Example 6.2: The maximum daily demand of water is 6 MLD. Design a plain sedimentation tank rectangle in shape, assuming the detention time of 4 hours and flow through velocity of 20 cm/minute. Also check up the flow rate.

Solution: Data: Daily demand = 6 MLD, Detention time = 4 hours.

1] Water to be treated per hour $= \ \frac{6 \times 10^6}{24} \ liters = \ 0.25 \ \times \ 10^6 \ lit/hr$

2] Capacity of sedimentation tank $= \ Hourly \ demand \ \times \ D.T.$

$= 0.25 \times 10^6 \times 4 = 10^6 \ lit \ = \ 1000 \ m^3$

3] Length of the tank $= \ flow \ velocity \ times \ D.T.$

$= 0.2 \times 4 \times 60 = 48 \ m$

4] Assume width of the tank = 10 m

5] Depth of the tank = $\frac{Capacity \ of \ the \ tank}{Length \ \times \ Width} = \frac{1000}{48 \times 10} = 2.08 \ m$

Assuming free board o 0.42 m, provide overall depth = 2.5 m

6] Tank dimension 48 m long x 10 m wide x 2.5 m deep.

7] Over flow rate = $\frac{Discharge \ rate \ per \ hour }{Surface \ area} = \frac{0.25 \times 10^6}{48 \times 10} = 520.83 \ lit/hr/day.$

Example 6.6: Design a rectangular sedimentation tank using the following data.

1] Population 1 lakh.

2] Overflow rate of 25 m/day.

3] Depth of tank = 3.2 m.

Solution: Data: Population = $10^5$, Overflow rate = 25 m/day

Tank depth = 3.2 m.

Assume the rate of water supply to be 200 lit/hr/day.

1] Daily demand = Rate of water supply x population

$= 200 \times 10^5 = 2 \times 10^7 = \frac{2 \times 10^7}{24} = 333.33 / m^3/hr$

2] Overflow rate = $\frac{Q}{A} = 25 \ m/day \ = \frac{25}{24} \ m/hr$

Here,

$\frac{Q}{A} = \frac{25 \times 100}{24 \times 60 \times 60} = V_s = \frac{25}{24 \times 36} = 0.029 \ cm/sec$

3] Detention time = $\frac{Depth \ of \ tank}{Overflow \ rate} = \frac{3.2}{25/24} = 3.072 \ hrs$

4] Capacity of the tank = $Hourly \ discharge \times D.T.$

$= 8.33.33 \times 3.072 = 2560 \ m^3$

5] Surface area of the tank = $\frac{Capacity}{Depth} = \frac{2560}{3.2} = 800 \ m^2$

Providing two units, surface area of one unit = $400 \ m^2$

6] Tank dimensions : Let length of tank = 4 x Width

$\therefore$ L = 4 B

Surface of the tank,$ A = L X B = 4 B X B = 4 B^2$

$\therefore$ $4 B^2 = 400$

$B^2 = 100$

$L = 4 \times B = 40 \ m$

$\therefore$ Provide 2 units, each of size 40 m x 10 x 3.5 (assuming free board = 0.3 m)

Example 6.4: Design horizontal flow type sedimentation tank of rectangular shape with the following data.

1] Diameter of the particle to be settled = 0.0001 cm

2] Temperature of water = 10°C

3] Design discharge = 1 MLD

4] Specific gravity of slit and clay = 2.65

5] Kinematic viscocity =$ 1.3101 \ \times 10^{-2} \ cm^2/sec$

Solution: Data : Particle diameter $=\mathrm{d}=0.001 \mathrm{cm}, s_{5}=2.65, \mathrm{t}=10^{\circ} \mathrm{C}$

$Q =1 \mathrm{MLD}, \quad \mathrm{v}=1.3101 \times 10^{-2} \mathrm{cm}^{2} / \mathrm{sec}$ .

1] Velocity of settling: $$ \begin{aligned} v_{s} &=\frac{g}{18}\left(s_{s}-1\right) \frac{d^{2}}{v}=\frac{981}{18}(2.65-1) \frac{(0.001)^{2}}{1.3101 \times 10^{-2}} \\ v_{s} &=6.8639 \times 10^{-3} \mathrm{cm} / \mathrm{sec} \end{aligned} $$

2] Surface loading:

Adopt surface loading of 864 $\mathrm{m}^{3} / \mathrm{m}^{2} / \mathrm{clay}$ for $\mathrm{V}_{5}=1 \mathrm{cm} / \mathrm{sec}$ . $\begin{aligned} \therefore \text { Surface loading for the velocity } 6.8639 \times 10^{-3} \\ &=864 \times 6.8639 \times 10^{-3} \\ &=5.9305 \mathrm{m}^{3} / \mathrm{m}^{2} / \mathrm{day} \end{aligned}$

3] Size and capacity of the tank: $$ \begin{aligned} Q &=1 \mathrm{MLD}=10^{3} \mathrm{m}^{3} / \mathrm{day} \\ \text { Surface area } &=\frac{\text { Daily demand }}{\text { Surface loading }}=\frac{10^{3}}{5.9305}=168.620 \mathrm{m}^{2} \end{aligned} $$

Adopt Length: Breadth $=2 : 1$

$A=2 B^{2}=168.62$

Solving, $B=9.2 \mathrm{m}$ and $\mathrm{L}=18.4 \mathrm{m}$

Assume effective depth $=1.5 \mathrm{m}$ and free board $=0.3 \mathrm{m}$

Tank dimensions 18.4 $\mathrm{m}$ long $\times 9.2 \mathrm{m}$ wide $\times 1.8 \mathrm{m}$ deep

$\begin{aligned} \therefore \quad \text { Capacity of the tank } &=18.4 \times 9.2 \times 1.5 \\ &=253.92 \mathrm{m}^{3} \end{aligned}$

4] Detention time:

$=\frac{\text { Capacity }}{\text { Discharge per hour }}=\frac{253.92}{10^{3} / 24}=6.09$ hours $=6.1$ hours

5] Weir loading:

$=\frac{Q}{\text { Width }}=\frac{10^{3}}{9.2}=108.7 \mathrm{m}^{3} / \mathrm{m} / \mathrm{day}$