---
title: "Crest level of fall, Q 30 m3/s"
subject: "Irrigation and Drainage Engineering"
chapter: 7
key: "semester_6/irrigation_and_drainage_engineering:fall-crest-level-30cumec-broad-crested"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/fall-crest-level-30cumec-broad-crested
site: BCT Engineering
---

# Crest level of fall, Q 30 m3/s

Irrigation and Drainage Engineering · Chapter 7: Regulating Structures

- Asked in: 2062 Kartik (old course), 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/fall-crest-level-30cumec-broad-crested
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7

## Question

Following data were observed in a canal fall: FSL of canal = 115 m; Bed level of canal = 112 m; Bed width of canal = 15 m; Design discharge = 30 m³/s; Side slope of canal = 2:1; Length of rectangular crest of the fall = 10 m (Broad crested weir). Calculate the crest level of the fall.

## Short answer

Crest level 113.55 m; head over crest H = 1.457 m, crest 1.55 m above canal bed.

## Answer

For a rectangular broad-crested weir with free flow, the discharge is $Q = 1.705\,L\,H^{3/2}$, where $H$ is the total head over the crest including the velocity head of approach. The crest level is the u/s energy level minus this head.

#### Data
$Q = 30$ m³/s, canal FSL 115.00 m, bed level 112.00 m (depth $D = 3.0$ m), bed width $B = 15$ m, side slope 2:1, crest length $L = 10$ m.

#### Step 1. Approach velocity
$$
\begin{aligned} A &= (B + mD)D = (15 + 2\times3)\times3 = 63\ \text{m}^2 \\ v &= \frac{Q}{A} = \frac{30}{63} = 0.476\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.0116\ \text{m} \end{aligned}
$$

#### Step 2. Head over the crest
$$
H = \left(\frac{Q}{1.705\,L}\right)^{2/3} = \left(\frac{30}{1.705\times10}\right)^{2/3} = (1.7595)^{2/3} = 1.457\ \text{m}
$$

#### Step 3. Crest level
$$
\text{Crest level} = FSL + h_a - H = 115.000 + 0.0116 - 1.457 = 113.55\ \text{m}
$$

The crest is 1.55 m above the canal bed. Check: the head over the crest measured from the FSL is $115 - 113.554 = 1.446$ m; adding the velocity head of 0.012 m gives $H = 1.457$ m.

**Answer:** crest level $= 113.55$ m (head over crest $H = 1.46$ m).

```text
 FSL 115.0 ~~~~~~~~~~~~~~~~~~~~~
                   ____ crest 113.55 m
 bed 112.0 _______|    |________
                  L = 10 m, broad crest
```

## Check yourself

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

1. Which relation gives the head over the broad crest here?
   A. Q = 1.705 L H^(3/2)
   B. Q = 1.705 L H^2
   C. Q = L H^(1/2)/1.705
   D. Q = 1.705 L H^(5/2)
   Answer: A. For a free-flow rectangular broad-crested weir Q = 1.705 L H^(3/2), so H = (Q/(1.705 L))^(2/3). (Key point: Broad-crested weir discharge formula)

2. What is the role of the approach velocity head h_a in finding crest level?
   A. It is subtracted from the FSL
   B. It is ignored because it is always zero
   C. It is added to FSL to give the u/s energy level before subtracting H
   D. It is multiplied by the crest length
   Answer: C. Total head H includes h_a, so crest level = FSL + h_a - H = 115 + 0.0116 - 1.457 = 113.55 m. (Key point: Velocity head of approach)

3. How is the approach velocity found for the trapezoidal canal?
   A. v = Q/(B x D) = 30/45
   B. v = Q/A with A = (B + mD)D = 63 m^2, giving 0.476 m/s
   C. v = sqrt(2gD)
   D. v = Q/L with L = 10 m
   Answer: B. Canal area for side slope 2:1 is (15 + 2x3)x3 = 63 m^2, so v = 30/63 = 0.476 m/s. (Key point: Approach velocity calculation)

4. Which result is correct for the crest level of the fall?
   A. 115.00 m
   B. 113.55 m
   C. 112.00 m
   D. 116.46 m
   Answer: B. Subtracting H = 1.457 m from the energy level of 115.0116 m gives about 113.55 m, 1.55 m above the bed. (Key point: Final crest level result)

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