Question

A sedimentation basin in a water treatment plant is designed for a flow rate of $0.2m^3/s$. The basin is rectangular with a length of $32m$, a width of 8m, and a depth of $4m$. Assume that the settling velocity of these particles is governed by Stoke's law. Given: density of the particles = $2.5g/c{m}^3$ density of water $=1g/c{m}^3$ dynamic viscosity of water = $0.01g/cm.s$; gravitational acceleration = $980cm/s^2$. If the incoming water contains particles of diameter $25\mu m$ (spherical and uniform) the removal efficiency of these particles is

• A. 78%
• B. 65%
• C. 100%
• D. 51%

Answer 1

The correct option is B 65%

Given:

Flow rate = $0.2m^3/sec$
Dimension of tank = $32m\times 8m\times 4m$
Density of particles= $2.5g/cc$
Density of water= $1g/cc$
Dynamic viscosity of water = $0.01g/cm.s$
Diameter of particle = $25\mu m$

We know that Over flow rate of tank

$\left(V_s\right)=\frac{0.2}{32\times 8}$

$=7.8125\times {10}^{-4}m/sec$

And settling velocity of particle $\left(v_s\right)$,
$v_s=\frac{({\gamma }_s-{\gamma }_w)d^2}{18\mu }$

$v_s=\frac{(2.5-1)\times 9.81\times (25\times {10}^{-6}{)}^2}{18\times 0.01\times {10}^4}$

$v_s=5.1094\times {10}^{-4}m/sec$

Now, % removal efficiency = $\frac{v_s}{V_s}\times 100$
$=\frac{5.1094\times {10}^{-4}}{7.8125\times {10}^{-4}}\times 100=65.4\%$

Hence option (b) is correct.