Question

A horizontal flow primary clarifier treats waste water in which 10%, 60% and 30% of particles have settling velocities of $0.1mm/s,0.2mm/s$ and $1.0mm/s$ respectively. What would be the total percentage of particles removed, if clarifier operates at a Surface Overflow Rate (SOR) of $43.2m^3/m^{2}d$?

• A. $100\%$
• B. $56\%$
• C. $86\%$
• D. $43\%$

Answer 1

The correct option is B $56\%$

If settling velocity $\left(V_s^{\prime }\right)$ is more than overflow rate $\left(V_s\right)$ then 100% removal will occur.

If $V_s^{\prime }<V_s$ then % removal

$=\left(\frac{V_s^{\prime }}{V_s}\right)\times 100$

Surface overflow rate,

$V_s=43.2m^3/m^{2}d$

$=43.2m/d$

$=\frac{4.32\times 1000}{86400}mm/s$

$=0.5mm/s$

30% particle have settling velocity more than (SOR) so they will be 100% removed. Hence total percentage of particle removed

$=30+10\times \frac{0.1}{0.5}+60\times \frac{0.2}{0.5}$
$=56\%$