Question

Water is flowing at the rate of 15 m³/s through a rectangular channel of 5 m width. If the acceleration due to gravity (g) is 9.81 m/s², the critical velocity for the flow is ______ m/s (rounded off to three decimal places).

Hint:

Find discharge per unit width q = Q/B, then use critical depth y_c = (q^2/g)^(1/3) and critical velocity V_c = sqrt(g y_c).

Answer 1

Correct Answer: 3.087

Instead of finding the critical depth first, this can be solved in one shot using a combined relation for a rectangular channel. At critical flow, $V_c = \sqrt{g y_c}$ and $y_c = (q^2/g)^{1/3}$; substituting one into the other gives the compact formula $V_c = (g q)^{1/3}$, where $q$ is the discharge per unit width. Here $q = Q/B = 15/5 = 3\ \text{m}^2/\text{s}$, so $V_c = (9.81 \times 3)^{1/3} = (29.43)^{1/3}$. Taking the cube root, $(29.43)^{1/3} \approx 3.087\ \text{m/s}$, since $3.087^3 \approx 29.42$, which matches closely. This single-formula route avoids computing the critical depth as an intermediate step and lands on the same numerical answer, confirming the result is consistent.

\[\boxed{V_c \approx 3.087\ \text{m/s}}\]