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

Step 1: Get the design flow in cubic metres per second.
The plant's demand is 12 million litres a day. Since $1 \ m^3 = 1000$ L, that is $Q = 12000 \ m^3/\text{day}$. There are $86400$ seconds in a day, so:
$Q = \dfrac{12000}{86400} = 0.1389 \ m^3/s$

Step 2: Find the tank length straight from the flow velocity.
A particle moving at the horizontal velocity $v$ takes exactly the detention period $t_d$ to travel the full length $L$ of the tank, so $L = v \times t_d$ gives the length before the width is even touched.
$t_d = 6$ hours $= 21600$ s, $v = 0.003$ m/s.
$L = 0.003 \times 21600 = 64.8$ m

Step 3: Get the required tank volume.
The volume held equals the flow rate times the detention time:
$V = Q \times t_d = 0.1389 \times 21600 = 3000 \ m^3$

Step 4: Solve for width from volume, length and depth.
$V = L \times W \times D$, so:
$W = \dfrac{V}{L \times D} = \dfrac{3000}{64.8 \times 4} = \dfrac{3000}{259.2}$
$W = 11.574$ m
Rounded to two decimal places this gives the same width as the direct continuity route, which confirms the answer.
\[ \boxed{W = 11.57 \ m} \]