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What is formula to calculate evaporation loss in cooling towers, with metric units?

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تم إضافة السؤال من قبل مستخدم محذوف‎
تاريخ النشر: 2016/09/19
مستخدم محذوف‎
من قبل مستخدم محذوف‎

Hi Dorothy,

 

It takes kJ to evaporate1kg of water (latent heat of vaporization Hv).

Heat load formula is Q = m Cp ΔT

with m = mass of water in kg

Cp = specific heat of water =4. kJ / kg / °C

ΔT = hot water T° - cold water T°

=> m = / (4, x  ΔT) ≈ kg/°C, it means that when kg of water is cooled down by1°C then1kg of water evaporates.

Generally, cooling water return is cooled down by°C therefore for m = kg of water,1kg is evaporated i.e.1/ ≈1,% of the circulating water flowrate.

 

If you want to include the circulating water flowrate C then the formula would be :

Evaporation loss = C x Cp x ΔT / Hv.

 

I assume that your final purpose is to calculate your make-up water flow rate so please be careful that there is not only evaporation losses to be taken into account:

Make-up = Evaporation losses + Windage (drift) losses + Draw-off (Blowdown to avoid salts accumulation).

 

 

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