Section 4: Soil and Water Conservation Engineering
Fluid Mechanics
Subtopic: Flow through orifices, weirs and notches
Question Statement:
In an experimental setup, the discharge through a triangular notch is 0.0074 m³ s⁻¹ at an operating head of 0.1 m. The coefficient of discharge for the notch is 0.7. If the required discharge is 0.1 m³ s⁻¹, the corresponding head in m is
Options:
(A) 0.00015
(B) 0.035
(C) 0.283
(D) 67.13
Official Answer Key:
C
Step-by-Step Solution & Explanation
Triangular (V-notch) weir formula: Q = (8/15)·C_d·√(2g)·tan(θ/2)·h^(5/2).
Step 1 — find tan(θ/2) from the calibration data (Q₁=0.0074 m³/s at h₁=0.1 m): coefficient (8/15)×0.7×√(2×9.81) = 1.6537; tan(θ/2) = 0.0074/(1.6537×0.1^2.5) = 0.0074/0.005230 = 1.415.
Step 2 — solve for h at Q₂=0.1 m³/s: 0.1 = 1.6537×1.415×h^2.5 ⇒ h^2.5 = 0.1/2.340 = 0.04274 ⇒ h = 0.283 m. Answer: (C).
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.