In a wide open channel of width B = 4 m, with depth y = 2 m, slope S = 0.001, Manning n = 0.030, compute discharge Q.

• A. 6.0 m^3/s
• B. 8.43 m^3/s
• C. 10.0 m^3/s
• D. 12.0 m^3/s

Answer: (B)

Open-channel discharge uses Manning’s equation: Q = (1/n) A R^(2/3) S^(1/2). For a rectangular channel, the area is A = B×y and the wetted perimeter is P = B + 2y, so the hydraulic radius is R = A/P.

Here, B = 4 m and y = 2 m, so A = 4×2 = 8 m^2. The wetted perimeter is P = 4 + 2×2 = 8 m, giving R = 8/8 = 1 m. The slope is S = 0.001, so S^(1/2) = sqrt(0.001) ≈ 0.03162. The Manning roughness is n = 0.030, so 1/n ≈ 33.333.

Plug these in: Q ≈ 33.333 × 8 × 1^(2/3) × 0.03162 ≈ 33.333 × 8 × 0.03162 ≈ 8.42 m^3/s, which rounds to 8.43 m^3/s.

This matches the listed value, so the discharge is about 8.43 m^3/s. The key steps are computing A, P, and R for the rectangular section and then applying Manning’s equation with the given slope and roughness.