Lining of canals is preferred to reduce percolation losses as well as seepage losses

1. CE 408 Irrigation Engineering
Lecture # 10
Lining of Irrigation Canals
Instructor: Engr. Prof. Dr. Daulat Khan

3. Lining of Canals
When the earthen channel surface is made impervious (lined) with stable materials like
concrete, bricks etc. such lining is known as canal lining. By lining the seepage losses are
controlled up to 100 % ,while evaporation losses also minimized because of high velocity.
Following are the main objectives / necessity of linings.
(a)To Control Seepage
The seepage is the maximum losses in unlined canals. Due to seepage the duty of canal
water is much more reduced. So, to control seepage losses through the bed and sides of the
canal, the lining of the canal is necessary.
(b) To Prevent Water-logging/ Anti water logging
There may be some low-lying areas on one side or on both sides of the canal. Due to seepage
of water through the sides and bottom of the canal these areas may get seepage water and
converted into marshy lands known water logging.

4. Lining of Canals
where crops can not be grown. Also waterlogged area may become the breeding place of
mosquitoes, which are responsible for many infectious diseases. Lining control it.
(c) To Increase the Capacity Of Canal
Due to the low velocity in unlined channel, the discharge capacity of the canal becomes low. If the
capacity of the canal is to be increased the cross-sectional area has to be increased which
involves more land width. So, the lining of the canal should be such that the velocity and the
discharge of the canal are more with minimum cross-sectional area.
( d) To Increase the Command Area
If the lining is provided in the canals the various losses can be controlled and ultimately the
command area of the project may be enhanced.
(e) To Protect the Canal From The Damage By Flood
The unlined canals may be severely damaged by scouring due to high velocity of flood water at
the time of heavy rainfall. So, to protect the canals from the damages, the lining should be provided.

5. Lining of Canals
(f) To Control the Growth of Weeds
The growth of various types of weeds along the sides of the canals is a common problem.
Again, some types of weeds are found to grow along the bed of the canals. These weeds
reduce the velocity of flow and the capacity of the canals. So, the unlined canals require
excessive maintenance works for clearing the weeds. If lining is provided in the canal, the
growth of weeds can be stopped, and velocity and the capacity of the canal may be increased.
Advantages of Canal Lining
Following are some of the Advantages of canal Lining
a). Lining of canals prevents seepage losses; hence the duty is enhanced, and more area will
be irrigated.
b). Lining control seepage losses and hence the adjoining area may not be water-logged.

6. Advantages of Canal Lining
c). It provides smooth surface, reduce the rugosity coefficient and hence the velocity of flow can be
increased.
d). Due to the increased velocity the discharge capacity of a canal is also increased.
e). Due to the increased velocity, the evaporation loss also can be reduced. Also silting can be minimized.
f). It eliminates the effect of scouring in the canal bed.
g). The increased velocity eliminates the possibility of weed growth on the sides and bed of canal
h). It provides the stable section of the canal in unstable soil.
i). It reduces the requirements of land width for the canal, because smaller section of the canal can
be used to produce greater discharge.
j). It prevents the sub-soil salt to come in contact with the canal water.
k). It reduces the maintenance cost for the canals.

7. Disadvantages of Canal Lining
1. The initial cost of lining is very high. So, it makes the project expensive with respect to the output.
2. It involves many difficulties for repairing the damaged section of lining.
3. It takes too much time to complete the project work.
4. It becomes difficult, if the outlets are required to be shifted or new outlets are required to be provided,
because the dismantling of the lined section is difficult.
Suitability of Canal Lining Materials
Canal lining materials must have the following properties.
i. The materials used for lining must be of complete watertight
ii. It must have low roughness coefficient, to make the section hydraulically more efficiently.
iii. The materials to be used must be strong and durable.
iv. The cost of materials used for lining must not be too expensive.
v. Similarly the materials must not be affected by animals, be withstand with high velocity and resists to
weed growth.

8. Types of Canal Lining
There are various types of lining. Like (a) Hard surface (b) Earth type © Buried and protected
membrane type.
Hard Surface linings: Concrete lining; shotcrete lining; precast concrete; cement mortar; Brick ;
stone block and asphaltic
Earth types: Soil cement ; clay puddle and Sodium Carbonate type
Buried and protected membrane types :These are prefabricated light membrane; Bentonite soil
and Road oil lining.
Concrete Lining
The cement concrete lining is widely accepted. It can resist the effect of scouring and erosion very
efficiently. The velocity of flow may be kept above 2.5 m/s. It can eliminate completely growth of
weeds.

9. Concrete Lining
The lining is done by the following steps;
(a) Preparation of sub-grade
The sub grade is prepared by ramming the surface properly with a layer of sand (about 15
cm). Then slurry of cement, gravel and sand (1:2:4) is spread uniformly over the prepared
bed.
(b) Laying of concrete
The cement concrete is spread uniformly according to the desired thickness, (generally the
thickness varies from 100 mm to 150 mm). After laying, the concrete is tapped gently until
the slurry comes on the top. The curing is done for two weeks. As the concrete is liable to
get damaged by the change of temperature, the expansion joints are provided at
appropriate places.

10. Lining of Canals
2. Brick Lining
This lining is prepared by the double layer brick flat soling laid with cement mortar (1:6) over the
compacted sub-grade. The first-class bricks should be recommended for the work. The surface of
the lining is finished with cement plaster (1:3). The curing should be done perfectly.
This lining is always preferred for the following reasons,
(a)This lining is economical.
(b)Work can be done very quickly.
(c) Expansion joints are not required.
(d) Repair works can be done easily.
(e)Bricks can be manufactured from the excavated earth near the site.

11. Lining of Canals
However, this lining has certain disadvantages,
(a) It is not completely impervious.
(b)It has low resistance against erosion.
(c)It is not so much durable.
3. Reinforced Cement Concrete Lining
Sometimes reinforcement is required to increase the resistance against cracks and shrinkage
cracks. The reduction in the cracks results in less seepage losses. However this
reinforcement does not increase the structural strength of the lining. This reinforcement adds
10 to 15 percent to the cost and for this reason steel reinforcement is usually omitted except
for very particular situations.

12. Design of Lining Canals
• Design of Lined Irrigation Channels
Irrigation channels should be aligned and laid out so that the velocity of flow is uniform.
High velocities of flow can be achieved due to hard wearing surface. While aligning the
channel, sharp curves should be avoided, as they not only reduce the velocity of flow, but
also required higher walls on the outside to retain the water as it rounds the curve.
• Channel Cross sections:
Generally two types of channel section are adopted
(i) Triangular channel section for smaller discharge (50 cumecs) and
(ii) Trapezoidal channel for larger discharge (more than 50)cumecs).
• For designing a lined section following data must be known
• The design discharge of the channel (Q)

13. Design Procedure of Lining Canals
The rugosity coefficient also known manning’s roughness(N) and Longitudinal slope (s)
The side slopes of the section Z:1 (1:1 to 1.5 :1)
The limiting velocity of flow (V) (1.5 m/sec to 2.5 m/sec)
• Equations to be used
V =
𝑅2/3× 𝑆1/2
𝑁
Q = A × 𝑣 and R =
𝐴
𝑃
A = Bh +h2(θ + cot θ) and
P = B + 2h (θ + cot θ)
From these equations the value of B and h can be calculated
A free board of 0.75 m is usually be provided.

14. Design of Lining Canals
Triangular Section: It is also known as curved section. When the discharge is
less than 50 cumecs such section can be used.
• Area of the section (A) = π × h2 (θ/ π ) +2 × (1/2 h. h Cot θ)
• = h2 (θ + cot θ )
• Wetted Perimeter ( P) = 2 π h × (θ/ π) + 2h cot θ = 2h (θ + cot θ )
• Hydraulic mean depth (R) = A/P = h2 (θ + cot θ )/ h (θ + cot θ ) = h/2

15. Design of Lining Canals
Trapezoidal section: when the discharge is greater than 50 cumecs then Trapezoidal channel has to be used
Area of the section (A) = B x h + 2 𝜋 ℎ2 𝜃
2𝜋
+ 2 ×
1
2
h × h Cot θ
Bh +h2 θ + h2cot θ = Bh + h2( θ +cot θ )
Wetted Perimeter ( P) =B + 2 × 2 π h × (
𝜃
2𝜋
) + 2h cot θ
= B +2h θ + 2 h cot θ )
= B +2h (θ + cot θ )
Hydraulic mean depth (R) = A/P =
Bh + h2
( θ +cot θ )
B +2h (θ + cot θ )

16. Example
• Design a lined channel to carry a discharge of 15 cumecs. The longitudinal slope is 1 in 9000. The side
slopes is 5/4 :1 and Manning’s roughness coefficient is 0.015.
• Given Information : Q = 15 cumecs; slope = 1/9000; Z:1 = 5/4:1 and n = 0.015
• Required: Design of channel
• Solution:
As the discharge is less than 50 cumecs so triangular channel will be designed.
So tan𝜃 =
1
1.25
=> 0.8
𝜃 = 𝑡𝑎𝑛−1
(0.8) = 𝐶𝑜𝑡 0.8 = 38.660
Converting 𝜃 𝑖𝑛𝑡𝑜 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 =
𝜃×𝜋
180
=
38.66×𝜋
180
= 0.675 radians
• A = Y2 (𝜃+ cot 𝜃) = Y2 (0.675+ 1.25) = 1.925Y2
• P = 2Y(𝜃+ cot 𝜃) = 2Y(0.675+1.25)= 3.85 Y
• R = Y/2 = 0.5Y
Using Continuity equation Q = A × 𝑉

17. Example
• Using Continuity equation Q = A × 𝑉
• Where Q = 15
𝑚3
𝑠𝑒𝑐𝑜𝑛𝑑
• A = 1.925Y2
• And V =
𝑅2/3×𝑆1/2
𝑛
plugging the values
Q =
𝐴×𝑅2/3×𝑆1/2
𝑛
=>15 =
1.925 𝑌2
×(0
.
5𝑌
)2
/
3
×0.000111/2
0.015
=
• 15 =
1.925 𝑌2
×(0
.
629𝑌
)2
/
3
×0.0105
0.015
=> 15 = 0.847Y8/3 or Y = [
15
0.847
]^3/8
=2.94 meter
Answer

18. Example
• Design a concrete lined channel to carry a discharge of 350 cumecs at a slope of 1 in 5,000. The side
slopes of the channel may be taken as 1.5:1.The value of n for lining is 0.014. Assume limiting velocity
in the channel as 2m/sec
Solution: As Q is greater than 50 cumecs so using trapezoidal channel
Using Manning’s Formula: V =
𝑅2/3×𝑆1/2
𝑁
2=
𝑅
2/3×(
1
5000)1/2
0.014
= 2 × 0.014 = 𝑅2/3
×0.01414
 R = [2 × 0.014/0.01414]
3/2
= 2.79 m
= 2 ×0.014 ×70.8 = R^2/3 or R = (1.98)^ 3/2 = 2.79 m
tan𝜃 =
1.0
1.50
= 0.667 => 𝜃 = 𝑡𝑎𝑛−1(0.667) = 33.700
Degree convert into Radian =
𝑐𝑜𝑡𝜃×𝜋
180
=>
33.700×𝜋
180
= 0.588

19. Example
Now area (A) = Bh + h2( θ +cot θ ) = h (B +h θ + hcot θ )
wetted Perimeter (P) = B +2h (θ + cot θ )
Also, A =
𝑄
𝑉
=
350
2
= 175 m2
Plugging the value of h and θ in Area
175 = h(B + 0.588h + 1.5h)=> h(B + 2.088h)
175
ℎ
= 𝐵 + 2.088 ℎ =>B =
175
ℎ
- 2.088 h
R =
𝐴
𝑃
= where P = B +2h (θ + cot θ )
plugging the value of R; A and P the equation become R = 2.79 =
𝐴
𝑃
=
175
𝐵+4.176ℎ

20. Example
2.79 =
175
175
ℎ
−2.09 ℎ+4.18 ℎ
=>
175
ℎ
− 2.09 ℎ + 4.18 ℎ =
175
2.79
= 62.7
= 175 – 2.09 ℎ2
+4.18 ℎ2
= 62.7 h
= 2.09 ℎ2 − 62.7h+175 = 0 divide by 2.09
= ℎ2-30h+83.7
h=
−𝑏± 𝑏2−4𝑎𝑐
2𝑎
=
30± 900−4×1×83.7
2×1
=
30−23.8
2
= 3.1 m and
30+23.8
2
=
26.9 m
Now B =
175
ℎ
- 2.088 h
B =
175
3.1
− 2.09 × 3.1 => 49.97m OR
B =
175
26.9
− 2.09 × 26.9 => - 49.72m (Negative value of width is not
possible ; so, B = 49.97 m