Question

The flow depth for a discharge of 10 m³/s in a wide rectangular channel is 2.0 m. Assume that the flow is uniform.

If the discharge is doubled, the flow depth (in m) in this channel is (rounded off to two decimal places).

Hint:

For a wide channel under uniform flow, discharge per unit width is proportional to $y^{5/3}$ (Manning's equation with hydraulic radius approx. equal to depth), so use that ratio directly.

Answer 1

Correct Answer: 3.03

Step 1: Understanding the Question.
The channel is wide, so the hydraulic radius is approximately equal to the flow depth y (the wetted perimeter is dominated by the wide bed, the sides barely matter). The slope and roughness of the channel do not change when the discharge changes, only the depth adjusts.

Step 2: Key Formula.
By Manning's equation, for a wide channel with $R \approx y$, the discharge per unit width q is:
\[ q = \frac{1}{n} y^{2/3} \cdot y \cdot S^{1/2} = \frac{S^{1/2}}{n} y^{5/3} \]
Since n and S stay the same, this gives the proportion:
\[ q \propto y^{5/3} \]

Step 3: Detailed Explanation.
Let $q_1 = 10$ m$^3$/s/m at $y_1 = 2.0$ m, and $q_2 = 20$ m$^3$/s/m (discharge doubled) at the unknown depth $y_2$. Using the proportion:
\[ \frac{q_2}{q_1} = \left(\frac{y_2}{y_1}\right)^{5/3} \]
\[ 2 = \left(\frac{y_2}{2}\right)^{5/3} \]
\[ \frac{y_2}{2} = 2^{3/5} = 2^{0.6} \]
\[ 2^{0.6} = 1.5157 \]
\[ y_2 = 2 \times 1.5157 = 3.031 \text{ m} \]

Step 4: Final Answer.
Rounded to two decimal places, the new flow depth is 3.03 m.
\[ \boxed{y_2 = 3.03 \text{ m}} \]