---
title: "Kennedy design: Q 7 cumec"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:kennedy-q7-n025-m11"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/kennedy-q7-n025-m11
site: BCT Engineering
---

# Kennedy design: Q 7 cumec

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2078 Baisakh, 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-q7-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 7 cumecs. Assume n = 0.025 and critical velocity ratio m = 1.1. The channel has a bed slope of 0.3 m per kilometer.

## Short answer

Depth 1.18 m, bed width 8.2 m, side slopes 0.5H:1V, bed slope 1 in 3333, velocity 0.67 m/s (discharge 7.0 m³/s).

## Answer

Given: $Q = 7\ \text{m}^3/\text{s}$, $n = 0.025$, $S = 0.0003$ (1 in 3333), $m = 1.1$, side slope 0.5H:1V. Bed slope $= 0.3$ m/km $= 0.0003$. 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) |
|---|---|---|---|---|---|---|
| 0.90 | 0.566 | 12.38 | 13.30 | 15.32 | 0.808 | 0.601 |
| 1.10 | 0.643 | 10.89 | 9.35 | 11.81 | 0.922 | 0.656 |
| 1.18 | 0.673 | 10.41 | 8.23 | 10.87 | 0.958 | 0.673 |
| 1.30 | 0.716 | 9.78 | 6.87 | 9.78 | 1.000 | 0.693 |

The velocities $V$ and $V_0$ become equal at $D \approx 1.18$ m, so this is the required depth.

#### Final section

Adopt $D = 1.18$ m, $B = 8.2$ m.

$$
\begin{aligned}
A &= (8.2 + 0.5\times 1.18)\times 1.18 = 10.37\ \text{m}^2 \\
P &= 8.2 + 2\times 1.18\sqrt{1+0.5^2} = 10.84\ \text{m},\quad R = 0.957\ \text{m} \\
V &= 0.673\ \text{m/s},\quad V_0 = 0.55\times 1.1\times 1.18^{0.64} = 0.673\ \text{m/s} \\
Q &= AV = 10.37\times 0.673 = 6.98\ \text{m}^3/\text{s}
\end{aligned}
$$

**Answer:** Depth $D \approx 1.18$ m, bed width $B \approx 8.2$ m, side slope 0.5H:1V, bed slope 1 in 3333, velocity $\approx 0.67$ m/s (discharge check 7.0 m³/s, i.e. within about 0.3% 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. A bed slope of 0.3 m per km is used in Manning's formula as which value?
   A. S = 0.003
   B. S = 0.03
   C. S = 0.0003 (about 1 in 3333)
   D. S = 0.3
   Answer: C. 0.3 m per 1000 m = 0.0003. (Key point: Slope conversion)

3. Given A = Q/V0 and a side slope z, how is the bed width found for an assumed D?
   A. B = A/D + zD
   B. B = A x D
   C. B = (A + zD^2)/D
   D. B = (A - zD^2)/D
   Answer: D. A = (B + zD)D, so B = A/D - zD = (A - zD^2)/D. (Key point: Bed width formula)

4. At which depth do V and V0 agree, and what is the bed width?
   A. D about 1.18 m, B about 8.2 m
   B. D about 0.90 m, B about 13.3 m
   C. D about 1.30 m, B about 6.9 m
   D. D about 1.18 m, B about 6.9 m
   Answer: A. At D = 1.18 m both velocities are 0.673 m/s and B = 8.2 m, giving Q = 6.98 m^3/s. (Key point: Final answer)

## Close variants set in other exams

- [Kennedy design: Q 5 cumec](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Akennedy-q5-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
- [Kennedy: Q 30, V 1.75 m/s](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Akennedy-q30-velocity-175) — Irrigation and Drainage Engineering, chapter 4
- [Kennedy design: Q 10, n 0.0245](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Akennedy-q10-n00245) — Irrigation and Drainage Engineering, chapter 4, asked 3×
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- [Kennedy design: Q 50, m 0.9](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Akennedy-q50-slope-15000-n00225-m09) — Irrigation and Drainage Engineering, chapter 4

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