---
title: "Discharge over overflow spillway"
subject: "Hydro Power"
chapter: 6
key: "semester_6/hydro_power:overflow-spillway-discharge"
source: https://www.bctengineer.com/q/semester_6/hydro_power/overflow-spillway-discharge
site: BCT Engineering
---

# Discharge over overflow spillway

Hydro Power · Chapter 6: Spillways

- Asked in: 2081 Chaitra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/overflow-spillway-discharge
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-6

## Question

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.

## 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.

## Answer

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

**Head over crest:**
$$
H = 135 - 133 = 2\ \text{m}
$$

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

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

**Substitution:**
$$
\begin{aligned}
Q &= \frac{2}{3} \times 0.74 \times \sqrt{2 \times 9.81} \times 60 \times 2^{3/2} \\
&= 2.185 \times 60 \times 2.828 \\
&= 370.8\ \text{m}^3/\text{s}
\end{aligned}
$$

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

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The head over the crest is
   A. 2 m
   B. 135 m
   C. 133 m
   D. 4 m
   Answer: A. 135 - 133 = 2 m. (Key point: Head)

2. The discharge formula used is
   A. Q = Cd L H
   B. Q = (2/3) Cd sqrt(2g) L H^(3/2)
   C. Q = (2/3) Cd L H^2
   D. Q = Cd sqrt(2g) L H^(1/2)
   Answer: B. Standard weir equation. (Key point: Formula)

3. The equivalent coefficient C = (2/3) Cd sqrt(2g) is about
   A. 0.74
   B. 4.43
   C. 2.19
   D. 9.81
   Answer: C. (2/3) x 0.74 x 4.429 = 2.185. (Key point: Coefficient)

4. The discharge is about
   A. 185 m^3/s
   B. 742 m^3/s
   C. 131 m^3/s
   D. 371 m^3/s
   Answer: D. 2.185 x 60 x 2.828 = 370.8. (Key point: Result)

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## Related in other subjects

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- [Reservoir routing through spillway, 6-hr inflow](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Areservoir-routing-spillway-6hr) — Engineering Hydrology, chapter 7
- [Gravity dam equilibrium and overtopping discharge coefficient](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-reservoir-capacity-curve-hrwl) — Hydropower Engineering, chapter 4
- [Minimum critical reservoir top elevation](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aspillway-min-critical-reservoir-elevation) — Hydropower Engineering, chapter 4