Section 4: Soil and Water Conservation Engineering
Fluid Mechanics
Subtopic: Flow in open channels
Question Statement:
A flow rate of 2 m³·s⁻¹ is to be carried out in an open channel at a velocity of 1 m·s⁻¹. The cross section of the channel is rectangular with depth equal to one-half of the width. Considering Manning’s roughness coefficient (n) as 0.02, the slope of the channel, in %, is __________. (Rounded off to 2 decimal places)
Official Answer Key:
0.09 to 0.11
Step-by-Step Solution & Explanation
Step 1 — flow area from continuity:
A = Q/v = 2/1 = 2 m²
Step 2 — channel dimensions. For a rectangular section of width B and depth d = B/2:
A = B × d = B × (B/2) = B²/2 = 2 ⇒ B² = 4 ⇒ B = 2 m, d = 1 m
Step 3 — hydraulic radius:
Wetted perimeter P = B + 2d = 2 + 2 = 4 m
R = A/P = 2/4 = 0.5 m
Step 4 — Manning’s equation, solved for slope:
v = (1/n) R^(2/3) S^(1/2)
1 = (1/0.02) × (0.5)^(2/3) × √S
1 = 50 × 0.62996 × √S
√S = 1/31.498 = 0.031748
S = 1.0079 × 10⁻³
Step 5 — express as a percentage:
S = 0.0010079 × 100 = 0.10 %
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