Triangular (V) notch discharge–head relationship Identify how discharge varies with head H for a sharp-crested triangular notch.

Difficulty: Easy

Correct Answer: directly proportional to H5/2

Explanation:


Introduction:
Flow measurement over notches and weirs relies on empirical–theoretical relations between discharge Q and head H. For triangular (V) notches, the exponent of H is a key discriminant. This question checks whether you know that Q varies with H to the power 5/2.


Given Data / Assumptions:

  • Sharp-crested, well-ventilated triangular notch.
  • Free, fully contracted flow without submergence.
  • Coefficient of discharge treated as approximately constant over small H ranges.


Concept / Approach:
Elementary derivation integrates elemental discharge strips across the varying width b(y) of the notch opening with local velocity proportional to sqrt(2gH_local). For a V-notch of angle theta, width b ∝ y, and integration yields Q = C_d * (8/15) * tan(theta/2) * sqrt(2g) * H^(5/2). Hence, the dependence is Q ∝ H^(5/2).


Step-by-Step Solution:

1) Express notch width versus depth: b = 2 * y * tan(theta/2).2) Local velocity v ∝ sqrt(2g * (H - y)).3) Integrate elemental discharge b * v dy from y = 0 to H to obtain Q ∝ H^(5/2).


Verification / Alternative check:
Standard hydraulics texts list Q = K * H^(5/2) for V-notches, distinguishing them from rectangular weirs where Q ∝ H^(3/2).


Why Other Options Are Wrong:

  • H^(3/2) (direct or inverse): characteristic of rectangular weirs, not V-notches.
  • Inverse proportionalities: unphysical for increasing head producing more flow.
  • Independent of H: contradicts basic weir theory.


Common Pitfalls:
Mixing up the H exponents between rectangular and triangular notches or forgetting the role of notch angle in the proportionality constant, not in the exponent.


Final Answer:
directly proportional to H5/2

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