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: Write Manning's equation out in full instead of jumping to a proportion.
For a wide rectangular channel per unit width, area $A = y \times 1$ and hydraulic radius $R \approx y$. Manning's formula for discharge per unit width is:
\[ q = \frac{1}{n} A R^{2/3} S^{1/2} = \frac{1}{n} y \cdot y^{2/3} \cdot S^{1/2} = \frac{\sqrt{S}}{n} y^{5/3} \]

Step 2: Use the first data point to fix the unknown constant.
Let $K = \sqrt{S}/n$ (this stays fixed since roughness and bed slope do not change). From the first condition, $q_1 = 10$ m$^3$/s/m at $y_1 = 2.0$ m:
\[ 10 = K (2.0)^{5/3} \]
\[ (2.0)^{5/3} = 2^{1.667} = 3.1748 \]
\[ K = \frac{10}{3.1748} = 3.1498 \]

Step 3: Use K to solve for the new depth.
With discharge doubled, $q_2 = 20$ m$^3$/s/m:
\[ 20 = 3.1498 \, y_2^{5/3} \]
\[ y_2^{5/3} = \frac{20}{3.1498} = 6.3496 \]
\[ y_2 = (6.3496)^{3/5} = (6.3496)^{0.6} \]

Step 4: Evaluate.
\[ y_2 = 3.031 \approx 3.03 \text{ m} \]
Finding K explicitly and then re-solving gives exactly the same result as taking the ratio of the two conditions directly, since K cancels out either way, but this route makes clear that the constant K itself never needs to be known if only the ratio is wanted.
\[ \boxed{y_2 = 3.03 \text{ m}} \]