Section 4: Soil and Water Conservation Engineering
Fluid Mechanics
Subtopic: Flow through orifices, weirs and notches
Question Statement:
A trapezoidal notch, placed over an emergency spillway, has the following details: Top width = 2 m, Bottom width = 1 m, Height = 0.5 m, Coefficient of discharge for the triangular portion = 0.65, Coefficient of discharge for the rectangular portion = 0.68. For a flow head of 0.4 m over the notch, the discharge in L s⁻¹ will be about
Options:
(A) 155
(B) 353
(C) 508
(D) 663
Official Answer Key:
D
Step-by-Step Solution & Explanation
• Side slope: half the width increase (2–1)/2 = 0.5 m occurs over the full notch height 0.5 m, giving a 1:1 side slope, i.e. equivalent triangular-weir slope factor z = 1 (total, both sides).
• Trapezoidal notch discharge = rectangular-notch component (base width b = 1 m) + triangular-notch component (side slope z = 1), both evaluated at head h = 0.4 m:
• Q_rect = (2/3) Cd,rect √(2g) b h^1.5 = (2/3)(0.68)√(19.62)(1)(0.4)^1.5 = 0.6667×0.68×4.429×0.2530 ≈ 0.508 m³/s = 508 L/s.
• Q_tri = (8/15) Cd,tri √(2g) z h^2.5 = (8/15)(0.65)(4.429)(1)(0.4)^2.5 = 0.5333×0.65×4.429×0.1012 ≈ 0.155 m³/s = 155 L/s.
• Total Q = 508 + 155 ≈ 663 L/s.
• Distractor logic: option (C) 508 L/s is only the rectangular-portion discharge — a common error is forgetting to add the triangular side contribution.
• Final Answer: (D) 663
Section 5 — Irrigation and Drainage Engineering
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