written 5.1 years ago by |
Example $8.1 :$ Design six units of slow sand for the following data
1] Population to be served $=50,000$ persons
2] Per capita water demand $=150$ lit/hr/day
3] Rate of filtration $=180 \mathrm{lit} / \mathrm{hr} / \mathrm{m}^{2}$
4] LIB=2
5] Maximum demand = $1.8 \times Average \ daily \ demand$
6] Out of six units, one unit will act as stand by. Draw its layout.
Solution:
$\begin{aligned} \text 1] { Average \ daily \ demand } &=\text { Population } \times \text { Rate of water supply } \\ &=50,000 \times 150=75 \times 10^{5} \text { lit/day } \end{aligned}$
$\begin{aligned} \text 2] { Maximum \ demand } &=1.8 \times \text { Average daily demand } \\ &=1.8 \times 75 \times 10^{5}=13.5 \times 10^{5} \mathrm{lit} / \mathrm{day} \end{aligned}$
3] Discharge per hour $=\frac{13.5 \times 10^{5}}{24}=562500$ lit/hour
4] Rate of filtration $=180 \mathrm{lit} / \mathrm{hr} / \mathrm{m}^{2}$
$\begin{aligned} \text 5] { Area \ (surface) \ required \ for \ filtration } & \\ &=\frac{562500}{180}=3125 \mathrm{m}^{2} \end{aligned}$
6] Six units of filter are to be provided, including 1 unit as stand by
7] Area per unit of filter $=\frac{3125}{5}=625 \mathrm{m}^{2}$
8] $\mathrm{L/B}=2 \quad \therefore \mathrm{L}=2 \mathrm{B}$ and Area $\mathrm{A}=625=\mathrm{L} \times \mathrm{B}=2 \mathrm{B} \times \mathrm{B}=2 \mathrm{B}^{2}$
Solving, $B=17.7 \mathrm{m}, \mathrm{L}=2 \mathrm{B}=2 \times 17.7=35.5 \mathrm{m}$
$\therefore$ Provide 6 units of size 35.5 $\mathrm{m}$ long $\times 17.7 \mathrm{m}$ wide.