Question Bank • QB_SWCE_FM_006 NAT • 2 Marks

Section 4: Soil and Water Conservation Engineering

Fluid Mechanics

Subtopic: Flow through orifices, weirs and notches

Question Statement:

A symmetrical $90^\circ$ V-notch weir is used for measuring small irrigation channel discharges. The discharge coefficient is $C_d = 0.60$. For a measured head $H = 0.30\text{ m}$ above the vertex of the notch, taking $g = 9.81\text{ m/s}^2$, the discharge through the notch in $\text{L/s}$ is ________ (round off to 1 decimal place).

Answer Key:

69.9

Step-by-Step Solution & Explanation

For a triangular V-notch with total included angle $2\theta = 90^\circ$ (so $\theta = 45^\circ$): $Q = \frac{8}{15} C_d \sqrt{2g} \tan(\theta) H^{5/2}$ Given: • $C_d = 0.60$ • $\sqrt{2 \times 9.81} = \sqrt{19.62} \approx 4.42945$ • $\tan 45^\circ = 1.0$ • $H = 0.30\text{ m} \implies H^{2.5} = (0.30)^{2.5} \approx 0.049295$ $Q = \frac{8}{15} \times 0.60 \times 4.42945 \times 1.0 \times 0.049295 = 0.32 \times 4.42945 \times 0.049295 = 0.06987\text{ m}^3\text{/s}$ $Q = 0.06987 \times 1000 = 69.87\text{ L/s} \approx 69.9\text{ L/s}$
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