Rectangular weir with velocity of approach—correct discharge expression For a sharp-crested rectangular weir where the approach velocity is not negligible, which expression correctly accounts for the velocity-of-approach correction (H is head over crest, h_a is velocity-head equivalent)?

Difficulty: Medium

Correct Answer: Cd x L x 2g [(H + ha)3/2 - ha3/2]

Explanation:


Introduction / Context:
For rectangular weirs, the classical discharge relation assumes negligible approach velocity. When approach velocity is significant, the upstream total head includes both the static head over the crest and the velocity head, requiring a correction to avoid overestimation of discharge.


Given Data / Assumptions:

  • Sharp-crested rectangular weir, fully aerated nappe.
  • H = static head above crest; h_a = v_a^2/(2 g) = velocity head in the approach channel.
  • Coefficient of discharge C_d and crest length L are known.


Concept / Approach:

The effective head driving flow is not just H but (H + h_a). However, the measured head gauge usually reads H referenced to a stilling point. The established correction subtracts the portion that would be counted if the approach section were also treated as a weir section (h_a term), leading to the standard form with a negative h_a^{3/2} correction.


Step-by-Step Solution:

Start from Q ∝ (H_total)^{3/2} with H_total = H + h_a.Correct to avoid double-counting approach velocity effect → subtract h_a^{3/2} term.Hence: Q = K · C_d · L · [ (H + h_a)^{3/2} − h_a^{3/2} ], where K contains √(2 g) and constants.


Verification / Alternative check:

When v_a → 0, h_a → 0 and the expression reduces to the classical rectangular weir formula proportional to H^{3/2}, as expected.


Why Other Options Are Wrong:

(a) adds h_a^{3/2} instead of subtracting; (c) and (d) incorrectly replace H with H − h_a, which is not the effective upstream total head.


Common Pitfalls:

Using gauge readings too close to the crest (velocity effects not dissipated); neglecting the velocity of approach for high specific discharges.


Final Answer:

Cd x L x 2g [(H + ha)3/2 - ha3/2]

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