---
title: "Discharge through chute spillway ogee crest"
subject: "Hydropower Engineering"
chapter: 4
key: "semester_7/hydropower_engineering:spillway-chute-discharge-ogee-250m"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/spillway-chute-discharge-ogee-250m
site: BCT Engineering
---

# Discharge through chute spillway ogee crest

Hydropower Engineering · Chapter 4: Headworks of Storage Plants

- Asked in: 2079 Baishakh, 4 marks
- Video: [Hydraulic Structures 15CV832 - Problems on Spillways](https://www.youtube.com/watch?v=FsjA3TThoGk) — sumanth Alekkady
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/spillway-chute-discharge-ogee-250m
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4

## Question

Determine the discharge through a chute spillway of 250.00 m long ogee crest, if the height of the spillway crest above the u/s approach channel is 10.50 m, the width of the approach channel is 2500 m, and the head over the crest is 4.50 m. Take $C_d = 0.85$.

## Short answer

Discharge over the ogee crest is computed as Q = (2/3) Cd sqrt(2g) L H^(3/2). With Cd=0.85, L=250 m, H=4.5 m, the discharge is approximately 5990 m³/s.

## Answer

**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**.

## Check yourself

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

1. Which expression gives the discharge?
   A. Q = (2/3) Cd sqrt(2g) L H^(3/2)
   B. Q = Cd L H
   C. Q = Cd L H^2
   D. Q = Cd sqrt(2g) H
   Answer: A. Weir flow over a crest of length L under head H follows the 3/2 power law. (Key point: Discharge formula)

2. Why can velocity of approach be neglected?
   A. Approach area is huge (2500 x 15 m), so Va = 0.16 m/s and ha is only 0.0013 m
   B. Approach velocity is zero by definition
   C. The crest is too long
   D. Cd already includes losses of 50%
   Answer: A. ha is 0.03% of H, so it hardly changes Q. (Key point: Velocity of approach)

3. What discharge results?
   A. About 5990 m3/s
   B. About 599 m3/s
   C. About 59,900 m3/s
   D. About 1500 m3/s
   Answer: A. Q = 627.5 x 4.5^1.5 = 5990 m3/s. (Key point: Result)

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