CHAPTER SIX:

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CHAPTER SIX:. DESIGN OF CHANNELS AND IRRIGATION STRUCTURES. 6.1 DESIGN OF CHANNELS FOR STEADY UNIFORM FLOW. Channels are very important in Engineering projects especially in Irrigation and, Drainage. Channels used for irrigation are normally called canals

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### CHAPTER SIX:

DESIGN OF CHANNELS AND IRRIGATION STRUCTURES

6.1 DESIGN OF CHANNELS FOR STEADY UNIFORM FLOW
• Channels are very important in Engineering projects especially in Irrigation and, Drainage.
• Channels used for irrigation are normally called canals
• Channels used for drainage are normally called drains.
6.1.1 ESTIMATION OF CANAL DESIGN FLOWS (Q)
• For Irrigation Canals, Design Flows are estimated Using the Peak Gross Irrigation Requirement
• For Example, in a Location with the Peak Gross Irrigation Requirement of 7.69 mm/day.
• Peak flow (Q) = 7.69/1000 m x 10000 x 1/3600 x 1/24 x 1000
• = 0.89 L/s/ha
• For a canal serving an area of 1000 ha, canal design flow is then 890 L/s or 0.89 m 3 /s.
• Typically, for humid areas, magnitude of discharges are in the range of 0.5 to 1.0 L/s/ha.
Definitions
• a) Freeboard: Vertical distance between the highest water level anticipated in the design and the top of the retaining banks. It is a safety factor to prevent the overtopping of structures.
• b) Side Slope (Z): The ratio of the horizontal to vertical distance of the sides of the channel. Z = e/d = e’/D
6.1.3 Estimation of Velocity in Channels
• The most prominent Equation used in the design is the Manning formula described in 6.1.3. Values of Manning's n can be found in standard texts (See Hudson's Field Engineering).
• 6.1.4 Design of Channels
• Design of open channels can be sub-divided into 2:
• a) For Non-Erodible Channels (lined)
• b) Erodible Channels carrying clean water
Design of Non-Erodible Channels

When a channel conveying clear water is to be lined, or the earth used for its construction is non-erodible in the normal range of canal velocities, Manning's equation is used. We are not interested about maximum velocity in design. Manning's equation is:

`

Q and S are basic requirements of canal determined from crop

water needs. The slope of the channels follows the natural

channel. Manning's n can also be got from Tables or estimated

using the Strickler equation: n = 0.038 d1/6 , d is the particle size diameter (m)

Design of Non-Erodible Channels Contd.
• LHS of equation (1) can be calculated in terms of A R2/3 termed section factor. For a trapezoidal section:
• A = b d + Z d2 ; P = b + 2 d (1 + Z)1/2
• The value of Z is decided (see Table 6.1) and the value of b is chosen based on the material for the construction of the channel.
• The only unknown d is obtained by trial and error to contain the design flow. Check flow velocity and add freeboard.
Example 6.1

Design a Non-Erodible Channel to convey 10 m3/s flow, the slope is 0.00015 and the mean particle diameter of the soil is 5 mm. The side slope is 2 : 1.

• Solution: Q = 1/n AR 2/3 S 1/2 ….. (1)
• With particle diameter, d being 5 mm, Using Strickler Equation, n = 0.038 d 1/6
• = 0.038 x 0.005 1/6 = 0.016
Solution of Example Contd.

Z = 2. Choose a value of 1.5 m for 'b‘

For a trapezoidal channel, A = b d + Z d2 = 1.5 d + 2 d2

P = b + 2 d (Z2 + 1)1/2 = 1.5 + 2 d 51/2 = 1.5 + 4.5 d

Try different values of d to contain the design flow of 10 m3/s

Soln of Example 6.1 Contd.

d(m) A(m2 ) P(m) R(m) R2/3Q(m3/s) Comment

2.0 11.0 10.5 1.05 1.03 8.74 Small flow

2.5 16.25 12.75 1.27 1.18 14.71 Too big

2.2 12.98 11.40 1.14 1.09 10.90 slightly big

2.1 11.97 10.95 1.09 1.06 9.78 slightly small

2.13 12.27 11.09 1.11 1.07 10.11 O.K.

The design parameters are then d = 2.13 m and b = 1.5 m

Check Velocity : Velocity = Q/A = 10/12.27 = 0.81 m/s

Note: For earth channels, it is advisable that Velocity should be above 0.8 m/s to inhibit weed growth but this may be impracticable for small channels.

Assuming freeboard of 0.2 d ie. 0.43 m, Final design parameters are:

D = 2.5 m and b = 1.51 m

Final Design Diagram

T = 11.5 m

D = 2.5 m

Z = 2:1

d = 2.13 m

b = 1.5 m

T = b + 2 Z d = 1.5 + 2 x2 x 2.5 = 11.5 m

Design of Erodible Channels Carrying Clean Water
• The problem here is to find the velocity at which scour is initiated and to keep safely below it. Different procedures and thresholds are involved including maximum permissible velocity and tractive force criteria.
• Maximum Permissible Velocities: The maximum permissible velocities for different earth materials can be found in text books e.g. Hudson's Field Engineering, Table 8.2.
Procedure For Design
• i)Determine the maximum permissible velocity from tables.
• ii) With the permissible velocity equal to Q/A, determine A.
• iii) With permissible velocity = 1/n S1/2 R2/3
• Slope, s and n are normally given.
• iv) R = A/P , so determine P as A/R
• v) Then A = b d + Z d and
• P = b+ 2 d (Z2 + 1)1/2 ,
• Solve and obtain values of b and d
Example 6.2:
• From previous example, design the channel using the maximum permissible velocity method.
• Solution: Given: Q = 10 m3 /s , Slope = 0.00015 , n = 0.016
• , Z = 2 : 1
• i) From permissible velocity table, velocity = 0.75 m/s
• A = Q/V = 10/0.75 = 13.33 m
• `
• iv) P = A/R = 13.33/0.97 = 13.74 m
• v) A = b d + Z d2 = b d + 2 d2
• P = b + 2 d (Z2 + 1)1/2 = b + 2 d 51/2 = b + 4.5 d
• ie. b d + 2 d2 = 13.33 m 2 ........(1)
• b + 4.5 d = 13.74 m ........ (2)

From previo

Solution of Equation 6.2 Contd.

From (2), b = 13.74 - 4.5 d .......(3)

Substitute (3) into (1), (13.74 - 4.5 d)d + 2 d2 = 13.33

13.74 d - 4.5 d2 + 2 d = 13.33

13.74 d - 2.5 d2 = 13.33

ie. 2.5 d2 - 13.74 d + 13.33 = 0

d = 1.26 m is more practicable

From (3), b = 13.74 - (4.5 x 1.26) = 8.07 m

Adding 20% freeboard, Final Dimensions are depth = 1.5 m and width = 8.07 m

6.1.5 Classification of Canals Based on Capacity:
• Canals can be classified as:
• (a)Main Canal: It is the principal channel of a canal system taking off from the headworks or a reservoir or tail of a feeder.
• It is a large capacity channel and usually there is no direct irrigation from it.
• Small capacity ditch distributaries running parallel to the canal are taken off from the main canal to irrigate adjoining areas.
• Main canals deliver supply to branch canal and main distributaries.
Canals Contd.
• (b)Branch or Secondary Canal:
• Branch canals take their supply from the main canal and convey to the distributaries.
• Very little direct irrigation is done from the branch canals.
• Sub-branch is a canal, which takes off from the branch canal but has capacity higher than a distributary.
Canals Contd.
• (c)Major Distributary:
• It is a distributing channel, which may take off from a main canal, branch canal or sub-branch and has discharge capacity less than that of a branch canal.
• It supplies water to another distributary.

Distributaries and minors take off from it.

Irrigation is done through outlets fixed along it.

Canals Contd.
• (d)Distributary:
• It is a channel receiving supply from branch canal or major distributary and has discharge less than that of major distributary.
• Minors take off from it, besides irrigation is done from it through outlets.
IRRIGATION STRUCTURES
• Structures are widely used in Irrigation, water conservation, flood alleviation, river works where water level and discharge regulation are required.
• Theseare hydraulic structures that are used to regulate, measure, and/or transport water in open channels.
• These structures are called control structures when there is a fixed relationship between the water surface elevation upstream or downstream of the structure and the flow rate through the structure.
• Hydraulic structures can be grouped into three categories:
Hydraulic Structures Contd.
• (i)Flow measuring structures, such as weirs
• (ii)Regulation structures such as gates and
• (iii)Discharge structures such as culverts.
Weirs
• Weirs: Weirs are elevated structures in open channels that are used to measure flow and/or control outflow elevations from basins and channels.
• There are two types of weirs in common use:
• Sharp-crested weirs and the broad-crested weirs.
• The sharp-crested weirs are commonly used in irrigation practice
Sharp-Crested Weirs
• Sharp-crested or thin plate, weirs consist of a plastic or metalplate that is set vertically across the width of the channel.
• The main types of sharp-crested weirs are rectangular, V-notches and the Cipolletti or the Trapezoidal weirs.
• The amount of discharge flowing through the opening is non-linearly related to the width of the opening and the depth of the water level in the approach section above the height of the weir crest.
Sharp Crested Weirs Contd.
• Weirs can be classified as being contracted or suppressed depending on whether or not the nappe is constrained by the edges of the channel.
• If the nappe is open to the atmosphere at the edges, it is said to be contracted because the flow contracts as it passes through the flow section and the width of the nappe is slightly less than the width of the weir crest (see figure).
• If the sides of the channel are also the sides of the weir opening, the streamlines of flow are parallel to the walls of the channel and there is no contraction of flow.
Figure 6.2: Rectangular Weirs

(b) Unsuppressed Weir

(Contracted)

(a) Suppressed Weir

Sharp Crested Weirs Contd.
• In this case, the weir is said to be suppressed. Some type of air vent must be installed in a suppressed weir so air at atmospheric pressure is free to circulate beneath the nappe. (See Figure 6.2 for suppressed and unsuppressed weirs).
Sharp Crested Weirs Contd.

The discharge, Q (m3/s) over a rectangular suppressed weir can be derived as:

Where: Cd is the discharge coefficient, b is the width of the weir crest, m (see Figure 6.2 above) and H is the head of water (m) above weir crest.

According to Rouse (1946) and Blevins( 1984),

………………..(2)

Where: Hw is the height of the crest of the weir above the bottom of the channel.

Weirs Contd

This equation is valid when H/Hw <5, and is approximated up to H/Hw = 10. If H/Hw < 0.4, Cd can be approximated as 0. 62 and equation (1) reduces to:

Q = 1.83 b H1.5 ………. (3)

This equation is normally used to compute flow over a rectangular suppressed weir over the usual operating range. It is recommended that the upstream head, H be measured between 4H and 5H upstream of the weir.

For the unsuppressed (contracted) weir, the air beneath the nappe is in contact with the atmosphere and venting is not necessary. The effect of side contractions is to reduce the effective width of the nappe by 0.1 H and that flow rate over the weir, Q is estimated as:

Q = 1.83 (b – 0.2 H) H1.5 ………………… (4)

This equation is acceptable as long as b is longer than 3 H

Cipoletti Weir
• A type of contracted weir which is related to the rectangular sharp-crested weir is the Cipoletti weir (see Figure 6.3 below) which has a trapezoidal cross-section with side slopes 1:4 (H:V). The advantage of a Cipolletti weir is that corrections for end contractions are not necessary.
Cipolletti Weir Contd.

The discharge formula can be written as:

Q = 1.859 b H1.5 …………….. (5)

Where: b is the bottom width of the Cipolletti weir. The minimum head on standard rectangular and Cipolletti weirs is 6 mm and at heads less than 6 mm, the nappe does not spring free of the crest.

Figure 6.3: A Trapezoidal of Cipolletti Weir

Example 6.3
• A weir is be installed to measure flows in the range of 0.5 to 1.0 m3/s. If the maximum depth of water that can be accommodated at the weir is 1 m and the width of the channel is 4 m, determine the height of a suppressed weir that should be used to measure the flow rate.
Solution to Example 6.3

The flow over the weir is shown in the Figure 6.4 below. The height of water is Hw and the flow rate is Q. The height of water over the crest of the weir, H is given by:

H = 1 – Hw

Assuming that H/Hw , 0.4, then Q is related to H by equation (3), where:

Q = 1.83 b H 1.5

Figure 6.4: Weir Flow

Solution to Example 6.3Concluded

Taking b = 0.4 m, Q = 1m3/s (the maximum flow rate will give the maximum head, H), then:

The height of the weir, Hw is therefore given by:

Hw = 1 – 0.265 = 0.735 m

And H/Hw = 0.265/0.735 = 0.36

The initial assumption that H/Hw < 0.4 is therefore validated, and the height of the weir should be 0.735 m.

V-Notch Weir

A V-notch weir is a sharp-crested weir that has a V-shaped opening instead of a rectangular-shaped opening. These weirs, also called triangular weirs, are typically used instead of rectangular weirs under low-flow conditions ( mainly < 0.28 m3/s), where rectangular weirs tend to be less accurate. It can be derived that the flow rate, Q over the weir is given by:

Parshall Flume
• Although weirs are the simplest structures for measuring the discharge in open channels, the high head losses caused by weirs and the tendency for suspended particles to accumulate behind weirs may be important limitations.
• The Parshall flume provides an alternative to the weir for measuring flow rates in open channels where high head losses and sediment accumulation are of concern.
• Such cases include flow measurement in irrigation channels.
• The Parshall flume (see Figures 6.7 and 6.8 below) consists of a converging section that causes critical flow conditions, followed by a steep throat section that provides for a transition to supercritical flow.
Parshall Flume Contd.
• The unique relationship between the depth of flow and the flow rate under critical flow conditions is the basic principle on which the Parshall flume operates.
• The transition from supercritical flow to subcritical flow at the exit of the flume usually occurs via a hydraulic jump, but under high tail water conditions the jump is sometimes submerged.
Parshall Flume Contd
• Within the flume structure, water depths are measured at two locations, one in the converging section, Ha and the other at the throat section, Hb. The flow depth in the throat section is measured relative to the bottom of the converging section as illustrated in the figure below.
• If the hydraulic jump at the exit of the Parshall flume is not submerged, then the discharge through the flume is related to the measured flow depth in the converging section, Ha by the empirical discharge relations given in Table 6.2, where Q is the discharge in ft3/s, W is the width of the throat in ft, and Ha is measured in ft.
Parshall Flume Contd
• Submergence of the hydraulic jump is determined by the ratio of the flow depth in the throat, Hb, to the flow depth in the converging section, Ha, and critical values for the Hb/Ha are given in Table 6.3.
• Whenever, the ratio exceeds the critical values in the table, the hydraulic jump is submerged and the discharge is reduced from the values given by the equations in Table 6.2.
• Corrections to the theoretical flow rates as a function of Ha and the percentage of submergence, Hb/Ha are given in the Figures 6.8 and 6.9 below for throat widths of 1 ft and 10 ft.
Parshall Flumes Contd.
• Flow corrections for the 1 ft flume are applied to larger flumes by multiplying the correction for the 1 ft flume by a factor corresponding to the flume size given in Table 6.4.
• Similarly, flow corrections for flume sizes greater than 10 ft. are applied to larger flumes by multiplying the correction for the 10 ft flume by a factor corresponding to the flume size given in Table 6.5.
• Parshall flumes do not reliably measure flow rates when the submergence ratio, Hb/Ha exceeds 0.95.
Example 6.4
• Example 6.4: Flow is being measured by a Parshall flume that has a throat width of 2 ft. Determine the flow rate through the flume when the water depth in the converging section is 2.00 ft and the depth in the throat section is 1.70ft.
Solution to Example 6.4

From the given data: W = 2 ft, Ha = 2 ft, and Hb = 1.7 ft. According to Table 6.2, Q is given by:

In this case: Hb/Ha = 1.7/2 = 0.85

Therefore, according to Table 6.3, the flow is submerged. Figure 6.8 gives the flow rate correction for a 1 ft flume as 2ft3/s, and Table 6.4 gives the correction factor for a 2 ft flume as 1.8. The flow rate correction, dQ for a 2 ft flume is therefore given by:

DQ = 2 x 1.8 = 3.6 ft3/s

And the flow rate through the Parshall flume is Q – dQ, where Q – dQ = 23.4 – 3.6

= 29.8 ft3/s

Gates
• Gates are used to regulate the flow in open channels.
• They are designed for either over-flow or underflow operation, with overflow operation appropriate for channels in which there is a significant amount of floating debris.
• The common types of gates are vertical and radial (Tainter) gates, which are illustrated below.
• Vertical gates are supported by vertical guides with roller wheels, and large hydrostatic forces usually induce significant frictional resistance to raise and lower the gate
Gates Contd.

Flow, Q through a gate could be established to be:

Cc = Cc = coefficient of contraction, = y2/yg = 0.61 for most vertical gates.

`

For For Tainter gates, Cc is generally greater than 0.61 and is commonly expressed as a function of the angle (degrees) shown in the diagram above.

Gates Concluded

It can be expressed as:

This equation applies as long as the angle is least than 900. All the equations apply where there is free flow through the gates. See texts for situations where the flows through the gates are submerged.

Drop Structures:
• Drop structures, typically constructed out of concrete, can accommodate a sudden change in elevation of the channel bottom while maintaining control of the flow.
• Drop structures are used in channels, which must be laid along relatively steep gradients to allow for dissipation of energy without causing scour in the channel itself.
• In such applications, the drop structure allows the main channel to be laid on subcritical slope while the excess potential energy of the flow due to the steep topography is absorbed in the drop structure. See Figure 6.12 of a drop structure below
Example 6.5
• An irrigation channel with a design discharge of 2.265 m3/s is to be laid along a terrain having an average slope of 0.005 m/m. To maintain subcritical flow in the channel section, the bottom of the channel must be limited to 0.001 m/m. The extra fall is to be absorbed by drop structures such as the one shown above in the diagram having a width of 3.048 m. Compute the number of structures required in a 16.09 km length of line if the drop height (dZ) is equal to 1.829 m.
Solution of Example 6.5
• Solution: The total drop to be absorbed by structures, ZT = (St - So) L
• Where St is the terrain slope, L is total distance, and So is the slope of the channel.
• ZT = ( 0.005 m/m - 0.001 m/m ) 16.09 km = 64.36
• The number of drop structures required,
• N = ZT/dZ = 64.36/1.829 = 36 Structures.