Oceans and Seas
Evaporation and Condensation

How fast does water evaporate?

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2017-03-19 12:23:07
2017-03-19 12:23:07

Water evaporation depends on a large number of factors, namely:

- Relative Humidity of the air

- Temperature of the air and the water

- Surface area of the water

- Velocity of the wind/air over the water

This is why sprays evaporate faster (large surface area), clothes dry faster on windy days (higher velocity of air), you feel hotter on a humid day (your sweat does not evaporate as easily) and you will see more steam from a hot cup of coffee than cold one (higher temperature).

Evaporation occurs by some molecules getting a higher amount of energy than is needed to turn into a gas, by collission with other molecules. When some molecules in an object turn into a gas even when the temperature is below boiling point, this is known as evaporation.

A typical equation from ASHRAE handbooks says that the rate of evaporation in kg/s/m2 is:

(Pw-Pa)x(0.089 + 0.0782V)/Y

Where Pw is the pressure exerted by water at the temperature of the air, Pa is the pressure of the vapour in the air at the air temperature and pressure, V is the velocity of the air over the water and Y is the latent heat of vapourisation of water. Pressures should be in kPa

So, the speed of evporation could be practically anything depending on these conditions.

As an example, let's take a cup of tap water in a kitchen at room temperature (21C).

So, the water from the tap will be at around 5C, at this air temperature of 21C the vapour pressure of water is 0.02486Bar = 2.486kPa.

Typical room conditions are around 60% humidity, so 0.6x0.02486 is the pressure of the water vapour in the air which will be 1.4916kPa.

(Remember humidity = Ps/Pg for given temperature, in this case 21C.

There is no wind in the kitchen, so V=0m/s, and Y for water is given as 2272kJ/kg.

So, ((2.486-1.4916)x0.089 + (0x0.0782V))/2272 = 0.000038953kg/s/m2

A typical cup is around 15cm high and with a diameter of 8cm, giving a surface area of 3.14x0.04x0.04 (circular surface area = pi x(r squared) = 0.00502, and so every second, 0.00502 x 0.000038953kg/s evaporate. This gives 0.000195799g/s. At this rate, a typical cup of water with 753g of water in it (calculated from above dimensions, water = 1000kg/m3) would take 1069hours to completely evaporate, or 44 days.

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