Question

35 MLD of water is flowing through a $5\text{km}$ long pipe of diameter $50\text{cm}$. Chlorine at the rate of 38 kg/day is applied at the entry of this pipe so that disinfected water is obtained at the exit. There is a proposal to increase the flow through this pipe to 55 MLD from 35 MLD. Assume the dilution coefficient, n =1. The minimum amount of chlorine (in kg per day) to be applied to achieve the same degree of disinfection for the enhanced flow is

• A. 82.32
• B. 103.48
• C. 93.84
• D. 74.23

Answer 1

The correct option is C 93.84

In the disinfection process we have the relationship,
$t{C}^n=k$
where t = time required to kill all the organisms
C = concentration of disinfectant
n = dilution coefficient
k = constant
$\therefore t_1{C}_1^n=t_2{C}_2^n$
In our case, n = 1
$\therefore t_1{C}_1=t_2{C}_2$
But $t_1=\frac{L}{V_1}$
where L = length of pipe ; $V_1$ = velocity of flow
$\therefore t_1=\frac{L}{Q/A}$
$\Rightarrow t_1=\frac{L.A}{Q_1}$
Also, $C_1=\frac{W_1}{Q_1}$
where $W_1$ = weight of disinfectant per day; $Q_1$ = discharge per day
$\therefore \frac{L.A}{Q_1}\times \frac{W_1}{Q_1}=\frac{L.A}{Q_2}\times \frac{W_2}{Q_2}$
$\Rightarrow W_2=\frac{Q_2^2}{Q_1^2}\times W_1$
$={\left(\frac{55}{35}\right)}^2\times 38\text{kg/day}=93.84\text{kg/day}$