A stable channel is to be designed for a discharge of Q with silt factor f as per Lacey's method. The mean flow velocity (m/s) in the channel is obtained by

Currently no explanation available

A. $[\frac{Qf^{2}}{140}]^{1/6}$

C. $[\frac{Q^{2}f^{2}}{140}]^{1/6}$

GATE CE · Irrigation Engineering

Design Of Stable Channels – GATE CE Irrigation EngineeringPractice Questions & PYQs

Generate GATE-level questions on Design Of Stable Channels in Irrigation Engineering. Focus on core concepts, previous year patterns, and numerical problem-solving techniques.

12 questions · 12 PYQs · 0 AI practice · GATE CE 2027

🎯 These are sample questions

Get the full Design Of Stable Channels experience — 100% free

✓ Attempt tracking & analytics
✓ Timer & exam-mode simulation
✓ Adaptive difficulty (Easy → Hard)
✓ Bookmarks, notes & mistake review

Just sign in to unlock everything. Free for all students.

🚀 Start Practice Mode Sign in free →

Q1GATE 2025MSQ

Lacey's regime equations, followed in India for making scour calculations while designing hydraulic structures across alluvial channels, are given below. Regarding these equations, which of the following statements is/are true: $D=0.470 \times \left[\frac{Q}{f_{s}}\right]^{1 / 3}$ $P=4.75 \times \sqrt{Q}$ $f_{s}=1.76 \times \sqrt{d}$ where, $Q$ is discharge and $f_{s}$ is silt factor

🚀 Solve in Practice Mode

📖 Explanation

Given Lacey's regime equations: $D = 0.47 \left( \frac{Q}{f_s} \right)^{1/3}$ $P = 4.75 \sqrt{Q}$ $f_s = 1.76 \sqrt{d}$ Where: $Q$ is discharge in $\text{m}^3/\text{s}$ $f_s$ is the silt factor $d$ is the average grain size diameter in millimetres $D$ is the regime flow depth (not scour depth) $P$ is the wetted perimeter, which in regime channels is approximately the channel width (A) $D$ is the depth of scour below the existing riverbed This is incorrect. $D$ represents the regime flow depth, not the scour depth. (B) $P$ is the Lacey's waterway width This is correct. In regime channels, the wetted perimeter $P$ approximates the width of the channel. (C) $d$ is the average grain size diameter of the bed material in centimetres This is incorrect. The silt factor equation uses $d$ in millimetres. (D) $D$ is the depth of scour below the design flood level This is incorrect. $D$ is the regime depth of flow, not a scour depth. Final correct option: $\boxed{(B)}$

Q2GATE 2025NAT

A symmetrical trapezoidal canal is 100 km long. The bottom width is 10 m and the side slope is 1 Horizontal : 1 Vertical. The average flow depth in the canal is 2.5 m throughout the month of April. The measurement from a Class-A evaporimeter in the vicinity of the canal indicated an average evaporation rate of 0.5 cm viay in April. The volume of water evaporated from the canal (in $\mathrm{m}^3$ ) in the month of April is close to _________ $\times 10^3$ (rounded off to 1 decimal place).

🚀 Solve in Practice Mode

📖 Explanation

In the month of April, number of days $=30$ Pan evaporation (in terms of depth) $=0.5 \mathrm{~cm} /$ day $\times 30$ day $=15 \mathrm{~cm}$ For trapezoidal canal of length 10 km , $\text { Area of free surface of canal }=15 \mathrm{~m} \times 100 \times 10^3 \mathrm{~m}$ $=15 \times 10^5 \mathrm{~m}^2$

\begin{aligned} \text { Pan evaporation } & =0.15 \mathrm{~m} \\ \text { Evaporation loss \in canal } & =C_P \times \text { Pan evaporation } \\ & =0.7 \times 0.15 \mathrm{~m}=0.105 \mathrm{~m} \end{aligned}

$\text { (For class A pan, } C_P=0.7 \text { ) }$ Evaporation loss from canal in terms of volurne of water $=0.105 \mathrm{~m} \times 15 \times 10^5 \mathrm{~m}^2$ $=157.5 \times 10^9 \mathrm{~m}^3$

Q3GATE 2024MCQ

A round-bottom trianglular lined canal is to be liad at a slope of $1 \mathrm{~m}$ in 1500 , to carry a discharge of $25 \mathrm{~m}^{3} / \mathrm{s}$ . The side slopes of the canal cross-section are to be kept at $1.25 \mathrm{H}: 1 \mathrm{~V}$ . If Manning's roughtness coefficient is 0.013 , the flow depth (in meters) will be in the range of

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \cot \theta & =1.25 \\ \theta & =0.675 \mathrm{rad} \\ \mathrm{n} & =0.013 \\ \mathrm{Q} & =25 \mathrm{~m}^{3} / \mathrm{s}, \mathrm{S}=\frac{1}{1500} \end{aligned}

Using manning's equation:

\begin{aligned} \text { Area } & =y^{2}(\theta+\cot \theta) \\ A & =1.925 y^{2} \\ R & =\frac{y}{2} \\ Q & =\frac{A}{n} R^{2 / 3}(S)^{1 / 2} \\ 25 & =\frac{1.925 y^{2}}{0.013}\left(\frac{y}{2}\right)^{2 / 3}\left(\frac{1}{1500}\right)^{1 / 2} \\ \Rightarrow \quad y & =2.40 \mathrm{~m} \end{aligned}

Q4GATE 2024NAT

A standard round bottom triangular canal section as shown in the figure has a bed slope of 1 in 200. Consider the Chezy's coefficient as 150 1/2/s. The normal depth of flow, $y[latex] (\in meters) for carrying a discharge of [latex] 20 \mathrm{~m}^{3} / \mathrm{s}$ is ____ (rounded off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

Discharge, $Q=A \times V$ Now, Cross-section area, $A=y^{2}\left(\theta_{\text {radian }}+\cot \theta\right)=y^{2}(0.588+1.5) =2.088 y^{2}$ Perimeter, $P=2 y\left(\theta_{\text {radian }}+\cot \theta\right)$ Hydraulic radius, $R=\frac{A}{P}=\frac{y^{2}\left(\theta_{\text {radian }}+\cot \theta\right)}{2 y\left(\theta_{\text {radian }}+\cot \theta\right)}=\frac{y}{2}$ Now, Velocity, $V=C \sqrt{R . S}=150 \sqrt{\left(\frac{y}{2}\right) \times \frac{1}{200}}=\frac{150}{20} y^{1 / 2}$ Now, $Q=A \times V$ $20=2.088 y^{2} \times \frac{150}{20} \cdot y^{1 / 2}$ $\Rightarrow \quad y=1.10 \mathrm{~m}$

Q5GATE 2023MCQ

Identify the cross-drainage work in the figure.

🚀 Solve in Practice Mode

Q6GATE 2022MSQ

In the context of cross-drainage structures, the correct statement(s) regarding the relative positions of a natural drain (stream/river) and an irrigation canal, is/are

🚀 Solve in Practice Mode

Q7GATE 2021NAT

An unlined canal under regime conditions along with a silt factor of 1 has a width of flow 71.25m. Assuming the unlined canal as a wide channel, the corresponding average depth of flow (in m, round off to two decimal places) in the canal will be ______________

🚀 Solve in Practice Mode

📖 Explanation

R = D (for wide rectangular channel)

\begin{aligned} &\begin{aligned} A f^{2} &=140\left(\frac{2}{5} f R\right)^{5 / 2} \\ (B D) f^{2} &=140\left(\frac{2}{5} f \times D\right)^{5 / 2} \\ (71.25 \times D) \times 1 &=140\left(\frac{2}{5} \times 1 \times D\right)^{5 / 2} \\ D \times 0.5089 &=\left(\frac{2}{5}\right)^{5 / 2} \times (D)^{5 / 2} \\ D^{3 / 2} &=5.029 \\ \text { or } \qquad \qquad \qquad \qquad D &=2.94 \mathrm{~m} \end{aligned} \end{aligned}

Q8GATE 2009MCQ

The depth of flow in an alluvial channel is 1.5 m. If critical velocity ratio is 1.1 and Manning's n is 0.018, the critical velocity of the channel as per Kenedy's method is

🚀 Solve in Practice Mode

📖 Explanation

The critical velocity as per Kennedy's method is given by,

\begin{aligned} V_{0} &=0.55 \mathrm{my}^{0.64}=0.55 \times 1.1 \times (1.5)^{0.64} \\ &=0.784 \mathrm{m} / \mathrm{s} \end{aligned}

Q9GATE 2008MCQ

A stable channel is to be designed for a discharge of Q $m^{3}/s$ with silt factor f as per Lacey's method. The mean flow velocity (m/s) in the channel is obtained by

🚀 Solve in Practice Mode

Q10GATE 2007MCQ

As per the Lacey's method for design of alluvial channels, identify the TRUE statement from the following :

🚀 Solve in Practice Mode

📖 Explanation

As per Lacey's method of design of alluvial channels, $P=475 \sqrt{Q}$

Q11GATE 2005MCQ

A launching apron is to be designed at downstream of a weir for discharge intensity of 6.5 $m^{3}$ /s/m. For the design of launching aprons the scour depth is taken two times of Lacey scour depth. The silt factor of the bed material is unity. If the tailwater depth is 4.4 m, the length of launching apron in the launched position is

🚀 Solve in Practice Mode

📖 Explanation

Lacey's scour depth,

\begin{aligned} R=1.35\left(\frac{q^{2}}{f}\right)^{1 / 3}&=1.35 \times \left(\frac{6.5^{2}}{1}\right)^{1 / 3}=4.7 \mathrm{m} \\ \text{Scour depth }&=2 R=9.4 \mathrm{m} \\ \therefore \quad D&=2 R-4.4=5 \mathrm{m} \end{aligned}

Q12GATE 2005MCQ

On which of the canal systems, R.G. Kennedy, executive engineer in the the Punjab Irrigation Department made his observations for proposing his theory on stable channels ?

🚀 Solve in Practice Mode

Want unlimited AI-generated Design Of Stable Channels questions?

Start practising for free →