Question

A water supply scheme transports 10 MLD (Million Litres per Day) water through a $450mm$ diameter pipeline for a distance of $2.5km$. A chlorine dose of $3.50mg/litre$ is applied at the starting point of the pipeline to attain a certain level of disinfection at the downstream end. It is decided to increase the flow rate from $10MLD$ to $13MLD$ in the pipeline. Assume exponent for concentration, $n=0.86$. With this increased flow, in order to attain the same level of disinfection, the chlorine dose (in mg/litre) to be applied at the starting point should be

• A. $4.75$
• B. $5.55$
• C. $3.95$
• D. $4.40$

Answer 1

The correct option is A $4.75$

As per chicks-Watson’s law,
In the disinfection process, for the constant degree of disinfection
We have the following relationships:

$t\times c^n=K$

where
t= time required
C= concentration of disinfectant
n= dilution coefficient = 0.86 (given)
k= constant

$\Rightarrow t_1\times C_1^n=t_2\times C_2^n$
$\Rightarrow t_1\times C_1^{0.86}=t_2\times C_2^{0.86}$
[as n= 0.86]

$t_1=\frac{L}{V_1}$

L= length of pipe
$V_1$= velocity of flow

$=\frac{Q_1}{A}$

A= cross-sectional area of pipe
$\Rightarrow t_1=\left(\frac{L\times A}{Q_1}\right),Q_1=10MLD$ (given)

$C_1=3.5mg/lt\left(given\right)$

Similarly, $t_2=\left[\frac{L\times A}{Q_2}\right]$

$Q_2=13MLDgiven$
$C_2=?$

$\Rightarrow \left(\frac{L\times A}{Q_1}\right)\times C_1^{0.86}=\left(\frac{L\times A}{Q_2}\right)\times C_2^{0.86}$

$\Rightarrow \frac{1}{10}\times (3.5{)}^{0.86}=\left(\frac{1}{13}\right)\times C_2^{0.86}$

$\Rightarrow C_2=\left[\frac{13\times (3.5{)}^{0.86}}{10}\right]mg/lt$

$=4.7485mg/lt$
$\approx 4.75mg/lt$