Question Bank • QB_SUB_SWCE_0124 NAT • 2 Marks

Section 4: Soil and Water Conservation Engineering

Fluid Mechanics

Subtopic: Flow through orifices, weirs and notches

Question Statement:

A symmetrical $90^\circ$ triangular V-notch weir operates under a head of $H = 0.30\text{ m}$. If the discharge coefficient is $C_d = 0.60$ and $g = 9.81\text{ m/s}^2$, the discharge $Q$ (in $\text{m}^3\text{/s}$, rounded to 4 decimal places) is:

Answer Key:

0.0699

Step-by-Step Solution & Explanation

Triangular V-notch weir discharge formula: $Q = \frac{8}{15} C_d \sqrt{2g} \tan\left(\frac{\theta}{2}\right) H^{5/2}$ For $\theta = 90^\circ$, $\tan(45^\circ) = 1.0$: $Q = \frac{8}{15} (0.60) \sqrt{2 \times 9.81} (1.0) (0.30)^{2.5} = 1.417 \times (0.30)^{2.5} = 0.0699\text{ m}^3\text{/s}$
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