Head–discharge relationship for a V-notch (triangular) weir How does the discharge Q over a sharp-crested V-notch vary with the measured head H above the crest (neglecting velocity of approach)?

Difficulty: Easy

Correct Answer: Q ∝ H^(5/2)

Explanation:


Introduction / Context:
Weirs are standard devices for measuring open-channel discharge. The triangular (V-notch) weir is especially sensitive at low flows, and its calibration follows a characteristic power law with head.



Given Data / Assumptions:

  • Sharp-crested V-notch, free and fully aerated nappe.
  • Head H measured at a standard upstream location.
  • Negligible velocity-of-approach correction for this conceptual relation.


Concept / Approach:
Integrating elemental discharge across the depth of the notch opening yields Q = C_d * (8/15) * √(2 g) * tan(θ/2) * H^(5/2). Thus, the fundamental dependence is Q ∝ H^(5/2). The stronger exponent compared with the rectangular weir (3/2) makes the V-notch more sensitive to small changes in H, aiding low-flow accuracy.



Step-by-Step Outline:

Express local width b(y) = 2 y tan(θ/2) at depth y below the head level.Elemental discharge dQ = C_d b(y) √(2 g (H − y)) dy.Integrate from y = 0 to H to obtain the H^(5/2) dependence.


Verification / Alternative check:
Calibration charts and ISO standards for V-notches reflect a 5/2 exponent in the governing formula, with C_d and angle terms as multiplicative factors.



Why Other Options Are Wrong:
H, 1/H, and H^(3/2) correspond to other devices or simplified relations; H^(7/2) overstates the head sensitivity and is not supported by the integration result.



Common Pitfalls:
Ignoring aeration (submergence reduces effective exponent), improper head measurement location, or using rectangular-weir exponents for triangular notches.



Final Answer:
Q ∝ H^(5/2)

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