Irrigation and Drainage Engineering
Dr. Tariq Hussein
3rd Year
2- Design of alluvial canals
It is a process of controlling the flow velocity in such away that the silt flowing in
the channel is not dropped in the bed without scouring the channel.
Some of theories towards the design of non-silting non-scouring channel section are
presented:
1) Lacey’s Theory
The fundamental requirements stated by Lacey for a channel are as follows:
The channel flows uniformly in incoherent alluvium. Incoherent alluvium
is the loose granular material which can scour or deposit with the same
ease.
The characteristics and the discharge of the sediment are constant.
The water discharge in the channel is constant.
The channel design procedure is as follows:
Step 1: Calculate the silt factor 𝑓 = 1.76 𝑚
Where f = silt factor
𝑚 = diameter of silt in mm, or it’s called mean particle size.
/
Step 2: Find the velocity 𝑉 =
Step 3: Determine the area A=Q/V
/
/
Step 4: Compute 𝑃 = 4.75 𝑄, and 𝑅 = 0.48 ( ) and 𝑆 = 0.0003 /
Ex: Design a stable channel for carrying a discharge of 30m3/s using Lacey’s method
assuming silt factor equal to 1.0.
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Irrigation and Drainage Engineering
Dr. Tariq Hussein
3rd Year
Ans.
𝑃 = 4.75 𝑄 = 4.75 √30 = 26.02m
/ /
𝑅 = 0.48 ( ) = 0.48 ( ) = 1.49m
/ /
𝑆 = 3 𝑥 10 /
= 3 𝑥 10 /
= 1.702 × 10–4
*Assume final side slope of the channel as 0.5H:1V (generally observed field value)
P = b+2d √1 + 𝑧 26.02= b+2d√1 + 0.5
b = 26.02 – 2.24d……………………….. (1)
and A= bd+z.d2 = P.R 26.02 × 1.49 = bd+0.5d2
38.77 = 26.02d – 2.24d2 + 0.5d2 1.74d2 – 26.02d + 38.77 = 0
Either d= 13.28m, then b would be a negative value!! Neglected.
Or d=1.68m, then b= 26.02 – 2.24 × 1.68 = 22.23m
Then add free board
H.W-1: Design an irrigation channel for a discharge of 50m3/s adopting the available
ground slope of 1.5 × 10–4. Then find the size of suitable river bed material.
(Ans. d= 1.99m, b= 29.13m, md=0.3mm), then compute the free board.
2) Kennedy’s Theory
The procedure of this theory is
23
Irrigation and Drainage Engineering
Dr. Tariq Hussein
3rd Year
Step 1: Assume (d) by trial and error, where d is the depth of water in (m).
.
Step 2: Calculate the critical velocity* in (m/sec) by using eq. 𝑉 = 0.55 𝑑
Which is defined as the mean velocity that will not allow scouring or silting in a
channel having depth of flow equal to (d)m.
Step 3: Find the area of the section 𝐴 =
Step 4: Since 𝐴 = 𝑏𝑑 + 𝑧. 𝑑 , then find b.
Step 5: Find P and R.
Step 6: Find (V) from Manning’s equation, which must equal to (Vo). If not, then
assume another (d).
*This critical velocity should be distinguished from the critical velocity of flow in
open channels corresponding to Froude number equal to 1.
Ex: Design a channel carrying a discharge of 30m3/s with critical velocity ratio and
Manning’s n equal to 1.0 and 0.0225, respectively. Assume that the bed slope is
equal to 1 in 5000.
Ans. Assume d=2m
𝑉 = 0.55𝑚 𝑑 .
= 0.55 (2)0.64 = 0.857 m/sec
A=Q/Vo = 30/0.857 = 35.01m2
For a trapezoidal channel with side slope 1H : 2V, or 0.5H:1V
A=bd+z.d2 35.01=2b+0.5(4) b= 16.51m
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Irrigation and Drainage Engineering
Dr. Tariq Hussein
3rd Year
.
𝑅= = = 1.67m
. ∗ ( . )
Therefore, from Manning’s eq.
/ / / /
𝑉= 𝑆 𝑅 = ( ) 1.67 = 0.885 𝑚/𝑠𝑒𝑐
.
Since the velocities obtained from Kennedy’s equation and Manning’s equation are
appreciably different, assume d = 2.25m and repeat the above steps.
.
𝑉 = 0.55 2.25 = 0.924 m/sec
A= 30/0.924 = 32.47 m2
b(2.25) + (0.5) (2.25)2 = 32.47 b= 13.31 m
R=1.77m
/ /
𝑉= 𝑆 𝑅 = 0.92 m/s
Since the two values of the velocities are matching, the depth of flow can be taken
as equal to (d = 2.25m) and the width of trapezoidal channel (b = 13.31m).
Then add the fee board. f1
To reduce the number of trials in finding (d), you can use the table below to use
recommended b/d values for assumption.
b/d
25
Irrigation and Drainage Engineering
Dr. Tariq Hussein
3rd Year
H.W-2: A stable channel is to be designed for a discharge of 40m3/s. Calculate the
dimensions of the channel using Kennedy’s method. Use a Manning’s coefficient of
0.0225 and bed slope of 1/5000.
Ex: A circular concrete culvert of diameter 100cm carries water of depth 40cm to
irrigate a 20 hectares field. What could be the discharge of this canal if the slope was
0.002? Use Manning’s n of 0.014.
Ans.
Length BC = (0.5) − (0.1) = 0.49m
∗ . ∗ .
Area BoA = = 0.049m2
.
Angle CoB = cos = 78.46o
.
؞Angle AoB = 2* 78.46 = 156.92o
=
.
, ؞circle section area = 0.342m2
؞water area = 0.342 – 0.049 = 0.293m2
=
.
؞wetted perimeter (P) = 1.37m
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Irrigation and Drainage Engineering
Dr. Tariq Hussein
3rd Year
=𝑄 ؞ 𝑅 /
𝑆𝑜 /
𝐴=
.
(
.
.
) (0.002) (0.293)
= 𝑄 ؞0.335 m3/sec.
H.W-3: Design the best cross-section of a trapezoidal canal to irrigate a field of
1000 hectare, where the water supply is 100 (m3/hectare/day). Knowing that the
side slope is 1:1 and bed slope of 10 cm/km. Use n = 0.02.
Ans. (d=1.133m, b=1.602m)
Hint: Q = water duty * area = (100 * 1000) / (24*3600) = 1.157 m 3/s
Find the relationship bet. b & d. then solve..
H.W-4: The slope of a channel in alluvium is 1/5000; Lacey’s silt factor is 0.9. Find
the channel section and maximum discharge which can be allowed to flow in it.
(Ans: b=7.45m, d=0.88m, Q=3.924m3/sec)
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Irrigation and Drainage Engineering
Dr. Tariq Hussein
3rd Year
Comparisons between Lacey’s and Kennedy’s theories
1- Kennedy introduced the Critical Velocity Ratio (m) but he did not give any
idea of how to measure m. While Lacey introduced the silt factor (f) and
suggested a method to determine(f) by relating it with particle size.
2- Kennedy assumed that silt is kept in suspension because of eddies generated
from the bed only and so he proposed a relation between V and d. Lacey
assumed that silt is kept in suspension because of eddies generated from the
entire perimeter and so he proposed a relation between V and R.
3- Kennedy gave no formula of determination of longitudinal slope of the
canal, where the slope is to be given based on experience. Lacey gave a
formula for the longitudinal slope of the canal.
4- Lacey’s theory does not involve any trial and error in the design procedure
whereas Kennedy’s theory involves a trial and error procedure.
5- The basic concept of both theories is the same that the silt remains in
suspension due to the force of vertical eddies.
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