---
title: "Kennedy design: Q 5 cumec"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:kennedy-q5-n025-m11"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/kennedy-q5-n025-m11
site: BCT Engineering
---

# Kennedy design: Q 5 cumec

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2079 Jestha, 6 marks
- Video: [Design of Channel by Kennedy's theory || Type 1 Kennedy's solved example with detail design steps](https://www.youtube.com/watch?v=aQtoJkNRygQ) — Professor Nidhi Trivedi
- 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/kennedy-q5-n025-m11
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4

## Question

Design an irrigation channel to carry a discharge of 5 cumecs. Assume n = 0.025 and critical velocity ratio m = 1.1. The channel has a bed slope of 0.5 m per kilometer.

## Short answer

By Kennedy's trial method: depth 1.69 m, bed width 2.7 m, side slope 0.5H:1V, bed slope 1/2000, velocity 0.85 m/s (discharge ~5.1 m³/s).

## Answer

Given: $Q = 5\ \text{m}^3/\text{s}$, $n = 0.025$, $S = 0.0005$ (1 in 2000), $m = 1.1$, side slope 0.5H:1V. Bed slope $= 0.5$ m/km $= 0.0005$. Side slope is not given, so 0.5H:1V is assumed; Manning's formula is used with the given $n$.

#### Method (Kennedy's silt theory, trial and error)

1. Assume depth $D$.
2. Kennedy's critical velocity: $V_0 = 0.55\,m\,D^{0.64}$.
3. Area $A = Q/V_0$, bed width $B = (A - zD^2)/D$, perimeter $P = B + 2D\sqrt{1+z^2}$, $R = A/P$.
4. Find $V$ from Manning's formula, $V = \frac{1}{n}R^{2/3}S^{1/2}$.
5. Repeat until $V = V_0$.

#### Trials

| D (m) | $V_0$ (m/s) | A (m²) | B (m) | P (m) | R (m) | V (m/s) |
|---|---|---|---|---|---|---|
| 1.30 | 0.716 | 6.99 | 4.72 | 7.63 | 0.916 | 0.843 |
| 1.50 | 0.784 | 6.38 | 3.50 | 6.85 | 0.930 | 0.852 |
| 1.69 | 0.846 | 5.91 | 2.65 | 6.43 | 0.919 | 0.845 |
| 1.90 | 0.912 | 5.48 | 1.93 | 6.18 | 0.886 | 0.825 |

The velocities $V$ and $V_0$ become equal at $D \approx 1.69$ m, so this is the required depth.

#### Final section

Adopt $D = 1.69$ m, $B = 2.7$ m.

$$
\begin{aligned}
A &= (2.7 + 0.5\times 1.69)\times 1.69 = 5.99\ \text{m}^2 \\
P &= 2.7 + 2\times 1.69\sqrt{1+0.5^2} = 6.48\ \text{m},\quad R = 0.925\ \text{m} \\
V &= 0.849\ \text{m/s},\quad V_0 = 0.55\times 1.1\times 1.69^{0.64} = 0.846\ \text{m/s} \\
Q &= AV = 5.99\times 0.849 = 5.09\ \text{m}^3/\text{s}
\end{aligned}
$$

**Answer:** Depth $D \approx 1.69$ m, bed width $B \approx 2.7$ m, side slope 0.5H:1V, bed slope 1 in 2000, velocity $\approx 0.85$ m/s (discharge check 5.1 m³/s, i.e. within about 1.7% of the design value).

## Check yourself

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

1. In Kennedy's method with a known slope and n, what is the condition that fixes the correct depth?
   A. The area A equals the wetted perimeter P
   B. The Manning velocity V equals Kennedy's critical velocity V0
   C. The depth D equals the bed width B
   D. The hydraulic mean radius R equals 1 m
   Answer: B. Assume D, find V0 = 0.55 m D^0.64 and the section from Q/V0, then compute V by Manning; repeat until V = V0. (Key point: Kennedy trial-and-error)

2. With m = 1.1 and D = 1.69 m, what is Kennedy's critical velocity?
   A. V0 = 0.55 x 1.69^0.64 = 0.77 m/s
   B. V0 = 1.1 x 1.69 = 1.86 m/s
   C. V0 = 0.55 x 1.1 x 1.69^0.64 = 0.846 m/s
   D. V0 = Q/B = 1.85 m/s
   Answer: C. The formula V0 = 0.55 m D^0.64 includes the critical velocity ratio m, giving 0.846 m/s. (Key point: Critical velocity)

3. What bed slope is used for 0.5 m per kilometre?
   A. S = 0.5 (1 in 2)
   B. S = 0.05
   C. S = 0.005 (1 in 200)
   D. S = 0.5/1000 = 0.0005 (1 in 2000)
   Answer: D. 0.5 m of fall in 1000 m is 0.0005. This is the slope used in Manning's formula. (Key point: Slope conversion)

4. Which section does the answer adopt?
   A. D about 1.69 m, B about 2.7 m, side slope 0.5H:1V
   B. D about 1.30 m, B about 4.7 m
   C. D about 1.90 m, B about 1.9 m
   D. D about 2.7 m, B about 1.69 m
   Answer: A. V and V0 become equal at D = 1.69 m, where B = 2.65, adopted as 2.7 m. (Key point: Final section)

## Close variants set in other exams

- [Kennedy design: Q 7 cumec](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Akennedy-q7-n025-m11) — Irrigation and Drainage Engineering, chapter 4, asked 2×

## Related in Irrigation and Drainage Engineering

- [Kennedy design: Q 40, slope 1 in 5000](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Akennedy-q40-slope-15000) — Irrigation and Drainage Engineering, chapter 4
- [Lacey and Kennedy bed width, Q 40](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Alacey-kennedy-q40-comparison) — Irrigation and Drainage Engineering, chapter 4
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