules, not the absolute temperature.
How do I pick a mass transfer coefficient?
There is no universal value — it normally comes from measurements or published figures for the situation at hand: still indoor air, a lightly ventilated room, or open outdoor wind. Indoor pools with poor air movement often sit near 0.03 to 0.06 kg/(h·m²·kPa); a breezy outdoor pool like the one in the worked example runs closer to 0.09; exposed tanks in steady wind can run several times higher. When in doubt, use a value taken from a measurement on a similar surface rather than a generic default.
What happens if the air is warmer than the water?
Evaporation can still continue, because Pws,air is multiplied by relative humidity before the subtraction — a warm room at moderate humidity often still has a lower effective vapor pressure than a heated pool surface. The arithmetic only goes negative, meaning condensation rather than evaporation, once the air's actual vapor pressure exceeds the water's saturation pressure — in effect, once the air is already more saturated than the surface itself.
Why use the Tetens formula instead of full steam tables?
Because it reproduces the saturation vapor pressure of water to within roughly 0.1 kPa across ordinary environmental temperatures using one exponential term, which is close enough for a mass-transfer estimate and far simpler than interpolating steam tables or solving the Clausius–Clapeyron relation directly. Meteorologists have used this and closely related fits since the 1930s for exactly that trade-off between accuracy and ease of calculation.
Does this formula include solar heating or splashing?
No — it only captures the vapor-pressure driven diffusion described by the mass transfer coefficient. Direct sunlight on the water, splashing, spray, and fountains all add evaporation the equation does not see on its own, which is why a coefficient calibrated against real measurements on a similar surface tracks reality better than one copied from a table built for a different setting.
Who actually uses a calculation like this?
Pool operators sizing a dehumidifier or budgeting make-up water and heating cost, HVAC engineers designing ventilation for indoor aquatic centers, and process or water-treatment engineers estimating losses from open tanks, cooling ponds, or reservoirs. In every case the same two numbers matter most: the vapor-pressure gap between water and air, and how hard the air is moving across the surface.