Example $8.3 :$ Calculate the dimensions of rapid sand filter for 1 lakh population with 150 lit/capita/day. Assume the filtration rate and mean size of sand 1.5 $\mathrm{mm}$ . Find the depth of sand bed for head loss of 2 $\mathrm{m}$ if break through index $\mathrm{B}=0.002 .$ Assume rate of filtration as 100 $\mathrm{lit} / \mathrm{min} / \mathrm{m}^{2}$ .

Solution:

1] Population $=1$ lakh $=10^{5}$

2] Water demand $=150$ lit/capita/day

3] Daily demand $=10^{5} \times 150=15 \times 10^{6}$ lit/day.

4] Hourly demand $=\frac{15 \times 10^{6}}{24}=0.625 \times 10^{6}$ lit/hr

5] Filter area, $A=\frac{0.625 \times 10^{6}}{100 \times 60}=104.16 \approx 105 \mathrm{m}^{2}$

6] Let the size of each filter is $7 \mathrm{m} \times 5 \mathrm{m},$ No. of units $=\frac{105}{35}=3$

Assuming one stand by provide 4 units, each of size $7 \mathrm{m} \times 5 \mathrm{m}$ .

7] Minimum depth of filter sand in metre is given by Hudson formula.

$\begin{aligned} L &=\frac{Q d^{3} h}{29323 \times B} \\ \text { where, } & Q=\text { Filtration rate in } m^{3} / h r / m^{2} \end{aligned}$

$Q=\frac{100}{1000} \times 60=6 \mathrm{m}^{3} / \mathrm{hr} / \mathrm{m}^{2}$

$\mathrm{d}=1.5 \mathrm{mm}, \mathrm{h}=$ Head loss $=2 \mathrm{m}$

$\begin{aligned} B &=0.002=2 \times 10^{-3} \\ &L =\frac{Q d^{3} h}{29323 \times B}=\frac{6 \times(1.5)^{3} \times 2}{29323 \times 2 \times 10^{-3}} \\ &=0.69 \mathrm{m}=69 \mathrm{cm} \end{aligned}$

Hence, provide 70 $\mathrm{cm}$ depth of filter sand.

8] Assume gravel depth $=45 \mathrm{cm},$ depth of under-drains $=60 \mathrm{cm}$

Water depth $=100 \mathrm{cm}$ and free board $=30 \mathrm{cm} .$

$\therefore$ Total depth of filter basin $=$ Depth of under drains $+$ Depth of gravel $+$ Depth of sand - Depth of water + Free board $\therefore \quad$ Total depth $=60+45+70+100+30=305 \mathrm{cm}=3.05 \mathrm{m} \approx 3.1 \mathrm{m}$ 9] Hence, provide 4 units of filters, each of size $7 \mathrm{m} \times 5 \mathrm{m} \times 3.1 \mathrm{m} .$