A rectangular chute spillway is designed for gully erosion control with a drop (H) of 5 m, peak flow of 1.81 m³ s⁻¹ and maximum inlet water level of 0.81 m. Frictional head loss over the apron is 20 % of H. The design height of chute blocks is equal to the depth of flow in the water way. Acceleration due to gravity (g) is 9.81 m s⁻². If coefficient of discharge of the weir section is 1.66, height of the chute blocks is _______ m.
Official Answer Key:
0.13 to 0.14
Step-by-Step Solution & Explanation
Weir (inlet) equation: Q = C L H_w^{3/2} ⇒ 1.81 = 1.66 × L × (0.81)^{1.5}.
(0.81)^{1.5} = 0.729 ⇒ L = 1.81/(1.66 × 0.729) = 1.496 m.
Unit discharge over the apron: q = Q/L = 1.81/1.496 = 1.210 m² s⁻¹.
Energy available at the toe after 20 % frictional loss: E = H − 0.2H = 0.8 × 5 = 4.0 m.
Energy equation: d + q²/(2g d²) = E, with q²/2g = (1.210)²/19.62 = 0.0746.
d + 0.0746/d² = 4.0 ⇒ by iteration d = 0.139 m (supercritical root).
Height of chute blocks = depth of flow = 0.139 m — within the official range 0.13–0.14.
Verification note: if the inlet head of 0.81 m is also added to the available energy (E = 4.81 m), d works out to 0.126 m, which falls below the official band. The accepted key therefore corresponds to E = 0.8 H only, i.e. the drop less friction loss, with the approach head neglected.
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