# Evaporation from Water Surfaces

## The amount of evaporated water from a water surfaces - like swimming pools or an open tanks - depends on the temperature in the water and in the air, and the humidity and velocity of the air above the surface - online calculator

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The evaporation of water from a water surface, as an open tank, a swimming pool or similar, depends the temperature in the water and the temperature in the air, the actual humidity of the air and the velocity of the air above the surface.

The amount of evaporated water can be expressed as:

g_{s}=ΘA (x_{s}- x) / 3600(1)

or

g_{h}=ΘA (x_{s}- x)

where

g_{s}= amount of evaporated water per second (kg/s)

g_{h}= amount of evaporated water per hour (kg/h)

Θ = (25 + 19 v) = evaporation coefficient (kg/m^{2}h)

v= velocity of air above the water surface (m/s)

A= water surface area (m^{2})

x_{s}= humidity ratio in saturated air at the same temperature as the water surface (kg/kg)(kg H_{2}O in kg Dry Air)

x= humidity ratio in the air (kg/kg)(kg H_{2}O in kg Dry Air)

**Note!** The units for *Θ* don't match since the this is an empirical equation - a result of experiments.

### Required Heat Supply

Most of the heat required for the evaporation is taken from the water itself. To maintain the water temperature heat must be supplied.

The heat supplied can be calculated as:

q = h_{we}g_{s}(2)

where

q= heat supplied (kJ/s, kW)

h_{we}= evaporation heat of water (2270 kJ/kg)

### Example - Evaporated Water from a Swimming Pool

For a swimming pool with water temperature *25 ^{o}C* the saturation humidity ratio is

*0.02 kg/kg*. With an air temperature of

*25*and

^{o}C*50%*relative humidity - the humidity ratio in air is

*0.0098kg/kg*- check the Mollier diagram.

For a *25 m x 20 m* swimming pool and *0.5 m/s* velocity of air above the surface, the evaporation can be calculated as:

g_{s}=( 25 + 19 (0.5 m/s)) ((25 m) (20 m)) ((0.02 kg/kg) - (0.0098kg/kg)) / 3600

= 0.049 kg/s

The heat supply required to maintain the temperature can be calculated as:

q =(2270 kJ/kg) (0.049 kg/s)

=111.2 kW

The heat supply can be reduced by

- reducing the air velocity above the water surface - limited effect
- reducing the size of the pool - not really practical
- reducing the water temperature - not a comfort solution
- reducing the air temperature - not a comfort solution
- increase the moisture content in the air - may increase the condensation and damage of building constructions
- remove the wet surface - possible with plastic blankets outside operation time. Very effective and commonly used

**Note!** During operation time the activity in a swimming pool may increase the evaporation and the need of heat supply dramatically.

To reduce the energy consumption and to avoid moisture damages in the building constructions, its more and more common to use heat recycling devices with heat pumps to move latent heat from the air to the water in the swimming pool.

### Surface Evaporation Calculator

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