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:
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:
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:
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
Discussion & Comments