Question

Width (in m) of a rectangular channel required to carry a discharge of 96 m³/s at a critical depth of 9.8 m is (rounded off to two decimal places).

Use acceleration due to gravity = 9.8 m/s²

Hint:

Use $y_c = (q^2/g)^{1/3}$ with $q$ as discharge per unit width, then divide the total discharge by $q$ to get the width.

Answer 1

Correct Answer: 1

Step 1: Use the Froude number condition instead of the direct formula.
Critical flow, by definition, is the depth at which the Froude number $Fr = 1$. For a rectangular channel this means:
\[ \frac{V_c^2}{g\,y_c} = 1 \implies V_c = \sqrt{g\,y_c} \]

Step 2: Find the critical velocity.
\[ V_c = \sqrt{9.8 \times 9.8} = 9.8 \text{ m/s} \]
This makes sense here only because $y_c$ was numerically equal to $g$; in general you would just substitute the given values.

Step 3: Use continuity to bring in the width.
For a rectangular channel, area $A = b\,y_c$, and continuity gives $Q = A\,V_c = b\,y_c\,V_c$. Solve for $b$:
\[ b = \frac{Q}{y_c\,V_c} = \frac{96}{9.8 \times 9.8} = \frac{96}{96.04} \]

Step 4: Compute the result.
\[ b = 0.9996 \text{ m} \approx 1.00 \text{ m} \]
This is the same numerical answer as the direct critical-depth-formula route, since both routes come from the same critical-flow condition, just written in terms of velocity and Froude number here instead of the compact $y_c$ formula.
\[ \boxed{b = 1.00 \text{ m}} \]