Question

A suspension of sand like particles in water with particles of diameter 0.10 mm and below is falling into a sedimentation tank at $0.1m^3/s,g=9.8m/s^2$, Specific gravity of particles = 2.65, and kinematic viscosity of water $=1.0105\times {10}^{-2}c{m}^2/s$. The minimum surface area (in $m^2$) required for this setting tank to remove particles of size 0.06 mm and above with 100% efficiency is

• A. 31.214

Answer 1

The correct option is A 31.214
$Q=0.1m^3/s$

Particle size $\to 0.1mm$ & below

$g=9.81m/s^2$
$G_s=2.65$
$V=1.0105\times {10}^{-2}c{m}^2/s$

$\frac{Q}{A}=\text{Overflow rate}$

$\le$ settling velocity of 0.06 mm particles

$\frac{Q}{A}\le V_s$

$A\ge \frac{Q}{V_s}$

$V_s=\frac{({\gamma }_s-{\gamma }_w)d^2}{18\mu }=\frac{(G_s-1){\gamma }_w{d}^2}{18{\rho }_{w}v}$

$V_s=\frac{(G_s-1)g{d}^2}{18v}$

$=\frac{(2.65-1)\times 9.81\times (0.06\times {10}^{-3}{)}^2}{18\times 1.0105\times {10}^{-6}{m}^2/s}m/s$

$=3.20366\times {10}^{-3}m/s$

$A\ge \frac{0.1m^3/s}{3.20366\times {10}^{-3}m/s}=31.214m^2$