Question

A rectangular open concrete drainage channel has base width of 3 m and flow depth of 1.5 m. The channel bed slope is 0.001. If the Manning's roughness coefficient of concrete used in this channel is 0.018, then the average flow velocity of this channel (in m/s) is (rounded off to two decimal places).

Hint:

Use Manning's equation V = (1/n) R^(2/3) S^(1/2) with R = A/P for the rectangular channel (top open to air, so P = b + 2d).

Answer 1

Correct Answer: 1.45

Step 1: Understanding the Question.
We need the average flow velocity in an open rectangular channel using Manning's equation, given the channel dimensions, bed slope, and roughness coefficient.

Step 2: Key Formula.
Manning's equation for the average velocity of flow in an open channel is
\[ V = \frac{1}{n} R^{2/3} S^{1/2} \]
where \(n\) is Manning's roughness coefficient, \(R\) is the hydraulic radius, and \(S\) is the bed slope.

Step 3: Find the flow area and wetted perimeter.
The channel is rectangular with base width \(b = 3\) m and flow depth \(d = 1.5\) m.
Flow area:
\[ A = b \times d = 3 \times 1.5 = 4.5 \text{ m}^2 \]
Wetted perimeter (the base plus the two vertical sides in contact with water; the top is open to air, so it is not wetted):
\[ P = b + 2d = 3 + 2(1.5) = 6 \text{ m} \]

Step 4: Find the hydraulic radius.
\[ R = \frac{A}{P} = \frac{4.5}{6} = 0.75 \text{ m} \]

Step 5: Apply Manning's equation.
With \(n = 0.018\), \(R = 0.75\) m, and \(S = 0.001\):
\[ R^{2/3} = (0.75)^{2/3} = 0.8256 \]
\[ S^{1/2} = (0.001)^{1/2} = 0.03162 \]
\[ V = \frac{1}{0.018} \times 0.8256 \times 0.03162 = 55.556 \times 0.8256 \times 0.03162 \]
\[ V = 1.45 \text{ m/s} \]

Final Answer:
The average flow velocity in the channel is 1.45 m/s. \[ \boxed{1.45 \text{ m/s}} \]