Error propagation for a triangular (V-notch) weir For a triangular notch (V-notch), discharge varies as H^(5/2), where H is the head over the crest. If there is a +1% error in measuring H, the approximate percentage error in computed discharge Q is:

Difficulty: Easy

Correct Answer: 2.50%

Explanation:


Introduction / Context:
Knowing how measurement errors propagate into discharge estimates helps set instrumentation accuracy. For a V-notch weir, Q depends strongly on head, so small head errors can produce larger percentage errors in Q.


Given Data / Assumptions:

  • Q ∝ H^(5/2) for a triangular notch under free flow.
  • Small relative error in head measurement: dH/H = +1%.
  • Linearized (differential) error propagation is adequate.


Concept / Approach:

For Q ∝ H^n, the relative error in Q is n times the relative error in H: dQ/Q ≈ n * dH/H. Here, n = 5/2 = 2.5.


Step-by-Step Solution:

Write Q = k * H^(5/2).Take logarithms and differentiate: dQ/Q = (5/2) * dH/H.For dH/H = 0.01 ⇒ dQ/Q ≈ 2.5 * 0.01 = 0.025 = 2.5%.


Verification / Alternative check:

More exact finite-difference check: ((H * 1.01)^(2.5) − H^(2.5)) / H^(2.5) ≈ 2.5% for small errors, confirming the differential result.


Why Other Options Are Wrong:

1.25% and 1.5% correspond to lower exponents; 2.0% assumes square dependence; 5.0% doubles the correct factor.


Common Pitfalls:

Using the rectangular-weir exponent (3/2) accidentally; mixing absolute and relative errors.


Final Answer:

2.50%

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