Section 4: Soil and Water Conservation Engineering
Soil and Water Erosion
Subtopic: Vegetative waterways
Question Statement:
A parabolic shaped grass-waterway is to be designed to carry a flow of 2.85 m³ s⁻¹ down the slope of 3%. The permissible velocity of water in the waterway is 1.78 m s⁻¹. If the freeboard depth is excluded, the most appropriate top width in m and depth in m, respectively are (Take Manning's roughness coefficient = 0.04)
Options:
(A) 4 and 0.6
(B) 6 and 0.4
(C) 7 and 0.5
(D) 5.5 and 0.6
Official Answer Key:
B
Step-by-Step Solution & Explanation
• Required flow area A = Q/V = 2.85/1.78 = 1.601 m²
• From Manning's equation, R^(2/3) = Vn/√S = (1.78×0.04)/√0.03 = 0.0712/0.1732 = 0.4111 ⟹ R = 0.4111^1.5 = 0.264 m
• For a parabolic channel, R ≈ (2/3)d ⟹ d = 0.264×1.5 = 0.40 m
• Parabolic area A = (2/3)Td ⟹ T = 3A/(2d) = 3×1.601/(2×0.3954) = 6.07 ≈ 6.0 m
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