Question

For a water treatment plant having a flow rate of 432 $m^3/hr$, what is the required plan area of a Type I settling tank to remove 90% of the particles having a setting velocity of 0.12 $cm/sec$

• A. $120m^2$
• B. $111m^2$
• C. $90m^2$
• D. $100m^2$

Answer 1

The correct option is C $90m^2$

Given, $Q=432m^3/hr$

$=\frac{432}{60\times 60}{m}^3/sec$

$=0.12m^3/sec$

$\eta =90\%$

$\eta =\frac{V_s}{V_0}\times 100$

$V_0=\frac{0.12cm/sec\times 100}{90}$

$=0.133cm/sec$

$\Rightarrow V_0=0.00133m/sec$

Also, $V_0=\frac{Q}{BL}$

Plan area, $BL=\frac{0.12}{0.00133}{m}^2$

$=90.225m^2$

$\simeq 90m^2$