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cooling towers losses 4
- Thread starter gbl62
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If you have 1 kg of water/sec circulating and the tower range is 50C, then heat to be removed is about 1*4.2*5 = 21kJ/sec. The latent heat of water is 2430kJ/kg. So amount of water to be evaporated is 21/2430 = 0.0086kgs. This becomes 0.86%.
Roughly 0.9% is considered for all practical calculations, nevertheless you can use the above method to find out evaporation losses.
With best quality drift eliminators, you consider 0.1% losses.
The blowdown can be calculated as ER/(C-1), where ER is rate of evaporation and C is cycles of concentration.
The link below gives you good concept about cooling towers.
Roughly 0.9% is considered for all practical calculations, nevertheless you can use the above method to find out evaporation losses.
With best quality drift eliminators, you consider 0.1% losses.
The blowdown can be calculated as ER/(C-1), where ER is rate of evaporation and C is cycles of concentration.
The link below gives you good concept about cooling towers.
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