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Simulation of the evaporation of a swimming pool
Everyone! I hope this site will allow those interested to get an idea of the evaporation rate of a body of water. Two formulas resulting from two different approaches of the phenomenon but giving appreciably similar results, most often, are used. It suffices to modify the data of the form.

The application belowallows the calculation of the water evaporation rate according to the average temperature of the ambient air, the temperature of the water and the relative humidity
for a surface S. The value of the surface can be calculated for different shapes(here).

We can take into account the presence or absence of swimmers and the corresponding duration of use.

The temperature must be between 0 and 60 ° C and the relative humidity between 1 and 100%. The corresponding thermal power is supplied.

Modifiable basic data | |||

m2 | bathers | hour/day | |

°C | °C | ||

°C | °C | ||

% | % | ||

m/s | m/s |

liter / hour | liter / hour | |

liter / hour | liter / hour | |

cm / 24 hours | cm / 24 hours | |

liter / 24 hours | liter / 24 hours | |

kW | kW |

The meteorological data useful for the calculation can be obtained on the site below for example:

Do not forget to modify the wind speed in this case to have the speed V2 above the pool according to the formula V2 = V1 * log (H2 / 0.0002) / log (H1 / 0.0002).

We generally take H2 = 0.3 m and H1 = 10 m.

(SEE THE ESTABLISHMENT OF FORMULA 1)

*sat index corresponds to saturated air at water temperature, index 0 corresponds to ambient air conditions*

Although this is not the main focus of the site, Formula 1 can be applied to large S surfaces (reservoirs or lakes).

Formula 2 based on the correlation of WH CARRIER (1918)

(f = 0.5 corresponds to an unoccupied swimming pool)__Calculation-measurement comparison__

Conclusion : Very good agreement between calculations and measurements if all possible measurement errors are taken into account, in particular on the wind speed in strong winds for a swimming pool sheltered by a wall of 1.70 m high, approximately 7 m from the swimming pool. The wind speed taken into account being that of the weather station. On a strong windy day (mistral at 20 km / h) it was appropriate in this example (sheltered swimming pool) to reduce the wind speed provided by the station by approximately 35%. Weather forecast to ensure the calculation-measurement coincidence.

__Approximate estimate of the annual water consumption for a 32 m2 pool:__

Applying the data provided in the default form to an outdoor 32 m2 pool occupied on average 4 hours per day by 2 bathers leads to:

__For a region with little wind __, V between 0 and 4 m / s (14.4 km / h), the evaporative loss is

obtained from 87.4 liters / day to 490.7 liters / day, i.e. an average annual consumption on the basis of 150 days

of use(cover by a tarpaulin the rest of the year) equal to 150 * (490.7 + 87.4) /2*0.001=43.3 m3.

The filling of the swimming pool represents 50 m3. We obtain a consumption for the filling and losses by evaporation of 93.3 m3 which leads to a cost of approximately

280 euros per year (based on 3 euros per m3 of water) .

__For a strongly windy region __, V between 0 and 8.7 m / s (31.3 km / h), we obtain a consumption for losses between 87.4 liters / day and

913 liters / day, i.e. following the same calculation than previously 76.5 m3 / year. The cost this time if we take into account the filling of about 380 euros.

The above values are probably overestimated because natural infill due to rain has not been taken into account; or even a simple gesture to further reduce consumption,

which is to use a tarpaulin at night.

__Drop in water temperature due to evaporation over 24 hours:__

The evaporation phenomenon causes the water to cool.If we take the previous example, formula 1 gives a thermal power linked to evaporation of 5.42 kW, wich corresponds to an energy loss equal to 24 hours at

Q = 5.42 * 1000 * 24 * 3600 = 4.68E8 joules (i.e. 130 kWh) .Then we obtain the cooling over a day equal to Dt = -Q / (M * Cp)

or M = mass of water = ro * V = 1000 * 50 = 50,000 kg, Cp = specific heat of water = 4180 J / kg.K This gives a drop in temperature Dt = -4.68E8 / (50,000 * 4180) = - 2.24 ° C.

The drop in water temperature per day is about 2.24 ° C due to evaporation in this case. Here again the wind speed will strongly influence this drop in temperature since the thermal

losses go from 5.42 kW to 16.42 kW when the wind speed goes from 2 to 8 m / s (28.8 km / h).

If you want to keep the water temperature constant, you can use A HEAT PUMP TO COMPENSATE FOR THE LOSSES. by evaporation With heat pump of efficiency eta = 4.5

the energy to besupplied for heating for 4 hours per day in the previous example is equal to Q / eta = 5.42 * 4 / 4.5 = 4.8 kWh If the energy source is electricity this represents

a cost = 4.8 *. 14 = 0.67 euro / day or about 101 euros for 150 days of use to compensate for losses by evaporation . To this are added essentially the losses (or gains) by

convection and radiation and in general the losses for the renewal of water. In the previous example where the air temperature is higher than the water temperature, we can evaluate

the gains by convection at 22.8 kWh and by radiation at 11.4 kWh per day. water is estimated -5.2 kWh. The balance is equal to: -130 + 22.

The average cost of heating by heat pump according to Distripool is 141 euros for a 32 m2 swimming pool in a temperate zone for heating from May to September if one benefits from

the peak hours / off-peak hours system.

__Simulation of a swimming pool cover__

The presence of a cover at night makes it possible to reduce evaporation. We can try to simulate the effect by taking a wind speed almost zero (0.1 m / s) and a humidity of 100 % the night.

In the example given in the form, the evaporation goes from 10.1 liters / hour to 6.1 liters / hour on average, ie a reduction of approximately 39.5%.

Water consumption drops from around 243 liters / day to 147 liters / day, i.e. a reduction of nearly 40% of the make-up water.

Evaporation has the effect of reducing the temperature of the water as well and therefore it will take more energy to maintain the temperature of the water.

In addition to saving energy and water consumption, swimming pool covers also have the following functions:

-reducing the consumption of pool chemicals from 35% to 60%. In fact, the evaporation of the products used is also reduced. for the treatment of the swimming pool.

-reduce cleaning time by keeping dirt and other debris out of the pool.

-allow to secure the basin according to the model.

Outdoor pools take advantage of the sun's heat and absorb 5% to 85% of solar energy. If you use the blanket, partly during the day, it will have the effect of reducing this contribution to heating

depending on the type of blanket you use.

When choosing a transparent bubble cover may be preferable to an opaque cover.

__Evaporation in the presence of bathers:__

evaporation. on the one hand because swimmers disturb the body of water and increase evaporation in a manner somewhat analogous to the wind; secondly, the wet bodies of swimmers and all the water that

is sprayed on the dry parts around the swimming pool increase losses It is as if we increased the evaporation surface. Empirical correlations for evaporation in the presence of bathers are generally

established as a function of the density of bathers per m ^ 2.

Applying the data from the form for a 32 m ^ 2 swimming pool used on average 4 hours per day by 4 bathers leads to an evaporation which goes from 10.1 liter / h to 12.1 liter / h, i.e. an increase

of l around 20%. This leads to a significant increase in the drop in the water level per day, which goes from 7.6 mm to 9.3 mm.

__Notes:__

The main factors that affect evaporation are surface area, air and water temperatures, presence or absence of wind, air humidity. the air is dry, for example, the greater the evaporation.

Other factors, which have not been taken into account here such as the level of the water in relation to the edge of the swimming pool, the presence or not of a wind breaker, the presence

or not of a cover, also play a role.

- Formula 1 seems to give better results in calm air (wind speed equal to 0) up to a light breeze (V less than 2 m / s or 7.6 km / h).Above V=2 m / s formula 2 seems more precise.

- Formula 2 is given for a residential pool free of occupant.

-To reduce your water consumption, a few precautions must be taken:

1) The wind greatly accelerates evaporation. It is therefore wise to protect the swimming pool from drafts by planting a hedge around it (or perpendicular to the prevailing winds). We can thus expect

a reduction of 30% to 50% of the wind speed depending on the density of the WINDBREAK .

To reduce evaporation, the windbreak must be high enough and close enough to the pool so as not to create turbulence on the pool. Turbulence has the effect of increasing evaporation, which is contrary

to the desired effect.

However, you also do not want the windbreak to be too close and / or too high and protect the pool from the sun's rays which contribute to heating.

A balance must be found between these two phenomena (turbulence and shadow) to have the right distance and the adequate height.

2) For an outdoor pool a night cover to reduce evaporation is also very useful especially if the water temperature is higher

than the air temperature which increases the evaporation.For the same for an indoor pool it is advisable to limit this phenomenon,

to keep the ambient air temperature 2 to 3 degrees above the water temperature.

3) If the water level is lower than the upper edge of the pool, a weakening of the convection and thus a reduction of the

evaporation can be expected. Do not fill to the brim!

4) The use of a blanket at night reduces the drop in temperature due to heat transfers to the outside but also reduces the

evaporation of water. heating), water savings.

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