Open-channel/Weir flow – discharge over a rectangular sharp-crested weir with velocity of approach Given H1 = H + Ha, where H is the head over the crest and Ha is the velocity-of-approach head, select the correct expression for discharge Q over a rectangular sharp-crested weir (Cd is coefficient of discharge, L is effective crest length).

Introduction / Context:Discharge over a sharp-crested rectangular weir depends on the effective head causing flow. When approach velocity is non-negligible, an additional head term appears. This question checks recognition of the correct functional dependence on head.

Given Data / Assumptions:

• Rectangular sharp-crested weir with coefficient Cd.
• H is static head over crest; Ha is velocity-of-approach head.
• H1 = H + Ha represents total effective head at approach section.
• Standard, free, fully aerated nappe; neglect end contractions beyond effective L.

Concept / Approach:For a sharp-crested weir, the theoretical discharge is proportional to the integral of velocity across depth, yielding the 3/2 power law in head. Accounting for velocity of approach, the effective head terms enter as H1^(3/2) − Ha^(3/2).

Step-by-Step Solution:

Verification / Alternative check:Dimensional analysis confirms Q has units of L^3/T when head is to the 3/2 power multiplied by L √(2g). Linear or quadratic head dependences (options a and c) do not match the classical derivation.

Why Other Options Are Wrong:

• (a) Linear in head: incorrect functional form.
• (c) Quadratic in head: inconsistent with weir theory.
• (d) 5/2 power: belongs to broad-crested or other profiles under different assumptions, not the sharp-crested rectangular case with this correction.

Common Pitfalls:Forgetting to subtract Ha^(3/2), which leads to overestimation of Q when approach velocity is significant.

Final Answer:Q = Cd · L · √(2g) · (H1^(3/2) − Ha^(3/2))