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Discharge over overflow spillway

Short answer

Discharge is computed using Q = (2/3) Cd sqrt(2g) L H^(3/2). With Cd=0.74, L=60 m, and head H=2 m, the result is 370.8 m³/s.

An overflow spillway is 60 m long, with a crest elevation of 133 m and a reservoir water surface elevation of 135 m. Calculate the discharge over the spillway, assuming a discharge coefficient of 0.74.

Given: crest length L=60L = 60 m, crest elevation =133= 133 m, reservoir water surface =135= 135 m, coefficient of discharge Cd=0.74C_d = 0.74.

Head over crest:

H=135−133=2 mH = 135 - 133 = 2\ \text{m}

Formula (flow over a weir/overflow spillway, approach velocity neglected):

Q=23 Cd 2g  L H3/2Q = \frac{2}{3}\,C_d\,\sqrt{2g}\;L\,H^{3/2}

Substitution:

Q=23×0.74×2×9.81×60×23/2=2.185×60×2.828=370.8 m3/s\begin{aligned} Q &= \frac{2}{3} \times 0.74 \times \sqrt{2 \times 9.81} \times 60 \times 2^{3/2} \\ &= 2.185 \times 60 \times 2.828 \\ &= 370.8\ \text{m}^3/\text{s} \end{aligned}

(The equivalent spillway coefficient C=23Cd2g=2.185C = \tfrac{2}{3}C_d\sqrt{2g} = 2.185, which is close to the usual value of about 2.2 for ogee crests in Q=CLH3/2Q = C L H^{3/2}.)

Answer: discharge over the spillway Q≈370.8Q \approx 370.8 m³/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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