| 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 temp0C: |
45 (113 0F ) |
| Cooled water temp0C: |
35 (95 0F) |
| Wet bulb temp0C: |
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 280C: |
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 M3/ hr |
| |
=18 M3/ hr |
| |
=18000 kg/hr |
|
| |
| Therefore 18000 kg of water will be required to be evaporated for cooling water from 450C to 350C 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/ M3 = 1.295 kg/M3 |
|
| (lb/2.2/cu.ft./35.28 = lb/cu. ft. = 35.28/2.2 = 16.03 kg/ M3 |
| |
| Therefore air required = 697674 / 1.295 = 538745 M3/hr. |
| |
| Now for 1000 M3/hr of water flow we have 555 nozzles |
| |
| Therefore air to be sucked per nozzle= 538745 / 555 =971 M3/hr |
| |
| Now water spray diameter is 300 mm and height of spray is 17’=5.18 mtr |
| |
| Therefore Volume of air = 0.3662 M3 / sec. |
| |
| = 1318 M3/hr (about 30 % more than required quantity of
971 M3hr as calculated above) |
| |
