Weirs and Notches MCQ with answer and solution

Concept

A Cipolletti weir has side slopes of 1H:4V. The extra flow through the sloping sides cancels the loss from end contractions, so it works like a rectangular weir without end contractions.

Formula

Q = (2/3) × C_d × √(2g) × L × H^3/2

Step-by-step solution

1. Data: L = 2.0 m, H = 1 m, C_d = 0.62, g = 9.81 m/s² 2. √(2g) = √(19.62) = 4.429 3. H^3/2 = 1^3/2 = 1 4. Q = (2/3) × 0.62 × 4.429 × 2.0 × 1 5. Q = 0.4133 × 4.429 × 2.0 = 3.66 m³/s

Explanation

In a Cipolletti weir, the sloping sides add flow equal to the flow lost by end contractions. So we use the simple rectangular weir formula with full crest length. This gives Q = 3.66 m³/s.

Why the other options are wrong

A (10.35): wrong use of constants or the head term. C (45.21) and D (75.68): far too large for L = 2 m and H = 1 m.

Shortcut

(2/3) × C_d × √(2g) = 1.83 for C_d = 0.62, so Q = 1.83 × L × H^3/2 = 1.83 × 2 = 3.66 m³/s

Common mistake

Applying the end-contraction correction (L - 0.2H). This is not needed for a Cipolletti weir.