Question

The maximum demand at a water purification plant has been estimated as 12 million litres per day. For the raw supplies, a rectangular sedimentation tank is to be designed with mechanical sludge removal arrangement. Consider depth of the tank as 4 m, detention period as 6 hours, and velocity of flow as 0.003 m/s.

The width (in m) of the detention tank is _________ (rounded off to two decimal places).

Hint:

Find the flow in \(m^3/s\), get the tank volume from the detention period, then use \(Q = v \times (W \times D)\) to solve for the width.

Answer 1

Correct Answer: 11.57

Concept: Sedimentation tank sizing.
A sedimentation tank is sized using two ideas together: the detention period fixes how much volume of water must sit in the tank at any instant, and the horizontal flow-through velocity fixes how big the cross-section (width times depth) must be to pass the design flow without going faster than that velocity.

Step 1: Convert the design flow to consistent units.
Maximum demand \(Q = 12\) million litres/day \(= 12 \times 10^6\) L/day. Since \(1 \, m^3 = 1000\) L:
\[ Q = \frac{12 \times 10^6}{1000} = 12000 \text{ m}^3/\text{day} \]
Convert this to \(m^3/s\) (1 day = 86400 s):
\[ Q = \frac{12000}{86400} = 0.13889 \text{ m}^3/\text{s} \]

Step 2: Find the required tank volume from the detention period.
The detention period is the average time a water particle spends inside the tank, so the tank volume equals the flow rate times the detention time. Detention period \(t_d = 6\) hours \(= 6 \times 3600 = 21600\) s.
\[ V = Q \times t_d = 0.13889 \times 21600 = 3000 \text{ m}^3 \]

Step 3: Relate volume to width using the given depth.
For a rectangular tank, \(V = L \times W \times D\), so the plan area works out to:
\[ L \times W = \frac{V}{D} = \frac{3000}{4} = 750 \text{ m}^2 \]

Step 4: Use the flow-through velocity to bring in width directly.
Water crosses the tank through a vertical cross-section of area \(W \times D\) (width times depth), and continuity gives \(Q = v \times (W \times D)\), where \(v\) is the horizontal flow velocity. So:
\[ W = \frac{Q}{v \times D} = \frac{0.13889}{0.003 \times 4} = \frac{0.13889}{0.012} \]
\[ W = 11.574 \text{ m} \]

Final Answer:
Rounded off to two decimal places, the width of the detention tank is 11.57 m. \[ \boxed{W = 11.57 \text{ m}} \]