Question

A water treatment plant treating 10 MLD of water requires 20 $mg/L$ of filter Alum $A{l}_2(S{O}_4{)}_3.18H_{2}0$. Water has 6 $mg/L$ of alkalinity as $CaC{O}_3$ .
( Al = 26.97, S = 32, 0 = 16, H = 1, Ca = 40, and C = 12)
Quantity of quick lime required (${10}^6$ mg per year as CaO) shall be

• A. $4132$
• B. $6132$
• C. $2132$
• D. $3000$

Answer 1

The correct option is B $6132$

In water treatment, water must have sufficient alkalinity for alum to be effective.
Reaction of alum with $HC{O}_3^{-}$ alkalinity is

$\underset{\left(666g\right)}{A{l}_2(S{O}_4{)}_3}.18H_{2}O+\underset{(3.\times 162g)}{3Ca(HC{O}_3{)}_2}\to$

$3CaS{O}_4+2Al(OH{)}_3\downarrow +6C{O}_2+\uparrow 18H_{2}O$

Discharge, $Q=10MLD$

$=10\times {10}^{6}L/d$

Alum required per day

$=10\times {10}^6\times 20mg/d$

$=200\times {10}^{6}mg/d$

666 g alum require alkalinity

$=3\times 162g$

$200\times {10}^{6}mg$ alum require alkalinity

$=\frac{3\times 162}{666}\times 200\times {10}^6$

$=145.95\times {10}^{6}mg/d$

Alkalinity expressed as

$CaC{O}_3=145.95\times {10}^6$

$\times \frac{\text{Equivalent weight of}CaC{O}_3}{\text{Equivalent weight of}Ca(HC{O}_3{)}_2}$

=$145.95\times {10}^6\times \frac{50}{81}$

$=90.09\times {10}^{6}mg/d$

as $CaC{O}_3=90({10}^6$ mg per day as $CaC{O}_3)$

Natural alkalinity available in water

$=6mg/L$
$=6\times 10MLD$
$=6\times 10\times {10}^{6}mg/d$
$=60\times {10}^{6}mg/d$

as $CaC{O}_3$ additional alkalinity required

$=90\times {10}^6-60\times {10}^6$
$=30\times {10}^{6}mg/d$

as $CaC{O}_3$ additional alkalinity required as $CaO$

$=30\times {10}^6\times \frac{Eq.wt.ofCaO}{Eq.wt.ofCaC{O}_3}$

$=30\times {10}^6\times \frac{28}{50}$

$=16.8\times {10}^{6}mg/d$ as CaO

$=16.8\times {10}^6\times 365$ mg/year as CaO

$=6132\times {10}^6$ mg/year as CaO

$=6132({10}^6$ mg per year as CaO)