Question

A water treatment plant having flow rate of $600m^3/hr$ has to remove 80% of the particles having a settling velocity of $0.18cm/sec$. The required plan area of a Type - I settling tank is

• A. $74.07m^2$
• B. $101.5m^2$
• C. $92.6m^2$
• D. $115.74m^2$

Answer 1

The correct option is A $74.07m^2$

$\eta =\frac{V_s}{V_o}\times 100$
$80=\frac{0.18}{V_o}\times 100$

$V_o=0.225cm/s$ $[\therefore V_o=\frac{Q}{\left(Planarea\right)}]$

$A=\frac{600/3600}{0.225\times {10}^{-2}}=74.07m^2$