Question Bank • QB_SWCE_ADV_092 NAT • 2 Marks

Section 4: Soil and Water Conservation Engineering

Fluid Mechanics

Subtopic: Flow through orifices, weirs and notches

Question Statement:

A standard $90^\circ$ V-notch (triangular weir) is installed to measure drainage runoff. The head over the vertex of the notch is $H = 0.35\text{ m}$ and the coefficient of discharge is $C_d = 0.60$. Taking $g = 9.81\text{ m/s}^2$, the measured runoff discharge $Q$ in liters per second ($\text{L/s}$) is ________ (round off to one decimal place).

Answer Key:

102.7

Step-by-Step Solution & Explanation

**Method 1: Canonical V-Notch Formula** $Q = \frac{8}{15} C_d \sqrt{2 g} \tan\left(\frac{\theta}{2}\right) H^{5/2}$ For a $90^\circ$ notch, $\theta = 90^\circ \implies \frac{\theta}{2} = 45^\circ \implies \tan(45^\circ) = 1.0$. $\sqrt{2 g} = \sqrt{2 \times 9.81} = \sqrt{19.62} = 4.4294$ $H^{5/2} = (0.35)^{2.5} = (0.35)^2 \times \sqrt{0.35} = 0.1225 \times 0.59161 = 0.072472\text{ m}^{2.5}$ $Q = \frac{8}{15} \times 0.60 \times 4.4294 \times 1.0 \times 0.072472 = 0.320 \times 4.4294 \times 0.072472 = 0.10272\text{ m}^3/\text{s}$ Converting to liters per second: $Q = 0.10272 \times 1000 = 102.72\text{ L/s} \approx 102.7\text{ L/s}$ **Method 2: Simplified Metric Empirical Factor** $Q = 1.4174 \times H^{2.5} = 1.4174 \times 0.072472 = 0.10272\text{ m}^3/\text{s} = 102.7\text{ L/s}$
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