The width of a settling tank with 2 hour detention period for treating sewage 378 cu m per hour, is A. 5 m B. 5.5 m C. 6.5 m D. 7 m

The correct answer is A. 5 m.

The width of a settling tank is calculated using the following formula:

$W = \frac{Q}{v}$

where:

• $W$ is the width of the tank (m)
• $Q$ is the flow rate of sewage (m3/h)
• $v$ is the velocity of the sewage (m/h)

The velocity of the sewage is calculated using the following formula:

$v = \frac{L}{T}$

where:

• $v$ is the velocity of the sewage (m/h)
• $L$ is the length of the tank (m)
• $T$ is the detention time (h)

The detention time is calculated using the following formula:

$T = \frac{Q}{V}$

where:

• $T$ is the detention time (h)
• $Q$ is the flow rate of sewage (m3/h)
• $V$ is the volume of the tank (m3)

The volume of the tank is calculated using the following formula:

$V = W \times L \times H$

where:

• $V$ is the volume of the tank (m3)
• $W$ is the width of the tank (m)
• $L$ is the length of the tank (m)
• $H$ is the depth of the tank (m)

Given:

• $Q = 378 \text{ m}^3/\text{h}$
• $T = 2 \text{ h}$
• $L = 10 \text{ m}$
• $H = 3 \text{ m}$

Therefore:

$W = \frac{Q}{v} = \frac{378 \text{ m}^3/\text{h}}{\frac{378 \text{ m}^3/\text{h}}{10 \text{ m} \times 3 \text{ m}}} = 5 \text{ m}$