---
title: "Regime canal for 90 cumecs"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:regime-canal-q90-d02cm"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/regime-canal-q90-d02cm
site: BCT Engineering
---

# Regime canal for 90 cumecs

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2065 Kartik (old course), 6 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/regime-canal-q90-d02cm
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4

## Question

Design a regime canal to carry a discharge of 90 cumecs. The average particle size of bed and bank material of the canal is 0.2 cm.

## Short answer

Lacey's silt factor f=2.489. Regime velocity 1.26 m/s, bed width 41.3 m, depth 1.70 m, side slope 0.5H:1V, bed slope 1 in 1547.

## Answer

Average particle size $d = 0.2$ cm $= 2$ mm. No side slope is given, so 0.5H:1V is assumed.

#### Silt factor

$$
f = 1.76\sqrt{d_{mm}} = 1.76\sqrt{2.0} = 2.489
$$

#### Lacey's regime values

Given: $Q = 90\ \text{m}^3/\text{s}$, $f = 2.489$, side slope 0.5H:1V.

$$
\begin{aligned}
V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{90\times 2.489^2}{140}\right)^{1/6} = 1.259\ \text{m/s} \\
A &= \frac{Q}{V} = \frac{90}{1.259} = 71.48\ \text{m}^2 \\
R &= \frac{5V^2}{2f} = \frac{5\times 1.259^2}{2\times 2.489} = 1.592\ \text{m} \\
P &= 4.75\sqrt{Q} = 4.75\sqrt{90} = 45.06\ \text{m} \\
S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{2.489^{5/3}}{3340\times 90^{1/6}} = 0.000647 = \frac{1}{1547}
\end{aligned}
$$

#### Trapezoidal section

For side slope $z = 0.5$: $A = (B + zD)D$ and $P = B + 2D\sqrt{1+z^2} = B + 2.2361D$.

Eliminating $B$:

$$
\begin{aligned}
1.7361\,D^2 - 45.06\,D + 71.48 &= 0 \\
D &= \frac{45.06 - \sqrt{45.06^2 - 4\times 1.7361\times 71.48}}{2\times 1.7361} = 1.697\ \text{m} \\
B &= P - 2.2361D = 45.06 - 2.2361\times 1.697 = 41.27\ \text{m}
\end{aligned}
$$

(The other root of the quadratic gives a very deep, narrow section, so the smaller root is taken.)

Check: $A = (41.27 + 0.5\times 1.697)\times 1.697 = 71.48\ \text{m}^2$ against required 71.48 $\text{m}^2$.

**Answer:** Bed width $B \approx 41.3$ m, depth of water $D \approx 1.70$ m, side slope 0.5H:1V, bed slope $S = 1$ in 1547 (0.000647), velocity $V = 1.26$ m/s, hydraulic mean radius $R = 1.59$ m.

## Check yourself

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

1. d = 0.2 cm in the silt factor formula is entered as
   A. 0.2 mm, giving f = 0.79
   B. 2 mm, giving f = 2.489
   C. 20 mm, giving f = 7.9
   D. 0.2 cm, giving f = 0.79
   Answer: B. f = 1.76 sqrt(d in mm) needs d in mm. (Key point: Unit conversion)

2. For Q = 90 m^3/s, what wetted perimeter does Lacey's P = 4.75 sqrt(Q) give?
   A. 22.53 m
   B. 67.59 m
   C. 55.06 m
   D. 45.06 m
   Answer: D. P = 4.75 x sqrt(90) = 45.06 m, independent of silt factor. (Key point: Wetted perimeter)

3. Using S = f^(5/3)/(3340 Q^(1/6)) with f = 2.489, Q = 90, the bed slope is about
   A. 1 in 1547
   B. 1 in 154
   C. 1 in 15470
   D. 1 in 773
   Answer: A. Slope is computed directly from Lacey's slope equation; a finer silt (smaller f) gives a flatter slope. (Key point: Regime slope)

4. After finding A and P, how is the depth D of the 0.5H:1V trapezoid obtained, and which root is taken (result D = 1.697 m, B = 41.27 m)?
   A. Assume D and iterate until V = V0
   B. Eliminate B using A = (B + 0.5D)D and P = B + 2.236D, solve the quadratic and take the smaller root
   C. Take D = R directly and find B from A
   D. Take the larger root, giving a deep narrow section
   Answer: B. The two equations in B and D give a quadratic; the smaller root is the sensible section, the larger gives a very deep narrow channel. (Key point: Trapezoidal section from A and P)

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