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: Set up the channel geometry, keeping in mind only the base and side walls are wetted.
For an open rectangular channel, only the base and the two side walls touch the water; the free water surface at the top is not a wetted boundary. Base $b = 3$ m, depth $d = 1.5$ m.

Step 2: Compute area and wetted perimeter.
$$A = b \times d = 3 \times 1.5 = 4.5 \text{ m}^2$$
$$P = b + 2d = 3 + 3 = 6 \text{ m}$$
$$R = A/P = 4.5/6 = 0.75 \text{ m}$$

Step 3: Work out $R^{2/3}$ using logarithms.
$$\log(0.75) = -0.1249$$
$$\frac{2}{3}\times(-0.1249) = -0.0833$$
$$R^{2/3} = 10^{-0.0833} = 0.8256$$

Step 4: Work out $S^{1/2}$.
$$S^{1/2} = \sqrt{0.001} = 0.0316$$

Step 5: Combine using Manning's formula.
$$V = \frac{1}{n}R^{2/3}S^{1/2} = \frac{1}{0.018}(0.8256)(0.0316)$$
$$V = 55.56 \times 0.02609 = 1.45 \text{ m/s}$$

Final Answer:
The average flow velocity works out to 1.45 m/s, the same result reached from the direct substitution method.