Calculation of Water Demand and Ca(OCl)Cl Required

Problem: The population of 2005 was 10000 and 2015 was 19500. What is the water demand in 2025? Given water demand is 90 lpcd. Chlorine content of 0.5 mg/l is to be added in water, how much bleaching powder s needed? Given 20%  Chlorine in bleaching powder.
Exam Name: Gas Transmission Company Ltd. 2016

Solution: 
We Know, P_f = P_p (1+r)ⁿ
  
Here, P_p  =19500
           r   = (P₁/P₂)^1/n -1
               = (19500/10000)^1/10 -1
               = 0.069
           n  = 10

∴ Water demand at 2025 = (38003* 90)
                                         =  3420270 liter
                                         =  3420.27 m³


Total amount of Cl₂ required
                 = (3420270*0.5) mg
                 = 1710135 mg
                 = 1.71 Kg

 
For 20 Kg Cl₂,  required Chlorine =100 Kg
 ∴  for 1.71Kg Cl₂,  required Chlorine ={(100/20)*1.71} Kg
                                                             = 8.55Kg 
Answer:  3420270 liter, 8.55Kg   

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