For a rectangular sharp-crested weir, discharge varies as H^(3/2). If the head H is measured with a +1% error, what is the approximate percentage error in the computed discharge?

Introduction / Context:Hydrometric structures like sharp-crested weirs convert a head measurement H into discharge Q using Q ∝ H^(3/2). Understanding error propagation through power relations is critical for estimating measurement uncertainty in field hydrology.

Given Data / Assumptions:

• Rectangular sharp-crested weir with standard flow conditions (aerated nappe, no submergence).
• Q ∝ H^(3/2) ignoring coefficient variations.
• Small relative error, linearized propagation applicable.

Concept / Approach:If Q ∝ H^n, then a small relative error dH/H produces dQ/Q = n * dH/H. For n = 3/2, a 1% head error gives 1.5% discharge error. This simple rule of thumb is widely used in discharge rating analyses.

Step-by-Step Solution:Write Q = k * H^(3/2), where k is constant for geometry and Cd.Differentiate: dQ/Q = (3/2) * dH/H.Substitute dH/H = 0.01 → dQ/Q = 0.015 → 1.5%.

Verification / Alternative check:Finite difference check: H → 1.01 H; Q_new/Q_old = (1.01)^(1.5) ≈ 1.015, confirming 1.5%.

Why Other Options Are Wrong:1.25%, 1.75%, 2.25%: do not match the 3/2 scaling; they over/understate sensitivity.

Common Pitfalls:

• Applying the 3/2 exponent to suppressed or drowned weirs without checking conditions.
• Ignoring additional uncertainty from Cd calibration and head measurement devices.

Final Answer:1.5%