Discharge through chute spillway ogee crest

Given: crest length $L=250$ m, head $H=4.5$ m, crest height above the approach channel $P=10.5$ m, approach channel width 2500 m, $C_d=0.85$.

Discharge over the ogee crest

$$Q=\frac{2}{3}C_d\sqrt{2g}\,L\,H^{3/2}$$ $$\begin{aligned} Q&=\frac{2}{3}(0.85)(4.429)(250)(4.5)^{3/2}\\ &=627.50\times 9.546 = 5990\ \text{m}^3/\text{s} \end{aligned}$$

Velocity of approach check

Approach channel depth $=P+H=10.5+4.5=15$ m, so area $=2500\times15=37{,}500$ m².

$$V_a=\frac{5990}{37500}=0.160\ \text{m/s},\qquad h_a=\frac{V_a^2}{2g}=0.0013\ \text{m}$$

This is only 0.03% of $H$, so its effect is negligible. Including it ($H_e=H+h_a$, iterated) gives $Q=5993$ m³/s, the same to three significant figures.

Answer: discharge through the spillway $\approx$ 5990 m³/s.