Head–discharge proportionality for notches Choose the correct statement(s) relating discharge Q to head H over a weir/notch under free-flow conditions.

Civil Engineering Hydraulics Difficulty: Easy
Choose an option
  • A
    For a triangular (V-notch), Q ∝ H^(5/2)
  • B
    For a rectangular sharp-crested weir/notch, Q ∝ H^(3/2)
  • C
    Both (a) and (b)
  • D
    Neither (a) nor (b)
  • E
    Only (b) is correct; (a) is false

Answer

Correct Answer: Both (a) and (b)

Explanation

Introduction / Context:Discharge measurement by notches and weirs is fundamental in hydraulic laboratories and field channels. The head–discharge exponent depends on the crest shape and flow contraction behavior.

Given Data / Assumptions:

  • Free, fully aerated overflow without submergence.
  • Standard coefficients near calibrated values; end-contraction corrections omitted for proportionality.
  • Sharp-crested geometry for the rectangular case; V-notch with fixed angle (e.g., 90°) for the triangular case.

Concept / Approach:

Classical results: rectangular sharp-crested weir Q ∝ H^(3/2); triangular (V-notch) Q ∝ H^(5/2). The larger exponent for the V-notch reflects the narrowing flow area with decreasing head, increasing sensitivity to H.

Step-by-Step Solution:

Rectangular: Q = C_d * b * √(2 g) * H^(3/2).Triangular: Q = (8/15) * C_d * √(2 g) * tan(θ/2) * H^(5/2).Hence proportionalities in (a) and (b) are both correct.

Verification / Alternative check:

Calibration charts and standard texts confirm these exponents under non-drowned, sharp-crested conditions.

Why Other Options Are Wrong:

Options denying either exponent contradict well-established discharge formulas.

Common Pitfalls:

Using rectangular exponents for V-notches; ignoring submergence or approach velocity corrections when high accuracy is needed.

Final Answer:

Both (a) and (b)

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