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Design a set of rapid sand filters for treating water required for a population
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Example $8.2 :$ Design a set of rapid sand filters for treating water required for a population of $80,000$ . Rate of water supply $=200$ lit/hr/day. The filters are rated to work at 5000 lit $/ \mathrm{hr} / \mathrm{m}^{2}$ . Show the arrangement of filter units.

Solution : 1] Daily demand $=80,000 \times 200=16 \times 10^{6}$ lit/day

2]. Assume 3$\%$ of filtered water is used for washing filter everyday. Also assume that 30 minutes are required for filter washing.

3] Quantity of filtered water per hour

$=\frac{1.03 \times 16 \times 10^{6}}{(24-0.5)}=0.7012766 \times 10^{6} \mathrm{lit} / \mathrm{hr}$

4] $\therefore$ Fiter area required $=A=\frac{0.7012766 \times 10^{6}}{5000}=140.25 \approx 140 \mathrm{m}^{2}$

5] Let the size of each filter unit be $9 \mathrm{m} \times 5.5 \mathrm{m}$

No. of units required $=\frac{140}{49.5}=2.82 \approx 3$

Provide one unit as stand by. Hence, provide 4 units each of size $9 \mathrm{m} \times 5.5 \mathrm{m}$

Design of Under-Drainage System

1] Assuming total area of perforations is 0.3$\%$ of entire filter area.

Total area of perforations $=0.003 \times 9 \times 5.5=0.1485 \mathrm{m}^{2}$

2] Assuming 12 mm diameter perforations.

Total cross section of laterals $=2 \times$ Area of perforation

$=2 \times$ Area of perforation

$=2 \times 0.1485 \mathrm{m}^{2}$

$=0.297 \mathrm{m}^{2}$

3] Keeping cross sectional area of mainfold $=1.5 \times$ Total cross section of laterals

$=1.5 \times 0.297=0.4455 \mathrm{m}^{2}$

$\therefore \quad$ Diameter of manifold $=\sqrt{\frac{0.4455 \times 4}{\pi}}=0.75 \mathrm{m}$

Provide 75 $\mathrm{cm}$ diameter manifold laid along the length of filter unit.

Assuming spacing of lateral of $20 \mathrm{cm},$ number of laterals $=\frac{9 \times 100}{20}=45$

4] Hence, provide 45 laterals on either side of manifold, total laterals $=90$ Length of each lateral $=\frac{(5.5-0.75)}{2}=2.375 \mathrm{m}$

5] Number and area of perforations Let, n be the total number of perforations, each of 12 $\mathrm{mm}$ diameter in all the 90 laterals.

$h \times \frac{\pi}{4}(12)^{2}=0.1485 \times(1000)^{2},$ solving $h=1313$

$\therefore$ Number of perforations on each lateral $=\frac{1313}{90}=14.58 \approx 15$

a] Hence provide 15 perforations per lateral

b] Area of perforations per lateral $=15 \times \frac{\pi}{4}(12)^{2}=1696.5 \mathrm{mm}^{2}$

c] \begin{aligned} \text { Area of each lateral } &=2 \times \text { Area of perforations on each lateral } \\ &=2 \times 1696.5=3393 \mathrm{mm}^{2} \end{aligned}

$\therefore$ Diameter of each lateral $=\sqrt{\frac{3393 \times 4}{\pi}}=65 \mathrm{mm}$

Hence, provide 65 $\mathrm{mm}$ diameter laterals at $20 \mathrm{cm} \mathrm{c} / \mathrm{c},$ each lateral having 15 perforations of 12 $\mathrm{mm}$ diameter.

Check: (i) As per design criteria, $\frac{\text { Length of lateral }}{\text { Diameter of lateral }}$ should not be greater than 60 .

$\therefore \quad \frac{2.375 \times 1000}{65}=36\lt60$

(ii) Spacing of perforation $=\frac{\text { Length of lateral in } \mathrm{cm}}{\text { No. of perforations per lateral }}$

$=\frac{2.375 \times 100}{15}$

$=16 \mathrm{cm}\lt20$

6] Minimum depth of filter sand, assuming mean diameter of sand $=1 \mathrm{mm}$

\begin{aligned} \text { and } B &=2 \times 10^{-3} \\ \text { Head } \operatorname{loss} &=2.5 \mathrm{m} \end{aligned}

\begin{aligned} L &=\frac{Q d^{3} h}{29323 \times B} \\ &=\frac{5 \times(1)^{3} \times 2.5}{29323 \times 2 \times 10^{-3}} \end{aligned}

$=0.213 \mathrm{m}=21.3 \mathrm{cm}$

$\therefore$ Provide 60 $\mathrm{cm}$ depth of sand.

7] \begin{aligned} \text { Depth of filter tank }=& \text { Depth of under-drains }+\text { Depth of gravel }+\text { Depth of sand } \\ &+\text { Depth of water }+\text { Free board } \end{aligned}

$=60+45+60+100+30$

$=295=300 \mathrm{cm}=3 \mathrm{m}$

$\therefore$ Provide 4 units of size $9 \mathrm{m} \times 5.5 \mathrm{m} \times 3 \mathrm{m}$ , one-unit being stand by.