Section 4: Soil and Water Conservation Engineering
Fluid Mechanics
Subtopic: Flow through orifices, weirs and notches
Question Statement:
A trapezoidal weir has a side slope of 1 horizontal unit to 4 vertical units. The length of the weir is less than the width of the channel. If the head over the weir is 70 cm and the discharge is 0.85 m³ s⁻¹, the length of the weir in m is _______. [round off to two decimal places] (Take coefficient of discharge = 0.62 and g = 9.81 m s⁻²)
Official Answer Key:
0.78 to 0.81
Step-by-Step Solution & Explanation
• A side slope of 1 horizontal : 4 vertical is the classical Cipolletti weir slope, specifically chosen so that the extra discharge through the two triangular end sections exactly offsets the loss of discharge due to end contraction of an equivalent suppressed rectangular weir. Consequently, a Cipolletti weir is designed to be computed using the simple rectangular-weir formula directly with the crest length L (no separate triangular term needed).
• Q = Cd (2/3)√(2g) L H^(3/2)
• √(2×9.81) = 4.429; (2/3)×0.62×4.429 = 1.831; H^1.5 = 0.7^1.5 = 0.5857
• Q = 1.831×0.5857×L = 1.0724L
• 0.85 = 1.0724L ⟹ L = 0.7926 m
• Initial computation attempted with the general trapezoidal-weir formula (rectangular term + separate triangular-end term) gave L ≈ 0.65 m, outside the official range; recognizing the 1:4 side slope as the defining Cipolletti geometry and switching to the simplified rectangular-equivalent formula resolved the discrepancy and matched the official range.
Computed value: 0.79 m (within official range 0.78–0.81), using the Cipolletti-weir simplified formula
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