In a rectangular open-channel with width B = 3 m, depth y = 1.2 m, discharge Q = 4 m^3/s, compute velocity V and specific energy E at the section.

• A. V = 1.111 m/s; E = 1.263 m
• B. V = 1.50 m/s; E = 1.400 m
• C. V = 0.833 m/s; E = 1.100 m
• D. V = 2.000 m/s; E = 2.000 m

Answer: (A)

In an open-channel of rectangular section, the flow velocity comes from the discharge divided by the cross-sectional area, and the specific energy is the depth plus the velocity head.

Compute the cross-sectional area: A = B × y = 3 m × 1.2 m = 3.6 m². The velocity is V = Q/A = 4 m³/s ÷ 3.6 m² = 1.111 m/s.

The velocity head is V²/(2g). With g ≈ 9.81 m/s², V² ≈ 1.111² ≈ 1.235, so V²/(2g) ≈ 1.235 ÷ 19.62 ≈ 0.063 m. The specific energy is E = y + V²/(2g) ≈ 1.2 m + 0.063 m ≈ 1.263 m.

So the velocity is about 1.111 m/s and the specific energy about 1.263 m, matching the given result.