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: Understanding the Question.
We are given the total discharge Q and the critical depth $y_c$ for a rectangular channel, and asked to find the channel width b that makes this depth critical.

Step 2: Key Formula or Approach.
For a rectangular channel, the critical depth is related to the discharge per unit width $q = Q/b$ by:
\[ y_c = \left(\frac{q^2}{g}\right)^{1/3} \]
which rearranges to:
\[ q = \sqrt{g\,y_c^3} \]

Step 3: Detailed Explanation.
Substitute $g = 9.8$ m/s$^2$ and $y_c = 9.8$ m:
\[ q = \sqrt{9.8 \times (9.8)^3} = \sqrt{(9.8)^4} = (9.8)^2 = 96.04 \text{ m}^3/\text{s per m width} \]
Now use $q = Q/b$ to find the width:
\[ b = \frac{Q}{q} = \frac{96}{96.04} = 0.9996 \text{ m} \]

Step 4: Final Answer.
Rounded to two decimal places, the required width is 1.00 m.
\[ \boxed{b = 1.00 \text{ m}} \]