---
title: "Kennedy design: Q 50, m 0.9"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:kennedy-q50-slope-15000-n00225-m09"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/kennedy-q50-slope-15000-n00225-m09
site: BCT Engineering
---

# Kennedy design: Q 50, m 0.9

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2068 Chaitra (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/kennedy-q50-slope-15000-n00225-m09
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4

## Question

Design an irrigation channel with side slope 0.5H:1V to carry a discharge of 50 m³/s at a slope of 1 in 5000. Take Kutter's n = 0.0225 and C.V.R (m) = 0.9.

## Short answer

Assume depth D, compute V0 = 0.55×0.9×D^0.64, then A = 50/V0, B, P, R, and Kutter’s V. Trial until V ≈ V0 yields D ≈ 3.67 m, bed width B ≈ 10.1 m, velocity ≈ 1.14 m/s.

## Answer

Given: $Q = 50\ \text{m}^3/\text{s}$, $n = 0.0225$, $S = 0.0002$ (1 in 5000), $m = 0.9$, side slope 0.5H:1V. Kutter's formula is used because Kutter's $n$ is specified. 

#### 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 Kutter's formula, $V = C\sqrt{RS}$ with $C = \dfrac{23 + \frac{1}{n} + \frac{0.00155}{S}}{1 + \left(23 + \frac{0.00155}{S}\right)\frac{n}{\sqrt R}}$.
5. Repeat until $V = V_0$.

#### Trials

| D (m) | $V_0$ (m/s) | A (m²) | B (m) | P (m) | R (m) | V (m/s) |
|---|---|---|---|---|---|---|
| 2.90 | 0.978 | 51.10 | 16.17 | 22.66 | 2.256 | 1.093 |
| 3.30 | 1.063 | 47.05 | 12.61 | 19.99 | 2.354 | 1.124 |
| 3.67 | 1.138 | 43.95 | 10.14 | 18.35 | 2.396 | 1.137 |
| 4.00 | 1.202 | 41.60 | 8.40 | 17.34 | 2.398 | 1.138 |

The velocities $V$ and $V_0$ become equal at $D \approx 3.67$ m, so this is the required depth.

#### Final section

Adopt $D = 3.67$ m, $B = 10.1$ m.

$$
\begin{aligned}
A &= (10.1 + 0.5\times 3.67)\times 3.67 = 43.80\ \text{m}^2 \\
P &= 10.1 + 2\times 3.67\sqrt{1+0.5^2} = 18.31\ \text{m},\quad R = 2.393\ \text{m} \\
V &= 1.137\ \text{m/s},\quad V_0 = 0.55\times 0.9\times 3.67^{0.64} = 1.138\ \text{m/s} \\
Q &= AV = 43.80\times 1.137 = 49.78\ \text{m}^3/\text{s}
\end{aligned}
$$

**Answer:** Depth $D \approx 3.67$ m, bed width $B \approx 10.1$ m, side slope 0.5H:1V, bed slope 1 in 5000, velocity $\approx 1.14$ m/s (discharge check 49.8 m³/s, i.e. within about 0.4% of the design value).

## Check yourself

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

1. Why is Kutter's formula used for velocity here?
   A. Kutter's n is specified in the data
   B. Manning's formula cannot be used for canals
   C. Kutter's formula needs no trial
   D. It gives V0 directly
   Answer: A. The problem gives Kutter's n, so V is found by Kutter's formula. (Key point: Why Kutter's formula)

2. How is Kennedy's critical velocity found?
   A. V0 = 0.55 D^0.64
   B. V0 = 0.55 m D^0.64
   C. V0 = m D^0.55
   D. V0 = 0.64 m D^0.55
   Answer: B. V0 = 0.55 m D^0.64, with m the critical velocity ratio (0.9 here). (Key point: Critical velocity)

3. What is the stopping condition of the trials?
   A. Q = 50 for any D
   B. B equals D
   C. V (Kutter) equals V0 (Kennedy)
   D. R equals D
   Answer: C. Depth is adjusted until the velocity from Kutter's formula equals Kennedy's V0. (Key point: Trial and error)

4. What final section does the answer adopt?
   A. D = 2.90 m, B = 16.2 m
   B. D = 4.00 m, B = 8.4 m
   C. D = 3.30 m, B = 12.6 m
   D. D = 3.67 m, B = 10.1 m
   Answer: D. V and V0 match at D about 3.67 m, with B about 10.1 m and velocity about 1.14 m/s. (Key point: Final result)

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