Now knowing the evaporation to be done for required delta ‘T’ we have to calculate quantity of atmospheric air required to pick up this much quantity of water. 

For example consider following parameter : 

Water flow M3/ hr: 
1000 
Hot water temp^{0}C: 
45 (113 0F ) 
Cooled water temp^{0}C: 
35 (95 0F) 
Wet bulb temp^{0}C: 
28 (82.4 0F) 


We can find out the humidity of air, as kg of water vapour carried by kg of dry air from sychometric chart. Air will pick up moisture till it becomes saturated, when it is allowed to be in contact with water. We can limit the absorption up to 90 % humidity 

Humidity at 28^{0}C: 
0.0192 kg of water/kg of dry air. 
Humidity at 90 %: 
0.0450 kg of water/kg of dry air. 
Therefore: 
0.0258 kg of water/kg of dry air is picked up by air. 
Evaporation losses: 
=1.8 % as calculated above 

=0.018 x 1000 M^{3}/ hr 

=18 M^{3}/ hr 

=18000 kg/hr 


Therefore 18000 kg of water will be required to be evaporated for cooling water from 45^{0}C to 35^{0}C in 1 hr. 

Quantity of air required 
= 18000 x 1/0.0258 

= 697674.41 kg/hr 
Now density of air 
= 0.0808 lb/cu. ft. 

= 0.0808x16.03 kg/ M^{3} = 1.295 kg/M^{3} 

(lb/2.2/cu.ft./35.28 = lb/cu. ft. = 35.28/2.2 = 16.03 kg/ M^{3} 

Therefore air required = 697674 / 1.295 = 538745 M^{3}/hr. 

Now for 1000 M3/hr of water flow we have 555 nozzles 

Therefore air to be sucked per nozzle= 538745 / 555 =971 M^{3}/hr 

Now water spray diameter is 300 mm and height of spray is 17’=5.18 mtr 

Therefore Volume of air = 0.3662 M^{3} / sec. 

= 1318 M3/hr (about 30 % more than required quantity of
971 M3hr as calculated above) 
