






DESIGN OF FANLESS INDUCED DRAFT COOLING TOWER: 

CALCULATION OF EVAPORATION LOSSES (1000 M3/HR): 

Latent heat of evaporation is 540.5 k.cal./kg of water say 550 k.cal./kg.
Heat load = m x cp x (th – tc )
Where cp = specific heat of water = 1
th = hot water temp. = 45^{0}C
tc = cooled water temp. =35^{0}C
m =weight of water kg.
550 k.cal. = m x 1 x ( 45 – 35 )^{0}C = evaporation of 1 kg of water
m = 550 = 55.0 kg of water.
10 x 1
i.e. 55.0 kg of water cools by ( 45 – 35 = 10^{0}C ), releases 550k.cal.of heat
Therefore when 55.0 kg of water is cooled from 45^{0}C to 350C it evaporates 1kg of water.
Therefore % of evaporation = 1 x 100 / 55.0 = 1.8 %
Therefore for 1000 M^{3}/ hr. water evaporated = 18.0 M^{3}/hr
Thus to cool 1000 M^{3} / hr of water from 45^{0}C to 350C, 18.0 M^{3}/ hr of water is to be evaporated
This can be achieved by supplying air with the help of fan or by nozzles.
Fans and the nozzles are the instrument to supply air to evaporate 18.0 M^{3} / hr of water and hence there is no difference in evaporation losses because of nozzles. For same delta ‘T’ based on the above we have to calculate the numbers of nozzles, which is equivalent to the fan as far as quantity of air, is concerned.














