Question

A water treatment plant is required to process 28800 $m^3/d$ of raw water (density = $1000kg/m^3$), kinematic viscosity $={10}^{-6}{m}^2/s$. The rapid mixing tank imparts a velocity gradient of 900 $s^{-1}$ to blend 35 $mg/L$ of alum with the flow for a detention time of 2 min. The power input (W) required for rapid mixing is

• A. $36$
• B. $32400$
• C. $324$
• D. $32.4$

Answer 1

The correct option is B $32400$

Velocity gradient

$G=\sqrt{\frac{P}{\mu V}}$

where

$G=900s^{-1}$

V= volume of mixing tank

= discharge $\times$ Detention time

$=28800m^3/d\times 2min$

$=\frac{28800}{86400}\times 2\times 60$

$=60m^3$

Dynamic viscosity,

$m=v\times \rho$

$={10}^{-6}\times 1000$

$={10}^{-3}Ns/m^2$

Power $P=G^2\mu V$

$=(900{)}^2\times {10}^{-3}\times 40$

$=32400watt$