Discharge over a Cipolletti weir A sharp-crested Cipolletti weir has crest length L = 2.185 m. If the measured head over the crest is H = 1.0 m, estimate the discharge (assume a typical sharp-crest coefficient C_d ≈ 0.62).

Difficulty: Medium

Correct Answer: 4.0 m^3/s

Explanation:


Introduction / Context:
The Cipolletti (trapezoidal) weir uses 1H:4V side slopes to cancel end-contraction effects so its discharge is approximated by the rectangular suppressed-weir formula using the measured crest length. This problem applies the standard sharp-crested relation with a representative C_d to estimate flow.


Given Data / Assumptions:

  • Crest length L = 2.185 m.
  • Head over crest H = 1.0 m (measured upstream at standard location).
  • Sharp-crested weir; use C_d ≈ 0.62.
  • g = 9.81 m/s^2; neglect velocity of approach and submergence.


Concept / Approach:

For a Cipolletti weir, the discharge is approximated by the rectangular suppressed-weir equation: Q = (2/3) * C_d * L * sqrt(2g) * H^(3/2). With H = 1, the power reduces to 1, simplifying evaluation.


Step-by-Step Solution:

Compute sqrt(2g) = sqrt(19.62) ≈ 4.429.Coefficient block: (2/3) * C_d = (2/3) * 0.62 ≈ 0.4133.Multiply by L: 0.4133 * 2.185 ≈ 0.902.Multiply by sqrt(2g): 0.902 * 4.429 ≈ 3.996 m^3/s ≈ 4.0 m^3/s.


Verification / Alternative check:

Reasonable magnitudes: a 2.2 m wide sharp-crest with 1 m head producing roughly 4 m^3/s aligns with standard tables. Minor deviations in C_d (0.60–0.62) change Q by only a few percent.


Why Other Options Are Wrong:

(a–c) Underestimate the computed discharge for H = 1 m. (e) Slight overestimate beyond typical C_d range with H = 1 m and L = 2.185 m.


Common Pitfalls:

Forgetting that Cipolletti geometry eliminates end-contraction correction; using effective length L_e incorrectly; neglecting submergence criteria or velocity-of-approach corrections when they are significant.


Final Answer:

4.0 m^3/s

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