hi,

read the solar pond literature. you will find mathematical equation to calculate heat loss from upper surface of pond/pool to atmosphere.

Dear Sitaram,

Generally, evaporation account for 30 to 50% of the total heat loss. In the middle of summer, mid-size pools typically lose about 50 mm of water per week in this way, the equivalent of 150 kilowatts-hours. This phenomenon is accelerating in dry or windy. The frequent use of a pool cover significantly reduce evaporation losses.

Here is the formula for evaporation losses

 Pevapo. = (25+19Vw) x S x (X - X') x Lv x t/1000

with:

Pevapo: evaporation losses (kWh)

25 + 19Vv: empirical formula giving the ratio of the heat exchange coefficient hconv by convection between the water and air (W / m²K) on the Cair specific heat of air (= 0,277Wh / kg of air.K)

Vw: wind speed (m / s)

S: the pool surface (m²)

X: specific humidity of saturated air at the temperature of the water (g water / kg air)

X ': specific humidity of ambient air at the temperature of the water (g water / kg air)

Lv: enthalpy (or latent heat) of vaporization of water: 0.625 Wh / g

t: time without solar cover (in hours)

With my best regards

Prof. Bachir ACHOUR

Hello dear, thank you for you help Muhammad Asim. 

Warm Regards 

Dear Prof Achour,

Thank you for your help Sir, and for the precise information and formula. I really appreciate it.

Warm Regards 

You can use the standard correlations for flat plate, find Sherwood number to get mass transfer coefficient.

Use this, area and water vapour pressure (or water vapour density) difference to get the rate of mass loss.  Ref: Heat and Mass Transfer by Incropera and De Witt.