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FAO 1992 Irrigation manual
SAFR/AGLW/DOC/007
Irrigation Manual
Planning, Development
Monitoring and Evaluation
of Irrigated Agriculture
with Farmer Participation
Developed by
Andreas P. SAVVA
Karen FRENKEN
Volume II
Module 7
Food and Agriculture Organization of the United Nations (FAO)
Sub-Regional Office for East and Southern Africa (SAFR)
Harare, 2002
The views expressed in this paper are those of the authors and do not necessarily reflect the views of the
Food and Agriculture Organization of the United Nations
The designations employed and the presentation of the material in this publication do not imply
the expression of any opinion whatsoever on the part of the Food and Agriculture Organization
of the United Nations concerning the legal status of any country, territory, city or area of its
authorities, or concerning the delimitation of its frontiers or boundaries
ISBN 0-7974-2315-X
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or
transmitted in any form or by any means, electronic, mechanical, photocopying or otherwise,
without the prior permission of the copyright owner
© FAO SAFR 2002
Design and Layout: Fontline Electronic Publishing, Harare, Zimbabwe
Printed by: Préci-ex, Les Pailles, Mauritius
ii
Foreword
The first edition of the Irrigation Manual was published in 1990 in two volumes by the “Smallholder Irrigation” Project
(UNDP/FAO/AGRITEX/ZIM/85/004). The authors of this first edition were FAO Staff on the project1. This edition of
one hundred copies ran out within two years from publishing.
Although the manual was written with Zimbabwe in mind, it soon became popular in several countries of the sub-region.
In view of the high demand, it was decided to proceed with a second edition. The experience gained from using the first
edition of the manual as the basic reference for the AGRITEX2 training programme of irrigation practitioners and the
University of Zimbabwe, was incorporated in the second edition which was published in 1994, in one volume by the
“Technical Assistance to AGRITEX” project (UNDP/FAO/AGRITEX/ZIM/91/005). This second edition was published
under the same authors as the first edition, with the assistance of a review committee from AGRITEX3. The two hundred
copies of this edition also ran out within two years of publishing.
In 1995, the FAO Sub-regional Office for East and Southern Africa (SAFR) was established in Harare, Zimbabwe, in
order to provide easy access to technical assistance and know-how for the countries of the sub-region4. In view of the
high demand for support in the field of smallholder irrigation by the countries of the sub-region, this office was
strengthened with four water resources management officers and a number of on-going programmes have been
developed to provide this support. One of these programmes is the publishing of a new regional edition of the irrigation
manual in support of the on-going national training programmes within several countries in the sub-region and to
provide the basic reference for another important programme, which is the sub-regional training on planning and design
of smallholder irrigation schemes.
This third edition aspires to further strengthen the engineering, agronomic and economic aspects of the manual and to
introduce new modules related to social, health and environmental aspects of irrigation development. The emphasis is
directed towards the engineering, agronomic and economic aspects of smallholder irrigation, in view of the limited
practical references in this area. This manual, being directed to the irrigation practitioner, does not provide an in-depth
analysis of the social, health and environmental aspects in irrigation development. It only attempts to introduce the
irrigation practitioner to these areas, providing a bridge between the various disciplines involved in irrigation
development.
The initiatives and efforts of the Water Resources Management Team of SAFR in publishing this Manual are considered
as a valuable contribution to the dissemination of knowledge and training of irrigation practitioners in the sub-region.
The material covered by this manual is expected to support both national and sub-regional training programmes in the
planning, design, construction, operation and maintenance and on-farm water management of irrigation schemes. This
will support the implementation of FAO’s mandate to increase food production through water control, intensification
and diversification, which are the basic components of the Special Programme for Food Security (SPFS).
The manual is the result of several years of field work and training irrigation engineers in the sub-region. The approaches
have been field tested and withstood the test of time.
1
A.P. Savva, Chief Technical Advisor; J. Stoutjesdijk, Irrigation Engineer; P.M.A. Regnier, Irrigation Engineer; S.V. Hindkjaer, Economist.
2
Agritex: Department of Agricultural Technical and Extension Services, Ministry of Lands and Agriculture, Zimbabwe.
3
Review committee: E. Chidenga, Acting Chief Irrigation Officer; P. Chipadza, Senior Irrigation Specialist; A. Dube, Senior Irrigation Specialist; L. Forichi, Irrigation
Specialist; L. Madhiri, Acting Principal Irrigation Officer; S. Madyiwa, Irrigation Specialist; P. Malusalila, Chief Crop Production; R. Mariga, Assistant Secretary, Economic
and Markets Branch; D. Tawonezvi, Agricultural Economist.
4
The following 21 countries are part of the FAO-SAFR region: Angola, Botswana, Burundi, Comoros, Eritrea, Ethiopia, Kenya, Lesotho, Madagascar, Malawi, Mauritius,
Mozambique, Namibia, Rwanda, Seychelles, South Africa, Swaziland, Tanzania, Uganda, Zambia, Zimbabwe.
iii
Irrigation manual
For ease of reference to the various topics covered by this Manual, the material has been divided into 14 modules,
covering the following:
Module 1:
Module 2:
Module 3:
Module 4:
Module 5:
Module 6:
Module 7:
Module 8:
Module 9:
Module 10:
Module 11:
Module 12:
Module 13:
Module 14:
Irrigation development: a multifaceted process
Natural resources assessment
Agronomic aspects of irrigated crop production
Crop water requirements and irrigation scheduling
Irrigation pumping plant
Guidelines for the preparation of technical drawings
Surface irrigation systems: planning, design, operation and maintenance
Sprinkler irrigation systems: planning, design, operation and maintenance
Localized irrigation systems: planning, design, operation and maintenance
Irrigation equipment for pressurized systems
Financial and economic appraisal of irrigation projects
Guidelines for the preparation of tender documents
Construction of irrigation schemes
Monitoring the technical and financial performance of an irrigation scheme
To those who have been waiting for so long for a practical irrigation engineering manual: here it is. I am sure that it will have
a lot to offer to both the new and experienced irrigation engineers.
Victoria Sekitoleko
FAO Sub-Regional Representative
for East and Southern Africa
iv
Irrigation Manual
Module 7
Surface Irrigation Systems
Planning, Design,
Operation and Maintenance
Developed by
Andreas P. SAVVA
and
Karen FRENKEN
Water Resources Development and Management Officers
FAO Sub-Regional Office for East and Southern Africa
In collaboration with
Simon MADYIWA, Irrigation Engineer Consultant
Patrick CHIGURA, Irrigation Engineer Consultant
Lee TIRIVAMWE, National Irrigation Engineer, Zimbabwe
Victor MTHAMO, Irrigation Engineer Consultant
Harare, 2002
v
Acknowledgements
The preparation of the third edition of the Irrigation Manual is an initiative of FAO's Sub-Regional Office for East and
Southern Africa (SAFR).
The whole project was managed and coordinated by Andreas P. Savva and Karen Frenken, Water Resources Development
and Management Officers at FAO-SAFR, who are considered as the main authors. Karen Frenken is also the main technical
editor.
The inputs by Simon Madyiwa, Patrick Chigura, Lee Tirivamwe and Victor Mthamo to this Module 7 are appreciated. The
preparation of several drawings by Solomon Maina for this Module is acknowledged.
Special appreciation is extended to Chris Pappas for his substantial contribution to the layout of the irrigation manual.
vi
Contents
Foreword
Acknowledgements
List of figures
List of tables
List of abbreviations
Unit conversion table
iii
vi
x
xiii
xv
xvii
1.
INTRODUCTION TO SURFACE IRRIGATION
1.1. Components of a surface irrigation system
1.1.1. The water source
1.1.2. The intake facilities
1.1.3. The conveyance system
1.1.4. The water storage facilities
1.1.5. The field canal and/or pipe system
1.1.6. The infield water use system
1.1.7. The drainage system
1.1.8. Accessibility infrastructure
1.2. The four phases of surface irrigation
1.2.1. The advance phase
1.2.2. The storage or ponding phase
1.2.3. The depletion phase
1.2.4. The recession phase
1.3. Infiltration and contact time
1.3.1. Estimation of the infiltration rate using the infiltrometer method
1.3.2. Estimation of the infiltration rate using the actual furrow method
1.3.3. Determination of optimum stream size and furrow length
1.3.4. Determination of optimum stream size and borderstrip length
1
1
1
1
1
1
1
3
3
3
3
3
3
4
4
4
5
7
9
10
2.
CRITERIA FOR THE SELECTION OF THE SURFACE IRRIGATION METHOD
2.1. Furrow irrigation
2.1.1. Furrow shape
2.1.2. Furrow spacing
2.1.3. Furrow length
2.2. Borderstrip irrigation
2.2.1. Borderstrip width
2.2.2. Longitudinal slope of the borderstrip
2.2.3. Borderstrip length
2.2.4. Guidelines for the determination of borderstrip width and length
2.3. Basin irrigation
2.3.1. Basin size
2.4. Efficiencies of surface irrigation systems and of the different surface irrigation methods
2.4.1. The different types of efficiencies in an irrigation system
2.4.2. Efficiencies of the different surface irrigation methods
2.5. Criteria for the selection of the surface irrigation method
2.5.1. Soil type
2.5.2. Type of crop
2.5.3. Required depth of irrigation application
2.5.4. Land slope
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13
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15
15
17
18
18
18
19
20
20
22
22
23
24
24
24
24
24
vii
2.5.5. Field shape
2.5.6. Labour availability
24
24
3.
DESIGN PARAMETERS FOR THE INFIELD WORKS
3.1. Crop and irrigation water requirements
3.2. Net and gross depth of water application
3.2.1. Net depth of water application
3.2.2. Gross depth of water application
3.3. Irrigation frequency and irrigation cycle
3.3.1. Irrigation frequency
3.3.2. Irrigation cycle
3.4. System capacity
25
25
25
25
26
26
26
26
27
4.
LAYOUT OF A SURFACE IRRIGATION SCHEME
4.1. General layout
4.2. Nabusenga irrigation scheme layout
4.3. Mangui irrigation scheme layout
29
29
31
34
5.
DESIGN OF CANALS AND PIPELINES
5.1. Design of canals
5.1.1. Calculation of the cross-section, perimeter and hydraulic radius of a canal
5.1.2. Factors affecting the canal discharge
5.1.3. Hydraulic design of canal networks using the chart of Manning formula
5.1.4. Canal section sizes used by Agritex in Zimbabwe
5.1.5. Longitudinal canal sections
5.1.6. Field canals for small irrigation schemes
5.1.7. Seepage losses in earthen canals
5.1.8. Canal lining
5.2. Design of pipelines
5.2.1. Design of the conveyance pipeline in Nabusenga irrigation scheme
5.2.2. Design of the piped system in Mangui irrigation scheme
5.2.3. Advantages and disadvantages of piped systems
37
37
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38
42
43
45
49
51
51
53
54
54
60
6.
HYDRAULIC STRUCTURES
6.1. Headworks for river water offtake
6.1.1. Headwork for direct river offtake
6.1.2. River offtake using a weir
6.1.3. River offtake using a dam
6.1.4. Scour gates for sedimentation control
6.2. Night storage reservoirs
6.2.1. Types of reservoirs
6.2.2. Reservoir components
6.3. Head regulators
6.4. Cross regulators
6.5. Drop structures and tail-end structures
6.5.1. Vertical drop structure
6.5.2. Chutes
6.5.3. Tail-end structures
6.6. Discharge measurement in canals
6.6.1. Discharge measurement equations
6.6.2. Weirs
6.6.3. Flumes
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62
63
70
71
73
74
75
77
80
80
83
85
86
86
86
89
96
viii
6.6.4. Orifices
6.6.5. Current meter
6.7. Discharge measurement in pipelines
6.7.1. Differential pressure flow meters
6.7.2. Rotating mechanical flow meters
107
108
110
110
110
7.
LAND LEVELLING
7.1. Profile method
7.2. Contour method
7.3. Plane method
7.4. The cut : fill ratio
7.5. Use of computers
111
111
111
112
119
119
8.
DESIGN OF THE DRAINAGE SYSTEM
8.1. Factors affecting drainage
8.1.1. Climate
8.1.2. Soil type and profile
8.1.3. Water quality
8.1.4. Irrigation practice
8.2. Determining hydraulic conductivity
8.3. Surface drainage
8.4. Subsurface drainage
8.4.1. Horizontal subsurface drainage
8.4.2. Vertical subsurface drainage
8.5. Salt problems
123
123
123
123
123
123
124
125
127
128
131
131
9.
BILL OF QUANTITIES
9.1. Bill of quantities for Nabusenga irrigation scheme
9.1.1. The construction of a concrete-lined canal
9.1.2. The construction of a saddle bridge
9.1.3. The construction of a diversion structure
9.1.4. The overall bill of quantities for Nabusenga irrigation scheme
9.2. Bill of quantities for Mangui irrigation scheme
133
133
133
135
138
139
141
10. OPERATION AND MAINTENANCE OF SURFACE IRRIGATION SYSTEMS
10.1. Operation of the irrigation system
10.1.1. Water delivery to the canals
10.1.2. Water delivery to the fields
10.1.3. Operational success determinants
10.2. Maintenance of the irrigation system
10.2.1. Special maintenance
10.2.2. Deferred maintenance
10.2.3. Routine maintenance
10.3. Operation and maintenance responsibilities
143
143
143
143
146
147
147
147
147
148
REFERENCES
149
ix
List of figures
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
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20.
21.
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23.
24.
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26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
x
Typical components of a surface irrigation system
Definition sketch showing the surface irrigation phases
Basic infiltration rate and cumulative infiltration curves
Cylinder infiltrometers
Analysis of the data of an infiltration test using an infiltrometer on a clay loam soil
Analysis of the data of an infiltration test using actual furrows on a clay loam soil
Time-advance graph for various stream flows in a furrow
Determining the head
Advance and recession of water on a borderstrip
Advance and recession curves for different borderstrip length needing different total volumes of water
to be applied
An example of a furrow irrigation system using siphons
Furrow shape depending on soil type
Soil moisture distribution on various soil types as a determinant of furrow spacing
Example of a borderstrip irrigation system
Cross-section of a borderstrip
Layout of basin irrigation
Typical layout of a surface irrigation scheme on uniform flat topography
The herringbone irrigation layout
Layout of Nabusenga surface irrigation scheme
Layout of Mangui piped surface irrigation scheme
Cumulative depth of irrigation versus time for different types of soil
Plot layout and hydrants
Flowchart for canal design calculations
Canal parameters
Different canal cross-sections
Hydraulic parameters for different canal shapes
Chart of Manning formula for trapezoidal canal cross-sections
Longitudinal profile of a field or tertiary canal
Longitudinal profile of a secondary or main canal
Longitudinal profile of a conveyance canal
Example of a longitudinal profile of a conveyance canal
Longitudinal canal profile generated by the Lonsec Programme
Methods commonly used to introduce water into the field
The longitudinal profile of the conveyance pipeline from Nabusenga dam to the night storage reservoir
Friction loss chart for AC pipes (Class 18)
Friction loss chart for uPVC pipes
Schemes irrigated from different water sources
Headwork with offtake structure only
Offtake possibilities in straight reach of river
2
3
4
5
6
8
9
10
10
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13
14
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21
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57
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62
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
51.
52.
53.
54.
55.
56.
57.
58.
59.
60.
61.
62.
63.
64.
65.
66.
67.
68.
69.
70.
71.
72.
73.
74.
75.
76.
77.
78.
79.
80.
81.
82.
Possible arrangements for offtakes based on site conditions
An example of an intake arrangement of a headwork
An example of a diversion structure
C1 coefficient for different types of weirs in relation to submergence, based on crest shape
C2 coefficient for different types of weirs in relation to crest shape
Types of weirs
Gabion weir
Typical parameters used in the design of a stilling basin
Schematic view of a weir and apron
Masonry weir and apron
Dam cross-section at Nabusenga
Gravity offtake with diversion dam
Scour sluice
Design of a typical earthen night storage reservoir
Courses in brick wall of a reservoir
A simple in-situ concrete proportional flow division structure
Precast concrete block division box
Timber division structures
Duckbill weir photograph
Duckbill weir design
Diagonal weir
Some drop structures used in open canals
Standard drop structure without stilling basin
A vertical drop structure
A chute structure
Static and velocity heads
Variation of specific energy with depth of flow for different canal shapes
Hydraulic jump over a concrete apron
The form of a hydraulic jump postulated in the momentum theory
Parameters of a sharp-crested weir
Trapezoidal (Cipoletti) weir
V-notch weirs
Broad-crested weir
Romijn broad-crested weir, hydraulic dimensions of weir abutments
Romijn broad-crested weir, sliding blades and movable weir crest
Approach velocity coefficient, Cv, as a function of the total head over the movable weir, Hcrt
Parshall flume
Discharge correction factors for Parshall flumes with different throat widths
Head loss through Parshall flumes
Trapezoidal flume
Cut-throat flume
Cut-throat flume coefficients
Examples of orifices
62
63
64
64
65
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66
67
68
70
70
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72
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78
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100-2
103
105
106
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107
xi
Irrigation manual
83.
84.
85.
86.
87.
88.
89.
90.
91.
92.
93.
94.
95.
96.
97.
98.
99.
100.
101.
102.
103.
104.
105.
106.
107.
xii
Free discharging flow through an orifice
Sluice gate under submerged conditions
Ott C31 propeller instrument
Depth-velocity integration method
Venturi flow meter
The profile method of land levelling: cut and fill and checking gradient levels with profile boards
The contour method of land levelling
Grid map showing land elevation and average profile figures
Average profile and lines of best fit
Part of the completed land levelling map for Nabusenga, assuming GX = 0.005
Irregular shaped field (elevations 0.0 are located outside the field)
Parameters for determining hydraulic conductivity
Cross-sections of drains
Rainfall-duration curve
Subsurface drainage systems at field level
Subsurface drainage parameters
Nomograph for the determination of equivalent sub-stratum depths
Nomograph for the solution of the Hooghoudt drain spacing formula
Salt accumulation in the root zone and the accompanying capillary rise
Cross-section of a concrete lined canal at Nabusenga
Saddle bridge for Nabusenga
Field canal bank breaching in order to allow the water to flow from the canal onto the field
Permanent outlet structure used to supply water from the canal onto the field
An example of a spile used to supply water from the canal onto the field
A siphon supplying water from a canal onto a field
107
108
109
109
110
111
112
113
116
118
122
124
125
126
127
128
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131
132
133
136
143
144
145
146
List of tables
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
Typical infiltration rates for different soils
Infiltration rate data from an infiltrometer test
Infiltration rate measurement in a 100 m long furrow
Discharge for siphons, depending on pipe diameter and head
Guidelines to determine when to stop the water supply onto a borderstrip
Measurement of water advance and recession distance and time on a borderstrip
Furrow lengths in metres as related to soil type, slope, stream size and irrigation depth
Practical values of maximum furrow lengths in metres depending on soil type, slope, stream size and
irrigation depth for small-scale irrigation
Typical borderstrip dimensions in metres as related to soil type, slope, irrigation depth and stream size
Suggested maximum borderstrip widths and lengths for smallholder irrigation schemes
Criteria for basin size determination
Basin area in m2 for different stream sizes and soil types
Approximate values for the maximum basin width
Selection of an irrigation method based on soil type and net irrigation depth
Design parameters for Nabusenga and Mangui surface irrigation schemes
Summary of the calculated design parameters for Nabusenga and Mangui surface irrigation schemes
Km and n values for different types of canal surface
Typical canal side slopes
Recommended b/d ratios
Maximum water velocity ranges for earthen canals on different types of soil
Canal capacities for standard Agritex canal sections
Longitudinal profile for field canal - output from the Lonsec computer programme
Seepage losses for different soil types
Hazen-Williams C value for different materials
Weighted-creep ratios for weirs depending on soil type
Reinforcement requirements in a clay brick wall of a reservoir
Cross-sectional areas of reinforcement steel rods
Discharge Q (m3/sec) for contracted rectangular weir, depending on h and b
Discharge Q (m3/sec) for Cipoletti weir, depending on h and b
Discharge Q (m3/sec x 10) for a 90° V-notch weir, depending on h
Standard dimensions of Parshall flumes
Discharge characteristics of Parshall flumes
Land levelling results
Input and output data types for computer land levelling programme LEVEL 4EM.EXE
Land levelling calculations with line of best fit and cut:fill ratio of 1.01
Land levelling calculations with 0.5% gradient in the X direction and cut:fill ratio of 1.01
Land levelling calculations with line of best fit and cut:fill ratio of 1.21
Computer printout of land levelling data for Mangui piped surface irrigation scheme
Values for runoff coefficient C in Equation 70
4
6
8
10
10
11
16
17
19
20
20
21
22
24
25
28
40
40
41
41
44
50
51
54
68
76
77
91
92
94
98
99
117
119
120
120
121
121-2
126
xiii
Irrigation manual
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
xiv
Concrete volume for different trapezoidal canal cross-sections
Summary of the bill of quantities for the construction of the 980 m long lined canal at Nabusenga
Summary of the bill of quantities for the construction of a saddle bridge
Summary of the bill of quantities for the construction of a diversion structure
Bill of quantities for Nabusenga scheme, downstream of the night storage reservoir
Summary of material requirements for Nabusenga (including 10% contingencies)
Bill of quantities for pipes and fittings and pumping plant at Mangui scheme
Discharge of permanent wooden field outlet structures
Rates of discharge through spiles (l/sec)
Discharge of siphons for different head and pipe diameter (l/sec)
Weed management and effectiveness
134
135
137
138
139
141
142
144
145
146
148
List of abbreviations
A
AC
ASAE
C
CI
J
D or d
d
dgross
dnet
E
EL
F
F
FC
Fr
g
G
GS
h
H
Hf100
HL
IC
IF
IT
Km
kPa
kW
L
n
NSR
P
P
P
PWP
q
Q
R
Area
Asbestos Cement
American Society of Agricultural Engineers
Cut
Cast Iron
Density of water
Diameter
Water depth
Gross depth of water application
Net depth of water application
Efficiency
Elevation
Freeboard
Fill
Field Capacity
Froude Number
Acceleration due to gravity
Regression coefficient
Galvanized Steel
water depth
Head
Friction losses per 100 m of pipe
Head Loss
Irrigation Cycle
Irrigation Frequency
Irrigation Time
Manning roughness coefficient
Kilopascal
kilowatt
Length
Roughness coefficient (=1/Km)
Night Storage Reservoir
Allowable moisture depletion
Wetted Perimeter
Pressure
Permanent Wilting Point
Discharge into one furrow or discharge per m width
Discharge
Hydraulic radius
xv
Irrigation manual
R
RZD
S
T
TDH
uPVC
V
V or v
z
xvi
Cut : Fill ratio
Effective Root Zone Depth
Slope or gradient
Irrigation time
Total Dynamic Head
unplasticized Polyvinyl Chloride
Volume
Water velocity
Elevation
Unit conversion table
Mass
Length
1 inch (in)
0.0254 m
1 ounce
28.3286 g
1 foot (ft)
0.3048 m
1 pound
0.4535 kg
1 yard (yd)
0.9144 m
1 long ton
1016.05 kg
1 mile
1609.344 m
1 short ton
907.185 kg
1 metre (m)
39.37 inches (in)
1 gram (g)
0.0353 ounces (oz)
1 metre (m)
3.28 feet (ft)
1 kilogram (kg)
1000 g = 2.20462 pounds
1 metre (m)
1.094 yards (yd)
1 ton
1 kilometre (km)
0.62 miles
1000 kg = 0.984 long ton
= 1.102 short ton
Pressure
Area
1 square inch
(in2)
6.4516 x
10-2
m2
1 pound force/in2
6894.76 N/m2
51.7 mm Hg
1 square foot (ft2)
0.0929 m2
1 pound force/in2
1 square yard (yd2)
0.8361 m2
1 Pascal (PA)
1 acre
4046.86 m2
1 N/m2
= 0.000145 pound force /in2
1 acre
0.4046 ha
1 atmosphere
760 mm Hg
= 14.7 pound force/in2
(lbf/in2)
1 square centimetre (cm2)
0.155 square inches (in2)
1 square metre (m2)
10.76 square feet (ft2)
1 atmosphere
1 bar
1 square metre (m2)
1.196 square yard (yd2)
1 bar
10 metres
1 square metre (m2)
0.00024 acres
1 bar
100 kpa
1 hectare (ha)
2.47 acres
Energy
Volume
1 B.t.u.
1055.966 J
1 cubic inch (in3)
1.6387 x 10-5 m3
1 foot pound-force
1.3559 J
1 cubic foot (ft3)
0.0283 m3
1 B.t.u.
0.25188 Kcalorie
1 cubic yard (yd3)
0.7646 m3
1 B.t.u.
0.0002930 KWh
1 cubic centimetre (cm3)
0.061 cubic inches (in3)
1 Joule (J)
0.000947 B.t.u.
1 cubic metre (m3)
35.315 cubic feet (ft3)
1 Joule (J)
0.7375 foot pound-force (ft.lbf)
1.308 cubic yards (yd3)
1 kilocalorie (Kcal)
4185.5 J = 3.97 B.t.u.
1 kilowatte-hour (kWh)
3600000 J = 3412 B.t.u.
1 cubic metre
(m3)
Capacity
1. imperial gallon
0.0045 m3
Power
1. US gallon
0.0037 m3
1 Joule/sec
0.7376 foot pound/sec
1. imperial barrel
0.1639 m3
1 foot pound/sec
1.3557 watt
1. US. barrel
0.1190 m3
1 cheval-vapor
0.9861 hp
1 pint
0.5681 l
1 Kcal/h
0.001162 kW
1 US gallon (dry)
0.0044 m3
1 watt (W)
1 litre (l)
0.22 imp. gallon
1 Joule/sec
= 0.7376 foot pound/sec (ft lbf/s)
1 litre (l)
0.264 U.S. gallon
1 horsepower (hp)
745.7 watt 550 ft lbf/s
1 litre (l)
0.0061 imperial barrel
1 horsepower (hp)
1.014 cheval-vapor (ch)
1 hectolitre (hl)
100 litres
1 kilowatt (kW)
860 Kcal/h
= 1.34 horsepower
= 0.61 imperial barrel
= 0.84 US barrel
1 litre (l)
1 cubic metre of water (m3)
1.760 pints
1000 l
= 227 U.S. gallon (dry)
1 imperial barrel
164 litres
Temperature
0C
(Celsius or centigrade-degree)
0F
(Fahrenheit degree)
K (Kelvin)
0F
0C
= 5/9 x (0F - 32)
= 1.8 x 0C + 0F
K = 0C + 273.15
xvii
Irrigation manual
xviii
Chapter 1
Introduction to surface irrigation
Surface irrigation is the oldest and most common method
of applying water to crops. It involves moving water over
the soil in order to wet it completely or partially. The water
flows over or ponds on the soil surface and gradually
infiltrates to the desired depth. Surface irrigation methods
are best suited to soils with low to moderate infiltration
capacities and to lands with relatively uniform terrain with
slopes less than 2-3% (FAO, 1974).
1.1. Components of a surface irrigation
system
Figure 1 presents the components of a surface irrigation
system and possible structures, which are described in
Chapter 6. The water delivery system, shown in Figure 1,
includes the conveyance system and the field canal system
described below. The water use system refers to the infield
water use system, showing one field in the block. The tail
water ditch and the water removal system are part of the
drainage system.
the conveyance canal itself does not need to be above
ground level all along the canal, but its starting bed level
should be such that there is sufficient command for the
lower order canals. Where possible, it could run quasiparallel to the contour line. Design aspects of canals and
pipelines are discussed in Chapter 5.
Although an open conveyance canal may be cheaper per
unit length than a pipeline, the latter would need to be
selected when:
Y
The water source is at lower elevation than the
irrigation area, and thus pumping is required
Y
The topography of the land is very uneven, such that
constructing an open canal could either be more
expensive or even impossible (for example when
crossing rivers and gullies)
A piped conveyance system also eliminates water losses
through evaporation and seepage. An added advantage is
that it does not provide the environment for water-borne
disease vectors along the conveyance.
1.1.1. The water source
The source of water can be surface water or groundwater.
Water can be abstracted from a river, lake, reservoir,
borehole, well, spring, etc.
1.1.2. The intake facilities
The intake is the point where the water enters into the
conveyance system of the irrigation scheme. Water may
reach this point by gravity or through pumping. Intake
facilities are dealt with during the design of headworks in
Chapter 6. Pumping units are discussed in detail in
Module 5.
1.1.3. The conveyance system
Water can be conveyed from the headworks to the inlet of
a night storage reservoir or a block of fields either by gravity,
through open canals or pipes, or through pumping into
pipelines. The method of conveyance depends mostly on
the terrain (topography and soil type) and on the difference
in elevation between the intake at the headworks and the
irrigation scheme. In order to be able to command the
intended area, the conveyance system should discharge its
water at the highest point of the scheme. The water level in
1.1.4. The water storage facilities
Night storage reservoirs (NSR) could be built if the
irrigation scheme is large enough to warrant such
structures. They store water during times when there is
abstraction from the water source, but no irrigation. In
Southern Africa it is common practice to have continuous
flow in the conveyance system combined with a NSR
located at the highest point of a block or the scheme.
Irrigation would then be practiced during daytime using the
combined flow from the conveyance system and the NSR.
Depending on the size of the scheme one could construct
either one reservoir located at the highest part of the
scheme or a number of reservoirs, each located at the
entrance of a block of fields. The conveyance system ends
at the point where the water enters the reservoir.
1.1.5. The field canal and/or pipe system
Canals or pipelines are needed to carry the water from
the conveyance canal or the NSR to a block of fields.
They are called the main canal or pipeline. Secondary
canals or pipelines supply water from the main canal or
pipeline to the tertiary or field canals or pipelines, which
1
Irrigation manual
Figure 1
Typical components of a surface irrigation system (Source: Walker and Skogerboe, 1987)
Parshall flume
Water supply
Check
Water
delivery
system
Gated pipe
Field canal
Drop
Head ditch
Water use
system
Division box
Tail water ditch
Water removal system (drain)
are located next to the field. Sometimes no distinction is
made between main and secondary and the canal or pipe
system from the reservoir to the tertiary canal is called
main canal or pipeline. The tertiary canals or the
pipelines with hydrants are used to supply water to the
furrows or borderstrips or basins. Where canals are used
to deliver irrigation water, they should be constructed
above ground level, as the water level in canals should be
2
above field level for siphoning to take place. At times,
water from the field canal is siphoned to a field earthen
ditch from where the furrows, borderstrips and basins
are supplied. When a piped distribution system is used,
the gated pipe is connected to the hydrant and water is
provided to the field from the gates of the gated pipe.
Alternatively, a hose is connected to the hydrant to supply
water to the field.
Module 7: Surface irrigation systems: planning, design, operation and maintenance
1.1.6. The infield water use system
This refers mainly to the method of water application to the
field, which can be furrow, borderstrip or basin irrigation.
These methods are described in detail in Chapter 2. It is
important to note that the method of conveyance and
distribution up to field level is independent of the selected
infield irrigation method.
In irrigation system design, the starting point is the infield
water use system as this provides information on the
surface irrigation method to use, the amount of water to be
applied to the field and how often it has to be applied. With
this information, we can then work backwards or upstream
to designing the field canal, distribution, storage,
conveyance system and ultimately the intake facilities, and
we can work forwards or downstream to determine the
capacity of the drainage facilities.
1.1.7. The drainage system
This is the system that removes excess water from the
irrigated lands. The water level in the drains should be
below the field level and hence field drains should be
constructed at the lower end of each field. These field or
tertiary drains would then be connected to secondary
drains and then the main drain, from where excess water is
removed from the irrigation scheme.
1.1.8. Accessibility infrastructure
The scheme is to be made accessible through the
construction of main roads leading to the scheme, and farm
roads within the scheme.
1.2. The four phases of surface irrigation
When water is applied to the soil surface by any of the three
surface irrigation methods (furrow, borderstrip or basin), it
will infiltrate into the soil to the required depth in order to
bring the soil back to field capacity. Using the borderstrip
and basin irrigation method, the entire soil surface is
wetted and the water movement through the soil is
predominantly vertical. Using the furrow irrigation
method, part of the soil surface is wetted and the water
movement through the soil is both vertical and lateral.
The surface irrigation event is composed of four phases, as
illustrated in Figure 2 and explained below.
1.2.1. The advance phase
The advance phase begins when water is applied onto the
field at the upstream end and ends when it reaches the
downstream end of the field. The stream size applied at the
head of the furrow, borderstrip and basin should be greater
than the soil infiltration rate. This means that part of the
water advances over the soil surface to the end of the field
and part of the water infiltrates into the soil. The time
between the start of irrigation and water advancement to
the end of the field is called the advance phase. The advance
curve in Figure 2 is the line showing the relationship
between the elapsed time (on y-axis) and the advance
distance (on x-axis).
1.2.2. The storage or ponding phase
When the water arrives at the tail end and the water
supply at the head is continued, water floods the whole
field. Some water continues infiltrating into the soil, some
water ponds on the field and some excess water is
collected as runoff. The time elapsed between the arrival
of the water at the tail end and the stopping of the inflow
at the top end is called the storage phase or ponding phase.
This phase ends when the inflow at the head of the field
is stopped.
Figure 2
Definition sketch showing the surface irrigation phases (Source: Basset et al., 1980)
3
Irrigation manual
1.2.3. The depletion phase
1.3. Infiltration and contact time
After stopping the inflow at the head end, water may
continue to pond on the soil surface for a while. Some
water still infiltrates the soil, with the excess being collected
as runoff. At a certain moment water will start receding
from the head end. The time between the stop of the inflow
at the head end and the appearance of the first bare soil that
was under water is called the lag time or depletion phase.
Infiltration, which is the movement of water into the soil, is
an important factor affecting surface irrigation in that it
determines the time the soil should be in contact with
water (the intake opportunity time or the contact time). It
also determines the rate at which water has to be applied to
the fields, thereby controlling the advance rate of the
overland flow and avoiding excessive deep percolation or
excessive runoff. The infiltration or intake rate is defined as
the rate at which water enters into the soil, usually
expressed in mm/hr.
1.2.4. The recession phase
After water starts receding from the head end, it continues
to the tail end. The time when water starts to disappear at
the head end until it eventually recedes from the whole field
is called the recession phase.
The time-difference between the recession and advance
curve is called the contact time or the intake opportunity time.
This is the time in hours or minutes that any particular
point in the field is in contact with water. Thus, by
increasing or decreasing the contact time, one can, within
limits, regulate the depth of water applied.
The following three basic principles are fundamental for
surface irrigation, though the possibility of applying them
depends a lot on the soil type:
i) The depth of infiltration varies in relation to contact
time
ii) The contact time can be increased by using flatter
slopes, increasing the length of run or reducing the
stream flow; any one or a combination of these factors
may be used
iii) The contact time can be decreased by steepening the
slope, shortening the length of run or increasing the
stream flow
Figure 3
Basic infiltration rate and cumulative infiltration curves
4
No matter where water infiltrates rapidly when it first
arrives, after which it slows down until it reaches a steady
state. This steady state is referred to as the basic infiltration
rate, which is close to the value of the saturated hydraulic
conductivity. When the basic infiltration rate is reached, the
cumulative infiltration curve becomes a straight line and the
basic infiltration rate curve becomes a horizontal line. This
phenomenon is shown using a graph in Figure 3.
The infiltration rates of soils are influenced, among others,
by the soil texture. Heavy soils have low infiltration rates by
virtue of their small pore sizes, while light soils have high
infiltration rates because of larger pore sizes. Some typical
infiltration rates for different soil types are given in Table 1.
Table 1
Typical infiltration rates for different soils
Soil Type
Infiltration rate mm/hr
Sand
> 30
Sandy Loam
30-20
Silty Loam
20-10
Clay Loam
10-5
Clay
<5
The infiltration rate is a difficult parameter to define
accurately, but it has to be determined in order to describe
the hydraulics of the surface irrigation event. When
planning a furrow irrigation scheme, one can determine
the infiltration rate by two methods: the infiltrometer
method and the actual furrow method. The former method
can also be used to determine the infiltration rate for
borderstrip and basin irrigation schemes.
Y
Fix a gauge (almost any type) to the inner wall of the inner
cylinder so that the changes in water level can be measured
Y
Fill the outer ring with water to a depth approximately
the same as will be used in the inner ring and also quickly
add water to the inner cylinder till it reaches 10 cm or
100 mm on the gauge
Y
Record the clock time immediately when the test begins
and note the water level on the measuring rod
1.3.1. Estimation of the infiltration rate using the
infiltrometer method
Y
The initial infiltration will be high and therefore regular
readings at short intervals should be made in the
beginning, for example every minute, after which they
can increase to 1, 2, 5, 10, 20, 30 and 45 minutes, for
example. The observation frequencies should be
adjusted to infiltration rates
Y
After a certain period infiltration becomes more or less
constant (horizontal line in Figure 3). Then the basic
infiltration rate is reached. After reading equal water
lowering at equal intervals for about 1 or 2 hours, the
test can stop.
Y
The infiltration during any time period can be
calculated by subtracting the water level measurement
before filling at the end of the period from the one
after filling at the beginning of that same period. For
example, the infiltration between 09.35 hr and
09.45 hr in Table 2 is 100 - 93 = 7 mm, which is
0.7 mm/min or 0.7 x 60 = 42 mm/hr
Y
After the tests the cylinders should be washed before
they become encrusted. This makes them easy to drive
into the soil, with minimal soil disturbance, next time
they are to be used
With this method, infiltration is measured by observing the
fall of water within the inner cylinder of two concentric
cylinders, with a usual diameter of 0.4 and 0.5 m and a
height of about 0.4 m, driven vertically into the soil surface
layer as illustrated in Figure 4. The outer ring acts as a
buffer preventing lateral seepage of water from the inner
one. This allows infiltration measurement from the inner
ring to be representative of infiltration from the actual
irrigation of a large area.
The procedure for installing the infiltrometer and for taking
measurements is as follows:
Y
Select possible locations for three to four infiltrometers
spread over the irrigation scheme and examine the sites
carefully for signs of unusual surface disturbance, animal
burrows, stones and so on, as they may affect the test
results
Y
Drive the cylinder into the soil to a depth of
approximately 15 cm by placing a driving plate over the
cylinder, or placing heavy timber on top, and using a
driving hammer. Rotate the timber every few pushes or
move the hammer equally over the surface in order to
obtain a uniform and vertical penetration
If the actual moisture conditions in the soil at the start of the
infiltration test are low, it will take longer to reach the basic
Figure 4
Cylinder infiltrometers
5
Irrigation manual
infiltration rate compared to the same soil wherein the
moisture is only slightly depleted. Preferably, tests should be
carried out at the expected depletion level during irrigation.
Results of an infiltrometer test on a clay loam soil are given
in Table 2.
Table 2
Infiltration rate data from an infiltrometer test
Site location: Nabusenga
Soil type: Clay Loam
Test date: 5 October 1990
Water level reading
Watch
reading
(hr:min)
Time
interval
Cumulative
time
(min)
(min)
before
filling
(mm)
after
filling
(mm)
Infiltration
Infiltration
rate
Infiltration
rate
Cumulative
infiltration
(mm)
(mm/min)
(mm/hr)
(mm)
3.0
1.50
90.0
3.5
1.17
70.0
4.5
0.90
54.0
7.0
0.70
42.0
5.5
0.55
33.0
7.0
0.35
21.0
5.0
0.25
15.0
7.0
0.175
10.5
7.0
0.175
10.5
7.0
0.175
10.5
Start = 0
09:25
0
100
2
09:27
2
97.0
100
3
09:30
5
96.5
10
96.5
20
93.0
30
93.5
50
93.0
70
95.0
110
93.0
150
94.0
40
12:35
190
35.5
101
40
11:55
30.5
100
40
11:15
23.5
100
20
10:35
18.0
100
20
10:15
11.0
99
10
09:55
6.5
100
10
09:45
3.0
101
5
09:35
Start = 0
42.5
100
49.5
93.0
Figure 5
Analysis of the data of an infiltration test using an infiltrometer on a clay loam soil
6
56.5
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Example 1
The net peak crop water requirement for a scheme is 6.0 mm/day. The available moisture for the clay loam is 130
mm/m and depletion is allowed up to around 46%. The root zone depth is 0.70 m. After how many days should
irrigation take place to replenish the soil moisture?
The moisture available to the crop in the root zone is 130 x 0.70 x 0.46 = 42 mm. The peak water requirement being
6.0 mm/day, after 42/6 = 7 days irrigation should take place to replenish the 42 mm soil moisture. This is equal to the
net irrigation requirement.
Example 2
Assuming a field application efficiency of 50% in the previous example, what is the time required to replenish the 42
mm soil moisture?
The net irrigation requirement being 42 mm and considering a field application efficiency of 50%, the gross irrigation
requirement is 42/0.50 = 84 mm. From the cumulative curve in Figure 5 it can be seen that the time required to
replenish this depth of water is approximately 384 minutes.
If the field application efficiency increased to 65%, due to improved water management, the gross irrigation requirement
would be 42/0.65 = 64.6 mm and the time required to replenish this depth would be reduced to 260 minutes.
It can be seen from Table 2, that the steady state has been
reached somewhere between 70 and 110 minutes after the
start of the test. From that moment on, the basic infiltration
rate curve (Figure 3) will be a horizontal line and the
cumulative infiltration curve will be a straight line. The
results of the above test are graphically presented in
Figure 5.
In this example, the furrow test starts at 14.00 hr with a
continuous uniform flow of 98 l/min (1.63 l/sec) being
discharged into the furrow. At the first recording of the
inflow at the top, there is no outflow at the bottom end of
the furrow. In this example, the outflow starts at 14.05 hr.
From a recording of zero outflow there is a sudden outflow
of 17 l/min (0.28 l/sec).
Examples 1 and 2 demonstrate the use of the intake rate in
estimating the time required to replenish the soil moisture
during irrigation.
While the inflow remains constant, the outflow increases
with time until the basic infiltration rate is reached at
15.10 hr. The infiltration is the difference between the
inflow and the average of the outflow during a given time
period. For example, between 14.20 hr and 14.30 hr, the
inflow is 98 l/min and the average outflow is 56.7 l/min
(={46 + 67.4}/2). Thus, the average infiltration rate over
this period is 98 - 56.7 = 41.3 l/min. The average
infiltration per period is calculated by multiplying the
infiltration per minute by the time period. The sum of the
infiltrations gives the cumulative infiltration.
The time required in Example 2 to replenish the required
depth of water is called the contact time, which is the time
the water should be in contact with the soil in order to have
the correct depth of water replenished in the soil.
1.3.2. Estimation of the infiltration rate using the
actual furrow method
With furrows, the infiltration rate and cumulative
infiltration curve can also be determined as follows. Three
adjacent furrows of a specific length, for example 30 m or
100 m, are wetted at the same time. Two measuring
devices, such as for example portable Parshall flumes, are
placed at the beginning and end of the middle furrow
respectively, and the inflow at the top end and outflow at
the tail end are measured simultaneously. The outer
furrows function in the same way as the outer ring of the
infiltrometer by preventing excessive lateral flow from the
middle furrow. The infiltration in l/min (volume/time) can
be converted into an infiltration in mm/hr (depth/time) by
dividing it by the area covered by furrow. Table 3 and Figure
6 show the result for the same clay loam soil.
The contact time can be determined from the cumulative
infiltration curve, as shown in Figure 6. Considering the
same data as used in Example 1 and 2, with a gross
irrigation requirement of 84 mm, and considering a furrow
length of 100 m and a furrow spacing of 0.75 m, the
volume of water required per furrow is:
V = 0.084 x 100 x 0.75 = 6.300 m3 = 6 300 litres
Figure 6 again gives a contact time of 384 minutes (x-axis)
for this volume of 6 300 litres on the right-hand y-axis
(cumulative infiltration).
7
Irrigation manual
Table 3
Infiltration rate measurement in a 100 m long furrow
Site location: Nabusenga
Soil type: Clay Loam
Test date: 5 October 1990
Water measurements
Watch
reading
Time
interval
Cumulative
time
(hr:min)
(min)
(min)
Inflow
(l/min)
Outflow
(l/min)
98.0
0
Intake rate
l/min
Intake
Cumulative
intake
over 100 m
(l/100 m)
(l/100 m)
98.00
490
67.00
335
52.50
525
Start = 0
14:00
0
5
14:05
5
98.0
0 to 17
5
14:10
10
98.0
20
98.0
46.00
30
98.0
67.40
10
14:30
20
14:50
50
98.0
70
98.0
110
98.0
84.88
150
98.0
84.88
40
16:30
40
17:10
190
98.0
1 350
41.30
413
26.25
525
17.51
375
13.12
525
1 763
2 288
84.88
40
15:50
825
76.10
20
15:10
490
45.00
10
14:20
0
2 638
3 163
13.12
525
13.12
525
3 688
84.88
Figure 6
Analysis of the data of an infiltration test using actual furrows on a clay loam soil
8
4 213
Module 7: Surface irrigation systems: planning, design, operation and maintenance
1.3.3. Determination of optimum stream size and
furrow length
In order to wet the root zone as uniformly as possible and
to have minimum percolation losses at the top end of the
field and minimum runoff at the bottom end of the field an
appropriate stream size has to be chosen. As explained
earlier, water flows from the top end of the field to the
bottom end. This is called the advance stream (Figure 2).
When water supply stops, the water moves away from the
top of the field, which is called the recession of the
waterfront. Usually the advance is slower than the recession
because water infiltrates quicker in dry soil. Therefore, the
top end of the field usually receives more water than the
bottom end of the field and water will be lost through deep
percolation. If the stream size is too small, it will take a long
time before the water reaches the end of the field, therefore
deep percolation will be high if the bottom end is also to
receive enough water. On the other hand, if the stream size
is too large, the waterfront will reach the bottom fast and
runoff losses will occur, unless the stream size is reduced.
Therefore the appropriate combination of stream flow size
and length of borderstrip or furrow has to be selected.
As a rule of thumb, one can say that the stream size must
be large enough to reach the end of the furrow in
approximately one quarter of the contact time. This is
called the one-quarter rule.
The optimum furrow length, or the optimum non-erosive
stream flow in existing schemes with known furrow length,
could be determined in the field with a test whereby the
advance of selected stream sizes is measured in furrows.
The results are plotted in a time-advance graph (Figure 7).
Following the above example, where the conditions would be
such that a furrow length of 100 m would be preferable, the
contact time is 384 minutes. Using the one-quarter rule of
thumb, the water should reach the end of the furrow in 96
minutes (= 1/4 x 384). Different flows are brought onto the
land and, using the plotted time-advance graph in Figure 7,
the optimum furrow length would be 60 m for a flow of 0.5
l/sec and 100 m for a flow of 0.7 l/sec. If 0.7 l/sec is a nonerosive flow, which depends on the soil type and the actual
state of the soil (which has to be checked in the field), the
100 m long furrow could be selected, which allows a more
cost-effective layout. If the field shape allows it, a furrow
should be as long as possible, in order to minimize the
number of field canals that have to supply water to the field.
This has a direct bearing on cost, since the cost increases with
the number of canals to be constructed.
If the flow is not reduced once the water reaches the end of
the furrow, a large runoff will occur. Therefore, the flow is
usually reduced once or twice during an irrigation, such
that runoff remains small. However, if the flow becomes
too small, deep percolation losses at the top of the field
might increase. The flow could be reduced by taking out
siphons from the furrow. Field tests are usually carried out
in order to make recommendations to farmers.
The discharge through siphons depends on the diameter of the
siphon and the head. For drowned or submerged discharge,
the head is the difference between the water level in the canal
and the water level in the field (Figure 8a). For free discharge,
the head is the difference between the water level in the canal
from where the siphon takes the water and the outlet from
the siphon (Figure 8b). Discharge can be altered by a change
in pipe diameter or a change in the head (Table 4).
Figure 7
Time-advance graph for various stream flow sizes in a furrow
9
Irrigation manual
Figure 8
Determining the head (Source: FAO, 1988)
Table 4
Discharge for siphons, depending on pipe diameter
and head (l/sec)
Pipe diameter
(cm)
5
Head (cm)
10
15
20
2
0.19
0.26
0.32
0.73
3
0.42
0.59
0.73
0.84
4
0.75
1.06
1.29
1.49
5
1.17
1.65
2.02
2.33
waterfront reaches between 2/3 and 3/4 of the borderstrip
length. On clayish soil, the inflow is usually stopped earlier
than on loamy soils, while on sandy soils the water could
almost cover the whole borderstrip length before the flow
is stopped. New irrigators can rely on the general guidelines
given in Table 5 to decide when to stop the flow. The actual
field cut-off times should then be decided through field
experience.
Table 5
1.3.4. Determination of optimum stream size and
borderstrip length
As in the case of furrow irrigation, it is important to use the
right stream size for the soil and land slope and to stop the
flow at the right time so that just enough water infiltrates
into the soil to satisfy the required irrigation depth. If the
flow is stopped too soon, there may not be enough water
on the borderstrip to achieve the required irrigation depth
at the bottom end of the borderstrip. If the water is left
running for too long, there may be large runoff losses. As a
rule of thumb, the water supply is stopped when the
Figure 9
Advance and recession of water on a borderstrip
10
Guidelines to determine when to stop the water
supply onto a borderstrip
Soil Type
Stop the flow when advance reaches the
following portion of borderstrip
Clay
Two thirds of total length
Loam
Three quarters of total length
Sand
Almost end of borderstrip
Where possible, it is recommended to carry out field tests
to determine the best borderstrip length. To do this, a
borderstrip is marked with pegs at 10 m interval along its
Module 7: Surface irrigation systems: planning, design, operation and maintenance
length. A selected discharge is then brought onto the strip
and the advance of the water is measured, which is the time
that it takes for the water to pass through pre-determined
distances along the borderstrip length (Figure 9). When the
desired volume of water has been delivered to the
borderstrip, the flow of water from the canal onto the
borderstrip is stopped. As explained above, usually this is
done before the water has reached the end of the border.
From that moment on, time is taken when the end of the
water flow passes through the pre-determined distances.
This is called the recession of the water. For a fixed
irrigation depth, the total volume desired depends on the
size of the borderstrip and thus is larger for a longer
borderstrip in cases where the width is the same.
Having measured the time and the distance of the advance
and the recession of the water, the advance and recession
curves can be drawn. If testing, for example, two different
borderstrip lengths, the total volumes of water to be applied
are different, leading to two different recession curves,
since the inflow is stopped later when the borderstrip is
longer. Table 6 shows the data for the advance stream,
which is the same for both recessions and the data for
recession 1 and recession 2.
The advance curve and the two recession curves have been
presented as graphs in Figure 10.
Table 6 shows the data for the advance stream, which is
similar for both recession streams. The same table also
shows the results from the two separately conducted
recession tests. From these data it can be seen that the
supply of test 2 was stopped earlier (at 404 minutes) when
compared to test 1 (at 464 minutes). The volume brought
onto the land during test 1 had completely infiltrated in the
soil at 744 minutes, covering the first 130 m of the border.
In test 2 all water disappeared after 624 minutes and the
required depth had been applied to the first 100 m.
Considering Example 2, where a contact time of 384
minutes was calculated, the ideal intake opportunity time
curve was drawn in Figure 10 to be parallel to the advance
curve derived from the data of Table 6. The same figure
presents the two recession curves derived from the data of
Table 6. The closer a recession curve is to the ideal intake
opportunity time curve the more efficient the water
application. Looking at the two recession curves, it is clear
that recession curve 2 is situated closer to the ideal intake
opportunity curve than recession curve 1. Recession curve
1 shows that over-irrigation took place, especially over the
first 100 m reducing later on for the remaining 30 m.
While some deep percolation is also indicated by the
position of recession curve 2, this is much less than that
demonstrated by recession curve 1. Hence the designs
should be based on 100 m length of border.
Table 6
Measurement of water advance and recession distance and time on a borderstrip
Advance
(m)
Time
(min)
Recession 1
(m)
Time
(min)
Recession 2
(m)
Time
(min)
0
0
0
464
0
404
10
20
10
476
10
422
20
38
20
494
20
444
30
57
30
524
30
464
40
80
40
560
40
480
50
100
50
586
50
502
60
120
60
608
60
524
70
143
70
629
70
554
80
170
80
649
80
582
90
205
90
671
90
604
100
240
100
692
100
624
110
278
110
710
110
120
315
120
726
120
130
350
130
744
130
11
Irrigation manual
Figure 10
Advance and recession curves for different borderstrips lengths, needing different total volumes of water
to be applied
12
Chapter 2
Criteria for the selection of the surface irrigation method
Surface irrigation methods refer to the technique of water
application over the soil surface in order to wet it, either
partially or completely. They do not bear any reference to
the conveyance and field canal or distribution system.
Three surface irrigation methods can be distinguished:
Y
Furrow irrigation
Y
Borderstrip irrigation
Y
Basin irrigation
Good surface irrigation practice calls for efficient water
management to be achieved through:
Y
Distributing the water evenly in the soil
Y
Providing adequate water to the crops (not too much,
not too little)
Y
Avoiding water wastage, soil erosion and salinity
2.1. Furrow irrigation
A furrow irrigation system consists of furrows and ridges.
The water is applied by means of small channels or furrows,
which follow a uniform longitudinal slope. The method is
best suited to row crops such as maize, potatoes, onions,
tomatoes, etc.
Water can be diverted from the field canal or the tertiary
canal into furrows by means of siphons placed over the side
of the ditch or canal bank and be allowed to flow
Figure 11
An example of a furrow irrigation system using
siphons (Source: Kay, 1986)
downstream along the furrow (Figure 11). The water level
in the canal must be raised to a sufficient height above the
level of the furrows by using a piece of wood, check plates,
or canvas filled with sand. This creates a head difference
between the water level in the field ditch and the furrow,
which is necessary for the water flow. Water can also be
diverted into furrows through gated pipes or hoses
connected to a hydrant fitted on buried pipes.
The water is gradually absorbed into the soil and spreads
laterally to wet the area between the furrows. The amount
of water that infiltrates the soil at any point along the
furrow depends on the soil type and the period during
which the water is in contact with the soil at that particular
point. This is known as the contact time or the intake
opportunity time (see Chapter 1). With furrow irrigation,
water is mainly lost by deep percolation at the head end of
furrow and runoff at the tail end. Furrows can be used on
most soil types, although coarse sands are not
recommended since percolation losses, especially at the top
end, would be high because of high infiltration rates. Soils
that crust easily are especially suited for furrow irrigation,
since the water does not flow over the ridge, which means
that the soil in which the plants grow remains friable.
Furrow design is an iterative process that should consider
the shape of the furrow, the spacing between furrows, with
the furrow length determined, amongst other factors, by
the stream size to apply and its application time, the soil
type and the slope.
As mentioned by Michael (1994), rational procedures for
predicting the water front advance and tail water recession
in furrows, which are applicable to field designs, have not
been developed. Various workers have proposed a number
of quasi-rational procedures with varying degrees of
adaptability. In the absence of more precise information on
predicting the water advance and recession in furrows,
general principles regarding stream size, furrow length and
furrow slope to obtain efficient irrigation are followed in
field design.
2.1.1 Furrow shape
The furrows are generally V-shaped or U-shaped in crosssection and are 15-30 cm deep and 25-40 cm wide at the
top. The shape of the furrow depends on the soil type and
13
Irrigation manual
Figure 12
Furrow shape depending on soil type
a. Deep, narrow furrow on sandy soil
b. Wide, shallow furrow on clay soil
Figure 13
Soil moisture distribution on various soil types as a determinant of furrow spacing (Source: FAO, 1985)
14
Module 7: Surface irrigation systems: planning, design, operation and maintenance
the stream size. Soils with low infiltration rates have usually
shallow wide parabolic or U-shaped furrows to reduce
water velocity and to obtain a large wetted perimeter to
encourage infiltration. Sandy soils, on the other hand,
require more or less V-shaped furrows to reduce the wetted
perimeter through which water infiltrates (Figure 12).
Shallow furrows are suited to fields that are graded to
uniform slope. The furrows should have uniform depth and
shape along the whole length to prevent overtopping. The
furrow should be large enough to carry the stream flow
and, in general, the larger the stream size the larger the
furrow must be to carry the desired flow.
2.1.3. Furrow length
Under mechanized agriculture, furrows should preferably be
as long as possible in order to reduce labour requirements
and system costs. However, they also should be short enough
to retain a reasonable application efficiency and uniformity.
Application efficiency and uniformity normally increase as
the furrow length decreases. Thus, when labour is not a
constraint or inexpensive and/or the water supply is limited,
short furrows may be most suitable. This does increase the
number of field canals and overall cost of the system. For
proper design of the furrow length, the following factors have
to be taken into account:
Y
Soil type
Y
Stream size
Y
Irrigation depth
Y
Slope
2.1.2. Furrow spacing
Y
Field size and shape
The spacing between furrows depends on the water
movement in the soil, which is texture related, on the crop
agronomic requirements as well as on the type of equipment
used in the construction of furrows. In practice a
compromise often has to be reached between these factors.
Y
Cultivation practices
Shallow-rooted crops require shallow furrows. Young crops
with shallow rooting depth should be wetted with shallow
furrows such that mainly the ridge is wetted. The furrow
can be deepened with increased crop growth.
When water is applied to a furrow, it moves vertically under
the influence of gravity and laterally by capillarity. Clay soils
have more lateral movement of water than sandy soils
because of their small pores, which favour capillary action
(Figure 13). In this regard, larger spacing can be used in
heavier soils than in light soils.
In general, a spacing of 0.3 m and 0.6 m has been proposed
for coarse soils and fine soils respectively. For heavy clay
soils up to 1.2 m has been recommended.
It should also be realized that each crop has its own
optimum spacing and the ridges should be spaced
according to the agronomic recommendations. In addition,
the equipment available on the farm determines the furrow
spacing, as this is adjustable only within limits. However, in
all instances the furrow spacing adopted should ensure a
lateral spread of water between adjacent furrows that will
adequately wet the entire root zone of the plants.
When water and labour are scarce, or when large areas must
be irrigated quickly, alternate furrows may be irrigated. In
this case, the lateral spread of water is only partial, and crop
yields may be reduced. An economic analysis should be
carried out to determine whether the advantages of this
practice outweigh any possible limitations on yield.
Soil type
Furrow lengths should be shorter in sandy soils than in clayish
soils, since water infiltrates faster into light soils than into
heavy soils. If furrow lengths are too long in sandy soils, too
much water will be lost as deep percolation at the top end of
the furrow. On heavier soils, water infiltrates slowly and
therefore furrows should be longer to allow sufficient time for
the water to infiltrate the soil to the desired depth. In Eastern
and Southern Africa designers are often confronted with the
fact that smallholders occupy land with high soil variability,
both vertically and horizontally, necessitating field tests to
establish the appropriate furrow length.
Stream size
On similar soils, and of the same slope and irrigation depth,
furrows can be longer when a larger stream size is used for
irrigation. This is because water will be advancing rapidly
down the furrow. However, the stream size should not
exceed the maximum non-erosive stream size determined
in field trials. The following equation provides guidance in
selecting stream sizes for field trials.
Equation 1
Qmax =
K
So
Where:
Qmax
=
Maximum non-erosive stream size (l/min)
So
=
Furrow slope in the direction of flow (%)
K
=
Unit constant (= 40)
15
Irrigation manual
Furrow stream sizes are sometimes selected on the basis of
the one-quarter rule. This rule states that the time required
for water to advance through a furrow till the end should be
one quarter of the total irrigation time (contact time).
However, it should be noted that only field experience will
show when to actually close the inflow, because of all the
different factors involved (see Section 1.3.3).
would facilitate uniform water delivery throughout the field
and scheme. It is therefore advisable that for new
developments this principle be adhered to and discussed in
detail with the farmers during the planning consultations.
Unfortunately, in practice in most cases no or insufficient
effort is made to discuss this matter with smallholders.
Where an area had been cultivated by farmers under dryland
conditions prior to the installation of the irrigation scheme,
farmers are often even more reluctant to change the shape
and borders of their individual fields. As a result the original,
irregular shapes of the fields are maintained and variable
lengths of run are used. This results in a complex operation
of the system and water shortages, and at the same time water
is wasted, leading to low irrigation efficiencies and higher
operation costs. For the same reason the number of field
canals increases, resulting in high development costs. This
shows the importance of involving farmers from the planning
stage onwards, so that they themselves also will become
convinced of the advantages and necessity of regular shaped
fields of equal size and thus are willing to change the original
borders of their individual fields.
Irrigation depth
A larger irrigation depth requires more contact time for
water to infiltrate to the desired depth than a shallow
irrigation depth, as explained in Chapter 1. The irrigation
depth can be increased by making the furrow longer in
order to allow more time for the water to reach the end of
furrow, which increases the contact time. Care should be
taken, however, to avoid too high percolation losses at the
top end.
Slope
Furrows should be put on proper gradients that allow water
to flow along them and at the same time allow some water
to infiltrate into the soil. Furrows put on steeper slopes can
be longer because water moves more rapidly. However, with
slopes steeper than 0.5% (0.5 m drop per 100 m length),
the stream sizes should normally be reduced to avoid
erosion, thus shorter furrows have to be used. Under
smallholder conditions the maximum slope of 0.5% should
not be exceeded (James, 1988).
Cultivation practices
When cultivation practices are mechanized, furrows should
be made as long as possible to facilitate working with
machinery. Short furrows also require a lot of labour, as the
flow must be changed frequently from one furrow to the
next.
Field size and shape
Guidelines for the determination of furrow lengths
Field size and shape provides challenges to designers. For the
system to operate at the level of efficiency earmarked by the
designer, the field shape of each farmer’s plot should be
regular and the length of run uniform for all farmers. This
Table 7 summarizes the main factors affecting the furrow
length and the suggested practical allowable furrow lengths
according to Kay (1986). The data given in this table are
appropriate for large-scale and fully mechanized conditions.
Table 7
Furrow lengths in metres as related to soil type, slope, stream size and irrigation depth (Source: Kay, 1986)
Soil type
Clay
Loam
Sand
Average irrigation depth (mm)
Furrow
slope
%`
Maximum
stream size
(l/sec)
75
150
50
100
150
50
0.05
3.0
300
400
120
270
400
60
90
150
0.10
3.0
340
440
180
340
440
90
120
190
0.20
2.5
370
470
220
370
470
120
190
250
0.30
2.0
400
500
280
400
500
150
220
280
0.50
1.2
400
500
280
370
470
120
190
250
1.00
0.6
280
400
250
300
370
90
150
220
1.50
0.5
250
340
220
280
340
80
120
190
2.00
0.3
220
270
180
250
300
60
90
150
16
75
100
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Table 8
Practical values of maximum furrow lengths in metres depending on soil type, slope, stream size and irrigation
depth for small-scale irrigation (Source: FAO, 1988)
Soil type
Clay
Loam
Sand
Net irrigation requirements (mm)
Furrow
slope
%`
Maximum stream
size per furrow
(l/sec)
50
75
50
75
50
75
0.0
3.0
100
150
60
90
30
45
0.1
3.0
120
170
90
125
45
60
0.2
2.5
130
180
110
150
60
95
0.3
2.0
150
200
130
170
75
110
0.5
1.2
150
200
130
170
75
110
Table 8 provides more realistic data for smallholder
irrigation.
The soil variability in most smallholders’ schemes,
combined with the small size of holdings, makes the
scheme more manageable when shorter furrows are used.
The figures in both tables should only be used as a guide in
situations where it is not possible to carry out field tests. As
much as practically possible, the furrow lengths should be
determined in the field based on the tests described in
Chapter 1.
2.2. Borderstrip irrigation
Borderstrips are strips of land with a downward slope but
are as horizontal as possible in cross-section (Figures 14
and 15). A horizontal cross-section facilitates an even rate
of water advance down the longitudinal slope. Borderstrips
can vary from 3-30 m in width and from 60-800 m in
length. They are separated by parallel dykes or border
ridges (levées).
Normally water is let onto the borderstrip from the canal
through intakes, which can be constructed with gates on
the wall of the canal or, when unlined canals are used, by
temporarily making an opening in the canal wall. The latter
is not recommended since it weakens the walls of the canal,
leading to easy breakage. Other means used for the same
purpose is the insertion of short PVC pipes into the canal
through the wall. The short pipes are usually equipped with
an end cup, which is removed when irrigation is practiced.
Some farmers use cloth or plastic sheet to close and open
the pipe. The most appropriate method of supplying water
from the canal to the field, however, is the use of siphons.
Figure 14
Example of a borderstrip irrigation system (Source: Kay, 1986)
17
Irrigation manual
Figure 15
Cross-section of a borderstrip
Leakage should be expected from all other techniques and,
as a consequence, waterlogging at the beginning of the
borderstrips. The use of siphons instead of one water inlet
also has the advantage that they can be spread over the
width of the borderstrip, thus facilitating an equal spread of
the water across the borderstrip width and then down the
slope towards the lower end. To ensure a proper lateral
spread, it is recommended that, longitudinally, the
borderstrips be horizontal for the first 10 m or so, with a
uniform downward slope thereafter. This irrigation method
is particularly suitable for pasture and close-growing crops
like wheat.
2.2.1. Borderstrip width
The borderstrip width depends on the topography of the
field, which determines the possible width that can be
obtained while keeping a horizontal cross-section without
requiring too much soil movement, and on the stream size.
The stream size also restricts strip width, as it should be
sufficient to allow complete lateral spreading throughout
the borderstrip width and length. The strip width also
depends on the cultivation practices, mechanized or nonmechanized for example. Borderstrips should not be wider
than 9 m on 1% cross-slopes (James, 1988).
on sandy soils ranges from 0.3-1% in humid areas and 1-2%
in arid areas on bare soils and soils with good crop cover
respectively. For clay soils the slope ranges from 0.5-2% in
humid areas and 2-5% in arid areas on bare soils and soils
with crop cover respectively (see also Module 1).
2.2.3. Borderstrip length
To determine the borderstrip length, the following factors
need to be considered:
Y
Soil type
Y
Stream size
Y
Irrigation depth
Y
Land slope
Y
Field size and shape
Y
Cultivation practices
Soil type
The borderstrip length can be longer on heavy soil than on
light soil because water infiltrates heavy soil more slowly. If
a border is made too long on a light soil, too much water
will infiltrate the soil at the top part, leading to too much
water loss due to deep percolation.
2.2.2. Longitudinal slope of the borderstrip
The slope should, wherever possible, be adapted to the
natural topography in order to reduce the need for land
grading, which may lead to the removal of too much
topsoil. On the one hand, slopes on sandy soils should be
greater than on heavy soils to avoid deep percolation losses
at the top of the field. Furthermore, the maximum slope
depends on the risk of soil erosion, which is greater in
sandy soils than in clayish soils. Crop cover can be a factor
that counts in determining the borderstrip slope. For
example, where the borderstrip will be covered by a
permanent crop, such as pasture, slopes can be up to 7%
for clay soils with stable aggregates (James 1988).
Borderstrips should have a minimum slope of 0.05-0.1% to
allow water to flow downstream over it. The maximum slope
18
Stream size
When a larger stream size is available, borderstrips can be
longer on the same soils, because water will spread more
rapidly across the soil surface. As a rule of thumb, the
stream size must be large enough to adequately spread
water across the width of border. However, it should not
exceed the maximum non-erosive stream size. The design
of the stream size must also result in rates of advance and
recession that are essentially equal (see Section 1.3.4).
Irrigation depth
A larger irrigation depth requires more contact time for
water to infiltrate to the desired depth than does a shallow
irrigation depth, as explained in Chapter 1. The irrigation
Module 7: Surface irrigation systems: planning, design, operation and maintenance
depth can be increased by making the borderstrip longer in
order to allow more time for the water to reach the end of
borderstrip, which increases the contact time. However,
soil type is a limiting factor.
farming. The width should preferably be a multiple of the
farm machinery used.
Slope
Table 9 provides some typical borderstrip widths and
lengths for various soil types, slopes, irrigation depths and
flows. It should be noted that in practice borderstrip
lengths are often shorter because of poor levelling.
Sometimes they have to be reduced after construction, if
the irrigation efficiencies turn out to be too low. It should
be noted that the figures for width and length in Table 9
apply to highly-mechanized agriculture on properly levelled
lands under good water management.
Borderstrips should be put on proper gradients that allow
water to flow downstream over the surface yet at the same
time to allow some water to infiltrate the soil. The
borderstrip can be longer on steeper slopes, since water
moves more rapidly. However, precautions should be taken
against erosion.
Field size and shape
Existing field size and shape are often practical limits to the
size of borders. However, the same remarks as mentioned
under furrow irrigation (see Section 2.1.3) are valid for
borderstrip irrigation.
Cultivation practices
2.2.4. Guidelines for the determination of
borderstrip width and length
The figures in Table 9 may not apply to small-scale
irrigators on small landholdings. Table 10 provides some
guidelines for determining borderstrip dimensions felt
suitable for smallholder irrigation in communal areas,
where farmers are responsible for the operation and
maintenance of these schemes.
Because borders are normally long so as to achieve good
water distribution, they are very suitable for mechanized
Table 9
Typical borderstrip dimensions in metres as related to soil type, slope, irrigation depth and stream size (Source:
Withers and Vipond, 1974)
Soil Type
Coarse
Medium
Fine
Slope
(%)
Depth applied
(mm)
Flow
(l/sec)
Strip width
(m)
Strip length
(m)
0.25
50
100
150
240
210
180
15
15
15
150
250
400
1.00
50
100
150
80
70
70
12
12
12
100
150
250
2.00
50
100
150
35
30
30
10
10
10
60
100
200
0.25
50
100
150
210
180
100
15
15
15
250
400
400
1.00
50
100
150
70
70
70
12
12
12
150
300
400
2.00
50
100
150
30
30
30
10
10
10
100
200
300
0.25
50
100
150
120
70
40
15
15
15
400
400
400
1.00
50
100
150
70
35
20
12
12
12
400
400
400
2.00
50
100
150
30
30
20
10
10
10
320
400
400
19
Irrigation manual
Table 10
Suggested maximum borderstrip widths and lengths for smallholder irrigation schemes
*
Soil
type
Borderstrip
slope
(%)
Unit flow per
metre width*
(l/sec)
Borderstrip
width
(m)
Borderstrip
length
(m)
Sand
(Infiltration rate greater than 25 mm/h)
0.2-0.4
0.4-0.6
0.6-1.0
10-15
8-10
5-8
12-30
9-12
6-9
60-90
80-90
75
Loam
(Infiltration rate of 10 to 25 mm/h)
0.2-0.4
0.4-0.6
0.6-1.0
5-7
4.6
2-4
12-30
9-12
6
90-250
90-180
90
Clay
(Infiltration rate less than 10 mm/h)
0.2-0.4
0.4-0.6
0.6-1.0
3-4
2-3
1-2
12-30
6-12
6
180-300
90-180
90
The flow is given per metre width of the border. The total flow into a border is equal to the unit flow multiplied by the border width (in metres)
However, in reality it seems that the widths of the
borderstrips in smallholder schemes are still less than the
figures given in Table 10. As an example, a typical
borderstrip in a smallholder irrigation scheme in
Zimbabwe can have a width varying between 2-4 m, which
is even less than half of the smallest width given in Table 10.
As much as practically possible, borderstrip lengths should
be determined in the field based on the tests described in
Chapter 1. However, the method is best suited to projects
at the planning stage. For projects that have passed through
this stage, and where the length and slope of the borderstrip
have been fixed by field shape and land topography during
the planning stage rather than by testing, a high uniformity
of water distribution can only be achieved by adjusting the
stream size and the time to stop inflow, on the condition
that the stream size is non-erosive. Such an arrangement,
however, makes the operation of the scheme very complex
for management by smallholders with limited or no past
experience with surface irrigation.
the contact time, so as to reduce difference in contact time
on the different sections of the basin. It may be used on a
wide variety of soil textures, though fine-textured soils are
preferred. As the area near the water inlet is always longer
in contact with the water, there will be some percolation
losses, assuming the entire root zone depth is filled at the
bottom of the field. Coarse sands are not generally
recommended for basin irrigation as high percolation losses
are expected at the areas close to water intake.
2.3.1. Basin size
The size of basin is critical in the design of this irrigation
method and, as for furrow and borderstrip irrigation,
depends on the following factors:
Y
Soil type
Y
Stream size
Y
Irrigation depth
Y
Field size and shape
2.3. Basin irrigation
Y
Land slope
A basin is a horizontal area of land surrounded by earthen
bunds and totally flooded during irrigation. Basin irrigation
is the most common type of surface irrigation. It is
particularly used in rice cultivation, where the fields are
submerged, but it is equally suitable for other crops like
cereals, fruit trees and pastures – as long as waterlogging
conditions do not last for too long. Ideally, the waterlogging
should not last longer than 24-48 hours. It is also used for
the leaching of salts by deep percolation in the reclamation
of saline soils. A basin irrigation system layout is illustrated
in Figure 16.
Y
Farming practices
Flooding should be done using a large stream size that
advances quickly in order for water to spread rapidly over
the basin. The advance time should not exceed a quarter of
20
Table 11 shows in summary the general criteria for selecting
a basin size, all of which will be discussed more in detail
below.
Table 11
Criteria for basin size determination
Criteria
Basin size small
Basin size large
Soil type
Sandy
Clay
Stream size
Small
Large
Irrigation depth
Small
Large
Land slope
Steep
Gentle or flat
Field preparation
Hand or animal
traction
Mechanized
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 16
Layout of basin irrigation (Source: FAO, 1985)
Direct method of water supply to the basins with a drainway midway between supply canals. “Basin a” is irrigated, then
“Basin b”, and so on.
Cascade method of water supply to the basins with a tier arrangement. Ideal on terraced land, where water is supplied to
the highest terrace, and then allowed to flow to a lower terrance and so on.
Soil type
Stream size
Water infiltrates heavier soils more slowly than lighter soils.
This means that there is more time for water to spread over
the soil surface on heavy soils than on light soils before
infiltrating. Therefore basins can be larger on heavier soils
than on lighter soils.
Basin size can increase with larger stream size, because
water will spread more rapidly over the basin. Table 12
below gives some guidelines on basin sizes in relation to
stream size and soil type.
Table 12
Basin area in m2 for different stream sizes and soil types (Withers and Vipond, 1974)
Stream size (l/sec)
Sand
Sandy loam
Clay loam
Clay
5
35
100
200
350
10
65
200
400
650
15
100
300
600
1 000
30
200
600
1 200
2 000
60
400
1 200
2 400
4 000
90
600
1 800
3 600
6 000
21
Irrigation manual
Irrigation depth
Cultivation practices
A deeper irrigation depth requires more contact time for
water to infiltrate to the desired depth than does a shallow
irrigation depth, as explained in Chapter 1. The irrigation
depth can be increased by increasing the size of the basin in
order to allow more time for the water to reach the end of
basin. This increases the contact time. However, soil type is
a limiting factor. If the soil is too light then too much water
will infiltrate and be lost by deep percolation next to the
water inlet and too little will infiltrate furthest away from
the water inlet, hence the unsuitability of basins for very
light soils. It would be better to wet the whole basin as soon
as possible and to leave the water ponding until the desired
volume of water has been applied.
Mechanized farming requires relatively large basins in order
to allow machines to turn round easily and to have long
runs without too many turns. In this case, the dimensions
of basins should be a multiple of the machine width. On
small farms, in general 1-2 ha or less in developing
countries, farming operations are done by hand or animal
power. Small basins are used on such lands. The basins are
often levelled by hand.
Land slope
The soil surface within each basin should be horizontal.
Basins can be as large as the stream size and soil type can
allow on level land. On steep slopes, the removal of the
topsoil and the associated land levelling costs may be limiting
factors for the basin size. Typically, terrace width varies from
about 2 m for 4% land slopes up to 150 m for 0.1 % land
slopes. Table 13 below provides some guidelines on the
possible width of a basin, in relation to the land slope.
Table 13
The types of crops grown may also influence the size of
basins. For example, small basins could be used per single
tree for orchards.
2.4. Efficiencies of surface irrigation
systems and of the different surface
irrigation methods
Surface irrigation schemes are designed and operated to
satisfy the irrigation water requirements of each field
while controlling deep percolation, runoff, evaporation
and operational losses. The performance of the system is
determined by the efficiency with which water is
conveyed to the scheme from the headworks, distributed
within the scheme and applied to the field, and by the
adequacy and uniformity of application in each field (see
also Module 1).
Approximate values for the maximum basin width (m)
Slope
(%)
Maximum width (m)
Average
Range
0.2
45
35-55
0.3
37
30-45
0.4
32
25-40
0.5
28
20-35
0.6
25
20-30
0.8
22
15-30
1.0
20
15-25
1.2
17
10-20
1.5
13
10-20
2.0
10
5-15
3.0
7
5-10
4.0
5
3-8
2.4.1. The different types of efficiencies in an
irrigation scheme
Conveyance efficiency (Ec)
Conveyance efficiency is the ratio of water received at the
inlet to a block of fields or a night storage reservoir to the
water released from the project headworks. Factors
affecting this efficiency include canal lining, evaporation of
water from the canal, technical and managerial facilities of
water control, etc. Conveyance efficiency is higher when
water is conveyed in a closed conduit than when it is
conveyed in an open one, since water in the latter is very
much exposed to evaporation as well as to ‘poaching’ by
people and to livestock watering.
Field size and shape
Field canal efficiency (Eb)
Basins are best adapted to regular field shapes (square or
rectangular). Irregular field shapes necessitate adapting
basins. This may lead to basin sizes being different from
what would be recommended based on other factors such
as soil types, etc. Although a regular shape is favourable,
basins can be shaped to follow contours. These are contour
basins or terraces, which are seen mainly on steep slopes
used for rice.
This is the ratio of water received at the field inlet to the
water received at the inlet to a block of fields or a night
storage reservoir. Among other factors, this efficiency is
affected by the types of lining in respect to seepage losses,
by the length of canals and by water management. Piped
systems have higher field canal efficiencies than do open
canal systems for reasons explained earlier.
22
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Field application efficiency (Ea)
This is the ratio of water directly available to the crop to
water received at the field inlet. It is affected, for example,
by the rate of supply, infiltration rate of soil, storage capacity
of the root zone, land levelling, etc. For furrow and
borderstrip irrigation, water is mostly lost through deep
percolation at the head end and through runoff at the tail
end, while for basin irrigation it is mostly through deep
percolation and evaporation, since the basin is closed.
managed. The value ranges from 50-70%. Losses will occur
through deep percolation at the top end of the field and
runoff at the bottom end. Properly designed and managed
borderstrips can reach a field application efficiency of up to
75%, although a more common figure is 60%. With basin
irrigation it is possible to achieve field application
efficiencies of 80% on properly designed and managed
basins, although a more common figure used for planning
varies between 60-65%. For more details on efficiencies
the reader is referred to Module 1.
Distribution system efficiency (Ed)
Example 3
Conveyance efficiency Ec and field canal efficiency Eb are
sometimes combined and called distribution system
efficiency Ed, expressed as:
At Nabusenga surface irrigation schemes, it is
assumed that Ec and Eb both are 90% for concretelined canals continuous flow and that Ea is 50%.
What is the overall irrigation efficiency?
Equation 2
Ep = Ec x Eb x Ea = 0.90 x 0.90 x 0.50 = 0.41 or 41%
Ed = Ec x Ef
Farm irrigation efficiency (Ef)
Field canal efficiency Eb and field application efficiency Ea
are sometimes combined and called farm irrigation
efficiency Ef, expressed as:
Equation 3
Ef = Eb x Ea
Overall irrigation efficiency (Ep)
The overall or project irrigation efficiency of an irrigation
scheme is the ratio of water made available to the crop to
that released at the headwork. It is the product of three
efficiencies, namely conveyance efficiency (Ec), field canal
efficiency (Eb) and field application efficiency (Ea), and is
expressed as:
Equation 4
Ep = Ec x Eb x Ea or:
Ep = Ed x Ea
In order to show the importance of contribution of Ea to
the overall irrigation efficiency while keeping the same Ec
and Eb but increasing the field application efficiency Ea
from 50% to 70%, an overall irrigation efficiency of 0.57
or 57% (0.90 x 0.90 x 0.70) can be achieved instead of
41%.
There are common problems that can reduce the field
application efficiency of the three surface irrigation
methods to a considerable extent. The most common ones
are discussed below:
Y
Poor land levelling can lead to waterlogging in some
places and inadequate water application in others. If
the cross slope is not horizontal for borderstrip
irrigation, water will flow to the lowest side causing
over-irrigation in that area.
Y
Different soil types along the furrows, borderstrips and
basins result in different infiltration rates.
Y
Too small an advance stream results in too long an
advance time, leading to over-irrigation at the top end
of the borderstrip and furrow. A small stream size
diverted into a basin will take too long to cover the
entire basin area, resulting in a contact time that is very
different at the various sections of the basin.
Y
Too large a stream size will result in water flowing too
fast down the borderstrip and furrow leading to a cutoff taking place before the root zone has been filled
with water. If the flow is allowed to continue under
these conditions there will be excessive runoff at the
end. A large stream size, on the other hand, can be
desirable for basins as this reduces the difference in
contact time on the various sections of the basin.
or:
Ep = Ec x Ef
The field application efficiency (Ea) is the one that
contributes most to the overall irrigation efficiency and is
quite specific to the irrigation method, as discussed below.
Any efforts that are made to improve on this efficiency will
impact heavily on the overall efficiency.
2.4.2. Efficiencies of the different surface irrigation
methods
Furrow irrigation could reach a field application efficiency
of 65% when it is properly designed, constructed and
23
Irrigation manual
Y
The use of irregular-shaped plots with variable lengths
of run complicates the operation of the system,
resulting in poor efficiencies.
most appropriate method of irrigation. Field experience
has shown that large application depths can be applied most
efficiently with basin irrigation.
As a rule of thumb, overall irrigation efficiencies used in
Zimbabwe for design purposes are in the range of 40-50%
when using concrete-lined canals and 55-65% for piped
systems.
In general, the gross irrigation depth is much larger than the
net irrigation depth for all three surface irrigation methods,
due to the lower irrigation efficiencies of surface irrigation
compared to pressurized systems (see Module 1). However,
of the three surface irrigation methods, basin irrigation can
have higher irrigation efficiency and use less water for the
same crop on the same soil than the other methods as water
is confined within bunds (see Section 2.4.2).
2.5. Criteria for the selection of the surface
irrigation method
Though it is not possible to give specific guidelines as to
which surface irrigation method to select under a given set
of conditions, each option usually has advantages and
disadvantages. The selection of a surface irrigation method
depends mainly on soil type, crops to be irrigated, the
irrigation depth, land slope, field shape, labour availability
and water source. Table 14 gives some guidelines on which
method would be most appropriate depending on soil type,
crop rooting depth and net depth of application.
2.5.1. Soil type
All three surface irrigation methods prefer heavy soils,
which have lower infiltration rates. A light soil with high
infiltration rates favours deep percolation losses at the top
of the fields, resulting in low field application efficiency.
2.5.4. Land slope
In general terms, all surface irrigation methods favour flat
land as steep slopes would necessitate excessive land
levelling in order to avoid erosion, which is expensive and
can lead to the removal of top soil.
Flat land with a slope of 0.1% or less is best suited for basin
irrigation (which needs a zero slope) since it requires
minimum land levelling. Borderstrip irrigation may be used
on steeper land, even up to 5%, depending upon other
limiting factors such as soil type. One has to be cautious
with furrow irrigation on such steep slopes. This is because
the flow is confined to a small channel (the furrow), which
could result in erosion.
2.5.5. Field shape
2.5.2. Type of crop
Furrow irrigation is particularly suitable for irrigating row
crops such as maize and vegetables. Furrows are also more
suitable for shallow-rooted crops. Borderstrip irrigation can
also be used for row crops or for close-growing crops that do
not favour water ponding for long durations, such as wheat
and alfalfa. Any crop, whether row or close-growing, that can
stand a very wet soil for up to 24 hours is best grown in basins.
2.5.3. Required depth of irrigation application
If the application depth is small, furrow irrigation is the
In general, furrow irrigation requires regular field shapes,
allowing the use of the same stream size for the same
furrow lengths. However, for all three types regularly
shaped fields are preferable, as explained in Sections 2.1,
2.2 and 2.3.
2.5.6. Labour availability
Basin irrigation requires less labour than the other two
methods and might have to be considered if there is a
critical labour shortage.
Table 14
Selection of an irrigation method based on soil type and net irrigation depth (Source: Jensen, 1983)
Soil type
24
Rooting depth
of crop
Net irrigation depth
per application (mm)
Surface irrigation method
Sand
Shallow
Medium
Deep
20-30
30-40
40-50
Short furrows
Medium furrows, short borders
Long furrows, medium borders, small basins
Loam
Shallow
Medium
Deep
30-40
40-50
50-60
Medium furrows, short borders
Long furrows, medium borders, small basins
Long borders, medium basins
Clay
Shallow
Medium
Deep
40-50
50-60
60-70
Long furrows, medium borders, small basins
Long borders, medium basins
Large basins
Chapter 3
Design parameters for the infield works
In order to calculate the design flow of the irrigation
system, a number of parameters must be taken into
consideration, notably the available moisture, the root zone
depth, the allowable moisture depletion, the net peak water
requirements, the irrigation frequency and cycle and the
irrigation efficiencies. The following surface irrigation
schemes will be used to demonstrate the process of using
and calculating design parameters:
Y
Y
Nabusenga irrigation scheme, which is a surface
irrigation scheme in Matabeleland North Province in
Zimbabwe using a concrete-lined canal system (see
Section 4.2 and Figure 19)
Mangui irrigation scheme, which is an imaginary
surface irrigation scheme using a piped system up to
field level (see Section 4.3 and Figure 20)
Table 15 shows the given design parameters.
3.2. Net and gross depth of water
application
3.2.1. Net depth of water application (dnet)
The net depth of water application (dnet) is the amount of
water in millimetres that needs to be supplied to the soil in
order to bring it back to field capacity. It is the product of
the available soil moisture (FC-PWP), the effective root
zone depth (RZD) and the allowable moisture depletion
(P), and is calculated as follows:
Equation 5
dnet = (FC - PWP) x RZD x P
Where:
dnet
=
Net depth of water application per
irrigation for the selected crop (mm)
FC
=
Soil moisture at field capacity mm/m)
PWP
=
Soil moisture at the permanent
wilting point (mm/m)
RZD
=
The depth of soil that the roots
exploit effectively (m)
P
=
The allowable portion of available
moisture permitted for depletion by
the crop before the next irrigation
3.1. Crop water and irrigation requirements
The calculation of the crop water and irrigation
requirements is discussed in detail in Module 4, to which
the reader is referred.
Table 15
Design parameters for Nabusenga and Mangui surface irrigation schemes
Parameter
Area
Soil type
Available soil moisture (= FC - PWP)
Nabusenga scheme
Mangui scheme
15 ha
2.4 ha
Clay loam
Sandy mixture
130 mm/m
80 mm/m
0.70 m for maize
0.75 m for maize
0.50
0.50
Assumed field application efficiency (Ea)
0.50
0.60
Assumed field canal/ efficiency (Eb)
0.90
1
Assumed conveyance efficiency (Ec)
0.90
1
0.60
Design root zone depth (RZD)
Allowable soil moisture depletion (P)
Farm irrigation efficiency (Ef = Eb x Ea)
0.45
Distribution system efficiency (Ed) (= Ec x Eb)
0.81
1
Overall irrigation efficiency (Ep) (= Ec x Eb x Ea)
0.41
0.60
6.0 mm/day
6.2 mm/day
Peak ETcrop
25
Irrigation manual
3.2.2. Gross depth of water application (dgross)
Equation 7
The gross depth of water per irrigation is obtained by
dividing the net depth of water (dnet) by efficiency, as in
Equation 6:
IF =
Equation 6
IF
=
Irrigation frequency (days)
dnet
=
Net depth of water application (mm)
ETcrop
=
Crop evapotranspiration (mm/day)
dgross =
dnet
ETcrop
Where:
dnet
E
Example 4
Based on the design parameters given in Table 14,
what are the net and gross depths of water
application for Nabusenga irrigation project?
It should be mentioned that for design purposes we are
particularly interested in the peak daily amount of water
used by the crop, which is the worst case scenario. The net
peak daily irrigation requirement (IRn) is determined by
subtracting the rainfall (if any) from the peak daily crop
water requirements.
dnet = 130 mm/m x 0.70 m x 0.50 = 45.5 mm
3.3.2. Irrigation cycle (IC)
The gross depths of water application at field and at
overall level would be:
dgross =
dgross =
45.5
0.50
45.5
0.41
= 91.0 mm at field level
= 111.0 mm at overall level
3.3. Irrigation frequency and irrigation cycle
3.3.1. Irrigation frequency (IF)
Irrigation frequency is the time it takes a crop to deplete
the soil moisture at a given depletion level and can be
calculated as follows:
The irrigation cycle is the time it takes to irrigate the entire
scheme. If, from an irrigation frequency of 7 days, for
example, we take the irrigation cycle to be 5 days, this
leaves us 2 days for other works and practices inside and
outside the scheme. The greater the difference between the
frequency and the cycle, the greater the flexibility to deal
with unforeseen situations such as breakdowns. Besides
this, it allows for the eventual expansion of the scheme,
utilizing the same conveyance and distribution system, once
water from the dam or river is found to be in surplus. On
the other hand, the greater the difference the more
expensive the scheme becomes, as the designer has to go
for larger water conveyance and distribution systems. As a
rule, the difference between the irrigation frequency and
Example 5
What is the irrigation frequency and what can be the irrigation cycle for Nabusenga scheme?
The irrigation frequency is equal to:
IF =
45.5
6.0
= 7.5 days
The system should be designed to provide 45.5 mm every 7.5 days. For practical purposes, fractions of days are not
used for irrigation frequency purposes. Hence, the irrigation frequency in our example should be 7 days, with a
corresponding dnet of:
dnet = 7 x 6.0 = 42 mm for an IF of 7 days
The dgross at field and overall level will be
42
0.50
= 84.0 mm and
42
0.41
= 102.4 mm respectively.
The adjusted allowable depletion for the 7 days (instead of 7.5 days) is equal to:
P =
7 x 6.0
130 x 0.7
= 0.46 or 46%
Based on the above, the Irrigation Cycle (IC) is fixed at 6 days.
26
Module 7: Surface irrigation systems: planning, design, operation and maintenance
the irrigation cycle should not exceed one day. This is
considered as a compromise between convenience and
cost.
3.4. System capacity (Q)
System capacity refers to the discharge that has to be
abstracted from the headwork during a given period per day
and it is used for the design of the headwork and the
conveyance system. It is determined by the following
equation:
Where:
V
=
Volume of water abstracted per day
(m3)
A
=
Area irrigated daily (ha)
dgross
=
Gross depth of application at overall
scheme level (mm)
10
=
Conversion factor to convert mm to
m3/ha
The area irrigated per day can be calculated as follows:
Equation 10
Equation 8
A =
Q =
V
At
IC
Where:
T
Where:
Q
= Discharge (m3/hr or l/sec)
V
= Volume of water to be abstracted per day
(m3 or l)
T
= Irrigation duration per day (hr or sec)
The volume of water to be abstracted per day is obtained as
follows:
A
= Area irrigated per day (ha)
At
= Total area (ha)
IC
= Irrigation cycle (days)
A summary of the calculated design parameters for
Nabusenga is given in Table 16. The design parameters for
Mangui should be calculated in the same way. The result is
also summarized in Table 16.
Equation 9
V = 10 x A x dgross
Example 6
What should be the system capacity for Nabusenga scheme, considering an irrigation duration of 10 hours per day?
The area irrigated per day is equal to:
A =
15
6
= 2.5 ha
The volume of water to be abstracted per day is equal to:
V = 10 x 2.5 x 102.4 = 2 560 m3/day
The system capacity, assuming 10 hours of irrigation per day, will be equal to:
Q =
2 560
10
= 256 m3/hr or 71.1 l/sec
If, however, this results in large conveyance dimensions, a night storage reservoir could be introduced so that
abstraction from the headworks could be continuous (24 hours/day) at peak demand. In such a case, the conveyance
system capacity would be 71.1 x (10/24) = 29.6 l/sec.
27
Irrigation manual
Table 16
Summary of the calculated design parameters for Nabusenga and Mangui surface irrigation schemes
Design parameter
Symbol
Nabusenga
Mangui
1.
Net depth of water application
dnet
45.5 mm
30.0 mm
2.
Calculated irrigation frequency
IF
7.5 days
4.8 days
3.
Adjusted irrigation frequency
IF
7 days
5 days
4.
Adjusted net depth of water application
dnet
42.0 mm
31.0 mm
5.
Adjusted allowable depletion
P
46%
52%
6.
Proposed irrigation cycle
IC
6 days
4 days
7.
Gross depth of water application (overall level)
dgross
102.4 mm
51.7 mm
8.
Gross depth of water application (field level)
dgross
84.0 mm
51.7 mm
9.
Area to be irrigated per day
A
2.5 ha
0.6 ha
10. Volume of water to be abstracted per day, assuming an
irrigation duration of 10 hours per day
V
2 560 m3/day
310.2 m3/day
11. System capacity, assuming an irrigation duration of 10 hours
day
Q
256.0 m3/hr
or 71.1 l/sec
31.02 m3/hr
or 8.62 l/sec
12. System capacity for 24 hours conveyance and night storage
Q
106.6 m3/hr
or 29.6 l/sec
By incorporating a night storage reservoir in the design of
Nabusenga scheme, the system capacity has been reduced
to less than half, thus allowing a smaller size conveyance
28
system to be used. The design of night storage reservoirs is
discussed in detail in Chapter 6.
Chapter 4
Layout of a surface irrigation scheme
Design of a surface irrigation system may be required for
either a planned new irrigation scheme or an existing
irrigation scheme where low performance requires
improvement by redesigning the system. In both cases, the
data required fall into six categories:
Y
The water resources to be used, including source of
water, flow rates and water quality
Y
The topography of the land surface
Y
The physical and chemical characteristics of the soil,
including infiltration rates, soil moisture holding
capacities, salinity
Y
The expected cropping pattern
Y
The economic and marketing situation in the area and
the availability of services, including the availability of
labour, maintenance and replacement services, energy,
availability of capital for the work
Y
The farming practices of the overall farming enterprise
Each surface irrigation and drainage system layout depends
on the local situation. In this chapter, some general rules
only will be discussed.
4.1. General layout
The main factors determining a surface irrigation scheme
layout are:
Y
Topography
Y
Farm size and degree of mechanization
Y
Possible lengths of furrows and borderstrips or possible
basin sizes
Two distinct irrigation layouts can be adopted in the design
of surface irrigation systems. The layout to choose is very
much dependent on the topography of the land amongst
other considerations.
The first layout is adaptable to flat lands with slopes of less
than 0.4%. In this layout, main canals/pipelines follow the
contour lines as much as possible. Secondary canals/pipelines
run perpendicular to the contour lines. If, in the case of
canals, the ground slope is steep in relation to the required
canal gradient, drop structures are incorporated into the
design in order to reduce the water velocity (see Section 6.5).
The tertiary canals/pipelines, which get their water from the
secondary canals, preferably run more or less parallel to the
contour lines while the furrows or borderstrips run in the
same direction as the secondary canals or perpendicular to the
contour lines. The distance between the field canals depends
on the design length of the furrows or borderstrips or on the
size of the basins. Where the natural slope is too steep, land
levelling should be carried out to reduce the slope. Furrows
or borderstrips could run at an angle from the tertiary canal
in order to reduce the longitudinal slope of the furrow or
borderstrip. Basins would require levelling to make them
horizontal. An example of this type of layout is given in
Figure 17.
The second layout is adaptable to lands with a gentle to steep
topography. In this layout, the tertiary canals/pipelines, which
get their water from the secondary ones, are constructed in
such a manner that they cross elevation contours almost at
right angles. Furrows and borders are then constructed along
the contours but slightly running away from them to create
some gradient for water flow.
Most land topography allows irrigation layouts with field
canals/pipelines irrigating only fields located on one side of
the canal/pipeline. Sometimes, however, the topography
might allow the field canals/pipelines to effectively irrigate
the fields located on both sides of the canal/pipeline. Where
this occurs, it is called the herringbone layout (Figure 18).
It is most adaptable to even slopes with contour lines
running more or less parallel to each other. The tertiary
canal/pipeline would then run perpendicular to the
contours and irrigate both sides.
The herringbone layout reduces costs for infield
developments, as fewer canals/pipelines need to be
constructed. However, the layout poses challenges to the
designer and the irrigator alike. The available command on
the two sides of the canal could be different, making it difficult
to apply correct volumes of water to each side. Precautions
should be taken to ensure that both sides are adequately under
command along the entire length of canal/pipeline.
For both layouts described above the angle between the field
canal and the furrow should not be too acute, otherwise a
very dense and expensive canal system has to be designed.
This also leaves large triangular pieces of land at the head of
tertiary canal/pipeline without command.
29
Irrigation manual
Figure 17
Typical layout of a surface irrigation scheme on uniform flat topography
30
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 18
The herringbone irrigation layout
Access to and within the field in the form of roads is
important and should be incorporated in the layout. The
access road to the scheme is very important for the delivery
of inputs as well the marketing of the produce. Roads within
the scheme facilitate access for carrying out adequate
maintenance works and the transportation of inputs and
produce to and from individual farmer plots. It is always
desirable to have a perimeter road surrounding the irrigation
project as well as a road next to higher order canals and
drains. Figures 17 and 19 show the roads network.
4.2. Nabusenga irrigation scheme layout
At Nabusenga, the farmers opted for furrow irrigation. In
order to keep the main supply or conveyance system from
the water source to the scheme as small as possible, a night
storage reservoir (NSR) was incorporated into the layout.
The reservoir had to be constructed at the highest point,
i.e. southwest of the small stream (Figure 19).
Nabusenga is located on a relatively flat area. Initially the
main canal, taking the water from the reservoir, can run
parallel to the contours (section a in Figure 19). As the area
served by the canals is relatively small, no distinction was
made between main and secondary canals. The canal starts
at the highest point, 92.30 m, and a short section runs in a
the northern direction, while another section will run in
southwest direction. To avoid this latter section running
into the ridge of high ground, it bends westward. The
direction of irrigation of section a is from south to west.
From the canal in section b no irrigation takes place,
therefore most of it could be in cut. The canal in section c
will run parallel to contour line 91.50 m.
As the contour lines change direction from contour line
91.00 m downwards, the canal also changes direction to
run parallel to that contour line. Water will be siphoned
from the canal in section d, thus there should be sufficient
command. From the end of section d, the canal bends in a
southwest direction and will run more or less parallel to the
river. From here, the highest point of each traverse is
located near the river.
Section f could irrigate from the canal immediately adjacent
to the drain of section d, but it can also be irrigated from
northwest to southeast direction. This latter option is
selected in order to save on canals. The furrow length
should be 125 m, so as to use as much land as possible.
Section g has a slightly steeper slope from northwest to
southeast than from northeast to southwest. As the slope is
close to 0.5 %, canal g will irrigate away from the river from
northwest to southeast. The anthill within this plot will be
destroyed using a bulldozer. A 90° bend is proposed
between sections i and j. The last three blocks, h, i and j, can
be very uniform. The topography is gently sloping and 3
blocks with furrow lengths of 100 m each can be designed
from northeast to southwest. Drains are planned parallel to
the field canals. The furrow lengths of 100 m fit in very well
with the available land.
On the whole furrow lengths were kept in the range of 75125 m. While the advance and recession test indicated an
ideal length of 100 m (see Section 1.3.3), the irregular
shape of the land combined with the topography allowed
only about 60% of the area to fully comply with the 100 m
length of run.
31
LEGEND
Irrigation canal
Drainage channel
Plot boundary
Fence and gate
Furrow direction and slope
Contour line
Road
lV
(C) DROP STRUCTURE;See drawing NABU/15
(B) SADDLE BRIDGE: See drawing NABU/12
(A) DIVERSION STRUCTURE: See drawing NABU/11
Road
Anthill
lll
a
l
ve
r
Ri
bu
Na
32
se
ng
a
0.25 ha
Irrigation manual
Figure 19
Layout of Nabusenga surface irrigation scheme
Road
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 20
Layout of Mangui piped surface irrigation scheme
33
Irrigation manual
4.3. Mangui irrigation scheme layout
Mangui irrigation scheme is located next to a river from
where water will be pumped to the scheme (Figure 20).
The difference in elevation between the highest point in the
field and the water level in the river during the low flow is
5.13 m.
Soil sampling and analysis has shown that the soil is light
with available moisture of 80 mm/m by volume. The
surveying team did not have access to an infiltrometer,
hence no infiltration data were available to the designer.
Similarly the design team could not have access to a pump
to run an advance and recession test. Therefore the only
data available for the designs were the topographic map, the
available moisture, climatic data from the nearest
meteorological station and the expressed wishes of the
farmers to grow specific crops among which maize was
prominent.
In order to proceed with the designs, the designers used an
infiltration curve from a similar soil (Figure 21), the
estimates of furrow length from Table 8 and the
CROPWAT software to estimate the crop water
requirements. Table 14 and 15 provide the design
parameters used. Farmers have requested that the scheme
operate for not more than 10 hours per day.
Referring to the cumulative infiltration curve (Figure 21),
the contact time required for the application of 51.7 mm of
water is estimated to be around 20-25 minutes.
According to Table 8, for light soils the maximum furrow
length for a slope of 0.4% should be about 75 m and the
maximum stream flow 1.6 l/sec. Looking at the general
slope of the land (Figure 20), it appears that it is not
possible to accommodate the 75 m length of furrow. It was
therefore decided to grade the land and provide a crest in
the middle, from east to west, allowing a furrow length of
45 m on each side. The furrow spacing is based on the row
spacing for maize, which is 0.9 m.
The irrigation duration for each furrow can be calculated
using the following equation:
Figure 21
Cumulative infiltration curves for different soil types (Source: Jensen, 1983)
34
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Equation 11
If the flow is reduced to 1.2 l/sec after 6 minutes, the
time it takes to apply the remaining depth of 37.5 mm
will be:
dgross x A
IT =
q
IT =
Where:
IT
=
Irrigation duration (seconds)
dgross
=
Gross depth of irrigation (m)
A
=
Area covered by one furrow (m2)
q
=
Discharge into one furrow (m3/sec)
Using the above figures gives:
IT =
0.0517 x 0.9 x 45
= 1 309 seconds
0.0016
or 22 minutes
This is more or less the same as the contact time according
to Figure 21, when considering a sand mixture soil.
Applying the one-quarter rule this means that the water
should reach the end of the furrow in about 6 minutes.
In order to irrigate the total area of 2.4 ha in 4 days, 0.6 ha
per day have to be irrigated, which equals:
6 000
45 x 0.9
= 148 furrows
0.0375 x 0.9 x 45
0.0012
= 1 266 seconds
or 21 minutes
This means that the total duration to irrigate one furrow
will be 21 + 6 = 27 minutes instead of 22 minutes.
The duration to irrigate 148 furrows = 148 x 27 =
3 996 minutes or 66 hours.
If six furrows are irrigated at the same time, the total
duration per day will be 66/6 = 11 hours, which is more
than the 10 hours which is the maximum duration per day
that the farmers are willing to irrigated.
While we retain a maximum flow of 1.6 l/sec per furrow
and 6 furrows to be irrigated at the same time (for our
design of the piped system of Mangui irrigation scheme in
this Module) field tests determining the size of the cutback
flow have to be carried out. This will then determine the
number of furrows to be irrigated at the same time and
thus the system capacity.
The duration to irrigate 148 furrows = 148 x 22 = 3 256
minutes or 54 hours.
When six furrows at a time will be irrigated, as calculated
above, the system capacity would then be 6 x 1.6 = 9.6
l/sec, or 34.56 m3/hr, instead of the originally estimated
capacity of 31.02m3/hr (Table 16).
If six furrows are irrigated at the same time, the total
duration per day will be 54/6 = 9 hours, which is a slightly
less than the 10 hours which is the maximum duration per
day that the farmers are willing to irrigate.
Looking at Figure 20, a field pipeline runs from west to east
in the middle of the field with hydrants installed along this
Figure 22
In case the flow has to be reduced or cut back once it
reaches the end of the furrow (by removing one or more
siphons), in order to avoid excessive runoff, the time to
finalize the irrigation increases and might even be more
than 10 hours, as shown in the calculations below.
Plot layout and hydrants
When the initial furrow stream of 1.6 l/sec reaches the end
of the furrow six minutes after the start, the depth of water
will be:
d =
(6 x 60) x 0.0016
0.9 x 45
= 0.0142 m or 14.2 mm
The remaining depth of water to be given is:
51.7 - 14.2 = 37.5 mm
35
Irrigation manual
line. Water can be supplied by hoses connected to each
hydrant. Along the field line are hydrant risers (bearing the
hydrants/gate valves) spaced at 30 m intervals. This line
would then be connected to the pumping unit with a supply
line. In order to provide some flexibility to the 18
smallholders participating in the scheme, each farmer will
be provided with one hydrant and a hose as shown in Figure
22. Each farmer has a plot of 30 m x 45 m. Each farmer
36
will irrigate one furrow at a time, meaning that 6 farmers
can irrigate at the same time (see Section 5.2.2).
Each hose will supply the furrows on either side of the
hydrant. The distance from the hydrant to the furthest
furrow is about 15 m. Therefore, the length of the hose
should be about 20 m to allow for loops and avoid sharp
bending by the hydrant.
Chapter 5
Design of canals and pipelines
5.1. Design of canals
The Manning Formula can be expressed as:
The canal dimensions and longitudinal slope, whether for
irrigation or drainage, can be calculated through trial and
error with the Manning formula. This formula is derived
from the continuity equation and the equation for unsteady
flow. These equations have been simplified by assuming
steady uniform flow in the canal (this assumes long canals
with constant cross-section and slope).
Equation 13
The Continuity equation is expressed as:
Equation 12
Q = Km x As x R2/3 x S1/2
or
Q =
V
x As x R2/3 x S1/2
Q
= Discharge (m3/sec)
Km = Manning roughness coefficient (m1/3/sec)
n
= Roughness coefficient; Km = 1/n or n =
1/Km (sec/m1/3)
As
= Wetted cross-sectional area (m2)
= Discharge (m3/sec)
P
= Wetted perimeter (m)
= Wetted cross-sectional area (m2)
R
= Hydraulic radius (m) (R=As/P)
S
= Canal gradient or longitudinal slope of the
canal
Where:
A
n
Where:
Q = AxV
Q
1
= Water velocity (m/sec)
A flowchart showing the various steps in applying the
Manning Formula is given in Figure 23.
Figure 23
Flowchart for canal design calculations
37
Irrigation manual
Figure 24
Canal parameters
5.1.1. Calculation of the cross-section, perimeter
and hydraulic radius of a canal
Figure 24 shows the different canal parameters. As and P,
and thus R in the Manning formula, can be expressed in d,
b and X, where:
The wetted perimeter is the sum of the bed width b and the
two sides from the water level to the bottom. The length of
a side, considering the formula c2 = a2 + b2, is:
—d2 + d2X2
Thus the wetted perimeter for the trapezoidal canal section is:
d
= Water depth (m)
b
= Bed width (m)
X
= Side slope = horizontal divided by vertical
Equation 17
P = b + {2(d2 + (dX)2}1/2 = b + 2d(1 + X2)1/2
For a trapezoidal canal, As is the sum of a rectangle and two
triangles.
The hydraulic radius R is:
The cross-sectional area of a rectangle is:
Equation 18
Equation 14
R =
Area of rectangle = (b x d)
Although the trapezoidal canal shape is very common,
other canal shapes, including V-shaped, U-shaped, semicircular shaped and rectangular shaped canals, can also be
designed as shown in Figure 25.
The cross-sectional area of a triangle is:
Equation 15
Area of triangle =
1
2
(base x height) =
1
2
(Xd x d)
Thus, the wetted cross-sectional area As of the trapezoidal
canal is:
Equation 16
As
5.1.2. Factors affecting the canal discharge
Canal gradient or longitudinal slope of the canal
The steeper the gradient, the faster the water will flow and
the greater the discharge will be. This is substantiated by the
Continuity Equation (Equation 12).
= b x d + 2(1/2 x Xd x d )
= b x d + Xd2
= d(b + Xd)
38
d(b + Xd)
b + 2d(1 + X2)1/2
Velocity increases with an increase in gradient or
longitudinal slope. It therefore follows that a canal with a
steeper gradient but with the same cross-section can
discharge more water than a canal with a smaller gradient.
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 25
Different canal cross-sections
As = 0.5Sd2
P = Sd
As = xd2
P = 2d(1 + x2)1/2
As = bd
P = b + 2d
The recommended maximum slope is 1:300 (that is 1 m
drop per 300 m canal length), which is equal to 0.33%.
Steeper slopes could result in such high velocities that the
flow would be super-critical (see Chapter 6). It would then
be difficult, for example, to siphon water out of the canal,
since an obstruction in a canal where super-critical flow
occurs tends to cause a lot of turbulence, which could result
in the overtopping of the canal. This is due to the change
from the super-critical state to the sub-critical state. The
state of flow could be checked using the Froude Number.
The Froude Number (Fr) is given by:
Equation 19
Fr =
V
(g x l)1/2
Where:
Canal roughness
The canal roughness, as depicted by the Manning roughness
coefficient, influences the amount of water that passes
through a canal. Unlined canals with silt deposits and weed
growth and lined canals with a rough finish tend to slow
down the water velocity, thus reducing the discharge
compared to that of a clean canal with a smooth finish.
Canals that slow down the movement of water have a low
Km or a high n (see Equation 13). It should be understood
that the higher the roughness coefficient Km, or the lower
n, the higher the ability of the canal to transport water,
hence the smaller the required cross-sectional area for a
given discharge.
The roughness coefficient depends on:
Y
The roughness of the canal bed and sides
Y
The shape of the canal
Y
Canal irregularity and alignment
V
= Water velocity (m/sec)
g
= Gravitational force (9.81m/sec2)
I
= Hydraulic depth of an open canal, defined
as the wetted cross-sectional area divided
by the width of the free water surface (m)
Y
Obstruction in the canal
Y
Proposed maintenance activities
Fr
= 1 for critical flow
Typical Km and n values are given in Table 17.
Fr
> 1 for super-critical flow
Fr
< 1 for sub-critical flow
It is important to maintain a Froude number of 1 or less so
that flow is at or below the critical level.
Manning coefficients Km often are assumed too high during
the design phase compared to what they actually will be
during scheme operation due to deterioration of the canals.
The result is an increased wetted cross-sectional area of the
39
Irrigation manual
Table 17
Km and n values for different types of canal surface (adapted from: Euroconsult, 1989)
Type of surface
Range of roughness coefficient
Km (= 1/n) in m1/3/sec
n (= 1/Km) in sec/m1/3
Pipes, precast and lined canals
Metal, wood, plastic, cement, precast concrete, asbestos, etc.
Concrete canal and canal structures
Rough concrete lining
Masonry
Corrugated pipe structures
65-100
65-85
40-60
30-40
40-45
0.010-0.015
0.012-0.016
0.017-0.025
0.025-0.035
0.023-0.025
Earthen canals, straight and uniform
Clean, recently completed
Clean, after weathering
With short grass, few weeds
50-65
40-55
35-45
0.016-0.020
0.018-0.025
0.022-0.027
Earthen canals, winding and sluggish
No vegetation
Grass, some weeds
Dense weeds or aquatic plants in deep channels
35-45
30-40
25-35
0.023-0.030
0.025-0.033
0.030-0.040
Canals, not maintained, weeds and brush uncut
Dense weeds, as high as flow depth
Clean bottom, brush on sides
8-20
10-25
0.050-0.120
0.040-0.080
canal during scheme operation with the danger of
overtopping the canal banks. This in turn means that the
canal discharge has to be reduced to below the design
discharge, in order to avoid overtopping. There is therefore
a need for regular and proper maintenance of canals.
Side slope
The side slope X (= horizontal/vertical) should be selected
depending on the type of canal, soil type and the expected
vegetation cover on the slopes.
Earthen canals
Canal shape
Canals with the same cross-sectional area, longitudinal
slope and roughness, but with different shapes, will carry
different discharges because of different wetted perimeters
and hydraulic radii (see Equation 13). The most efficient
geometry is when the wetted perimeter is minimal for a
given discharge. Under these circumstances, the crosssectional area for a given discharge will also be minimal.
The optimum canal shape, hydraulically, also tends to be
the cheapest to construct as the amount of surface lining
material required will be minimized.
The semi-circle is the canal section that has the lowest
wetted perimeter for a given cross sectional area, but semicircular canals are difficult to construct. The closest canal
section to a semi-circle is the trapezoid. This is a quite
common cross-section as it is relatively easy to construct.
Figure 26 shows rectangular and trapezoidal canals with
different hydraulic radii. Canals with narrower beds and
higher water depths have a smaller wetted perimeter, and
thus a higher discharge, than canals with larger beds and
lower water depths, for the same cross-sectional area. This
is due to the fact that the hydraulic radius R (= As/P)
increases if the wetted perimeter decreases, while keeping
the wetted cross-sectional area the same (see Equation 13).
40
If the side slopes are very steep (low X) there is high risk of
banks collapsing, especially after heavy rainfall. Therefore, a
compromise has to be reached between loss of land (due to
larger width of canal surface) and bank safety. Table 18 gives
suggested side slopes for canals in different soil types.
Table 18
Typical canal side slopes
Soil type
Side slope X
(=horizontal/vertical)
Stiff clay or earth with concrete lining
Heavy, firm clay or earth for small
ditches
Earth, with stone lining or earth for large
canals
Fine clay, clay loam
Sandy clay or loose sandy earth
Fine sand or sandy loam
Coarse sand
1 to 2
1 to 1.5
1
1.5 to 2
2
2 to 3
1.5 to 3
Concrete-lined canals
There are no strict rules for the side slopes of concretelined canals. A major consideration is ease of construction
and the fact that the concrete should stay in place during
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 26
Hydraulic parameters for different canal shapes
As = 0.5 x 1.0 = 0.5 m2
P = (2 x 0.5) + 1.0 = 2.0 m
R = 0.5 / 2.0 = 0.25 m
As = 0.25 x 2.0 = 0.5 m2
P = (2 x 0.25) +2.0 = 2.5 m
R = 0.5 / 2.5 = 0.2 m
As = (0.5 x 0.75) + (0.75 x 0.75) = 0.94 m2
P = 0.5 + 2 x 0.75 x (1 + 12)½ = 2.62 m
R = 0.94 / 2.62 = 0.36 m
As = (1.5 x 0.475) + (0.475 x 0.475) = 0.94 m2
P = 1.50 + 2 x 0.475 x (1+12)½ = 2.84 m
R = 0.94 / 2.84 = 0.33 m
construction, thus the side slope should not be too steep.
Side slopes of around 60º should be easy to construct.
Table 20
Maximum water velocity ranges for earthen canals on
different types of soil (Source: Peace Corps
Information Collection and Exchange, undated)
Bed width / water depth ratio for trapezoidal canals
Soil type
The recommended bed width/water depth (b/d) ratios for
earthen trapezoidal canals are given in Table 19.
Table 19
Recommended b/d ratios
Water depth
b/d ratio
Small (d < 0.75 m)
1 (clay) - 2 (sand)
Medium (d = 0.75-1.50 m)
2 (clay) - 3 (sand)
Large (d > 1.50 m)
>3
The bed width should be wide enough to allow easy
cleaning. A bed width of 0.20-0.25 m is considered to be
the minimum, as this still allows the cleaning of the canal
with small tools such as a shovel. Lined trapezoidal canals
could have similar b/d ratios as given above.
Maximum water velocities
The maximum permissible non-erosive water velocity in
earthen canals should be such that on the one hand the
canal bed does not erode and that on the other hand the
water flows at a self-cleaning velocity (no deposition). A
heavy clay soil will allow higher velocities without eroding
than will a light sandy soil. A guide to the permissible
velocities for different soils is presented in Table 20. In
winding canals, the maximum non-erosive velocities are
lower than in straight canals.
Maximum flow velocity (m/sec)
Sand
0.3 - 0.7
Sandy loam
0.5 - 0.7
Clayish loam
0.6 - 0.9
Clay
0.9 - 1.5
Gravel
0.9 - 1.5
Rock
1.2 - 1.8
Lined canals can manage a range of velocities, as erosion is
not an issue. However, for easy management of water, the
permissible velocity should be critical or sub-critical.
Freeboard
Freeboard (F) is the vertical distance between the top of the
canal bank and the water surface at design discharge. It
gives safety against canal overtopping because of waves in
canals or accidental raising of the water level, which may be
a result of closed gates.
The freeboard can be calculated using Equation 20:
Equation 20
F = C x h1/2
Where:
C
= 0.8 for discharges of up to 0.5 m3/sec up to
1.35 for discharges in excess of 80 m3/sec
h
= Water depth (m)
41
Irrigation manual
Example 7
What is the water depth for a trapezoidal canal with the following known parameters:
= 0.09 m3/sec
Km = 55 (rough concrete lining)
S
= 0.001 (0.1%)
X
= 1 (45°)
b
= d
V
= < 0.75 m/sec
Q
The cross-sectional area of a trapezoidal canal is given by Equation 16:
= d(b + Xd)
As
Substituting the above given data for b, d, and X gives:
= d(d + d) = 2d2
As
The wetted perimeter is given by Equation 17:
= b + 2d(1 + X2)1/2
P
Substitution again of the above given data for b, d, and X gives:
= d + 2d(1 + 12)1/2 = d +2d(1.414) = d + 2.83d = 3.83d
P
The hydraulic radius is given by Equation 18:
R
As
=
P
=
2d2
3.83d
= 0.52d
The Manning formula is given by Equation 13:
= Km x As x R2/3 x S1/2
Q
This gives:
= 55 x 2d2 x (0.52d)2/3 x 0.0011/2 = 110d2 x (0.52d)2/3 x 0.0316 = 1.807d2.66
0.09
2.66
—0.09/1.807 Ÿ d = 0.30 m
d
=
As
= 2d2 = 2 x 0.302 = 0.18 m2
V
=
Q
As
= 0.09 / 0.18 = 0.50 m/sec
This means that the water velocity is less than the maximum allowable velocity given of 0.75 m/sec, which is acceptable.
However, the Froude Number should be calculated using Equation 19 to make sure the flow is sub-critical:
Fr =
V
(g x l)1/2
Where:
l =
As
Width of free water surface
Thus, Fr =
0.50
(9.81 x 0.20)1/2
=
As
b + 2d
=
0.18
(0.30 +2 x 0.30)
= 0.20
= 0.36 which is < 1.
This means that the flow is sub-critical.
For lined canals, F ranges from 0.40 m for discharges less
than 0.5 m3/sec up to 1.20 m for discharges of 50 m3/sec
or more. For very small lined canals, with discharges of less
than 0.5 m3/sec, the freeboard depths could be reduced to
between 0.05-0.30 m.
42
5.1.3. Hydraulic design of canal networks using the
chart of Manning formula
The hydraulic design of canal networks for irrigation and
drainage requires the following steps (Euroconsult, 1989):
1. Design water surface levels in relation to natural
ground slope and required head for irrigation of fields
or for drainage to outlet, taking into account head
losses for turnouts and other structures.
Module 7: Surface irrigation systems: planning, design, operation and maintenance
2. Calculate corresponding hydraulic gradients.
3. Divide network into sections of uniform slope (S) and
discharge (Q).
4. Determine required design (maximum) discharge per
section.
5. Select roughness coefficient (Km or n)
– side slopes
– preferred minimum velocity and permissible
maximum velocity
– bottom width/water depth ratio
6. Calculate hydraulic section dimensions and
corresponding velocity, using:
– nomograph series, if available
– the nomograph presented in Figure 27 (chart of
Manning formula)
– basic equations and calculator
7. Check calculated velocities against preferred and
maximum velocity values; if V is too high, reduce
hydraulic gradient and corresponding bottom slope.
The gain in head should preferably be used in upstream
and downstream canal sections but, if this is
impossible, it must be absorbed by drop structures.
The chart presented in Figure 27 can be used to determine
the optimum canal parameters for trapezoidal canal
sections through trial and error.
5.1.4. Canal section sizes commonly used by Agritex
in Zimbabwe
The Irrigation Branch of Agritex in Zimbabwe has adopted
a 60º trapezoidal canal. The following standard size sections
have been recommended: a flow depth of 0.30 m plus
freeboard of 0.05 m, with bed widths of 0.25 m, 0.30 m,
0.375 m and 0.50 m depending upon gradient and capacity
Figure 27
Chart of Manning formula for trapezoidal canal cross-sections
(m2)
A
Qn
(m-3s-1)
n
A1/3S1/2
X+p
Qn
S1/2
(m1m-1)
S
(m3s-1) (m1s-1)
Q
V
p =
b
h
A= h2(x + p)
Example
x = 2
p = 3
Q = 1 m3s-1
n = 0.02
S = 10-4
n
S1/2
– Draw lines 1, 2, 3, 4, 5,
– Read A = 3.1m2, V = 0.32m1s-1
– Calculate h = 0.78m1, b = 2.34m1
43
Irrigation manual
or discharge required. The total depth of 0.35 m (water
depth + freeboard) is easily reached by construction gangs
while placing concrete. It provides an adequate siphon head
and gives efficient flows within range. The narrowest bed
width used, 0.25 m, is still easy to clean out with a shovel.
By varying the bed width only and not the depth, transition
from one section to another is simplified. This involves no
loss of head and also overcomes the need to make an
allowance when pegging the canal.
The capacities for the above types of Agritex canal sections
have been worked out and are presented in Table 21.
Table 21
Canal capacities for standard Agritex canal sections
Canal
gradient &
hydraulic
data
Canal bottom widths in mm
250
300
375
500
Velocity
(m/sec)
Capacity
(l/sec)
Velocity
(m/sec)
Capacity
(l/sec)
Velocity
(m/sec)
Capacity
(l/sec)
Velocity
(m/sec)
Capacity
(l/sec)
1 : 300
0.79
100.0
0.875
124
1.02
168
1.09
205
1 : 500
0.62
78.0
0.675
96
0.78
128
0.85
172
1 : 750
0.50
63.5
0.595
85
0.68
112
0.69
140
1 : 1 000
0.43
54.5
0.475
68
0.55
92
0.60
126
AS (m2)
0.127
0.142
0.165
0.202
P (m)
0.946
0.996
1.071
1.196
R (m)
0.135
0.143
0.163
0.17
55
55
55
55
Km
Example 8
What is the bed width for a trapezoidal canal with a side slope angle of 60° and a water depth of 0.3 m, assuming Km
= 55 and that the canal has to discharge 78.30 l/sec at a gradient of 0.001 (0.1%) and 0.002 (0.2%) respectively?
In order to calculate X, one has to determine the tangent as follows:
Tan 60° =
1
X
, therefore X =
1
tan 60°
=
1
1.73
= 0.58
Substituting the value for X and the water depth d = 0.30 m in Equations 16 and 17 respectively gives:
As = 0.30(b + 0.58 x 0.30) = 0.30(b + 0.174) = 0.3b + 0.05
P = b + 2d(1 + X2)1/2 = b + 2(0.30)(1 + 0.582)1/2 = b + 0.6(1.156) = b + 0.69
The hydraulic radius, using Equation 18, is:
R =
As
P
=
(0.30b + 0.05)
(b + 0.69)
Substituting the data in the Manning Formula gives:
Q = 55 x (0.30b + 0.05) x
(0.30b + 0.05)
(b + 0.69)
2/3
x 0.0011/2
Substituting values of bed widths in the formula by trial and error will result in a bed width that suits the design
discharge, fixed at 0.0783.
Try b = 0.20 m. The result of the calculation is a flow Q = 0.049 m3/sec when the gradient is 0.001. This means that
the canal with a bed width b = 0.20 m and a water depth d = 0.30 m will not be able to discharge the design flow of
0.0783 m3/s.
After a few runs of trial and error, we get Q = 0.0783 m3/sec, when b =0.35 m, for the 0.001 gradient and 0.24 m for
the 0.002 gradient, with water velocities of 0.50 m/s and 0.64 m/s respectively.
The engineer in Zimbabwe designing the canals would simply use Table 21 to choose a canal section with 250 mm
bed width at a gradient of 1 : 500 and 350 or 375 mm bed width at a gradient of 1 : 1 000.
Calculation of the Froude Number according to Equation 19 gives a value of 0.26, implying that the flow is sub-critical.
44
Module 7: Surface irrigation systems: planning, design, operation and maintenance
5.1.5. Longitudinal canal sections
The best way to present canal design data for construction
is to draw a longitudinal profile of the canal route and to
tabulate the data needed for construction. The longitudinal
profile shows the chainage or distance along the canal at the
horizontal or x-axis and the elevations of the natural
ground, the ground after levelling and the canal bed at the
vertical or y-axis. The data are tabulated under the graph,
showing the elevation of ground and canal bed in figures at
each given distance. Water depths could also be shown. The
chainage starts from a reference point, which is usually the
beginning of the canal. Where possible the survey results of
the topographic survey are used. If these are not sufficient,
a detailed survey of the proposed alignments should be
made. The following are guidelines for the presentation of
longitudinal profiles.
Y
Direction: water flow is always given from left to right.
Y
Horizontal scale:
1 : 1 000 for short canals (1 cm = 10 m)
1 : 5 000 for long canals (1 cm = 50 m)
Y
Vertical scale:
1 : 20 for small canals and low gradient
(1 cm = 0.2 m)
1 : 100 for larger canals and higher gradient
(1 cm = 1 m)
(Note: the vertical scale should be chosen in such a way
that the water depth is clearly visible)
Y
The profile should show the ground level, the bed level
and eventually the water level at design discharge
Y
Structures should be marked by a vertical line at the
place of the structure, with the structure identification
written along the x-axis
Y
Distance is measured in metres from the canal inlet,
with intervals depending on length to be covered (5,
10, 50 m etc.). For very long canals, it can be measured
in kilometres. The distance to structures or major
change of direction is always measured and added to
the tabulated data
Y
Ground levels are tabulated from survey data
Y
Bed levels and eventually water levels are tabulated at
the end of each reach, which means upstream and
downstream of each structure where the water level
changes. A single value can be given at structures when
there is no fall
Figures 28, 29 and 20 show longitudinal sections of a field,
secondary and conveyance canal respectively.
The field or tertiary canals should have sufficient command
over the whole length in order to allow the correct
discharge to be supplied to the field. Figure 28 shows an
example of a field canal with sufficient command over its
full length after land levelling. For these canals the ground
elevations after land levelling have to be taken into account
in deciding the slope of the canal. As normal practice, the
water depth should be more or less 10-15 cm above the
levelled ground surface in order to maintain a good
siphoning head.
Secondary and main canals can be designed in cut at places
where there are no offtakes. The designer should ensure
that there is sufficient command at field canal offtakes.
Figure 29 illustrates this. Ideally, an offtake should be
placed before a drop.
Figures 30 and 31 show examples of longitudinal profiles of
a conveyance canal. The starting bed elevation of the
conveyance canal should be high enough to give sufficient
command to the lower order canals. The conveyance canal
itself does not necessarily need to have a water level above
ground level since no water will be abstracted from it. It is
in fact preferable to design them in cut as much as possible.
Where possible, it could run quasi-parallel to the contour
line as shown in Figure 19. Drops should be incorporated,
when the canal goes in fill, but the command required
should be maintained.
45
Irrigation manual
Figure 28
Longitudinal profile of a field or tertiary canal
46
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 29
Longitudinal profile of a secondary or main canal
47
Irrigation manual
Figure 30
Longitudinal profile of a conveyance canal
48
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 31
Example of a longitudinal section of a conveyance canal
5.1.6. Field canals for small irrigation schemes
Field canals (tertiary canals and sometimes secondary
canals) usually run at an average gradient of 1:500 (0.0020
or 0.2%) to 1:300 (0.0033 or 0.33%). When the existing
land slope exceeds the proposed canal gradient, drop
structures can be used in order to avoid the canal being
suspended too much above the ground level, which would
require too much fill.
A common drop in small canals is 0.15 m. Such small drops
do not require stilling basins because of their short fall (see
Chapter 6). In order to have a minimum of 0.15-0.20 m
command, the drop is constructed when the bed level of
the canal reaches the ground level after land levelling. A
small Cipoletti weir (see Chapter 6) is constructed at every
drop in order to allow for support for the check plate.
A problem often encountered is that field canals in
irrigation schemes lack command, making siphoning onto
adjacent land difficult or even impossible. One main reason
for the lack of command is the use of the original ground
level to site the drops. Once the canal bed level has reached
the original ground level, it is dropped by 0.15 m. However,
when land levelling is done afterwards, it might result in fill
near the canal, thus reducing or eliminating the command.
Computer programmes for calculating the location and
elevation of the standard drop structures are available
nowadays. As an example a short description of the Lonsec
programme, which is such kind of programme, is given
below. It is written in Quick Basic.
The programme requires the input of:
1. The chainage at the beginning and at the end of the
field canal
2. The ground levels after land levelling at these two
chainages
3. The canal bed level at the beginning of the canal
4. The canal gradient (it assumes a uniform canal gradient)
Figure 32
Longitudinal canal profile generated by the Lonsec Programme
49
Irrigation manual
The output consists of the ground level, the canal bed level
immediately before and after a standard drop and the
chainage where a drop occurs. Furthermore the ground and
canal bed levels are calculated at 10 m intervals, independent
of the fact whether there is a drop structure or not. Thus,
the output is suitable for use during construction.
The programme can also show and print a visual impression
of the longitudinal section of the field canal. Table 22 shows
an example of output data, while Figure 32 gives the visual
impression, which could for example be included in
feasibility reports together with the output tables.
Table 22
Longitudinal profile for field canal – output from the Lonsec computer programme
50
Chainage
(m)
Ground level
(m)
Canal level
(m)
0.0
10.0
20.0
99.920
99.828
99.737
99.820
99.787
99.754
25.6
99.686
99.736
30.0
40.0
50.0
99.645
99.554
99.462
99.572
99.539
99.506
51.1
99.452
99.502
60.0
70.0
99.371
99.279
99.323
99.290
76.7
99.218
99.268
80.0
90.0
100.0
99.188
99.096
99.004
99.108
99.075
99.042
102.2
98.984
99.035
110.0
120.0
98.913
98.821
98.859
98.826
127.7
98.751
98.801
130.0
140.0
150.0
98.730
98.638
98.547
98.644
98.611
98.578
153.3
98.516
98.567
160.0
170.0
98.455
98.364
98.395
98.362
178.8
98.283
98.333
180.0
190.0
200.0
98.272
98.180
98.089
98.179
98.147
98.114
204.3
98.050
98.100
210.0
220.0
97.997
97.906
97.931
97.898
229.9
97.815
97.866
230.0
240.0
250.0
97.814
97.723
97.631
97.715
97.683
97.650
255.4
97.582
97.632
260.0
270.0
280.0
97.540
97.448
97.357
97.467
97.434
97.401
280.9
97.348
97.398
290.0
292.0
97.265
97.248
97.219
97.212
Canal level after drop
(m)
99.586
99.352
99.118
98.885
98.651
98.417
98.183
97.950
97.716
97.482
97.248
Module 7: Surface irrigation systems: planning, design, operation and maintenance
1. The chainage at the beginning of the section = 0.0 m
and the chainage at the end of the section = 292.0 m
inflow and outflow measurements will not only
represent seepage losses, but evaporation losses as well.
2. The ground level after levelling at the beginning of the
section = 99.92 m and the ground level after levelling
at the end of the section = 97.24 m
2. Measurement of the rate of fall of the water level in a
canal stretch that has been closed and where the water is
ponding. From these losses the estimated evaporation
should be subtracted to get the seepage losses.
3. The canal bed level at the beginning of the section =
99.82 m
4. The canal gradient = 0.0033 (1:300)
5.1.7. Seepage losses in earthen canals
Unlined earthen canals are the most common means of
conveying irrigation water to irrigated lands. Farmers
prefer them because they can be built cheaply and easily
and maintained with farm equipment. Unlined canals are
also flexible, as it is easy to change their layout, to increase
their capacity or even to eliminate or rebuild them the next
season. However, unlined canals have many disadvantages
that make them less desirable compared to lined canals or
underground pipes. These are:
Usually, seepage losses are expressed in m3 of water per m2
of the wetted surface area of a canal section (P x L) per day.
If a field test cannot be carried out, seepage can be
estimated from Table 23, which gives average seepage losses
for different types of soil.
Table 23
Seepage losses for different soil types
Type of soil
Seepage (m3 water/
m2 wetted surface area
per day)
Impervious clay loam
0.07 - 0.10
Clay loam, silty soil
0.15 - 0.23
Sandy loam
0.30 - 0.45
They usually lose more water due to seepage, leakage
and spillage
Sandy soil
0.45 - 0.55
Sandy soil with gravel
0.55 - 0.75
Y
Rodents can cause leakage
Pervious gravelly soil
0.75 - 0.90
Y
Frequent cleaning is needed because of weed growth
Y
Earth ditches can erode and meander, creating
problems in maintaining straight or proper alignments
Seepage could be localized where a portion of highly
permeable material has been included in the bank or where
compaction has been inadequate during canal construction.
Y
Labour costs of maintenance of unlined canals are
normally higher than of lined canals and pipelines
5.1.8. Canal lining
Y
Y
They provide an ideal environment for the vector of
bilharzia
When designing earthen canals, it is important to ensure
that the slope is such that the bed does not erode and that
the water flows at a self-cleaning velocity (see Section
5.1.2). From all standpoints, relatively flat lands on soils
with a high percentage of silt and clay are the most suitable
for canal construction, because of low infiltration rates.
In earthen canals, seepage occurs through the canal bed and
sides. In areas where relatively permeable soils are used to
construct canals, high seepage can be expected. The higher
the seepage losses in the canals the lower the distribution
system (conveyance and field canal) efficiencies, since much
less water than that diverted at the headworks reaches the
fields.
Seepage always occurs, even if the canals are constructed
with clay soils. If there is abundant water available that can
be diverted under gravity, one might accept the water losses
without resorting to lining. In fact, worldwide, unlined
canals are the most common as they are the cheapest and
easiest type of canal to construct. However, if water has to
be used more efficiently, due to its scarcity or if it has to be
pumped, it usually becomes economical to line the canals.
Another consideration in analyzing the economics is the
health-related cost (of medicines and time lost by
smallholders due to poor health).
Seepage is difficult to predict. Two simple ways to estimate
seepage losses are:
Canal lining is generally done in order to reduce seepage
losses and thus increase the irrigation efficiencies. It also
substantially reduces drainage problems and canal
maintenance as well as water ponding, thus reducing the
occurrence of vector-borne diseases. Also, smooth surface
linings reduce frictional losses, thereby increasing the
carrying capacity of the canals.
1. Measurement of inflow into and outflow from the
canal at selected points. The difference between the
Below different lining methods are briefly explained. The
actual construction is dealt with in detail in Module 13.
51
Irrigation manual
Example 9
An earthen canal with a 1:1000 gradient, constructed in and using sandy loam, is designed to convey 78.3 l/sec for
24 hours per day over a distance L of 2 km. The Manning coefficient for the canal Km is 30, the side slope X is 1.5
and the b/d is 1.5. What are the seepage losses as a percentage of the daily discharge?
The canal cross sectional area is calculated from the Manning Formula as follows:
Q
= Km x As x R2/3 x S1/2
Where:
As
= 1.5d2 + 1.5d2 = 3d2
P
= 5.10d
Substitution, of As and P in the equation gives:
0.0783 = 30 x 3d2 x
Therefore:
3d2
2/3
5.10d
x (0.001)1/2 Ÿ d = 0.30 m
As = 0.27 m2 and P = 1.53 m
The total wetted surface area over the 2 km stretch is:
Wetted surface area = P x L = 1.53 m x 2 000 m = 3 060 m2
The seepage loss through a sandy loam is estimated at 0.40 m3/m2 per day (Table 23). Thus, the total estimated
seepage loss from the canal is:
Total seepage loss per day = 3 060 x 0.40 = 1 224 m3/day
The total volume of water supplied per day = 0.0783 x 24 x 60 x 60 = 6 765 m3
This means that approximately
1 224
6 765
x 100 = 18% of the supplied water is lost to seepage.
Material used for lining:
Y
Clay
Y
Polyethylene plastic (PE)
Y
Concrete
Y
Sand-cement
Y
Brick
Y
Asbestos cement (AC)
Furthermore, tools such as shovels and slashers can easily
damage it during maintenance works. Weed growth and soil
erosion could also cause problems in the canal.
Concrete
The selection of a lining method depends mainly on the
availability of materials, the availability of equipment, the
costs and availability of labour for construction.
The materials required for concrete lining are cement, fine
and coarse aggregates. Concrete lining is an expensive but
very durable method of lining. When properly constructed
and maintained, concrete canals could have a serviceable
life of over 40 years. This durability is an important aspect
to consider, more so for small-scale self-run schemes in
remote areas. Details on the preparation of concrete lining
are given in Module 13.
Clay
Sand-cement
If a sufficient volume of clay soil can be found in the vicinity
of the scheme, clay lining might be the cheapest method to
use to reduce seepage losses. One has to ensure that the clay
is well spread in the canal and well compacted. However, clay
lining is susceptible to weed growth and possible soil erosion.
If coarse aggregates are not available for the preparation of
concrete, the method of sand-cement lining could be
considered. A strong mixture is either placed in-situ on the
canal sides and bed or is precast (thickness 5-7 cm). A mix
of 1:4 (cement : river sand) is recommended. More details
are given in Module 13.
Polyethylene plastic
Polyethylene plastic sheeting can be used for lining canals.
The sheets have to be covered with well-compacted soil,
since the plastic deteriorates quickly when exposed to light.
52
Brick
If good clayish soils, suitable for producing good quality
burnt bricks, are found near the scheme area, brick lining
Module 7: Surface irrigation systems: planning, design, operation and maintenance
could be considered. The construction however is
laborious. Cement is required for mortar and plastering. A
disadvantage of this lining method is the large amount of
firewood needed to burn the bricks. It could, however, be
justified if the scheme area had to be cleared of trees, which
could then be used for burning the bricks.
Asbestos cement
Precast asbestos cement flumes can be used as lining
materials. The flumes are easy to place and join. A
disadvantage is usually the high unit cost and the health risk
of working with asbestos.
5.2. Design of pipelines
In piped surface irrigation systems, water is transported in
closed conduits or pipes in part or all of the distribution
system from the headwork up to the field inlet. The pipes
can be all buried, with outlets in the form of hydrants
protruding above ground level on field pipes. Or only the
conveyance and supply lines can be buried with field pipes
being portable and laid above ground. In the latter case, the
above ground pipes are made of aluminium fitted with
adjustable gate openings (Figure 33).
Piped systems for surface irrigation, unlike piped systems
for sprinkler irrigation, do not require a lot of head at the
hydrant outlet. The head should only be sufficient to push
water through the irrigation hose that takes the water from
the hydrant to the soil. In view of the low head
requirements for the systems, it is possible to employ
gravity flow where there is sufficient head to overcome the
frictional losses in pipes. In situations where the head is not
adequate, small power pumps would be used with low
operational costs. Pipes with low-pressure rating are also
used for these systems as they operate at reasonably low
pressures. At times when the pressure in the system is very
low, buried PVC pipes rated at two bar can be used with
these systems, if available.
If the water level at the headwork is higher than the water
level required at scheme level, the water can be transported
through the pipes by gravity. If the water level at the
headwork is lower than the water level required at scheme
level, then the water needs to be pumped through the pipe
Figure 33
Methods commonly used to introduce water into the field
Takeout pipe
Field lateral
Secondary ditch
Portable siphon
Lath pipe or spile
a)
Field lateral or
secondary ditch
c)
b)
d)
53
Irrigation manual
to arrive at the scheme at the required elevation necessary
to be able to irrigate by gravity from the field inlet onwards.
application efficiency. Also the hose diameter would be too
large for the farmer to move around.
5.2.1. Design of the conveyance pipeline in
Nabusenga irrigation scheme
Allowable pressure variation and head losses in the
hose
The friction losses of the outlet pipe and the conveyance pipe
should not exceed the difference in elevation between the
lowest drawdown level in Nabusenga dam and the top of the
scheme or the block of fields. To ensure this, there is a need
to draw a longitudinal profile of the alignment of the pipeline.
The profile will show the elevations of the pipeline
corresponding to distances from a reference point (also called
chainages) along the pipe alignment. Figure 34 shows the
longitudinal section of a pipeline from the Nabusenga dam to
the top of the scheme. Figures 35 and 36 can be used to
calculate the friction losses in AC and uPVC pipes respectively.
Before proceeding with the calculations of the hydraulics, it
should be pointed out that the system should be designed
for equity in water supply. Therefore, each hose should
provide about the same amount of water ± 5%. For this
reason, the pressure variation within the system should not
exceed 20% of the head losses in the hose.
The Hazen-Williams equation will be used for this purpose.
Equation 21
Kx
Q
1.852
C
High points along the proposed alignment should be
carefully checked in order to ensure that there is enough
head available to discharge the required flow over these
points. Measures to be taken to ensure this include:
Where:
Hf100
=
Y
Friction losses over a 100 m distance
(m)
K
=
Constant 1.22 x 1 012, for metric units
Excavation of a deep trench. This may not always be
feasible for huge elevation differences due to the nature
of the underlying bedrock and the distance over which
the digging has to be done
Y
Taking a new route altogether for the pipeline
Y
Changing the pipe size diameter with the hope that the
friction losses would be reduced sufficiently to
overcome the problem
Hf100 =
D4.87
Q
=
Flow (l/s)
C
=
Coefficient of retardation based on
type of pipe material (C = 140 for
plastic)
D
=
Inside diameter (mm)
Table 24 gives C values for different materials.
Table 24
The pressure is generally lower at high points along
pipelines and air or other gases tend to be released from
solution forming an air pocket that interrupts the flow of
water. It is imperative that air-release valves be fitted at
these points to let air out of the system when it forms.
Along our pipeline, an air-release valve would be fitted at
chainage 880 m.
5.2.2. Design of the piped system in Mangui
irrigation scheme
Based on the layout discussed earlier (see Section 4.3,
Figure 20 and Figure 22), each farmer’s plot will be
equipped with one hydrant and one hose irrigating one
furrow at a time. A total of eighteen hydrants (gate valves)
have been provided for the system. One option would be
allowing six hydrants to operate at a time. Another option
would be that all water is delivered to one hydrant and that
thus one farmer would irrigate at a time. Such an option,
while technically feasible, would increase the cost of the
system in addition to requiring more labour per plot to
manage the water to the level required for 60% field
54
Hazen-Williams C value for different materials
Material
Constant C
uPVC
140 - 150
Asbestos cement (AC)
140
Cast iron (new) (CI)
130
Galvanized steel (new) (GS)
120
Note:When aging, the roughness of cast iron and galvanized steel pipes
increases. For example, for a year old cast iron pipe the C might be reduced
to 120 and to 100 for a 20-year-old cast iron pipe.
Assuming that a 50 mm inside diameter and 20 m long
hose is used, the friction losses for a flow of 1.6 l/s will be
as follows:
1.22 x 1012 x
Hf100 =
1.6
1.852
140
504.87
= 1.64 m per 100 m
For the 20 m hose the head losses HL will be:
HL = 1.64 x (20/100) = 0.32 m
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 34
The longitudinal profile of the conveyance pipeline from Nabusenga dam to the night storage
reservoir (NSR)
55
Irrigation manual
Figure 35
Friction loss chart for AC pipes (Class 18)
Delivery-litres per second
Frictional head-metres per 100 metres of pipe(on hydraulic gradient x 100)
56
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 36
Friction loss chart for uPVC pipes (Source: South African Bureau of Standards, 1976)
Frictional head-metres per 100 metres of pipe(on hydraulic gradient x 100)
Frictional head-metres per 100 metres of pipe(on hydraulic gradient x 100)
57
Irrigation manual
Example 10
An existing Class 6 AC pipeline, with a diameter of 225 mm and a length of 464 m, has to be extended by 1739 m
(2 193 - 454) of uPVC pipe in order to irrigate an additional area. Figure 34 shows that the minimum water level in
the dam is 103.38 m. The outlet level of the existing pipeline is at 99.70 m. The ground is high at certain points along
the pipeline. The highest point is at chainage 880 (1334 - 454), where the elevation is 99.72 m. The design maximum
water level of the night storage reservoir is 94.0 m. A flow of 32.6 l/sec (117.4 m3/hr) has to be discharged through
the pipeline. What is the best pipe to use for conveyance?
Using the friction loss charts (Figure 35 and 36) for AC and uPVC pipes, the friction headlosses per 100 m are drawn
up for the different pipe sizes and presented below.
Pipe size (mm)
Friction losses (m per 100 m)
AC 225 (Class 18)
0.29
uPVC 200 (Class 4)
0.42
uPVC 250 (Class 4)
0.15
Thus, the friction losses (HL) in the existing AC pipe, inclusive of 20% extra for losses in fittings are:
HL = (0.29/100) x 464 x 1.20 = 1.62 m
The head loss line will run from 103.38 m, being the minimum water level in the dam, to 101.76 m (103.38 - 1.62) at
the end of the AC pipe. The designed maximum water level of the night storage reservoir leaves 7.76 m (101.76 94.0) for friction losses within the remaining 1 739 m of pipeline.
In selecting the pipe sizes to be used, it is possible to use different sizes of pipe along the sections of the pipeline.
The level of the pipe should be below the head loss line along its length, so that the pipe can pass the design
discharge. Therefore, high points should be checked to ensure that the design discharge passes.
If a 200 mm diameter uPVC pipe is selected from that point to the night storage dam it would give a head loss of:
HL = 0.42 x 1.20 = 0.504 m per 100 m, including 20% extra. The high point at chainage 880 m should be checked.
At that point the head loss line would be at elevation:
101.76 m – (0.504/100) x (1 739 – 880) = 101.76 – 4.33 = 97.43 m
This is lower than the ground level elevation of 99.72 m at the high point at chainage 880 m. Therefore, the pipe
should be laid at a depth below 97.43 m at that point. Figure 34 shows that the trench is dug to elevation 97.24 m,
thus the depth is adequate.
Where the night storage reservoir is located, the head loss line would be at elevation:
101.76 – (0.504/100) x 1 739 = 101.76 – 8.76 = 93.0 m.
The head loss between the minimum water level in the dam and chainage 0 is 8.76 m, which is more than the 7.76
m limit. The head loss line of 93.0 m is below the design water level of the night storage reservoir, meaning that there
is insufficient head available in order to deliver the discharge required. A different combination of pipes that reduces
head losses needs to be selected.
As a second option, a 225 mm diameter AC pipe (same as the existing one) is used from chainage 880 m to 1 739 m.
The friction losses for this section would be:
(0.29/100) x 1.2 x (1 739 – 880) = 2.99 m
If, for the remaining 880 m, a 200 mm diameter Class 4 uPVC pipe is used, the friction loss for this section would be:
(0.42/100) x 1.2 x 880 = 4.44 m.
Therefore, the total friction loss of the 1 739 m pipe section is 2.99 + 4.44 = 7.43 m, which is less than the 7.76 m
limit. The head loss line at the night storage reservoir is 94.33 m (101.76 m - 7.43 m), giving an excess head of
58
Module 7: Surface irrigation systems: planning, design, operation and maintenance
In order to reduce costs and ease operation the option of a
32 mm inside diameter hose will also be looked at. For this
hose the head losses will be:
1.22 x 1012 x
Hf100 =
32
1.6
1.852
140
4.87
= 14.4 m per 100 m
For the 20 m hose the head losses HL will be:
HL = 14.4 x (20/100) = 2.94 m
Hence, the 32 mm inside diameter hose is adopted and the
allowable pressure variation would be 20% of the head losses
of this hose, which is 2.94 x 0.2 = 0.59 m. This implies that
the head losses in the field line, including elevation difference
along this line, should not exceed 0.58 m.
This is above the allowable pressure variation of 0.59 m.
The difference in elevation within the hydraulic unit,
from the first to the last hydrant, is 0.22 m (= 10.13 9.91). However, this is down slope hence the negative
difference in elevation, so when added to the total head
losses, they drop to 0.40 m (= 0.62 - 0.22) and are thus
within the 0.58 limit. This also implies that we can
reduce the diameter of part of the 180 m length pipeline
from 160 mm to 140 mm and redo the calculations as
follows:
Q1
=
34.56 m3/hr
D1
=
160 mm PVC class 4
L1
=
100 m
HL1
=
0.19 x 1 = 0.19 m
Q1
=
34.56 m3/hr
D2
=
140 mm PVC class 4
L2
=
80 m
There are three options in operating the system:
HL2
=
0.35 x 0.8 = 0.28 m
Y
The last six hydrants (gate valves)* operate at the same
time
Q2
=
23.04 m3/hr (= 34.56 - 11.52)
D3
=
110 mm PVC class 4
Y
The first six hydrants operate at the same time
L3
=
30 m
Y
The middle six hydrants operate at the same time
HL3
=
0.56 x 0.3 = 0.17 m
Q3
=
11.52 m3/hr
D4
=
90 mm PVC class 4
L4
=
30 m
HL4
=
0.36 x 0.3 = 0.11 m
HLtotal
=
HL1 + HL2 + HL3
Head losses in field pipeline
The worst case scenario would be when the last six hydrants
operate at the same time, hence the adopted calculations.
The flow per hydrant will be 34.56/6 = 5.76 m3/hr. Using
Figure 36 the head losses are determined as follows:
Q1
=
34.56 m3/hr
D1
=
160 mm PVC class 4
=
0.19 + 0.28 + 0.17
L1
=
180 m
=
0.75 m
HL1
=
0.19 x 1.8 = 0.34 m
Q2
=
23.04 m3/hr (= 34.56 - 5.76 - 5.76 for
two hydrants)
D2
=
110 mm PVC class 4
L2
=
30 m
HL2
=
0.58 x 0.3 = 0.17 m
Q3
=
11.52 m3/hr (= 23.04 - 5.76 - 5.76)
D3
=
90 mm PVC class 4
L3
=
30 m
HL3
=
0.36 x 0.3 = 0.11 m
HLtotal
=
HL1 + HL2 + HL3
=
0.34 + 0.17 + 0.11
=
0.62 m
If we include the difference in elevation of -0.22 m the
HLtotal becomes 0.53 m (= 0.75 - 0.22), which is within
the allowable pressure variation of 0.59 m.
Head losses in supply pipeline
The head losses in supply pipeline from the pumping
station to the first set of hydrants are as follows:
Qsp
=
34.56 m3/hr
Dsp
=
160 mm PVC class 4
Lsp
=
90 m
HLsp
=
0.19 x 0.9 = 0.17 m
* A hydrant in this case is a gate valve, fitted on a riser, and there are two of them on each riser. Therefore, six hydrants operating at the same time implies that three hydrant
risers are operating at once.
59
Irrigation manual
Head losses in galvanized risers
Using Equation 21 the head losses are as follows for the 1.5
m, 75 mm inside diameter riser, using a C = 80 for old
steel pipes:
3.2
1.22 x 1012 x
Hf100 =
5.2.3. Advantages and disadvantages of piped
systems
Following are some advantages of the use of piped systems:
Y
The cost of medium and small diameter PVC pipes
compares very favourably with the cost of constructing
smaller canals
Y
Seepage and evaporation losses are eliminated
Y
There are no stilling boxes required or other places
where stagnant water can collect and become a
breeding ground for mosquitoes and snails.
Furthermore, there is no weed growth in pipelines
Y
Pipelines are normally safer than open channels since
humans and equipment cannot fall into the water
stream
Y
With only hydrants protruding above ground, it is
possible to undertake land levelling and other
mechanical cultivation after the scheme has been
installed
Y
The system can be installed faster than canal systems
Y
Pipelines permit the conveyance of water uphill against
the normal slope of the land over certain distances to
overcome obstacles
Y
Very little land is lost at the headlands of each plot as
the crops can be planted right up to or even over the
pipeline. Also, the use of buried pipes allows the use of
most direct routes from the water source to the field
Y
The farmer has control over the water supply to the
plot, and since water can be available “on demand” in
case no pumping is required, there is some flexibility in
when to irrigate and it is less important to adhere to
strict rotation
Y
The underground pipes form a closed system and as a
result the conveyance losses are negligible. There are
also no incidences of water poaching as could occur
with canal conveyance systems
1.852
80
754.87
x
1.5
100
= 0.035 m = 0.04 m
Total head requirements
The total head requirements are composed of the suction
lift (assumed to be 2 m), the head losses in the supply line,
the head losses in the field line, the head losses in the
hydrant riser and hose, and miscellaneous losses for fittings,
plus the difference in elevation between the water level and
the highest point in the field.
They are calculated as follows:
Suction lift
2.00 m
Supply line
0.17 m
Field line
0.62 m
Riser
0.04 m
Hose
2.90 m
Miscellaneous 10%
0.57 m
Difference in elevation
Total
5.13 m (= 10.13 - 5.00)
11.43 m
Power requirements
The following equation is used:
Equation 22
kW =
QxH
360 x Ep
x 1.2
Where:
kW
=
Power requirements (kW)
Q
=
Discharge (m3/hr)
H
=
Head (m)
Ep
=
Pump efficiency (obtained from the
pump performance chart)
Amongst the disadvantages, the following can be
mentioned:
360
=
Conversion factor for metric untis
Y
1.2
=
20% derating (allowance for losses in
transferring the power to the pump)
The system can be expensive to install, especially when
large diameter pipes are to be used and when the
trenching requires blasting in some areas
Y
Some skill is required to fix a hydrant when it gets
broken at the bottom. However, these incidences are
rare when the hydrants are properly protected
kW =
34.56 x 11.43
360 x 0.5
x 1.2 = 2.63kW
Depending on the availability in the market place, the
closest size to 2.7 kW should be selected. However no unit
smaller than 2.7 kW should be purchased.
60
Chapter 6
Hydraulic structures
Hydraulic structures are installed in open canal irrigation
networks to:
e. Drop structures
f.
Tail-end structures
Y
Control and measure discharge
g. Canal outlets
Y
Control water levels for command requirements
h. Discharge measurement structures
Y
Dissipate unwanted energy
i.
Y
Deliver the right volume of water to meet crop water
requirements
Y
Incorporate recycled tail water, if available
Crossings, like bridges, culverts, inverted siphons
Depending on the size and complexity of the irrigation
scheme, some or all of the above-mentioned structures
could be incorporated in the design.
The most common structures are:
6.1. Headworks for river water offtake
a. Headworks for river water offtake
Abstraction and/or diversion of water from its source to the
scheme is often difficult and can be quite costly, depending
on its complexity. Figure 37 presents a sketch of schemes
irrigated from different water sources.
b. Night storage reservoirs
c. Head regulators
d. Cross regulators
Figure 37
Schemes irrigated from different sources (Source: FAO, 1992)
Irrigation from a reservoir
Simple river
diversion (1)
Pumping from
a river (3)
River diversion
using a weir (2)
Groundwater
pumping
61
Irrigation manual
Figure 38
Headwork with offtake structure only
The function of a headwork is to divert the required
amount of water at the correct head from the source into
the conveyance system. It consists of one or more of the
following structures:
Y
Offtake at the side of the river
Y
Regulating structure across the river or part of it
Y
Sediment flushing arrangement
This section concentrates mainly on the headworks for
direct river offtake and offtakes using a weir. Some
attention is paid to important dam and reservoir aspects,
such as the outlet pipe diameter. However, for detailed dam
design, the reader is referred to other specialized literature.
6.1.1. Headwork for direct river offtake
In rivers with a stable base flow and a high enough water
level throughout the year in relation to the bed level of the
intake canal, one can resort to run-off-river water supply
(Figure 38 and Example 1 in Figure 37). A simple offtake
structure to control the water diversion is sufficient.
The offtake should preferably be built in a straight reach of
the river (Figure 39). When the water is free from silt, the
centre line of the offtake canal could be at an angle to the
centre line of the parent canal. When there is a lot of silt in
the system, the offtake should have a scour sluice to
discharge sediments or should be put at a 90° angle from
the parent canal.
If it is not possible to build the offtake in a straight reach of
the river, one should select a place on the outside of a bend,
as silt tends to settle on the inside of bends. However,
erosion usually takes place on the outside of the bend and
therefore protection of the bank with, for example,
concrete or gabions might be needed. The offtake can be
perpendicular, at an angle or parallel to the riverbank,
depending on site conditions, as illustrated in Figure 40.
62
Figure 39
Offtake possibilities in straight reach of river
Figure 40
Possible arrangements for offtakes based on
site conditions
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 41
An example of an intake arrangement of a headwork
The functions of the offtake structures are:
Y
To pass the design discharge into the canal or pipeline
Y
To prevent excessive water from entering during flood
Considering these functions, the most important aspect of
the structure is the control arrangement, which can be a
gate, stop logs, or other structures. When the gate is fully
opened, the intake behaves like a submerged weir (Figure
41) and its discharge is given by equation 23.
that the gate opening to the offtake structure will be at a
higher elevation than the normal base flow water level. To
abstract the required discharge in these situations, one
could consider the options below:
Y
Select an offtake site further upstream. However, site
conditions, the increased length of the conveyance
canal, and other factors have to be considered carefully.
Y
Build a cheap temporary earthen dam and temporary
diversion structure. This method is especially suitable
in unstable rivers, where high expenses for a
permanent structure are not warranted because of the
danger of the river changing its course.
Y
Construct a permanent diversion dam or structure
(weir or gate) across the river, where the design
elevation of the weir should relate to the design water
level in the conveyance canal, similar to the previous
example.
Equation 23
Q = C x B(h + hd)3/2
Where:
(m3/sec)
Q
= Discharge in intake
C
= Weir coefficient
B
= Width of the intake (m)
h
= Difference between river water level and
canal design water level (m)
hd
= Difference between canal design water
level and sill level of the intake (m)
In some instances, the base flow water level fluctuates
greatly over the year and the water level can become so low
6.1.2. River offtake using a weir
Figure 42 shows an example of a river diversion structure,
in this case a weir (Example 2 in Figure 37).
Example 11
A discharge of 1.25 m3/sec has to be abstracted from a river, into an open conveyance canal. The base flow water
level of the river is 125.35 m. The design water level in the canal is 124.90 m and the water depth is 0.60 m. The weir
coefficient is 1.60. The width of the intake is 1.50 m and the length of the weir is 0.50 m (Figure 41). What will be the
sill level?
Q
= 1.25 m3/sec
C
= 1.60
h
= 125.35 - 124.90 = 0.45 m
B
= 1.50 m
The next step would be to substitute these values in Equation 23:
1.25 = 1.60 x 1.50(0.45 + hd)2/3 Ÿ hd = 0.20 m
Thus the sill level should be at an elevation of 124.90 - 0.20 = 124.70 m
63
Irrigation manual
Figure 42
An example of a diversion structure
Structures constructed across rivers and streams with an
objective of raising the water level are called cross regulators
(see Section 6.4).
A weir should be located in a stable part of the river where
the river is unlikely to change its course. The weir has to be
built high enough to fulfil command requirements. During
high floods, the river could overtop its embankments and
change its course. Therefore, a location with firm, welldefined banks should be selected for the construction of
the weir. Where possible, the site should have good bed
conditions, such as rock outcrops. Alternatively, the weir
should be kept as low as possible. Since weirs are the most
common diversion structures, their design aspects will be
discussed below.
Design of a weir for flood conditions
The weir height has to be designed to match the design
water level in the conveyance canal. The weir length has to
be designed to allow the design flood to safely discharge
over the weir.
After deciding upon the location of the weir, the design
flood, which is the maximum flood for which the weir has
to be designed, has to be determined. If data are available,
a flood with a return period of 50 or 100 years for example
could be selected. If sufficient data are not available, flood
marks could be checked, upon which the cross-sectional
area can be determined and used, together with the
gradient of the river, to calculate the flood discharge. Some
formulae have been developed for this purpose, based on
peak rainfall intensity and catchment characteristics.
Figure 43
C1 coefficient for different types of weirs in relation to submergence, based on crest shape
64
Module 7: Surface irrigation systems: planning, design, operation and maintenance
The general equation for all weir types is:
Figure 44
Equation 24
C2 coefficient for different types of weirs in
relation to crest shape
Q = C1 x C2 x B x H3/2
Where:
Q
=
Discharge (m3/sec)
C1
=
Coefficient related to condition of
submergence and crest shape
(Figure 43)
C2
=
Coefficient related to crest shape
(Figure 44)
B
=
Weir length, i.e. the weir dimension
across the river or stream (m)
H
=
Head of water over the weir crest (m)
Three general types of weirs are shown in Figure 45. The
choice depends, among other aspects, on:
Y
Availability of local materials
Y
Available funds
Y
Local site conditions and floods
As an example, a broad-crested weir would be selected if
gabion baskets were available as construction material
(Figure 46). Gabion baskets are made of galvanized steel
and look like pig netting (see Module 13). However, for the
filling of gabion baskets large quantities of stones are
required as well as plenty of cheap labour, since the
construction method is labour intensive. Stone size is
critical for a gabion weir, as large stones leave big spaces
between them that allow water to quickly flow through,
while too small stones may pass through the mesh.
Figure 45
Types of weirs
a. Broad-(long)crested weir
b. Short-crested weir
c. Sharp-crested weir
65
Irrigation manual
Figure 46
Gabion weir
Example 12
In Example 11 the weir coefficient C, which is the product of C1 and C2, was assumed to be 1.60. Can this be
confirmed by calculating C1 and C2 respectively?
From Example 11 the difference between the water level in the river and the sill elevation can be calculated as follows:
hu = 125.35 - 124.70 = 0.65m
The weir length L is 0.50 m, thus
L
L
0.50
in Figure 44 is:
=
= 0.77
hu
hu
0.65
This relates to a weir type between 1c and 1d in Figure 44. By interpolation, C2 is approximately 1.8.
The difference between the canal design water level and the sill elevation hd = 0.20 m.
Thus
hd
h
0.20
, which is the y-axis in Figure 43, is: d =
= 0.31
hu
hu
0.65
Using the curve for weir type 1b-d in Figure 43, gives a value for C1 of approximately 0.9.
Thus C = C1 x C2 = 0.9 x 1.8 = 1.62, which is almost the same as the weir coefficient 1.60 used in Example 11.
Example 13
A broad-crested weir is to be constructed with gabion baskets. The top width L, which is the dimension of the weir in
the direction of the river, is 1.50 m. There will be non-submerged conditions, which means that the water level
downstream of the weir will be below the weir crest. The design discharge is 37 m3/sec. Due to local site conditions,
the head of water over the crest should not exceed 0.75 m. The freeboard (F), which is the distance between the
design level of the water and the top of the river bank is 0.70 m (Figure 46). What should be the weir length or the
dimension of the weir across the river?
The first step is to determine the values of C1 and C2 from Figures 43 and 44 respectively:
hd is the distance from the crest of the weir to the design water level downstream of the weir (Figure 41). Since there
will be non-submerged conditions, hd will be below the crest of the weir. This means that hd is 0.
As a result
hd
= 0, thus C1 = 1 (Figure 43)
hu
L = 1.50 m and hu = 0.75 m, thus
L
= 2, which means that C2 = 1.5, which is weir type 1b in Figure 44.
hu
Substituting the above data in Equation 24 gives:
37 = 1.0 x 1.5 x B x 0.753/2 Ÿ B = 38 m.
66
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 47
Typical parameters used in the design of a stilling basin
The downstream side of the weir has to be protected using
a stilling basin to dissipate the energy of the dropping water.
It could be constructed using masonry, concrete, gabions
or Reno mattresses.
Design of a stilling basin
The length of the stilling basin should be correctly
determined in order to avoid bed scour and the subsequent
undermining of the structure. The parameters used in the
design of a stilling basin are shown in Figure 47.
Equation 27
d1
z
= 0.54 x D0.425
Equation 28
d2
z
= 1.66 x D0.27
Equation 29
Lj = 6.9 x (d2 - d1)
The empirical formulae to use for the design of a stilling
basin (apron) are:
Where:
D
= Drop number (no limit)
Equation 25
q
= Discharge per metre length of the weir (m2/sec)
g
= Gravitational force (9.81 m/sec2)
D =
q2
z
= Drop (m)
(g x z3)
Ld
= Length of apron from the drop to the point
where the lowest water level d1 will occur
(hydraulic jump) (m)
d1
= Lowest water level after the drop (m)
d2
= Design water level after the apron (m)
Lj
= Length of apron from the point of lowest
water level to the end of the apron (m)
Equation 26
Ld
z
= 4.30 x D0.27
Example 14
A weir with a length B of 38 m across the river and a design discharge Q of 37 m3/sec, has a design drop z of 1.25 m.
What will be the apron length?
The unit discharge is
37
= 0.974 m3/sec per metre length of weir.
38
Substituting this value and the drop z in Equations 25 to 29 for the design of a stilling basin dimension gives:
D =
d1
1.25
0.9742
= 0.05
(9.81 x 1.253)
= 0.54 x 0.050.425 Ÿ d1 = 0.19 m
Ld
= 4.30 x 0.050.27 Ÿ Ld = 2.40 m
1.25
d2
1.25
= 1.66 x 0.050.27 Ÿ d2 = 0.92 m
Lj = 6.9 x (0.92 - 0.19) = 5.04 m
Thus the total apron length is (Ld + Lj) = 2.40 + 5.04 = 7.44 m
67
Irrigation manual
Apron floors should have sufficient thickness to counterbalance the uplift hydrostatic pressure and should be
sufficiently long to prevent piping action. This is
responsible for the removal of the bed material from under
the floor, thereby causing its collapse.
Bed material that allows uplift is liable to piping. Piping
could be avoided by using sheet piling, which is a method
whereby metal or wooden posts are driven vertically into
the ground until they reach an impermeable sub-layer.
However, this is expensive. Alternatively, horizontal,
impermeable layers could be provided. By applying Lane’s
weighted-creep theory, which is an empirical, but simple
and proven method, the length can be determined. This is
defined by the following terms:
Y
The weighted-creep distance Lw of a cross-section of a
weir or a dam is the sum of the vertical creep distances
(steeper than 45°) plus one-third of the horizontal
creep distances (Equation 30).
Y
The weighted-creep ratio is the weighted-creep
distance (Lw) divided by the effective head on the
structure, which in this case is the drop (z)
(Equation 31).
Y
The upward pressure may be estimated by assuming
that the drop in pressure from headwater to tail water
along the line of contact of the foundation is
proportional to the weighted-creep distance
(Equation 32).
Figure 48 shows the different heights and lengths to be used
in determining the weighted-creep ratios.
The weighted-creep distance is as follows:
Equation 30
Lw = h1 + h2 + h3 + h4 + h5 +
1
3
(W1+ L1 +L2 + W2)
Figure 48
Schematic view of a weir and apron
68
The weighted-creep ratio is formulated as follows:
Equation 31
Lw
z
For designing the floor thickness, the uplift pressure P has
to be estimated. The uplift pressure at point B of Figure 48
is calculated as follows:
Equation 32
h1 + h2 + h3 +
Pb = z -
1
3
(W1 + L1)
Lw
The thickness of a floor can be determined using the
following equation:
Equation 33
t = Uplift pressure x
Unit weight of water
Unit weight of submerged mansonry
The recommended weighted-creep ratios are given in
Table 25.
Table 25
Weighted-creep ratios for weirs, depending on soil
type
Bed materials
Weighted-creep ratio
Medium sand
6
Coarse sand
5
Fine gravel
4
Medium gravel
3.5
Coarse gravel
3.0
Boulders with gravel
2.5
Medium clay
2
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Thus, for a given type of bed material, the weighted-creep
ratio can be brought within the recommended value by
selecting a suitable combination of the floor lengths and
vertical cut-offs, as given in Table 25. The materials being
used often determine the cut-off walls, though the apron
length should also be long enough to dissipate the energy.
Example 15
A masonry weir (Figure 49) has to be built in a coarse sand bed material. The proposed dimensions are as follows
(Figure 48):
z
= 1.25 m
h5
= 1.00 m
h1
= 1.00 m
W1
= 1.50 m
h2
= 0.25 m
L1
= 3.50 m
h3
= 0.50 m
L2
= 3.00 m
h4
= 0.50 m
W2
= 1.00 m
Would this structure be safe against piping?
The weighted-creep distance Lw is:
1
Lw = 1.00 + 0.25 + 0.50 + 0.50 + 1.00 +
3
(1.50 + 2.50 + 1.00) = 6.25 m
The weighted creep ratio is:
Lw
z
=
6.25
1.25
= 5.0
Comparing this value to the recommended one given for coarse sand in Table 25 shows that the selection of the
structure dimensions is acceptable and that no piping should be expected.
Example 16
Using the same data as given in the previous example, calculate the floor thickness at points B and D (Figure 48).
Assuming a masonry floor, the unit dry weight can be taken as 2 400 kg/m3, while the unit weight of water is 1 000
kg/m3. The submerged weight of masonry is the difference between the unit dry weight of the masonry minus the unit
weight of water, thus: (2 400 – 1 000) = 1 400 kg/m3.
The uplift pressure at point B is:
1.00 + 0.25 + 0.50 +
Pb = 1.25 -
1
3
(1.50 + 3.50)
= 0.70 m
6.25
Thus the required floor thickness at point B is:
t = 0.70 x
1 000
1 400
= 0.50 m
Similarly for point D, the uplift pressure is:
1.00 + 0.25 + 0.50 + 0.50 +
Pb = 1.25 -
1
3
6.25
(1.50 + 3.50 + 3.00)
= 0.46 m
The required floor thickness at point D is:
t = 0.46 x
1 000
1 400
= 0.33 m
As there is a drop in pressure from the head to the tail of the structure, the floor thickness can be less at point D,
which is further away from the head of the structure.
69
Irrigation manual
Figure 49
Masonry weir and apron
6.1.3. River offtake using a dam
down level one would like to abstract the design discharge.
Design of dams will not be discussed in this Irrigation
Manual. For this the reader is referred to other specialized
literature available. In this section only those aspects of
dams that affect irrigation designs will be discussed. A
typical dam cross-section is given in Figure 50.
A simple relation between the discharge, the difference in
height between lowest drawdown level and outlet pipe
invert and outlet pipe diameter can be obtained using
Equation 34:
Equation 34
Lowest drawdown level
The lowest drawdown level is the minimum water level in
the reservoir that can be abstracted into the irrigation
system. The water remaining below the lowest drawdown
level is called dead storage. This could be used as drinking
water for human beings and animals. The lowest drawdown
level often coincides with the latter part of the dry season,
when water requirements are high. Even at the lowest draw
Figure 50
Dam cross-section at Nabusenga
70
Q = C x A x —2gh
Where:
Q
= Discharge (m3 /sec)
C
= Discharge coefficient, approximately 0.5
A
= Cross-sectional area of pipe (m2)
g
= Gravitational force (9.81 m/sec2)
h
= Available head (m)
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Example 17
The lowest drawdown level of Nabusenga dam is 3.38 m above the outlet pipe (Figure 50). The pipe has a diameter
of 225 mm. Would this outlet pipe be able to deliver a discharge of 78.3 l/sec?
Substituting these data in Equation 34 gives:
78.3 x 10-3 = 0.5 x (
1
4
x S x 0.2252) x (2 x 9.81 x h)1/2 Ÿ hrequired = 0.79 m
The minimum required head of 0.79 m is much less than the havailable of 3.38 m. Thus no reduction in discharge
should be expected when the water level in the dam reaches its minimum level.
Example 18
Using the data from Example 17, what would be the outlet pipe diameter if a discharge of 78.3 l/sec has to be
abstracted at a minimum available head of 1.25 m instead of 3.38 m?
Substituting the data in Equation 34 gives:
78.3 x 10-3 = 0.5 (
1
x S x d2) x (2 x 9.81 x 1.25)1/2 Ÿ d = 200 mm
4
Friction losses in outlet pipe
There should be sufficient head available to overcome
friction losses in the outlet pipe as well as in the conveyance
pipeline in case the conveyance system is a pipe and not an
open canal. The available head refers to the water height
above the outlet pipe. The friction losses (HL) through a
pipe can be calculated using the Hazen-Williams equation,
which was given in Equation 21 (see Section 5.2.2):
Kx
Hf100 =
Q
The value of the material constant C depends on the
smoothness of the material (Table 24). If the pipe size is
small in relation to the discharge, high friction losses are
expected, which means that the water head above the pipe
outlet should be large. In such a case, the discharge would
be reduced at a lower drawdown level than for a larger pipe.
6.1.4. Scour gates for sedimentation control
1.852
Many rivers carry substantial sediment loads, especially
during the rainy season, in the form of sand, silt, weeds, moss
and tree leaves. Approximately 70% of all suspended and bed
load sediments travel in the lower 25% of the flow profile.
C
D4.87
Example 19
The Nabusenga dam has a 70 m long AC outlet pipe with a diameter of 225 mm. What are the friction losses for
discharges of 78.3 l/sec and 32.6 l/sec, including 20% extra for minor losses?
Substituting the above data in Equation 21 gives for Q = 78.3 l/sec:
1.22 x
Hf100 =
78.3
1.852
140
= 1.46 m per 100 m or HL = 1.02 m per 70 m + 20% = 1.22 m
2254.87
For Q = 32.6 l/sec:
1.22 x
Hf100 =
32.6
140
2254.87
1.852
= 0.29 m per 100 m or HL = 0.20 m per 70 m + 20% = 0.24 m
Since the minimum head available is 3.38 m (Example 17 and Figure 50) no reduction in discharge is expected, even
if the full discharge of 78.3 l/sec has to be delivered.
71
Irrigation manual
Example 20
What are the friction losses for a discharge of 78.3 l/sec through a 70 m long galvanized steel pipeline with a 200 mm
diameter? The minimum available head is 1.25 m.
1.22 x
Hf100 =
78.3
140
2004.87
1.852
= 3.44 m per 100 m or HL = 2.41 m per 70 m + 20% = 2.89 m
Already at 1.64 m (2.89 - 1.25 m) above the minimum drawdown level, the discharge will be reduced as the water
head is insufficient to overcome the friction losses of the design discharge.
It should be noted that with aging the C for galvanized steel pipes drops to 80. This will further increase the head
losses in the pipe.
While suspended silt can be beneficial to the scheme by
adding nutrients to the farmland, coarse sediments usually
cause problems once they are blocked by a weir or other
diversion structure. Headworks have to be adapted to these
sediment loads to avoid silting of canals and structures. A
properly-designed intake should divert only the relatively
clean upper part of the water flow into the canal and
dispose of the lower part down the river. A sluice should
therefore be incorporated into the diversion structure
design. It should be placed in line with the weir near the
canal intake (Figure 51). Its seal level is generally placed at
the river bed level while the floor to the intake gate should
be located higher (Figure 52).
The control arrangement in the scour sluice generally
consists of a series of stop logs (timber, concrete) or a sluice
gate. This arrangement allows the water to be raised when
there are very few or no sediments in the water. During the
flood season, the sluice is permanently open or opened at
regular intervals so that depositions of sediments can be
flushed away. The guide wall prevents lateral movement of
sediments deposited in front of the weir and separates the
flow through the sluice and the flow over the weir.
Figure 52
Scour sluice
72
Figure 51
Gravity offtake with diversion dam
a Desilting section
b Flushing sluice
c Conveyance sluice
d Flushing canal
e Gravel and sand trap
f
Flushing gate
g Regulation basin
Module 7: Surface irrigation systems: planning, design, operation and maintenance
6.2.
Night Storage Reservoirs (NSR)
Night storage reservoirs (NSR) store water during times
when there is abstraction from the headwork but no
irrigation. Depending on the size of the scheme one could
construct either one reservoir, located at the top of the
scheme as shown in Figure 19, or more than one to
command sections of the scheme they are serving.
Night storage reservoirs could be incorporated in the
design of a scheme when:
i.
The distance from the water source to the field is
very long, resulting in a long time lag between
releasing water from the source and receiving it in
the field.
ii
The costs of constructing the conveyance canal or
pipeline are very high because of the large discharge it
has to convey without a NSR. Incorporating a reservoir
means that a smaller size conveyance system can be
built.
iii. The discharge of the source of the water is smaller than
would be required for the area without storing the
water during times of no irrigation.
The following examples illustrate scenarios i to iii.
Example 21
A discharge of 78.30 l/sec has to be delivered through a 7 km long canal with a wetted cross-section of 0.19 m2.
When should the headwork gate be opened, if water has to reach the field at 07.00 hours?
The water velocity (V) was given by the Continuity Equation 12:
V =
Q
A
Substituting the values in Equation 12 gives:
V =
0.0783
0.19
= 0.41 m/sec
The time (t) it takes for the water to reach the top of the field is given:
t =
distance
velocity
=
7 000
0.41
= 17 073 seconds or 4 hours and 45 minutes.
This would mean that the head gate should be opened at 02.15 hours if irrigation is to start at 07:00 hours. If this is
unsuitable for proper management, one should incorporate a night storage reservoir.
Example 22
Water is abstracted from a river with a base flow of less than 78.3 l/sec (required if the delivery period is 10 hours per
day), but more than 32.6 l/sec (required if the delivery period can take place 24 hours per day). If abstraction only
takes place during daytime the area under irrigation would have to be reduced. Determine the size of the reservoir
for the scheme in order to be able to irrigate the whole area.
With a night storage reservoir, one could collect the required discharge of 32.6 l/sec from the water source. At an
abstraction rate of 32.6 l/sec, the volume of water accumulated during the 14 hours when there is no irrigation should
be stored in the night storage reservoir. Thus the volume (V) to be stored is:
V =
32.6 x 3 600 x 14
1 000
= 1 643 m3
If 20% is added to cater for evaporation and seepage losses, a night storage reservoir with a capacity of 1 970 m3
could be proposed.
73
Irrigation manual
Example 23
If the friction losses in a conveyance or supply pipeline, delivering 78.3 l/sec for a period of 10 hours per day, are kept
at around 0.30 m per 100 m, then a 300 mm diameter AC pipe (Class 18) could be used (Figure 35). What pipe size
could be used, if a night storage reservoir were built allowing a water flow 24 hours per day?
If abstraction could take place for 24 hours per day, then the discharge would reduce to 32.6 l/sec (= 78.3/(24/10))
and subsequently a pipe size of 225 mm could be selected, considering the same friction loss of 0.30 m per 100 m.
The need for a night storage reservoir should be carefully
considered, weighing advantages, such as money saving in
water delivery works, against disadvantages, such as cost of
reservoir construction, maintenance, seepage and
evaporation losses and disease vector control costs.
Example 24
If the reservoir in Example 23 has water depth, h, of
2.0 m (the maximum recommended depth for brick
reservoirs), what would be the required diameter for
the reservoir?
6.2.1. Types of reservoirs
Using Equation 35:
Reservoirs can be classified on the basis of:
1 970 =
Y
The material used in construction, such as bricks,
concrete or earth
Y
Their shape, which can be circular, square or
rectangular
1
4
.
x 3.14 x d2 x 2 Ÿ d = 35.42 m
The best site for a reservoir is on a flat area with firm,
uniform soils. It is not recommended to build a reservoir
on made-up ground, unless the compaction is extremely
well done.
Earthen reservoirs
Earthen reservoirs are the most common, as they are
usually cheaper to construct. Figure 53 shows a design of a
typical square earthen reservoir, including the inlet, the
outlet and the spillway. The embankments should be well
compacted. If the original soils are permeable, a core
trench should be dug and filled up with less permeable
soils.
Circular reservoirs
A circular reservoir is the common shape of a concrete or
brick reservoir. It is the most economical, as the perimeter
of a circle is smaller than the perimeter of a square or
rectangle for the same area. It also does not need heavy
corner reinforcement to resist the water pressure, as do
square or rectangular reservoirs. The formula for the
calculation of the volume (V) of a circular reservoir is:
6.2.2. Reservoir components
Foundation and floor
A foundation of 450-600 mm in width and 225-300 mm
in depth for a 250 mm wall thickness should be adequate
for circular reservoirs on firm, solid ground. Normally,
foundations do not need reinforcement, except when
placed on unstable soils.
A floor thickness of 100 mm should be adequate. Often a
reinforcement grid with 200-300 mm interval is placed in
the concrete floor. Joints, meant to control cracking, are
placed in the reservoir floor and the reinforcement grid
should not cut across these joints. The concrete panels
should not exceed 6 m in either length or width. Long
narrow panels should be avoided.
Reinforcement
Equation 35
V =
1
4
Sd2h
Where:
V
= Volume of reservoir (m3)
d
= Diameter (m)
h
= Water depth (m)
74
The pressure of the water in a circular reservoir produces
tension in the wall. Pressure exerted by the water is directly
proportional to the water head (depth) from the surface to
the depth considered. Tension produced in the wall of a
circular reservoir is directly proportional to the water depth
and the diameter of the reservoir.
The tension is taken up, to some degree, by the material of
the wall. However, concrete and bricks are weak in
tension, therefore reinforcement should be provided. In a
Notes: –
–
–
–
Exact position of inlet and outlet to be determined on site
Levels given are approximate, details to be determined on site
Levels related to local benchmarks
All dimensions in metres, unless otherwise stated
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 53
Design of a typical earthen night storage reservoir
75
Irrigation manual
brick or concrete block wall the reinforcement rods are
best placed within the mortar of the horizontal joints
between courses. A course is a continuous level line of
bricks or stones in a wall (Figure 54). A simple formula to
calculate the cross-sectional area (A) of reinforcement
needed per course is:
Figure 54
Courses in brick wall of a reservoir
Equation 36
A = 44.6 x d x H x h
Where:
A
=
Cross-sectional area of
reinforcement needed for the course
or band under consideration (mm2)
d
=
Diameter of the reservoir (m)
H
=
Distance down from the design water
level to the bottom of the course (m)
hcourse =
Height of a course (m)
hband
Height of band (m)
=
The band for a concrete wall is assumed to be 300 mm. The
course for a brick wall is 90 mm (including 15 mm for the
mortar) and the course for a concrete block is assumed to
Example 25
What are the reinforcement requirements for a brick reservoir of 2 m high and 36 m in diameter, with a wall as shown
in Figure 54?
The calculations of the steel rod requirements for courses 1, 5 and 21 are given below. Calculations for all other courses
are similar and summarized in Table 26. Cross-sectional areas of the different rod sizes are given in Table 27.
Using Equation 36:
Course 1:
A = 44.6 x 36 x (1 x 0.09) x 0.09 = 13 mm2
Ÿ 1 steel rod of 4 mm diameter is required
Course 5:
A = 44.6 x 36 x (5 x 0.09) x 0.09 = 65 mm2
Ÿ 3 steel rods of 6 mm diameter are required
Course 21: A = 44.6 x 36 x (21 x 0.09) x 0.09 = 273
mm2
Ÿ 6 steel rods of 8 mm diameter or 4 steel rods
of 10 mm diameter are required
Table 26
Reinforcement requirements in a clay brick wall of a reservoir
Course
Reinforcement
(mm2)
Number of rods
and diameter
Course
Reinforcement
(mm2)
Number of rods
and diameter
1
13.0
1 x 4 mm
12
156.1
3 x 8 mm
2
26.0
1 x 6 mm
13
169.1
4 x 8 mm
3
39.0
2 x 6 mm
14
182.1
4 x 8 mm
4
52.0
2 x 6 mm
15
195.1
4 x 8 mm
5
65.0
3 x 6 mm
16
208.1
5 x 8 mm
6
78.0
3 x 6 mm
17
221.1
5 x 8 mm
7
91.0
4 x 6 mm
18
234.1
5 x 8 mm
8
104.0
4 x 6 mm
19
247.1
5 x 8 mm
9
117.0
5 x 6 mm
20
260.1
6 x 8 mm
10
130.1
5 x 6 mm
21
273.1
6 x 8 mm
11
143.1
5 x 6 mm
Note :It is not recommended to have more than 8 steel bars in a course for a wall thickness of 250 mm. If need be, then it would be necessary to use smaller
diameter bars.
76
Module 7: Surface irrigation systems: planning, design, operation and maintenance
be 160 mm high (including 20 mm for the mortar). The
height could differ, depending on the actual block or brick
used and should be confirmed on site.
Table 27
Cross-sectional areas of reinforcement steel rods
Diameter of rod (mm)
Cross-sectional area
= 1/4 S d2 (mm2)
4
12.6
6
28.3
8
50.3
10
78.5
12
113.1
a screen as a precaution against blockage. The diameter of
the outlet pipe depends on the design discharge and the
available head. This pipe should also have a gate valve to be
able to shut it off.
An overflow pipe should be installed in the wall with its
bottom at the same height as the full supply level of the
reservoir. A 100 mm diameter pipe usually suffices.
A scour pipe should be provided at a level slightly below
floor level, so that the reservoir can be regularly cleaned of
sediments. It could have a 100 mm diameter and should be
fitted with a gate valve.
6.3. Head regulators
Pipe requirements
The main pipe requirements are:
Y
A supply pipe for filling the reservoir
Y
An outlet pipe
Y
An overflow pipe
Y
A scour pipe
The supply pipe generally discharges into the reservoir over
the wall from the outside, although it could also be brought
under the foundation and up through the floor. The pipe
should have a gate valve so that supplies can be shut off
when necessary. The diameter depends on the design
discharge.
The outlet pipe may be installed into the reservoir wall
about 150 mm above floor level or under the foundation
and up through the floor. By positioning the pipe above
floor level, sludge and sediments are prevented from
entering the delivery system. The pipe should be fitted with
A head regulator is a structure used to control, and usually
also to measure, the discharge of water into the irrigation
system. It should be designed in such a way that head losses
are kept as low as possible.
On large schemes needing large quantities of water, head
regulators can be very large and would usually be built in
concrete. The use of concrete will result in strong
structures, but can be expensive. The thickness of the floor
and the walls should be between 10 cm and 15 cm. A
cheaper structure could be a concrete block or brick
structure, which would be suitable for smaller structures
with low walls. Wooden diversion structures could be used
where discharges are less than 200 l/sec. In small schemes,
concrete blocks, bricks or even stones could be used to
build the regulators. In this case, they have manual lifting
gates or moveable weirs.
In addition to the headwork described in Section 6.1, head
regulators could also be located at the top of any canal in
the scheme, for example a secondary canal or even a
Figure 55
A simple in-situ concrete proportional flow division structure (Source: Jensen, 1983)
77
Irrigation manual
tertiary canal. In these cases, the head regulator is usually
called a diversion structure. Weir-type diversion structures
have been discussed in Section 6.1. Below diversion
structures as regulating structures in general will be
discussed.
A diversion structure regulates the flow from one canal into
one or more other canals. It normally consists of a box with
vertical walls in which controllable openings are provided.
The minimum dimensions of the structure depend on its
performance in the fully open position. The width of the
outlet is usually proportional to the division of water flow
to be made. Figures 55-57 show some examples of
diversion structures (in-situ concrete, pre-cast concrete
and timber structures respectively). The walls can be made
either of concrete (10-15 cm thick precast or in-situ) or
masonry or even wood.
Large backup of water upstream of the structure, which
would result in overtopping of the canal, should be avoided.
Since a lined canal is designed to carry water at relatively
high velocities, a full gate-opening at the intake to the box,
covering approximately the same area as the canal section,
should be provided. In earthen canals, gate-opening
dimensions can be based on assuming velocities of less than
1.0-1.5 m/sec.
Figure 56
Precast concrete block division box (Source: FAO, 1975a)
All dimensions are in metres
78
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 57
Timber division structures (Source: FAO, 1975a)
79
Irrigation manual
Example 26
What should be the minimum width of the opening b
of a diversion structure, if the discharge Q is equal to
78.3 l/sec and if the water depth in the opening is not
to exceed 0.30 m?
Using Equation 12:
Q = A x V = 0.30 x b x V
Assuming a Vmax = 1.50 m/sec for concrete lining
and substituting it into Equation 12 gives:
0.00783 = 0.30 x b x 1.50 Ÿ bmin = 0.18 m
Example 27
What should be the water depth over a weir crest, if
the discharge Q is equal to 78.3 l/sec, the weir length
B is equal to 0.40 m and C is equal to 1.75?
minimum so as to maintain command in the canals. This
would be achieved by making a wider and larger diversion
structure.
6.4. Cross regulators
A cross regulator is a structure built across the canal to
maintain the water level at the command level required to
irrigate the fields. Cross regulators could be simple timber
stop logs, check plates, weirs or expensive automatically
operated gates, which automatically control a constant
water level.
In Section 6.1, weirs have been discussed as headwork
structures. In the context of cross regulation, examples of
common weirs are duckbill and diagonal weirs, which
control the water level at a given height, (Figure 58, 59 and
60). Detailed explanations of weirs as discharge
measurement structures are given in Section 6.6.
Using Equation 24:
Q = C x B x H3/2 ŸH = 0.23 m
The structure should be designed in such a way that the
water velocity will not cause erosion in the earthen canal.
Thus, the water velocity should reduce to its canal design
value after the opening and before the water re-enters the
earthen canal.
There is a relationship between the width of the opening of
the gate and the head loss. Hydraulic losses through a
properly designed structure are small. When the ground
slope is very gentle, head losses should be kept to a
Figure 58
Duckbill weir photograph (Source: FAO, 1975b)
80
6.5. Drop structures and tail-end structures
Drop structures and chutes are flow control structures
that are installed in canals when the natural land slope is
too steep compared to the design canal gradient (see
Section 5.1) to convey water down steep slopes without
erosive velocities. If a canal were allowed to follow a steep
natural gradient, the velocities would be too high. This in
turn would cause erosion and make water management
difficult. For this, the canal is divided into different
reaches over its length. Each reach follows the design
canal gradient. When the bottom level of the canal
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 59
Duckbill weir design (Source: FAO, 1975b)
81
Irrigation manual
Figure 60
Diagonal weir (Source: FAO, 1975b)
Range of suitable dimensions for capacities
up to 500 l/s
B
= 0.20 to 1.00
f
= 0.20 to 1.00
y1
= 0.10 to 0.70 (upstream water depth)
H(crt) = 0.05 to 0.15 (difference between upstream
water level and crest level)
s
82
= 0.10 to 0.60
c
= 0.15 (thickness of weir)
l
= (width of available upstream water surface)
B(t)
1
= (crest length) = l x cos
v
v
= (angle between weir crest and cross-section
of channel)
m
= 1.5f - 1.5s + 0.20
k
= Lsinv
p
= fn
t
= f
Module 7: Surface irrigation systems: planning, design, operation and maintenance
becomes too high compared to the natural ground level,
drop structures are installed. Vertical drops are normally
used for the dissipation of up to 1 m head for unlined
canals and up to 2 m head for lined canals. For larger
drops, chutes are usually used.
For canals that do not require command, the position of
drops is determined by considering the cost of canal
construction, including balancing the cuts and fills and the
cost of the structure. Where there is need for command,
the drops should be located in such a way that the canal
banks are not too high, but still keeping enough command
at the same time.
Figures 61 and 62 show examples of drop structures built
with different materials.
6.5.1. Vertical drop structure
An important aspect of a drop is the stilling basin,
required to avoid downstream erosion. The floor of the
stilling basin should be set at such a level that the
hydraulic jump occurs at the upstream end of the basin
floor in order to avoid erosion at the unprotected canal
bed downstream. A common straight drop structure is
shown in Figure 63.
Figure 61
Some drop structures used in open canals (Source: James, 1988)
83
NOTES: All dimensions in metres unless otherwise stated.
Earthworks and surrounding grounds not shown.
Design based on standard drop structure
implemented in Mushandike by hydraulics research.
–
–
–
Concrete mix is 1 : 2 : 3.
Mortor/plaster mix is 1 : 4.
All brickwalls to be plastered on both sides.
84
‡ 8 mm steel bar in 0.15 mm grid
Reinforced concrete
Drawing: NABU/15
Scale as shown
Irrigation manual
Figure 62
Standard drop structure without stilling basin
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 63
A vertical drop structure
Example 28
Given a discharge of 0.0783 m3/sec, a drop height of 0.50 m and a drop width of 0.30 m, what would be (Figure 63
and 47):
–
The length of the apron from the drop to the hydraulic jump, where the lowest water level will occur (Ld)?
–
The height of the jump or the lowest water level after the drop (d1)?
–
The design water level after the drop (d2)?
–
The total length of the apron (LB)?
Q = 0.0783 m3/sec Ÿ Q = 0.261 m3/sec per 0.30 m width
Using Equation 25:
D =
0.2612
(9.81 x 0.053)
Ÿ D = 0.0556
Substituting the data in Equations 26 to 29 respectively gives:
Ld
= 0.50 x 4.30 x 0.05560.27 = 0.99 m
d1
= 0.50 x 0.54 x 0.05560.425 = 0.15 m
d2
= 0.50 x 1.86 x 0.05560.27 = 0.38 m
Lj
= 6.9 x (0.38 - 0.15) = 1.59 m Ÿ LB = 0.99 + 1.59 = 2.58 m
Due to the impact of the water flow on the basin floor and
the turbulent circulation, an amount of energy ('HL) is
lost. Further energy is lost in the hydraulic jump
downstream of the section U in Figure 63. Experiments
have shown that the energy head (H2) is equal to about
2.5 x d1 (that is 2.5 times the critical depth). This provides
a satisfactory basis for design.
An upward step is often added at the end of the basin floor
in order to be sure that the hydraulic jump occurs
immediately below the drop. This step has the disadvantage
of retaining standing water when the canal is not in use,
thereby posing a danger to health.
6.5.2. Chutes
An example of a chute structure is given in Figure 64.
Chutes are normally rectangular, although they are also
made in a trapezoidal shape. They have an inlet, a steepsloped section of a lined canal, a stilling basin or some other
energy dissipating devices, baffle blocks, and an outlet. The
energy dissipation is usually effected by the creation of a
85
Irrigation manual
Figure 64
A chute structure
hydraulic jump at the toe of the steep-sloped section of the
structure. Baffle blocks could be used to facilitate the
creation of a hydraulic jump.
The slope of the downstream face (steep-sloped section)
usually varies between 1 in 4 and 1 in 6. The length of the
stilling basin (often called cistern), Lj, can be estimated
with the following equation:
Equation 37
Lj = 5 x d2
6.5.3. Tail-end structures
Most canals need some way of getting rid of water. A tailend structure should be provided at the end of the canal so
that excess water can flow safely into the drain. It normally
consists of a drop structure to bring the water level from a
command canal level to the drain level from where it will be
taken to the main drainage system of the project.
that upstream and downstream flow are independent, can be
used as a measuring device, provided that it can be calibrated.
Standard structures, which have already been accurately
described and calibrated, exist. Weirs, flumes and orifices are
the devices that are normally used for discharge
measurement.
6.6.1. Discharge measurement equations
The three fundamental equations used to solve discharge
problems in canals are based on the principles of
conservation of mass, energy and momentum. For our
purposes, only the conservation of mass and energy
equations will be dealt with.
Conservation of mass
Conservation of mass leads to the Continuity Equation 12
to be constant:
Q = A x V = Constant
6.6. Discharge measurement in canals
Conservation of energy
Discharge measurement in irrigation schemes is important
for the following reasons:
Conservation of energy applied along a streamline results in
the Bernoulli Equation:
Y
To ensure the maintenance of proper delivery schedules
Equation 38
Y
To determine the amount of water delivered for water
pricing, where it is applicable
Y
P
J
+
V2
2g
+ z = Constant
To detect the origin of water losses and to estimate the
quantity
Where:
P
= Pressure (kgf/m2)
Y
To ensure efficient water distribution
J
= Density of water (kg/m3)
Y
To conduct applied research
V
= Water velocity (m/sec)
g
= Gravitational force (9.81 m/sec2)
z
= Elevation above reference line (m)
Almost any kind of obstacle that partially restricts the flow of
water in an irrigation canal and provides a free fall, to ensure
86
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 65
Static and velocity heads
Equation 38 sums up the pressure head, velocity head and
gravitational head to give the total head. For an open
canal, the pressure head equals the water depth h (Figure
65).
When there is frictional loss along the flow path, an
expression for frictional head loss must be included. Thus
applying the Bernoulli Equation to two successive crosssections along a flow path results in:
Equation 40
E =
h
2g
+ V2
Where:
E
= Specific energy (m)
h
= Depth of flow (m)
g
= Gravitational force (9.81 m/sec2)
V
= Water velocity (m/sec)
Equation 39
h1 =
V12
2g
+ z1 = h2 +
V22
2g
+ z2 + HL
The numbers 1 and 2 refer to the first and second crosssection in Figure 65. HL is the frictional head loss.
Assuming a uniform velocity distribution, the specific
energy is constant across the section. Combining the above
equation and the Continuity Equation gives:
Equation 41
E =
h
2g
+
Q
2
A
Specific energy
The concept of specific energy is used in the analysis of
critical flow. At any cross-section of a canal, the energy with
respect to the canal bed is referred to as specific energy. It
is derived from the Bernoulli Equation according to the
following equation:
The cross-sectional area varies with the depth of flow only
if the geometry of the canal is constant. Therefore, for a
given discharge the specific energy is a function of depth
alone.
87
Irrigation manual
Figure 66
Variation of specific energy with depth of flow for different canal shapes
Specific energy can be determined for different structures:
Rectangular canal
A = bxh
E =
h
2g
+
Q
2
bxh
Trapezoidal canal (with a side slope of 60°)
A = b x h + 0.58 x h2
E =
h
2g
+
Q
For critical flow, the specific energy is a minimum for a
given discharge. In this case, a relationship exists between
the minimum specific energy and the critical depth. This
relationship is found by differentiating Equation 41 with
respect to h, while Q remains constant. This gives:
2
b x h + 0.58 h2
Plotting E against h for different values of (Q/b) gives curves
as shown in Figure 66.
The curves show that, for a given discharge and specific
energy, there are two alternate depths of flow, which
coincide at a point where the specific energy is a minimum
for a given discharge. Below this point, flow is physically not
possible. At this point flow is critical and it occurs at critical
depth and velocity. At a greater depth, the velocity is low
and flow is sub-critical. At the lesser depth, the velocity is
high and flow is super-critical.
For sub-critical flow, the mean velocity is less than the
velocity of propagation of stream disturbances such as
waves. Thus, stream effects can be propagated both
88
upstream and downstream. This means that downstream
conditions affect the behaviour of flow. When flow is supercritical, the velocity of flow exceeds the velocity of
propagation. Consequently, stream effects (for example,
waves) cannot be transmitted upstream, and downstream
conditions do not affect the behaviour of the flow.
Equation 42
Vc =
g x Ac
1/2
bc
Froude Number
The Froude Number is calculated according to Equation 19
(see Section 5.1.2):
Fr =
v
(g x h)1/2
Where:
Fr
= 1 for critical flow
Fr
= > 1 for super-critical flow
Fr
= < 1 for sub-critical flow
Module 7: Surface irrigation systems: planning, design, operation and maintenance
If a structure is built in a canal which has sub-critical flow,
it may cause the flow to pass through the critical to the
super-critical state. This means that the state upstream of
the structure becomes independent of the state
downstream. This can either be achieved if the structure
narrows the canal, which means increasing the (Q/b)ratio without altering the specific energy, or if it raises the
canal bed, which means reducing the specific energy
without altering the discharge per unit width. That is how
critical flow is obtained with a measuring device. A
control section in a canal is a section that produces a
definitive relationship between water depth and
discharge.
Hydraulic jump
If, through a structure, super-critical flow is introduced in
a canal where the normal flow is sub-critical, flow adjusts
back to the sub-critical state through a hydraulic jump in
which the water level rises over a short distance with much
visible turbulence. This situation occurs, for example,
downstream of a sluice gate or a flume. It is undesirable to
have a hydraulic jump in an unlined canal because of the
risk of scour. In such cases, a jump is usually induced over
a concrete apron by means of a sill or baffle blocks set in the
floor, as shown in Figure 67.
The relationship between depths just upstream and
downstream of a hydraulic jump is found by the application
of the momentum theory to the simplified situation shown
in Figure 68. It is assumed that boundary frictions are
negligible over the length of the jump. For a rectangular
canal it can be shown that:
Equation 43
h2 = -
h1
2
+ 0.5 x h12 + 8 x V12 x
h1
1/2
9
6.6.2. Weirs
The weir is the most practical and economical device for
water measurement. Weirs are simple to construct, easy to
inspect, robust and reliable. Discharge measurement weirs
can either be sharp-crested (Figure 69, 70, 71) or broadcrested (Figure 72).
Sharp-crested weirs
Sharp-crested weirs, also called thin plate weirs, consist of
a smooth, vertical, flat plate installed across the channel and
perpendicular to the flow (Figure 69). The plate obstructs
flow, causing water to back up behind the weir plate and to
Figure 67
Hydraulic jump over a concrete apron
Figure 68
The form of a hydraulic jump postulated in the momentum theory
89
Irrigation manual
Figure 69
Parameters of a sharp-crested weir
flow over the weir crest. The distance from the bottom of
the canal to the weir crest, p, is the crest height. The depth
of flow over the weir crest, measured at a specified distance
upstream of the weir plate (about four times the maximum
h1), is called the head h1. The overflowing sheet of water is
known as the nappe.
Thin plate weirs are most accurate when the nappe springs
completely free of the upstream edge of the weir crest and
air is able to pass freely around the nappe. The crest of a
sharp-crested weir can extend across the full width of
channel or it can be notched. The most commonly used
notched ones are:
Y
Rectangular contracted weir
Y
Trapezoidal (Cipoletti) weir (Figure 70)
Y
Sharp sided 90° V-notch weir (Figure 71)
case be less than 30 cm. This will allow the water to fall
freely, leaving an airspace under and around the jets.
ii. At a distance upstream of about four times the
maximum head a staff gauge is installed on the crest
with the zero placed at the crest elevation, to measure
the head h1.
iii. For the expected discharge, the head (h1) should not
be less than 6 cm and should not exceed 60 cm.
iv. For rectangular and trapezoidal weirs, the head (h1)
should not exceed 1/3 of the weir length.
v.
The weir length should be selected so that the head for
the design discharge will be near the maximum, subject
to the limitations given in (ii) and (iii).
vi. The thickness of the crest for sharp-crest weirs should
be between 1-2 mm.
The type and dimensions of the weir chosen are based on
the expected discharge and the limits of its fluctuation. For
example, a V-notch weir gives the most accurate results
when measuring small discharges and is particularly
adapted to the measurement of fluctuating discharges.
Calibration curves and tables have been developed for
standard weir types.
In sediment-laden canals, a main disadvantage of using
weirs is that silt is deposited against the upstream face of the
weir, altering the discharge characteristics. Weirs also
cannot be used in canals with almost no longitudinal slopes,
since the required difference in elevation between the water
levels upstream and downstream side of the weir is not
available.
The conditions and settings for standard weirs are as
follows:
Discharge equations for weirs are derived by the application
of the Continuity and Bernoulli Equations (Equation 12
and 38 respectively). In each case, a discharge coefficient is
used in order to adjust the theoretical discharge found by
laboratory measurements.
i.
90
The height of the crest from the bottom of the
approach canal (p) should preferably be at least twice
the depth of water above the crest and should in no
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Equation 45
Rectangular contracted weir
A rectangular contracted weir is a thin-plate weir of
rectangular shape, located perpendicular to the flow. To
allow full horizontal contraction of the nappe, the bed and
sides of the canal must be sufficiently far from the weir
crest and sides.
Many practical formulae have been developed for
computing the discharge, amongst which are the
following:
Equation 44
Hamilton-Smith formula:
Q = 0.616 x 1 -
0.1h
b
x
2
3
Francis formula:
Q = 1.838 x (b - 2h) x h3/2
Where:
Q
= Design discharge over weir (m3/sec)
b
= Length of weir crest (m)
h
= Design water depth measured from the top
of the weir crest (m)
Table 28 gives discharge data related to length of crest, b,
and water head, h, over a weir.
Trapezoidal (Cipoletti) weir
x
(2g)1/2
xbx
h3/2
The trapezoidal weir has a trapezoidal opening, the base
being horizontal. The Cipoletti weir is a trapezoidal weir
Example 29
A rectangular contracted weir has to be placed in a lined canal. The design discharge is 0.0783 m3/sec and the
maximum allowable water depth, h, at the measuring gauge can be 0.15 m. What should be the minimum weir crest
length, b, calculated using the Francis formula?
Using Equation 45:
Q = 0.0783 = 1.838 x (b - 0.2 x 0.15) x 0.153/2 = 0.1068 x b - 0.0032 Ÿ b = 0.76 m.
Table 28
Discharge Q (m3/sec) for contracted rectangular weir, depending on h and b
Head h (m)
Length of crest b (m)
0.30
0.40
0.50
0.75
1.00
1.25
1.50
0.0025
0.0001
0.0001
0.0001
0.0002
0.0002
0.0003
0.0003
0.015
0.0010
0.0013
0.0017
0.0025
0.0034
0.0042
0.0051
0.030
0.0028
0.0038
0.0047
0.0071
0.0095
0.0119
0.0143
0.045
0.0051
0.0069
0.0086
0.0130
0.0174
0.0218
0.0262
0.060
0.0078
0.0105
0.0132
0.0199
0.0267
0.0335
0.0402
0.075
0.0108
0.0145
0.0183
0.0278
0.0372
0.0466
0.0561
0.090
0.0140
0.0190
0.0239
0.0363
0.0487
0.0612
0.0736
0.105
0.0175
0.0237
0.0300
0.0456
0.0612
0.0769
0.0925
0.12
0.0211
0.0287
0.0364
0.0555
0.0746
0.0937
0.1128
0.15
0.0288
0.0395
0.0502
0.0769
0.1036
0.1303
0.1570
0.0511
0.18
0.0651
0.1002
0.1353
0.1704
0.2055
0.21
0.0810
0.1253
0.1695
0.2137
0.2580
0.24
0.0977
0.1517
0.2058
0.2598
0.3139
0.27
0.1795
0.2440
0.3085
0.3730
0.30
0.2084
0.2840
0.3595
0.4350
0.36
0.2692
0.3685
0.4678
0.5671
0.42
0.4584
0.5835
0.7086
0.48
0.5527
0.7055
0.8584
0.54
0.8331
1.0155
0.60
0.9655
1.1791
91
Irrigation manual
with the sides having an outward sloping inclination of 1
horizontal to 4 vertical (Figure 70). This side slope is such
that the water depth-discharge relationship is the same as
that of a full width rectangular weir.
Figure 70
Trapezoidal (Cipoletti) weir
The discharge equation for a Cipoletti weir is:
Equation 46
Q = 1.859 x b x h3/2
Where:
Q
= Design discharge over weir (m3/sec)
b
= Length of weir crest (m)
h
= Design water depth measured from the top
of the weir crest (m)
Table 29 shows discharge data, related to the design water
depth, h, and weir length, b.
Example 30
A Cipoletti weir has to be placed in a lined canal. The design discharge is 0.0783 m3/sec and the maximum allowable
head, h, at the measuring gauge is 0.15 m. What should be the minimum weir crest length, b?
Using Equation 46:
0.0783 = 1.859 x b x 0.153/2 = 1.108b Ÿb = 0.73 m
Table 29
Discharge Q (m3/sec) for Cipoletti weir, depending on h and b
Head h
(m)
0.30
0.40
0.50
0.75
1.00
1.25
1.50
0.0025
0.0001
0.0001
0.0001
0.0002
0.0002
0.0003
0.0003
0.015
0.0010
0.0014
0.0017
0.0026
0.0034
0.0043
0.0051
0.030
0.0029
0.0039
0.0048
0.0072
0.0097
0.0121
0.0145
0.045
0.0053
0.0071
0.0089
0.0133
0.0177
0.0222
0.0266
0.060
0.0082
0.0109
0.0137
0.0205
0.0273
0.0341
0.0410
0.075
0.0115
0.0153
0.0191
0.0286
0.0382
0.0477
0.0573
0.090
0.0151
0.0201
0.0251
0.0376
0.0502
0.0627
0.0753
0.105
0.0199
0.0253
0.0316
0.0474
0.0632
0.0791
0.0949
0.12
0.0232
0.0309
0.0386
0.0580
0.0773
0.0966
0.1159
0.15
0.0324
0.0432
0.0540
0.0810
0.1080
0.1350
0.1620
0.0568
0.18
0.0710
0.1065
0.1420
0.1774
0.2129
0.21
0.0894
0.1342
0.1789
0.2236
0.2683
0.24
0.1093
0.1639
0.2186
0.2732
0.3278
0.27
0.1956
0.2608
0.3260
0.3912
0.30
0.2291
0.3054
0.3818
0.4582
0.36
0.3011
0.4015
0.5019
0.6023
0.48
0.3060
0.6325
0.7590
0.54
0.6182
0.7727
0.9273
0.9220
1.1065
1.0799
1.2959
0.60
92
Length of crest b (m)
Module 7: Surface irrigation systems: planning, design, operation and maintenance
V-notch weir
A V-notch weir has two edges that are symmetrically
inclined to the vertical to form a notch in the plane
perpendicular to the direction of flow. The most commonly
used V-notch weir is the one with a 90° angle. Other
common V-notches are the ones where the top width is
equal to the vertical depth (1/2 x 90° V-notch) and the one
where the top width is half of the vertical depth
(1/4 x 90° V-notch) (Figure 71). The V-notch weir is an
accurate discharge-measuring device, particularly for
discharges less than 30 l/sec, and it is as accurate as other
types of sharp-crested weirs for discharges from 30 to 300
l/sec (U.S. Department of Interior, 1975).
To operate properly, the weir should be installed so that the
minimum distance from the canal bank to the weir edge is
at least twice the head on the weir. In addition, the distance
from the bottom of the approach canal to the point of the
weir notch should also be at least twice the head on the weir
(U.S. Department of Interior, 1975).
The general and simple discharge equation for a V-notch
weir is:
Equation 47
Q = 1.38 x tan(
1
2
x T) x h5/2
Where:
Q
= Design discharge over the weir (m3/sec)
T
= Angle included between the sides of the
notch (degrees)
h
= Design water depth (m)
Table 30 gives discharge data for the three common Vnotches related to water depth (head) and angle°.
Example 31
A design discharge of 0.0783 m3/sec has to pass through a V-notch weir with an angle T of 90°. What will be the
water depth over the weir?
Substituting the above data in Equation 47:
0.0783 = 1.38 x tan(
1
2
x 90) x h5/2 Ÿ h5/2 = 0.0783 Ÿ h = 0.317 m.
Figure 71
V-notch weirs
93
Irrigation manual
Table 30
Discharge Q (m3/sec x 10) for a 90° V-notch weir, depending on h
Head
(m)
Discharge
(m3/sec x 10)
Head
(m)
Discharge
(m3/sec x 10)
Head
(m)
Discharge
(m3/sec x 10)
0.050
0.008
0.160
0.142
0.270
0.523
0.055
0.010
0.165
0.153
0.275
0.548
0.060
0.012
0.170
0.165
0.280
0.573
0.065
0.015
0.175
0.177
0.285
0.599
0.070
0.018
0.180
0.190
0.290
0.626
0.075
0.022
0.185
0.203
0.295
0.653
0.080
0.025
0.190
0.217
0.300
0.681
0.085
0.029
0.195
0.232
0.305
0.710
0.090
0.034
0.200
0.247
0.310
0.739
0.095
0.039
0.205
0.263
0.315
0.770
0.100
0.044
0.210
0.279
0.320
0.801
0.102
0.050
0.215
0.296
0.325
0.832
0.110
0.056
0.220
0.313
0.330
0.865
0.115
0.062
0.225
0.332
0.335
0.898
0.120
0.069
0.230
0.350
0.340
0.932
0.125
0.077
0.235
0.370
0.345
0.966
0.130
0.084
0.240
0.390
0.350
1.002
0.135
0.093
0.245
0.410
0.355
1.038
0.140
0.102
0.250
0.432
0.360
1.075
0.145
0.111
0.255
0.454
0.365
1.113
0.150
0.121
0.260
0.476
0.370
1.152
0.155
0.131
0.265
0.499
0.375
1.191
0.380
1.231
Broad-crested weir
A broad-crested weir is a broad wall set across the canal
bed. The way it functions is to lower the specific energy and
thus induce a critical flow (Figure 72).
One of the most commonly used broad-crested weirs for
discharge measurements is the Romijn broad-crested weir,
which was developed in Indonesia for use in relatively flat
areas and where the water demand is variable because of
different requirements during the growing season (FAO,
1975b). It is a weir with a rectangular control section, as
shown in Figure 73.
The Romijn weir consists of two sliding blades and a movable
weir crest, which are mounted in one steel guide frame
(Figure 74). The bottom blade, which is locked under
operational conditions, acts as the bottom terminal for the
movable weir. The upper blade, which is connected to the
bottom blade by means of two steel strips placed in the frame
grooves, acts as the top terminal for the movable weir. Two
steel strips connect the movable weir to a horizontal lifting
beam. The horizontal weir crest is perpendicular to the water
flow and slopes 1:25 upward in the direction of the flow. Its
upstream nose is rounded off in such a way that flow
separation does not occur. The operating range of the weir
equals the maximum upstream head (Hcrt) which has been
selected for dimensioning the regulating structure.
Figure 72
Broad-crested weir
1 = vc2/2g
2 = hc
3 = h1
94
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 73
Romijn broad-crested weir, hydraulic dimensions of weir abutments (Source: FAO, 1975b)
Figure 74
The discharge equation for the Romijn broad-crested weir
is written as:
Romijn broad-crested weir, sliding blades
and movable weir crest (Source: FAO, 1975b)
Equation 48
Q =
2
3
x Cd x Cv x
2
3
1/2
xg
x Bt x Hcrt3/2
Where:
Q
=
Design discharge over the weir
(m3/sec)
Cd
=
Discharge coefficient
Cv
=
Approach velocity coefficient
g
=
Acceleration due to gravity
(= 9.81 m/sec2)
Bt
=
Width (or breadth) of the weir across
the direction of flow (m)
Hcrt
=
Design upstream water depth over
the weir (m)
The value of the discharge coefficient, Cd, has been
determined in laboratory tests. For field structures with
concrete abutments, it is advisable to use an average value
of Cd = 1.00. The value of the approach velocity
coefficient, Cv, ranges between 1.00 and 1.18, depending
on Hcrt (Figure 75).
95
Irrigation manual
Figure 75
Approach velocity coefficient, Cv, as a function of the total head over the movable weir crest, Hcrt
(Source: FAO, 1975b)
Where both Cd and Cv are considered to be 1.00,
substituting these values and the value for g in Equation 48
gives Equation 49:
6.6.3. Flumes
Equation 49
Y
Can be used under almost any flow condition
Y
Have smaller head-losses than weirs, thus are more
accurate over a large flow range
Y
Are insensitive to the velocity of approach
Y
Are relatively less susceptible to sediment and debris
transport
q = 1.7 x Bt x Hcrt3/2
More details on the Romijn weir can be found in FAO
(1975b).
Discharge measurement flumes are extensively used in
irrigation schemes mainly because they:
Example 32
A Romijn broad-crested weir has to discharge 0.0783 m3/sec. The maximum allowable water depth over the weir can
be 0.15 m. What should be the minimum width of weir?
Considering a Cd value of 1.00 and an average Cv value of 1.04 (Figure 75), Equation 48 gives:
0.0783 =
2
3
x 1.00 x 1.04 x
2
3
1/2
x 9.81
x Bt x 0.153/2 Ÿ Bt = 0.76 m
Using the simplified Equation 49 would give:
0.0783 = 1.7 x Bt x 0.153/2 Ÿ Bt = 0.79 m
96
Module 7: Surface irrigation systems: planning, design, operation and maintenance
However, major disadvantages of flumes include the relative
large sizes and the accurate manufacturing/construction
workmanship required for optimum performance (James,
1988).
A canal section that causes flow to pass from sub-critical
through critical to the super-critical state forms a control and
the discharge is a single valued function of the upstream water
level. Critical flow can be achieved by raising the canal bed,
thereby reducing the specific energy, or by decreasing the
canal width, thereby increasing the discharge per unit width
(see Section 6.6.1). This latter technique is the one used by
flumes.
A flume has:
Y
A convergent section, in which the flow accelerates
Y
A throat, in which critical flow occurs
Y
A divergent section, in which the flow returns to normal
Super-critical flow passing from the throat will return to
sub-critical flow downstream of the flume. This occurs due
to the development of a hydraulic jump, which is induced
within the divergent section by a sill or other barrier. Where
there is sufficient head available, the divergent section of the
flume could be avoided as the flow could fall freely in a
stilling basin. In this case, weirs could also be used.
However, if canals are expected to carry a lot of sediment,
the flume should be the better choice.
Flumes are most commonly rectangular or trapezoidal in
cross-section. The former type is the most simple to
construct, but if the canal cross-section is not rectangular
there is a risk that unpredictable flow patterns will result
from an abrupt change of cross-section.
The most commonly used flumes are:
Y
Parshall flume
Y
Trapezoidal flume
Y
Cut-throat flume
Figure 76
Parshall flume
97
Irrigation manual
Table 31
Standard dimensions of Parshall flumes (the letters are shown in Figure 76) (Adapted from: FAO, 1975b)
b
A
a
B
C
D
E
‘+“
mm
1”
25.4
363
242
356
93
167
229
2”
50.8
414
276
406
135
214
3”
76.2
467
311
457
178
259
6”
152.4
621
414
610
394
9”
228.6
879
587
864
1”
304.8 1372
914
134
1’6”
457.2 1448
965 1419
L
G
H
K
206
19
M
N
P
R
X
Y
Z
mm
2’
609.6 1524 1016 1495
3’
914.4 1676
76
203
-
29
-
-
8
13
3
254
114
254
257
22
-
43
-
-
16
25
6
457
152
305
309
25
-
57
-
-
25
38
9
397
610
305
610
-
76 305
114
902 406
51
76
-
381
575
762
305
-
76 305
114
1080 406
51
76
-
610
845
914
610
914
-
76 381
229
1492 508
51
76
-
762
1026
914
610
914
-
76 381
229
1676 508
51
76
-
914
1206
914
610
914
-
76 381
229
1854 508
51
76
1118 1645
1219
1572
914
610
914
-
76 381
229
2222 508
51
76
-
4’ 1219.2 1829 1219 1794
1524
1937
914
610
914
-
76 457
229
2711 610
51
76
-
5’ 1524.0 1981 1321 1943
1829
2302
914
610
914
-
76 457
229
3080 610
51
76
-
6’ 1828.8 2134 1422 2092
2134
2667
914
610
914
-
76 457
229
3442 610
51
76
-
7’ 2133.6 2286 1524 2242
2438
3032
914
610
914
-
76 457
229
3810 610
51
76
-
8’ 2438.4 2438 1626 2391
610
4172 610
2743
3397
914
914
-
76 457
229
51
76
-
10’
3048
- 1829 4267
3658
4756
1219
914 1829
-
76
-
343
-
- 305
229
-
12’
3658
- 2032 4877
4470
5607
1542
914 2438
-
152
-
343
-
- 305
229
-
15’
4572
- 2337 7620
5588
7620
1829 1219 3048
-
152
-
457
-
- 305
229
-
20’
6096
- 2845 7620
7315
9144
2134 1829 3658
-
305
-
686
-
- 305
229
-
25’
7620
- 3353 7620
8941 10668
2134 1829 3962
-
305
-
686
-
- 305
229
-
30’
9144
- 3861 7925 10566 12313
2134 1829 4267
-
305
-
686
-
- 305
229
-
40’
12192
- 4877 8230 13818 15481
2134 1829 4877
-
305
-
686
-
305
229
-
50’
15240
- 5893 8230 17272 18529
2134 1829 6096
-
305
-
686
-
- 305
229
-
Parshall flume
The Parshall flume is a widely-used discharge measurement
structure. Figure 76 shows its general form. The
characteristics of Parshall flumes are:
Y
Small head losses
Y
Free passage of sediments
Y
Reliable measurements even when partially submerged
Y
Low sensitivity to velocity of approach
The Parshall flume consists of a converging section with a
level floor, a throat section with a downward sloping floor
and a diverging section with an upward sloping floor. Flume
sizes are known by their throat width.
Care must be taken to construct the flumes accurately if the
calibration curves have to be used. Each size has its own
characteristics, as the flumes are not hydraulic scale models
of each other. In other words, each flume is an entirely
different device (see Table 31).
The flow through the Parshall flume can occur either under
free flow or under submerged flow conditions. Under free
flow the rate of discharge is solely dependent on the throat
width and the measured water depth, ha. The water depth
is measured at a fixed point in the converging section.
98
The upstream water depth-discharge relationship,
according to empirical calibrations, has the following
general form:
Equation 50
Q = K x (ha)u
Where:
Q
= Discharge (m3/sec)
ha
= Water depth in converging section (m)
K
= A fraction, which is a function of the throat
width
u
= Variable, lying between 1.522 and 1.60.
Table 32 gives the values for K and u for each flume size.
When the ratio of gauge reading hb to ha exceeds 60% for
flumes up to 9 inches, 70% for flumes between 9 inches
and 8 feet and 80% for larger flume sizes, the discharge is
reduced due to submergence. The upper limit of
submergence is 95%, after which the flume ceases to be an
effective measuring device because the head difference
between ha and hb becomes too small, such that a slight
inaccuracy in either head reading results in a large discharge
measurement error.
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Table 32
Discharge characteristics of Parshall flumes
Throat width
b
Discharge range
Minimum
feet + inches
Maximum
(m3/sec x 10-3)
Equation
Q = K x ha
Head range
u
Minimum
(m3/sec)
1.55
Modular limit
Maximum
(m)
hb/ha
(m)
1”
0.09
5.4
0.0604 ha
0.015
0.21
0.50
2”
0.18
13.2
0.1207 ha1.55
0.015
0.24
0.50
1.55
3”
0.77
32.1
0.1771 ha
0.030
0.33
0.50
6”
1.50
111
0.3812 ha1.58
0.030
0.45
0.60
9”
2.50
251
0.5354 ha1.53
0.030
0.61
0.60
1.522
1’
3.32
457
0.6909 ha
0.030
0.76
0.70
1’6”
4.80
695
1.056 ha1.538
0.030
0.76
0.70
2’
12.1
937
1.428 ha1.550
0.046
0.76
0.70
1.566
3’
17.6
1 427
2.184 ha
0.046
0.76
0.70
4’
35.8
1 923
2.953 ha1.578
0.060
0.76
0.70
5’
44.1
2 424
3.732 ha1.587
0.060
0.76
0.70
1.595
6’
74.1
2 929
4.519 ha
0.076
0.76
0.70
7’
85.8
3 438
5.312 ha1.601
0.076
0.76
0.70
8’
97.2
3 949
6.112 ha1.607
0.076
0.76
0.70
8.28
7.463 ha1.60
0.09
1.07
0.80
1.60
m3/sec
10’
0.16
12’
0.19
14.68
8.859 ha
0.09
1.37
0.80
15’
0.23
25.04
10.96 ha1.60
0.09
1.67
0.80
20’
0.31
37.97
14.45 ha1.60
0.09
1.83
0.80
1.60
25’
0.38
47.14
17.94 ha
0.09
1.83
0.80
30’
0.46
56.33
21.44 ha1.60
0.09
1.83
0.80
40’
0.60
74.70
28.43 ha1.60
0.09
1.83
0.80
93.04
1.60
0.09
1.83
0.80
50’
0.75
35.41 ha
The discharge under submerged conditions is:
Equation 51
Step 2: List flumes capable of taking the given discharge,
using Table 32.
Then, for free flow at the maximum canal discharge:
Qs = Q - Qc
Where:
Qc
= Reduction of the modular discharge due to
submergence.
Figure 77 gives the corrections Qc for submergence for
flumes with 6 inch, 9 inch and 1 foot throat width. The
correction for the 1 foot flume is made applicable to other
sizes by multiplying the correction Qc for the 1 foot by the
factors given in Figure 77 (1 foot flume).
Usually the smallest practical size of flume is selected
because of economical reasons. In general the width should
vary between 1/3 to 1/2 of the canal width. Often the head
loss across the flume is the limiting factor.
The procedure for selecting the appropriate flume is as
follows:
Step 1: Collect site information: maximum and minimum
canal discharges, corresponding normal flow depths and
canal dimensions.
i) List values of ha for the maximum canal discharge
passing through the flumes.
ii) Apply the submergence limit appropriate to the flume
to find the value of hb corresponding to the values of ha
(Table 32).
iii) Subtract hb from the normal flow depth at maximum
discharge to give the vertical distance from the canal
bed to the flume crest level. This assumes that at
maximum submergence the downstream stage is the
same as that at hb, and that the flow downstream of the
flume is not affected by it.
iv) Find the head loss across the flume at maximum
discharge (Figure 78). Add this to the downstream
water depth to obtain the water depth upstream of the
flume.
v) Select the smallest size of flume for which the upstream
stage is acceptable.
99
Irrigation manual
Figure 77
Discharge corrections due to submergence for Parshall flumes with different throat width
a. Parshall flume with a throat width b of 6 inch or 15.2 cm
100
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 77
Discharge corrections due to submergence for Parshall flumes with different throat width
b. Parshall flume with a throat width b of 9 inch or 22.9 cm
101
Irrigation manual
Figure 77
Discharge corrections due to submergence for Parshall flumes with different throat width
c. Parshall flume with a throat width b of 1 foot or 30.5 cm
102
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 78
Head loss through Parshall flumes
103
Irrigation manual
Example 33
Select the most appropriate flume to be placed in a canal with the following characteristics:
Maximum discharge = 0.566 m3/sec
Canal water depth = 0.77 m
Canal banks at 3 m apart
The freeboard of the canal = 0.15 m
a.
Consider the flumes with a throat width of 3 and 4 foot. Table 32 gives discharge equations for the different flume
sizes.
The discharge equation for the 3 foot flume is:
Q = 2.184 x ha1.566
Ÿ
0.566 = 2.184 x ha1.566
Ÿ
ha = 0.43 m
Ÿ
ha = 0.35 m
The discharge equation for the 4 foot flume is:
Q = 2.953 x ha1.578
Ÿ
0.566 = 2.953 x ha1.578
b.
Assume that the submerging of 70% must not be exceeded. This means that hb = 0.70 x ha (Table 32). Thus for
the 3 foot flume the water depth hb = 0.30 m and for the 4 foot flume hb = 0.25m.
c.
The elevation of the crest above the bottom of the canal (K in Figure 76) equals the design water depth minus
hb. Thus K = 0.77 m - 0.30 m = 0.47 m for the 3 foot flume and K = 0.77 m - 0.25 m = 0.52 m for the 4 foot flume.
d.
From Figure 78 it can be seen that the head loss is 0.16 m for the 3 foot flume and 0.13 m for the 4 foot flume.
Thus the upstream water depth becomes 0.77 m + 0.16 m = 0.93 m and 0.77 m + 0.13 m = 0.90 m for the 3 foot
and 4 foot flume respectively.
e.
The upstream water depth of the 3 foot flume just exceeds the sum of the normal water depth and freeboard,
thus overtopping would result. The 4 foot flume is just within the available limit of depth. Thus this flume could
be selected for implementation. If there was sufficient freeboard available for either of the flumes, considering
the rise in water level upstream of the flume, one should select the 3 foot flume because this is cheaper.
Trapezoidal flume
Cut-throat flume
Whenever the canal section is not rectangular, trapezoidal
flumes such as those shown in Figure 79, are often
preferred, especially for measuring smaller discharges. A
typical trapezoidal flume has an approach, a converging
section, a throat, a diverging and an exit section. A
minimum transition will be required. An additional
advantage is the flat bottom, which allows sediment to pass
through fairly easily. Furthermore, the loss in head may be
less for comparable discharges.
The cut-throat flume has a converging inlet section, throat
and diverging outlet section. The flume has a flat bottom
and vertical walls (Figure 80).
Trapezoidal flumes are particularly suited for installation in
concrete-lined canals. The flume should normally be put
on top of the lining, thus constricting the flow section to the
extent required for free flow conditions over a whole range
of discharges up to the canal design discharge. As a rule of
thumb, one can say that the lower the canal gradient the
higher the elevation of the flume above the canal bed level.
Equation 52
The flow characteristics of the flume can be determined
experimentally. This allows for the calibrations of the flume.
As an example, a flume with dimensions such as those given
in Figure 79 can be located in a canal with a bed width of
0.30 m (1 foot), having side slopes of 1:1. The range of
calibrated water depth is 6-37 cm and the range of
calibrated discharge is 1.4-169 l/sec. This will suit most
conditions in a typical small-scale irrigation canal.
104
It is preferable to have the cut-throat flume operating under
free flow conditions. This facilitates measurements and
ensures a high degree of accuracy. Free flow conditions
through the cutthroat flume are described by the following
equations:
Q = C x (ha)n
Equation 53
C = K x W1.025
Where:
Q
= Discharge (m3/sec)
C
= Free flow coefficient
ha
= Upstream water depth (m)
K
= Flume length coefficient
W
= Throat width (m)
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 79
Trapezoidal flume (Source: FAO, 1975b)
For a given flume length, the values of n and K are obtained
from Figure 81. In order to ensure free flow conditions, the
ratio between the water depths ha and hb should not exceed
a certain limit, which is called the transition submergence, St.
Example 34
A cut-throat flume is to be installed with a length L = 1.22 m and throat width W = 0.36 m. The maximum discharge
through the structure is 0.20 m3/sec. How should it be installed in order to operate under free flow conditions?
From Figure 81, it follows that for a flume length L = 1.22 m:
St = 68.2%
K
= 3.1
n
= 1.75
Using Equations 53 and 52 respectively:
C = 3.1 x 0.361.025 = 1.088
Q = 1.088 x ha1.75 = 0.200 Ÿ ha = 0.38 m
St =
hb
= 0.682 Ÿ hb = 0.682 x 0.38 = 0.26 m
ha
Therefore the floor of the flume should be placed not lower than 0.26 m below the normal water depth, in order to let
pass the maximum discharge of 0.20 m3/sec. under free flow conditions.
105
Irrigation manual
Figure 80
Cut-throat flume (Source: FAO, 1975b)
Figure 81
Cut-throat flume coefficients (Source: FAO, 1975b)
106
Module 7: Surface irrigation systems: planning, design, operation and maintenance
6.4.4. Orifices
Orifices, such as gates and short pipes, are also used as
water measuring devices (Figure 82). However, they do not
offer any advantage over the use of weirs or flumes.
Furthermore, their calibrations are not as accurate nor as
stable as other types of measuring devices.
For weirs the discharge is proportional to the head above
the crest raised to the power 3/2 (Equations 44, 45, 46,
48). Therefore, they are sensitive to the fluctuations in the
upstream water level. For orifices, including gates and short
pipes, the discharge is proportional to the head of water
above the crest raised to the power 1/2, as shown by
Equation 34 (see Section 6.1.3). Therefore, they are less
sensitive to small fluctuations of the upstream water level.
Under submerged conditions both the upstream and
downstream sides of the structure need water level
recordings. For free flow conditions, the discharge is a
function of the upstream water depth alone.
Figure 82
Examples of orifices
Figure 83
Free discharging flow through an orifice
Example 35
A circular orifice is placed in a canal, which discharges 0.0783 m3/sec. The maximum allowable water depth over the
centre of the orifice is 0.25 m. What should be the opening of the orifice?
Substituting the above data in Equation 34 gives:
0.0783
1
= 0.075 Ÿ d = 0.27 m.
0.0783 = 0.6 x ( x S x d2) x (2 x 9.81 x 0.25)1/2 Ÿ d2 =
4
1.0437
Thus the diameter of the orifice should be 0.27 m
107
Irrigation manual
The general discharge equation for a free flow orifice is
(Equation 34):
Q = C x A x (2gh1
)1/2
Equation 54
Q = C x A (2g[h1 - h2])1/2
Where:
Where:
(m3/sec)
Q
= Design discharge through orifice (m3/sec)
C
= Discharge coefficient, which is 0.63 for
sluice gates and submerged orifices and
0.85 for short pipes
Q
= Design discharge through orifice
C
= Design coefficient (approximately 0.60)
A
= Cross-sectional area of the orifice (m2)
g
= Gravitational force (9.81 m/sec2)
A
= Cross-sectional area of the orifice (m2)
h1
= Water depth upstream of orifice over
reference level (m) (Figure 83)
g
= Gravitational force (9.81 m/sec2)
h1
= Water depth upstream of orifice over
reference level (m)
h2
= Water depth downstream of the structure (m)
Partially-opened sluice gates could be used for discharge
measurements, in which case they will be acting like
submerged orifices (Figure 84).
For partially-opened sluice gates and submerged orifices
the discharge equation reads:
6.6.5. Current meters
Current meters are used to measure the velocity in a canal,
from where the discharge can be calculated using the
Continuity Equation 12 (see Section 5.1). Most current
meters have a propeller axis in the direction of the current.
The flowing water sets the propeller turning. On a meter,
Figure 84
Sluice gate under submerged conditions
Example 36
A sluice gate is installed in a canal with a design water depth of 0.30 m. The canal discharges 0.0783 m3/sec. The
maximum allowable rise in water level upstream of the sluice gate is 0.25 m. The width of the gate opening is 0.40
m. What should be the height d of the opening?
h2 being 0.30 m and the allowable rise in water level upstream of the gate being 0.25 means that:
h1 = 0.30 m + 0.25 m = 0.55 m.
Substituting the above data in Equation 54 gives:
0.0783 = 0.63 x (0.40 x d) x (2 x 9.81 x [0.55 - 0.30])1/2 Ÿ d = 0.14 m.
108
Module 7: Surface irrigation systems: planning, design, operation and maintenance
forming part of the equipment, the number of revolutions
per time unit can be read and, by means of a calibrated
graph or table, the velocity can be determined. A wellknown type of current meter is the Ott instrument C31 for
velocities up to 10 m (Figure 85). Propeller meters are
reliable and accurate, but rather expensive.
In measuring the velocities, the number of points per
vertical and the number of verticals per cross-section
should be determined. For this purpose, the quantity of
work and the time required should be weighed against the
degree of accuracy (Euroconsult, 1989). For example,
measurements can be taken at 10 cm horizontal distance
over the cross-section and at 0.2h and 0.8h depth at each
10 cm (h is the water depth). The velocity is the average of
the velocity at 0.2h and 0.8h depth. If the water depth is
less than 0.5-0.6 m, one reading can be done at 0.6h. Then,
for each vertical the flow per unit width can be calculated
according to q = vaverage x h (Figure 86a). These qs are
distributed over the total width (Figure 86b) and the area
between the q-line and the water surface gives the total
discharge. It is also possible to establish the discharge per
section and to consider the sum of the discharges in the
sections as the total discharge.
Figure 85
Ott C31 propeller instrument
Figure 86
Depth-velocity integration method
109
Irrigation manual
6.7. Discharge measurement in pipelines
Several types of devices can be used to measure the
discharge in pipelines. This section will discuss differential
pressure and rotating mechanical meters, as they are the
ones commonly used.
6.7.1. Differential pressure flow meters
Differential pressure flow meters create a pressure
difference that is proportional to the square of the
discharge. The pressure difference is created by causing
flow to pass through a contraction. Manometers, bourdon
gauges, or pressure transducers are normally utilized to
measure the pressure difference. One good example of a
differential pressure flow meter is the Venturi tube (Figure
87).
Venturi tube
The pressure drop between the inlet and throat is created
as water passes through the throat. In the section
downstream of the throat, the gradual increase in crosssectional area causes the velocity to decrease and the
pressure to increase. The pressure drop between the
Venturi inlet and the throat is related to the discharge, as
follows:
Equation 55
Q =
Cd2K(P1 - P2)1/2
[1 - (d/D)2]1/2
Figure 87
Venturi flow meter
110
Where:
Q
= Discharge (l/min)
C
= Flow coefficient
D
= Diameter of upstream section (cm)
d
= Diameter of contraction (cm)
P1
= Pressure in upstream section (kPa)
P2
= Pressure in contraction (kPa)
K
= Unit constant (K is 6.66 for Q in l/min, d
and D in cm, and P1 and P2 in kPa)
The flow coefficient C for a Venturi metre is 0.97.
6.7.2. Rotating mechanical flow meters
There are many types of rotating mechanical flow meters
used in pipelines. These flow meters normally have a rotor
that revolves at a speed roughly proportional to the
discharge and a device for recording and displaying the
discharge and total volume. The rotor may be a propeller or
axial flow turbine, or a vane-wheel with the flow impinging
tangentially at one or more points.
Calibration tests are usually needed to accurately relate
rotor revolutions to the flow. The lowest discharge that can
be accurately measured by a rotating mechanical flow meter
depends on the amount of bearing friction that can be
tolerated while the occurrence of cavitation often
establishes the largest flow rate that can be measured (see
Module 5). Head loss through most rotating mechanical
discharge meters is moderate.
Chapter 7
Land levelling
Proper land levelling is important for efficient surface
irrigation. It involves moving soil in order to have level
fields for basin irrigation or uniform sloping fields for
furrow or borderstrip irrigation.
When levelling or grading land, one should avoid large
volumes of cut and fill. Besides being expensive, too much
soil movement tends to leave shallow topsoil in areas of cut,
which is not ideal for crop production.
A detailed topographic survey, preferably grid, is needed to
calculate the most economic land levelling requirements.
Based on the spot heights of the grid points and the required
gradient of the land, the cut and fill can be calculated. The
total volume of cut should preferably exceed the total volume
of fill by 10-50% depending on the total volume to be moved
and the compressibility of the soil.
The three most widely used methods for calculating the
amounts of soil cuts and fills are:
Y
Profile method
Y
Contour method
Y
Plane or centroid method
The plane method is the most popular of the three and will
be described more in detail in Section 7.3.
7.1. Profile method
The grid points following the proposed direction of slope
are used to represent a strip of land. The ground level
elevation points are plotted to show the existing profile.
The required gradient is superimposed and the gradient
line moved through trial and error until the volume of cut
equals the volume of fill. In general, the greater the amount
of fill required the greater should be the over-cut in
earthwork balances. For the purpose of over-cut the line of
equal cut and fill is lowered. After levelling, the work can be
checked using a level instrument or profile boards as shown
in Figure 88.
7.2. Contour method
The contour method requires an accurate contour map. A
new set of contour lines is chosen by visually balancing the
areas indicating cut and those indicating fill. Figure 89
shows a layout for the contour method.
The cut and fill areas are measured using a planimeter.
Approximate volumes of cut and fill between successive
contours are found by multiplying the average of the top
and bottom areas by the contour interval. As an example, if
the area of cut in zone 1 is 3.75 m2 and that of cut in zone
2 is 2.25 m2, the average cut area between contours 98 and
97 m is (3.75 + 2.25)/2 = 3.00 m2. If the distance
between the contour lines is 125 m, the volume of cut
between these lines is 3.00 m2 x 125 m = 375 m3.
All volumes of cut and fill are summed up and checked to
ascertain that they balance according to the cut to fill ratio.
If this is not correct, the new contours have to be adjusted
and the procedure repeated.
Figure 88
The profile method of land levelling: cut and fill and checking gradient levels with profile boards
111
Irrigation manual
Figure 89
The contour method of land levelling
7.3. Plane method
Step 1
The plane method is a least-squares fitting of field
elevations to a two-dimensional plane with subsequent
adjustments for variable cut-fill ratios. The aim is to grade
the surface of a field to a uniformly inclined plane. Grid
point elevations are used for the calculation. Each grid
point is taken to be representative of the square of a grid
size of which it is the centre. It is possible to calculate the
inclination and direction of the slope for minimum cut and
fills, although often a slope suited to the designed irrigation
system is selected.
The initial step is to determine the weighted average
elevations of each grid point in the field. The purpose of the
weighting is to adjust for any boundary stakes that
represent larger or smaller areas than given by the standard
grid dimension. The weighting factor is defined as the ratio
of actual area represented by a grid point to the standard
area. The grid point area is assumed to be the proportional
area surrounding the stake or other identification of the
grid point elevation.
Giving the field a basic X-Y orientation, the plane equation
is written as follows:
Equation 56
EL(X,Y) = (GX x X) + (GY x Y) + C
Where:
EL (X,Y)
=
Elevation of the (X,Y)
coordinate (m)
GX and GY
=
Regression coefficients
X and Y
=
Distance from origin to grid
point (m)
C
=
Elevation of the origin (m)
The calculation of the regression coefficients GX and GY
and the elevation of the centroid can be accomplished using
a four-step procedure.
112
Figure 90 shows a portion of an irrigation layout with a field
irrigation canal planned on the grid line i = 5 and the
drainage channel on grid line i = 1. The grid points on the
canal and drain alignment and the plot boundaries have to
be adjusted as they represent a smaller area than the
standard grid dimension of 25 m x 25 m. In the example
of Figure 90 the edge points only count for either 25% or
50%, thus the weighting factors are respectively 0.25 and
0.50. The weighting factors, other than those that are 1.00,
have been indicated between brackets in Figure 90.
The figures between brackets on the X-axis and the Y-axis
represent the distance.
The weighted average elevation has to be determined in
both field directions. Using the grid map, the elevations are
added by horizontal rows and by vertical columns, taking
the weighting factors into account, after which the average
of each row and column is calculated.
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 90
Grid map showing land elevation and average profile figures
The average elevation of column i (ELi in Figure 90) is
calculated by:
Where:
Equation 57
4ij
N
¦4ij x ELij
ELi =
ELij = Elevation of the (i,j) coordinate, found from
field measurements (m)
= Weighing factor of the (i,j) coordinate,
which is the ratio of actual area
represented by grid point (i,j) to the
standard grid area
j=1
N
¦4ij
j=1
113
Irrigation manual
Similarly, the average elevation of row j (Elj) is expressed by:
The elevation of the centroid is the average of the average row
or the average column elevations and is calculated as follows:
Equation 58
Equation 61
M
¦4ij x ELij
ELj =
M
¦4i x ELaverage, i
i=1
M
¦4ij
i=1
ELcen =
M
¦4i
i=1
For example, the average elevation of row EL1 (j=1) is:
(0.25 x 88.837) + (0.5 x 89.159) + (0.5 x 89.057)
+ (0.5 x 89.098) + (0.25 x 89.478)
0.25 + 0.50 + 0.50 + 0.50 + 0.25
= 89.118 m
i=1
Where:
ELcen
=
Elevation of the centroid (m)
ELaverage, i
=
Average elevation of column i (m)
In Figure 90, ELcen is:
Step 2
(3.5 x 88.678) + (7 x 88.846) + (7 x 88.919) + (3.5x 89.126)
Locate and calculate the elevation of the centroid of the
field with respect to the grid system. Usually, an origin is
located one grid spacing in each direction away from the
first grid position. The origin could, however, be related to
any corner of the field. The final results will be the same,
irrespective of the origin location. The distance from the
origin to the centroid in the i direction is found by:
Equation 59
M
¦4i x Xi
Xcen =
i=1
M
¦4i
3.5 + 7 + 7 + 7 + 3.5
= 88.905 m
Step 3
Calculate the best fitting straight line through the average
row and column elevations using the least squares method.
This is called linear regression, which is a statistical method
to calculate a straight line that best fits a set of two or more
data pairs. Thus, using this method the calculated slope line
fits the average profile best. These slopes, GX and GY, can
be calculated with the following formulae:
Equation 62
i=1
¦Xi x ELaverage,i -
Where:
Xcen
=
Distance from origin to centroid (m)
Xi
=
Distance in x direction from origin to
i-th grid position (m)
Gx =
j=1
M
M
2
¦Xi
M
¦
2
X -
¦4ij
=
M
i=1
i=1
N
4i
M
i=1
¦Xi x ¦ELaverage,i
M
i
i=1
i=1
M
Where:
GX
=
Slope in the x direction
Similarly, the distance from the origin to the centroid in the
j direction is:
Xi
=
Distance of average grid point
elevation ELaverage from the
origin (m)
Equation 60
ELaverage,i
=
Average elevation of column i (m)
M
=
Number of grid points in the Xdirection
N
¦4j x Yj
Ycen =
j=1
N
¦4j
j=1
114
Module 7: Surface irrigation systems: planning, design, operation and maintenance
The formula for the calculation of GY is:
Figure 91 gives a graphical impression of the lines of best
fit.
Equation 63
N
N
j=1
j=1
¦Yj x ELaverage,j -
The final step involves defining the best-fit plane (Equation
56) and requires the determination of C, which is the
elevation of the origin. As the lines of best fit go through the
centroid, the elevation of that point can be used to calculate
C as follows:
N
j=1
GY =
Step 4
¦Yj x ¦ELaverage,j
N
N
2
¦Yj
N
¦Yj2 j=1
j=1
N
C = ELcentroid - (GX x Xcen) - (GY x Ycen)
GX and GY can be calculated with a normal standard
calculator, although this is a very laborious method. A
programmable calculator, or one with linear regression
functions, could be used. Also, a number of land levelling
programmes have been written for use by computer.
Examples are given in Section 7.5.
In the above example:
C = 88.905 - (0.0039 x 75) - (-0.0015 x 112.50)
= 88.781 m
Example 37
For the example of figure 90 the value for GX can be calculated as follows:
We substitute M = 5 in the following equations:
5
¦Xi x ELaverage,i
= (25 x 88.678) + (50 x 88.846) + (75 x 88.955) + (100 x 88.919) + (125 x 89.126) = 33 363.525 m2
i=1
5
¦Xi
= 25 + 50 + 75 + 100 + 125 = 375 m
i=1
5
¦ELaverage,i
= 88.678 + 88.846 + 88.919 + 89.126 = 444.524 m
i=1
5
¦Xi)2
= (252 + 502 + 752 + 1002 + 1252 = 34 375 m2
i=1
5
¦Xi
2
= (25 + 50 + 75 + 100 + 125)2 = 140 625 m2
i =1
Substitution of the above data in the Equation 62 gives:
33 363.525 GX =
34 375 -
(375 x 444.524)
5
140 625
=
24.225
6 250
= 0.0039
5
This means that the line of best fit will rise from the origin at 0.39 cm per metre distance (0.39 m/100 m).
A similar calculation for GY would give a value -0.0015. This means that the line of best fit would drop from the origin
(because of the minus sign) at 0.15 cm per metre distance.
It should be noted that if the origin had been selected at the bottom right side of the field, the GX would have a
negative sign and the GY a positive one. The values would, however, remain the same.
115
Irrigation manual
Figure 91
Average profile and lines of best fit
Thus the equation for computing the elevation at any grid
point will be (Equation 56):
EL(X,Y) = (0.0039 x X) - (0.0015 x Y) + 88.781
The value of each grid point elevation can now be calculated
by substituting the distances of each point from the origin.
As an example, the elevation at the point with (X,Y) =
(25,25) coordinate is:
EL(25,25) = (0.0039 x 25) - (0.0015 x 25) + 88.781
= 88.841 m
Table 33 gives the results of all calculations. The differences
in elevation (3rd row in Table 33) are the necessary cuts,
where the calculated EL is lower than surveyed grid point
elevation, or fills, where the calculated EL is higher than
surveyed grid point elevation.
The volumes of cut and fill can be calculated by multiplying
the depth of cut or fill at each grid point with the grid area,
in this case an area of 625 m2 (= 25 m x 25 m) per grid
point, except for points with a weighing factor smaller than
1. The cut and fill volumes of our example of Table 33 are
764 m3 and 757 m3 respectively. The fourth row (adjusted
cut or fill) will be discussed later.
If the slopes GX and/or GY of the lines of best fit are too
steep or too flat to suit the irrigation method, they can be
changed. The slopes should still pass through the centroid,
which means that the volume of earth to be moved will
normally increase. The adjusted slopes are entered in the
equation to calculate C. If, for example, the slope in the Xdirection is changed to 0.005, the C-value becomes
88.698 m. Thus the equation for computing the elevation
at any grid point becomes:
EL(X,Y) = (0.005 x X) - (0.0015 x Y) + 88.698
116
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Table 33
Land levelling results
Surveyed ground level
Elevation after levelling
Cut or Fill
Adjusted cut or Fill
X:Y
88.837
88.841
+0.004
-0.004
25 : 25
89.159
88.939
-0.220
-0.228
50 : 25
89.057
89.036
-0.021
-0.029
75 : 25
89.098
89.134
+0.036
+0.028
100 : 25
89.478
89.231
-0.470
-0.255
125 : 25
Surveyed ground level
Elevation after levelling
Cut or Fill
Adjusted cut or Fill
X:Y
88.520
88.804
+0.284
+0.276
25 : 50
89.017
88.901
-0.116
-0.124
50 : 50
89.108
88.999
-0.109
-0.117
75 : 50
88.976
89.096
+0.120
+0.112
100 : 50
89.181
89.194
+0.013
+0.005
125 : 50
Surveyed ground level
Elevation after levelling
Cut or Fill
Adjusted cut or Fill
X:Y
88.731
88.766
+0.035
+0.027
25 : 75
88.814
88.864
+0.050
+0.042
50 : 75
89.043
88.961
-0.082
-0.090
75 : 75
89.027
89.059
+0.032
+0.024
100 : 75
89.264
89.156
-0.108
-0.116
125 : 75
Surveyed ground level
Elevation after levelling
Cut or Fill
Adjusted cut or Fill
X:Y
88.983
88.729
-0.254
-0.262
25 : 100
88.908
88.826
-0.082
-0.090
50 : 100
88.775
88.924
+0.149
+0.141
75 : 100
88.722
89.021
+0.299
+0.291
100 : 100
89.066
89.119
+0.053
+0.045
125 : 100
Surveyed ground level
Elevation after levelling
Cut or Fill
Adjusted cut or Fill
X:Y
88.654
88.691
+0.037
+0.029
25 : 125
88.802
88.789
-0.013
-0.021
50 : 125
88.905
88.886
-0.019
-0.027
75 : 125
88.846
88.984
+0.138
+0.130
100 : 125
89.026
89.081
+0.055
+0.047
125 : 125
Surveyed ground level
Elevation after levelling
Cut or Fill
Adjusted cut or Fill
X:Y
88.623
88.654
+0.031
+0.023
25 : 150
88.768
88.751
-0.017
-0.025
50 : 150
88.957
88.849
-0.108
-0.116
75 : 150
88.864
88.946
+0.082
+0.074
100 : 150
89.039
89.044
+0.005
-0.003
125 : 150
Surveyed ground level
Elevation after levelling
Cut or Fill
Adjusted cut or Fill
X:Y
88.591
88.616
+0.025
+0.017
25 : 175
88.697
88.714
+0.017
+0.009
50 : 175
88.918
88.811
-0.107
-0.115
75 : 175
88.981
88.909
-0.072
-0.080
100 : 175
89.024
89.006
-0.018
-0.026
125 : 175
Surveyed ground level
Elevation after levelling
Cut or Fill
Adjusted cut or Fill
88.450
88.579
+0.129
+0.121
88.668
88.676
+0.008
+0.000
88.900
88.774
-0.126
-0.134
88.940
88.871
-0.069
-0.077
89.081
88.969
-0.112
-0.120
X:Y
25 : 200
50 : 200
If the same calculations on volumes of cut and fill are done
again using the above equation, they result in a total volume
of cut of 822 m3 and a total volume of fill of 829 m3.
If the change in slope would give unsatisfactory results,
such as an excessive cut, it could be more beneficial to
irrigate at an angle to the canal.
This method of calculating the cut and fill volumes assumes
that the elevation of a grid point is representative for a full
grid area. This assumption is, of course, not always true. A
more accurate, but also more laborious, method to calculate
the cut and fill volumes is the Four-Corners method. This
method takes the depth of cut or fill at each corner of a
square into account. For boundaries, where complete grid
spacings are not present, the procedure is to assume that the
elevations of the field boundaries are the same as those of the
nearest grid point, while the actual edge area is taken into
account.
75 : 200
100 : 200
125 : 200
Equation 64
Vc =
L2 x C2
4 x (C + F)
Equation 65
Vf =
L2 x F2
4 x (C + F)
Where:
Vc
= Volume of cut (m3)
Vf
= Volume of fill (m3)
L
= Grid spacing (m)
C
= Sum of cut depth at grid points (m)
F
= Sum of fill depth at grid points (m)
117
Irrigation manual
Figure 92
Part of the completed land levelling map for Nabusenga, assuming GX = 0.005
118
Module 7: Surface irrigation systems: planning, design, operation and maintenance
As the calculations are very elaborate, they should
preferably be carried out with a programmable calculator
or a computer.
Where:
R
= Cut : fill ratio
Figure 92 shows part of the completed land levelling map
for Nabusenga surface irrigation scheme, assuming GX =
0.005.
Vf
= Volume of fill (m3)
Vc
= Volume of cut (m3)
Ai
= Total grid area which requires cut (m2)
More often than not, one tends to get a variety of slopes
within a scheme or a block of fields or even a field. To level
it as an entity will result in a lot of compromises as far as the
depths of cuts and fills are concerned. To avoid this, the
scheme or block of fields or fields can be divided into
sections. A section could be taken as a piece of land with a
uniform slope and can be treated as an area commanded by
a field canal or pipeline. The sections are levelled separately
with different parameters being used.
7.4. The cut : fill ratio
As explained above, the volume of cut (Vc) should exceed
the volume of fill (Vf) since the disturbance of the soil
reduces its density. The ratio is called the cut : fill ratio (R)
and should be in the range of 1.1 to 1.5, depending on soil
type and its condition. Selecting a cut : fill ratio remains a
matter of judgement and is therefore subjective.
As an example, if the volume of cut should exceed the
volume of fill by 20%, the cut : fill ratio is 1.20. The depth
required in order to lower the surface plane to achieve a
cut : fill ratio of 1.20 can be estimated with the following
formula:
Equation 66
d =
(R x Vf) - Vc
d
= Depth by which the surface plane has to be
lowered (m)
Following the example of Table 33, where 10 full grid areas,
7 half grid areas and 2 quarter grid areas have cuts (negative
values in 3rd row):
d =
(120 x 757) - 764
((10 + (0.5 x 7) +(0.25 x 2)) x 25 x 25) x 2.20
= 0.0075 m
Thus, in order to achieve a cut : fill ratio of approximately
1.20, the plane has to be lowered by 7.5 mm (the 4th row
in Table 33). This results in a final cut volume of 836 m3
and a final fill volume of 689 m3.
7.5. Use of computers
As already indicated previously, a number of programmes
have been written to calculate the land levelling
requirements by computer. One such programme, written
by E.C. Olsen of Utah State University, is called LEVEL
4EM.EXE. It calculates land-grading requirements based
on the least squares analysis for both rectangular and
irregularly shaped fields. The inputs required are given in
Table 34 below.
Some results of the use of computer for land levelling
calculations for different cut : fill ratios are given in Tables
35, 36 and 37.
I
¦(Ai x (1 + R))
i=1
Table 34
Input and output data types for computer land levelling programme LEVEL 4EM.EXE
INPUTS
OUTPUTS
The minimum and maximum acceptable cut : fill ratios
Elevations after grading
The units, either metric or imperial
Grade in horizontal direction
Number of grid points in horizontal direction
Grade in vertical direction
Grid distance in horizontal direction
Cut or fill required
Grid distance in vertical direction
Centroid elevation
Weighing factors other than 1
Cut : fill ratio
Number of grid points in vertical direction
Area levelled
Elevations of all grid points
Volume of excavation
119
Irrigation manual
Table 35
Table 36
Land levelling calculations with line of best fit and
cut:fill ratio of 1.01
Land levelling calculations with 0.5% gradient in the
X direction and cut:fill ratio of 1.01
Location
N M
Elevation
(m)
Ground elevation
(m)
Operation
(m)
Location
N M
Elevation
(m)
Ground elevation
(m)
Operation
(m)
1
1
1
1
1
1
2
3
4
5
88.84
89.16
89.06
89.10
89.48
88.84
88.94
89.04
89.14
89.24
C
C
C
F
C
0.00
0.22
0.02
0.04
0.23
1
1
1
1
1
1
2
3
4
5
88.84
89.16
89.06
89.10
89.48
88.79
88.91
89.04
89.16
89.29
C
C
C
F
C
0.05
0.25
0.02
0.06
0.19
2
2
2
2
2
1
2
3
4
5
88.52
89.02
89.11
88.98
89.18
88.80
88.90
89.00
89.10
89.21
F
C
C
F
F
0.28
0.12
0.11
0.13
0.02
2
2
2
2
2
1
2
3
4
5
88.52
89.02
89.11
88.98
89.18
88.75
88.87
89.00
89.12
89.25
F
C
C
F
F
0.23
0.14
0.11
0.15
0.07
3
3
3
3
3
1
2
3
4
5
88.73
88.81
89.04
89.03
89.26
88.76
88.86
88.96
89.06
89.17
F
F
C
F
C
0.03
0.05
0.08
0.04
0.10
3
3
3
3
3
1
2
3
4
5
88.73
88.81
89.04
89.03
89.26
88.71
88.84
88.96
89.09
89.21
F
F
C
F
C
0.02
0.02
0.08
0.06
0.05
4
4
4
4
4
1
2
3
4
5
88.98
88.91
88.78
88.72
89.07
88.72
88.82
88.92
89.03
89.13
C
C
F
F
F
0.26
0.09
0.15
0.30
0.06
4
4
4
4
4
1
2
3
4
5
88.98
88.91
88.78
88.72
89.07
88.67
88.80
88.92
89.05
89.17
C
C
F
F
F
0.31
0.11
0.15
0.33
0.11
5
5
5
5
5
1
2
3
4
5
88.65
88.80
88.90
88.85
89.03
88.68
88.78
88.89
88.99
89.09
F
C
C
F
F
0.03
0.02
0.02
0.14
0.06
5
5
5
5
5
1
2
3
4
5
88.65
88.80
88.90
88.85
89.03
88.64
88.76
88.89
89.01
89.14
F
C
C
F
F
0.02
0.04
0.02
0.17
0.11
6
6
6
6
6
1
2
3
4
5
88.62
88.77
88.96
88.86
89.04
88.64
88.75
88.85
88.95
89.05
F
C
C
F
F
0.02
0.02
0.11
0.09
0.01
6
6
6
6
6
1
2
3
4
5
88.62
88.77
88.96
88.86
89.04
88.60
88.72
88.85
88.97
89.10
F
C
C
F
F
0.02
0.04
0.11
0.11
0.06
7
7
7
7
7
1
2
3
4
5
88.59
88.70
88.92
88.98
89.02
88.60
88.71
88.81
88.91
89.01
F
F
C
C
C
0.01
0.01
0.11
0.07
0.01
7
7
7
7
7
1
2
3
4
5
88.59
88.70
88.92
88.98
89.02
88.56
88.69
88.81
88.94
89.06
F
F
C
C
C
0.03
0.01
0.11
0.04
0.04
8
8
8
8
8
1
2
3
4
5
88.45
88.67
88.90
88.94
89.08
88.57
88.67
88.77
88.87
88.97
F
F
C
C
C
0.12
0.00
0.13
0.07
0.11
8
8
8
8
8
1
2
3
4
5
88.45
88.67
88.90
88.94
89.08
88.52
88.65
88.77
88.90
89.02
F
F
C
C
C
0.07
0.02
0.13
0.04
0.06
Final grade in M direction
Final grade in N direction
Final centroid elevation
Final ratio of cuts/fill
Area levelled
Final volume of excavation
120
:
:
:
:
:
:
0.41 m/100 m
-0.15 m/100 m
88.905 m
1.01
1.750 ha
768.301 m3
Final grade in M direction
Final grade in N direction
Final centroid elevation
Final ratio of cuts/fill
Area levelled
Final volume of excavation
:
:
:
:
:
:
0.50 m/100 m
-0.15 m/100 m
88.905 m
1.01
1.750 ha
841.959 m3
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Table 37
Table 38a
Land levelling calculations with line of best fit and
cut:fill ratio of 1.21
Computer printout of land levelling data for the area
south of the main pipeline in Mangui piped surface
irrigation scheme
Location
N M
Elevation
(m)
Ground elevation
(m)
Operation
(m)
1
1
1
1
1
1
2
3
4
5
88.84
89.16
89.06
89.10
89.48
88.83
88.93
89.03
89.13
89.24
C
C
C
F
C
0.01
0.23
0.03
0.04
0.24
2
2
2
2
2
1
2
3
4
5
88.52
89.02
89.11
88.98
89.18
88.79
88.89
88.99
89.10
89.20
F
C
C
F
F
0.27
0.13
0.11
0.12
0.02
3
3
3
3
3
1
2
3
4
5
88.73
88.81
89.04
89.03
89.26
88.75
88.85
88.95
89.06
89.16
F
F
C
F
C
0.02
0.04
0.09
0.03
0.11
4
4
4
4
4
1
2
3
4
5
88.98
88.91
88.78
88.72
89.07
88.71
88.81
88.92
89.02
89.12
C
C
F
F
F
0.27
0.09
0.14
0.30
0.05
5
5
5
5
5
1
2
3
4
5
88.65
88.80
88.90
88.85
89.03
88.67
88.78
88.88
88.98
89.08
F
C
C
F
F
0.02
0.03
0.03
0.13
0.06
6
6
6
6
6
1
2
3
4
5
88.62
88.77
88.96
88.86
89.04
88.64
88.74
88.84
88.94
89.04
F
C
C
F
F
0.01
0.03
0.12
0.08
0.00
7
7
7
7
7
1
2
3
4
5
88.59
88.70
88.92
88.98
89.02
88.60
88.70
88.80
88.90
89.00
F
F
C
C
C
0.01
0.00
0.12
0.08
0.02
8
8
8
8
8
1
2
3
4
5
88.45
88.67
88.90
88.94
89.08
88.56
88.66
88.76
88.86
89.97
F
F
C
C
C
0.11
0.01
0.14
0.08
0.11
Final grade in M direction
Final grade in N direction
Final centroid elevation
Final ratio of cuts/fill
Area levelled
Final volume of excavation
:
:
:
:
:
:
0.41 m/100 m
-0.15 m/100 m
88.897 m
1.21
1.750 ha
841.988 m3
Location
N M
Elevation
(m)
Ground elevation
(m)
Operation
(m)
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
9.70
10.02
10.03
10.06
9.96
10.08
9.94
10.03
9.97
10.04
9.81
9.87
9.83
9.75
10.15
10.13
10.11
10.09
10.07
10.06
10.04
10.02
10.00
9.98
9.97
9.95
9.93
9.91
F 0.45
F 0.11
F 0.08
F 0.03
F 0.11
C 0.02
F 0.10
C 0.01
F 0.03
C 0.06
F 0.16
F 0.08
F 0.10
F 0.16
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
9.99
10.04
10.03
10.00
10.09
10.00
10.05
9.87
9.96
9.78
9.48
9.97
9.61
9.98
10.07
10.05
10.03
10.01
9.99
9.98
9.96
9.94
9.92
9.90
9.89
9.87
9.85
9.83
F
F
F
F
C
C
C
F
C
F
F
C
F
C
0.08
0.01
0.00
0.01
0.10
0.02
0.09
0.07
0.04
0.13
0.41
0.10
0.24
0.15
3
3
3
3
3
3
3
3
3
3
3
3
3
3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
10.06
10.03
10.09
10.30
10.14
10.35
10.01
10.01
10.17
10.24
10.23
9.83
9.83
9.80
9.99
9.97
9.95
9.93
9.91
9.90
9.88
9.86
9.84
9.82
9.81
9.79
9.77
9.75
C
C
C
C
C
C
C
C
C
C
C
C
C
C
0.07
0.06
0.14
0.37
0.23
0.45
0.13
0.15
0.33
0.41
0.42
0.04
0.06
0.05
Final grade in M direction
Final grade in N direction
Final centroid elevation
Final ration of cut/fills
Area levelled
Final volume of excavation
=
=
=
=
=
=
-0.09 m/100 m
-0.40 m/100 m
9.950m
1.48
1.680 ha
1 396.401 m3
The computer programme has also been used to calculate
the land levelling requirements for the gross area of Mangui
piped surface irrigation scheme and the results are shown
in Table 38 and in Figure 20. The slope along the pipeline
has been maintained as fairly level, while the slope
perpendicular to the pipeline, which is the furrow slope,
has been fixed at 0.4%.
121
Irrigation manual
Table 38b
Computer printout of land levelling data for the area
north of the main pipeline in Mangui piped surface
irrigation scheme
Location
N M
Elevation
(m)
Ground elevation
(m)
Operation
(m)
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
9.70
10.02
10.03
10.06
9.96
10.08
9.94
10.03
9.97
10.04
9.81
9.87
9.83
9.75
10.01
9.99
9.97
9.95
9.94
9.92
9.90
9.88
9.86
9.85
9.83
9.81
9.79
9.77
F 0.31
C 0.03
C 0.06
C 0.11
C 0.02
C 0.16
C 0.04
C 0.15
C 0.11
C 0.19
F 0.02
C 0.06
C 0.04
F 0.02
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
9.75
10.07
9.94
10.00
9.97
10.23
9.94
9.27
9.83
9.48
9.85
9.84
10.01
9.83
9.93
9.91
9.89
9.87
9.86
9.84
9.82
9.80
9.78
9.77
9.75
9.73
9.71
9.69
F 0.18
C 0.16
C 0.05
C 0.13
C 0.11
C 0.39
C 0.12
F 0.53
C 0.05
F 0.29
C 0.10
C 0.11
C 0.30
C 0.14
3
3
3
3
3
3
3
3
3
3
3
3
3
3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
9.92
9.77
9.88
9.75
9.82
9.74
9.82
9.68
9.68
9.58
9.56
9.51
9.39
9.49
9.85
9.83
9.81
9.79
9.78
9.76
9.74
9.72
9.70
9.69
9.67
9.65
9.63
9.61
C 0.07
F 0.06
C 0.07
F 0.04
C 0.04
F 0.02
C 0.08
F 0.04
F 0.02
F 0.11
F 0.11
F 0.14
F 0.24
F 0.12
Final grade in M direction
Final grade in N direction
Final centroid elevation
Final ration of cut/fills
Area levelled
Final volume of excavation
122
=
=
=
=
=
=
-0.09 m/100 m
-0.40 m/100 m
9.810m
1.30
1.680 ha
1 163.194 m3
For irregularly shaped fields, zero elevations are given to
grid points that fall outside the field boundary as shown in
Figure 93. These points will not be included in the
calculation.
Figure 93
Irregular shaped field (elevations 0.0 are located
outside the field)
Chapter 8
Design of the drainage system
Good water management of an irrigation scheme not only
requires proper water application but also a proper
drainage system. Agricultural drainage can be defined as the
removal of excess surface water and/or the lowering of the
groundwater table to below the root zone in order to
improve plant growth. The common sources of the excess
water that has to be drained are precipitation, overirrigation and the extra water needed for the flushing away
of salts from the root zone. Furthermore, an irrigation
scheme should be adequately protected from drainage
water coming from adjacent areas.
8.1.2. Soil type and profile
Drainage is needed in order to:
In soils with a layered profile drainage problems may arise,
when an impermeable clay layer exists near the surface for
example.
Y
Maintain the soil structure
Y
Maintain aeration of the root-zone, since most
agricultural crops require a well aerated root-zone free
of saturation by water; a notable exception is rice
Y
Assure accessibility to the fields for cultivation and
harvesting purposes
Y
Drain away accumulated salts from the root zone
A drainage system can be surface, sub-surface or a
combination of the two.
The rate at which water moves through the soil determines
the ease of drainage. Therefore, the physical properties of
the soil have to be examined for the design of a subsurface
drainage system. Sandy soils are easier to drain than heavy
clay soils.
Capillary rise is the upward movement of water from the
water table. It is inversely proportional to the soil pore
diameter. The capillary rise in a clay soil is thus higher than
in a sandy soil.
8.1.3. Water quantity
The quantity of water flowing through the soil can be
calculated by means of Darcy’s law:
Equation 67
Q = kxAxi
Q
= Flow quantity (m3/sec)
k
= Hydraulic conductivity (m/sec)
8.1. Factors affecting drainage
A
= Cross-sectional area of the soil through
which the water moves (m2)
8.1.1. Climate
i
= Hydraulic gradient
An irrigation scheme in an arid climate requires a different
drainage system than one in a humid climate. An arid
climate is characterized by high-intensity, short-duration
rainfall and by high evaporation throughout the year. The
main aim of drainage in this case is to dispose of excess
surface runoff, resulting from the high-intensity
precipitation, and to control the water table so as to prevent
the accumulation of salts in the root zone, resulting from
high evapotranspiration. A surface drainage system is most
appropriate in this case.
In a humid climate, that is a climate with high rainfall
during most of the year, the removal of excess surface and
subsurface water originating from rainfall is the principal
purpose of drainage. Both surface and subsurface drains are
common in humid areas.
The hydraulic conductivity, or the soils’ ability to transmit
water, is an important factor in drainage flow. Procedures
for field measurements of hydraulic conductivity are
discussed below.
8.1.4. Irrigation practice
The irrigation practice has a bearing on the amount of
water applied to the soil and the rate at which it is removed.
For example, poor water management practices result in
excess water being applied to the soil, just as heavy
mechanical traffic results in a soil with poor drainage
properties due to compaction.
123
Irrigation manual
8.2. Determining hydraulic conductivity
Hydraulic conductivity is very variable, depending on the
actual soil conditions. In clear sands it can range from 11 000 m/day, while in clays it can range from 0.001-1
m/day. Several methods for field measurement of hydraulic
conductivity have been established. One of the best-known
field methods for use when a high water table is present is
Hooghoudt’s single soil auger hole method (Figure 94).
A vertical auger hole is drilled to the water table and then
drilled a further 1-1.5 m depth or until an impermeable layer
or a layer with a very low permeability is reached. The water
level in the hole is lowered by pumping or by using buckets.
The rate of recharge of the water table is then timed.
For the calculation of the hydraulic conductivity the
following formula has been established:
Equation 68
3 600 x a2
k =
(d + 10a) x 2 -
y
d
x
+y
= Average distance from the static
groundwater table to the groundwater table
during the measurement:
y = 0.5 x (y1 + y2) (m)
Note that this is an empirical formula and the units should
be as explained above.
Example 38
An auger hole with a radius of 4 cm is dug to a depth
of 1.26 m below the static groundwater table. The
rise of the groundwater table, measured over 50
seconds, is 5.6 cm. The distance from the static
groundwater table to the groundwater table is 0.312
m at the start of the measurement and 0.256 m at the
end of the measurement. What is the hydraulic
conductivity?
a
= 0.04 m
d
= 1.26 m
'H = 5.6 cm
'H
't
't
= 50 sec
y
= 0.5 x (0.312 + 0.256) = 0.284 m
Substituting the above data in Equation 68 gives:
Where:
k
= Hydraulic conductivity (m/day)
a
= Radius of the auger hole (m)
d
= Depth of the auger hole below the
static ground water table (m)
'H = Rise in groundwater table over a time (t)
(cm)
't
y
= Time of measurement of rise in
groundwater table (sec)
k =
3 600 x 0.042
(1.26 + 10 x 0.04) + 2 -
124
x 0.284
5.6
50
If the water table is at great depth, the inverted auger hole
method can be used. The hole is filled with water and the rate
of fall of the water level is measured. Refilling has to continue
until a steady rate of fall is measured. This figure is used for
determination of k, which can be found from graphs.
Parameters for determining hydraulic conductivity using the auger hole method
y2
1.26
x
Ÿ k = 0.77 m/day
Figure 94
y1
0.0284
Module 7: Surface irrigation systems: planning, design, operation and maintenance
8.3. Surface drainage
When irrigation or rainfall water cannot fully infiltrate into
the soil over a certain period of time or cannot move freely
over the soil surface to an outlet, ponding or waterlogging
occurs. Grading or smoothening the land surface so as to
remove low-lying areas in which water can settle can partly
solve this problem. The excess water can be discharged
through an open surface drain system. Examples of a layout
of a drainage system are given in Figure 17 and 19, the
latter representing Nabusenga surface irrigation scheme.
The drainage water can flow directly over the fields into the
field drains. Drains of less than 0.50 m deep can be Vshaped. In order to prevent erosion of the banks, field
drains often have flat side slopes, which in turn allow the
passage of equipment. The side slopes could be 1:3 or
flatter. Larger field drains and most higher orders drains
usually have a trapezoidal cross-section as shown in
Figure 95.
The water level in the drain at design capacity should
ideally allow free drainage of water from the fields. The
design of drain dimensions should be based on a peak
discharge. It is, of course, impractical to attempt to
provide drainage for the maximum rainfall that would
likely occur within the lifetime of a scheme. It is also not
necessary for the drains to instantly clear the peak runoff
from the selected rainfall because almost all plants can
tolerate some degree of waterlogging for a short period.
Therefore, drains must be designed to remove the total
volume of runoff within a certain period. If, for example,
12 mm of water (= 120 m3/ha) is to be drained in 24
hours, the design steady drainage flow of approximately
1.4 1/sec per ha (= (120 x 103)/(24 x 60 x 60)) should
be employed in the design of the drain.
If rainfall data are available, the design drainage flow, also
called the drainage coefficient, can be calculated more
precisely for a particular area. The following method is
usually followed for flat lands. The starting point is a
rainfall-duration curve, an example of which is shown in
Figure 96. This curve is made up of data that are generally
available from meteorological stations. The curve
connects, for a certain frequency or return period, the
rainfall with the period of successive days in which that
rain is falling. Often a return period of 5 years is assumed
in the calculation. It describes the rainfall which falls in X
successive days as being exceeded once every 5 years. For
design purposes involving agricultural surface drainage
systems X is often chosen to be 5 days. Thus from Figure
96 it follows that the rainfall falling in 5 days is 85 mm.
This equals a drainage flow (coefficient) of 1.97 l/sec per
ha (= (85 x 10 x 103)/(5 x 24 x 60 x 60)).
Figure 95
Cross-sections of drains
125
Irrigation manual
rainfall. The design discharge could then be calculated with
empirical formulas, two of the most common ones being:
Figure 96
Rainfall-duration curve
Y
The rational formula
Y
The curve number method
The rational formula is the easier of the two and generally
gives satisfactory results. It is also widely used and will be
the one explained below. The formula reads:
Equation 70
Q =
CxIxA
360
Where:
The design discharge can be calculated, using the following
equation:
Equation 69
Q =
Where:
= Design discharge (m3/sec)
q
= Drainage flow (coefficient) (l/sec per ha)
A
= Drainage area (ha)
= Design discharge (m3/sec)
C
= Runoff coefficient
I
= Mean rainfall intensity over a period equal
to the time of concentration (mm/hr–)
A
= Drainage area (ha)
The time of concentration is defined as the time interval
between the beginning of the rain and the moment when
the whole area above the point of the outlet contributes to
the runoff. The time of concentration can be estimated the
following formula:
qxA
1 000
Q
Q
Equation 71
Tc = 0.0195 x K0.77
It would seem contradictory to take 5 days rainfall, when
the short duration storms are usually much more intensive.
However, this high intensity rainfall usually falls on a
restricted area, while the 5 days rainfall is assumed to fall on
the whole drainage area under consideration. It appears
from practice that a drain designed for a 5 days rainfall is,
in general, also suited to cope with the discharge from a
short duration storm.
Having said this, the above scenario is not necessarily true
in small irrigation schemes, especially on sloping lands
(with slopes exceeding 0.5%), which may cover an area that
could entirely be affected by an intense short duration
Where:
Tc
= Time of concentration (minutes)
K
=
L
—S
and S =
H
L
L
= Maximum length of drain (m)
H
= Difference in elevation over drain length (m)
The runoff coefficient represents the ratio of runoff
volume to rainfall volume. Its value is directly dependent
on the infiltration characteristics of the soil and on the
retention characteristics of the land. The values are
presented in Table 39.
Table 39
Values for runoff coefficient C in Equation 70
Slope (%)
Sandy loam
Clay silty loam
Clay
Forest
0-5
5-10
10-30
0.10
0.25
0.30
0.30
0.35
0.50
0.40
0.50
0.60
Pastures
0-5
5-10
10-30
0.10
0.15
0.20
0.30
0.35
0.40
0.40
0.55
0.60
Arable land
0-5
5-10
10-30
0.30
0.40
0.50
0.50
0.60
0.70
0.60
0.70
0.80
126
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Example 39
An irrigation scheme of 100 ha with sandy loam soils and a general slope of less than 5% has a main drain of 2.5 km
long with a difference in elevation of 10 m. What is the time of concentration?
S =
K =
H
L
L
—S
=
=
10
2 500
2 500
—0.004
= 0.004 or 0.4%
= 39 528
Substituting this value of K into Equation 71 gives:
Tc = 0.0195 x 39 5280.77 = 68 minutes
The rainfall intensity can be obtained from a rainfallduration curve, such as shown in Figure 96. For short
duration rainfall, it is necessary to make a detailed rainfallduration curve for the first few hours of the rainfall.
Example 40
In Example 39, the 68 minutes rainfall with a return
period of 5 years is estimated at 8.5 mm. What is the
design discharge of the drain?
The mean hourly rainfall intensity is (60/68) x 8.5 =
7.5 mm/hour.
The runoff coefficient for sandy loam arable land with
a slope of less than 5% is 0.30 (Table 39).
Thus the design discharge for the scheme is:
Q =
CxIxA
360
=
0.30 x 7.5 x 100
360
Ÿ Q = 0.625 m3/sec or 6.25 litres/sec per ha
Once the design discharge has been calculated, the
dimensions of the drains can be determined using the
Manning Formula (Equation 13). It should be noted that
higher order canal design should not only depend on the
design discharge, but also on the need to collect water from
all lower order drains. Therefore, the outlets of the minor
drains should preferably be above the design water level of
the collecting channel.
8.4. Subsurface drainage
Subsurface drainage is used to control the level of
groundwater so that air remains in the root zone. The
natural water table can be so high that without a drainage
system it would be impossible to grow crops. After
establishing the irrigation system the groundwater table
might rise into the root zone because of percolation of
water to the groundwater table. These situations may
require a subsurface drainage system.
Figure 97
Subsurface drainage systems at field level
127
Irrigation manual
A subsurface drainage system at field level can consist of any
of the systems shown in Figure 97:
Y
Y
Horizontal drainage by open ditches (deep and
narrowly-spaced open trenches) or by pipe drains
Vertical drainage by tubewells
8.4.1. Horizontal subsurface drainage
Open drains can only be justified to control groundwater if
the permeability of the soil is very high and the ditches can
consequently be spaced widely enough. Otherwise, the loss
in area is too high and proper farming is difficult because of
the resulting small plots, especially where mechanized
equipment has to be used.
Instead of open drains, water table control is usually done
using field pipe drains. The pipes are installed underground
(thus there is no loss of cultivable land) to collect and carry
away excess groundwater. This water could be discharged
through higher order pipes to the outlet of the area but,
very often, open ditches act as transport channels.
The materials used for pipe drains are:
Y
Clay pipes (water enters mainly through joints)
Y
Concrete pipes (water enters mainly through joints)
Y
Plastic pipes (uPVC, PE, water enters through slots)
Plastic pipes are the most preferred choice nowadays,
because of lower transport costs and ease of installation,
although this usually involves special machinery
The principal design parameters for both open trenches
and pipe drains are spacing and depth, which are both
shown in Figure 98 and explained below Equation 72.
The most commonly used equation for the design of a
subsurface drainage system is the Hooghoudt Equation:
Equation 72
S2 =
(4 x k1 x h2) + (8 x k2 x d x h)
q
Where:
S
= Drain spacing (m)
k1
= Hydraulic conductivity of soil above drain
level (m/day)
k2
= Hydraulic conductivity of soil below drain
level (m/day)
h
= Hydraulic head of maximum groundwater
table elevation above drainage level (m)
(Figure 98)
q
= Discharge requirement expressed in depth
of water removal (m/day)
d
= Equivalent depth of substratum below
drainage level (m) (from Figure 99)
Figure 98
Subsurface drainage parameters
k1
k2
128
Module 7: Surface irrigation systems: planning, design, operation and maintenance
It should be noted that the Hooghoudt Equation is a steady
state one, which assumes a constant groundwater table with
supply equal to discharge. In reality, the head losses due to
horizontal and radial flow to the pipe should be considered,
which would result in complex equations. To simplify the
equation, a reduced depth (d) was introduced to treat the
horizontal/radial flow to drains as being equivalent to flow
to a ditch with the impermeable base at a reduced depth,
equivalent to d. The equivalent flow is essentially horizontal
and can be described using the Hooghoudt formula. The
average thickness (D) of the equivalent horizontal flow zone
can be estimated as:
D = d+
h
2
Nomographs have been prepared to determine the
equivalent depth more accurately (Figure 99).
Nomographs have also been developed to determine the
drain spacing (Figure 100). From example 41:
(4 x k1 x h2)
q
=
(4 x 0.80 x 0.62)
0.002
= 576
and
(8 x k2 x h)
q
=
(8 x 0.80 x 0.6)
0.002
= 1 920
Drawing a line from 576 on the right y-axis to 1 920 on the
left y-axis in Figure 100, gives an S of about 90 m at the
point where D = 5 m. Note that results obtained from the
nomographs could differ slightly from the ones calculated
with trial and error as above, because of reading
inaccuracies.
Example 41
A drain pipe of 10 cm diameter should be placed at a depth of 1.80 m below the ground surface. Irrigation water is
applied once every 7 days. The irrigation water losses, recharging the already high groundwater table, amount to 14
mm per 7 days and have to be drained away. An average water table depth, z of 1.20 m below the ground surface
,has to be maintained. k1 and k2 are both 0.8 m/day (uniform soil). The depth to the impermeable layer D is 5 m. What
should be the drain spacing?
q = 14/7 = 2 mm/day or 0.002 m/day
and
h = 1.80 – 1.20 = 0.60 m
The calculation of the equivalent depth of the substratum d is done through trial and error. Initially the drain spacing
has to be assumed (Figure 99). After determining d, the assumed S should be checked with the calculated S from
the Hooghoudt Equation.
Lets assume S = 90 m. The wetted perimeter of the drain pipe, u, is 0.32 m (= 2 x S x r = 2 x 3.14 x 0.05). Thus
D/u = 5/0.32 = 15.6.
From Figure 99 it follows that d = 3.65 m. This has been determined as follows:
–
Draw a line from D = 5 on the right y-axis to D/u = 15.6 on the left y-axis
–
Determine the intersection point of the above line with the S = 90 line
–
Draw a line from this point to the right y-axis, as shown by the dotted line
–
The point where it reaches the right y-axis gives the d value
Substitution of all known parameters in the Hooghoudt Formula (Equation 72) gives:
S2 =
(4 x 0.8 x 0.62) + (8 x 0.8 x 3.65 x 0.6)
0.002
= 7 584 m2
Thus S = 87 m, which means that the assumed drain spacing of 90 m is quite acceptable.
129
Irrigation manual
Figure 99
Nomograph for the determination of equivalent sub-stratum depths
130
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Figure 100
Nomograph for the solution of the Hooghoudt drain spacing formula
Graph A: S = 5 - 25 m
8k2h
q
Graph B: S = 10 - 100 m
4k1h2
q
8k2h
q
4k1h2
q
For pipe drains with ro = 0.04 - 0.10 m, u = 0.30 m
8.4.2. Vertical subsurface drainage
Where soils are of high permeability and are underlain by
highly permeable sand and gravel at shallow depth, it may
be possible to control the water table by tubewells located
in a broad grid, for example at one well for every 2-4 km2.
Tubewells minimize the cost of and disturbance caused by
field ditches and pipe drains and they require a more sparse
drainage disposal network. If the groundwater is of good
quality, it could be re-used for irrigation.
8.5. Salt problems
Salt problems in the root-zone occur mainly in arid
countries. Drainage systems installed for the purpose of
salinity control aim at removing salts from the soils so that
a salinity level that would be harmful to plants is not
exceeded. Irrigation water always contains salts, but to
varying degrees. When the water is applied to the soil
surface, some of it evaporates or is taken up by the plants,
leaving salts behind in the root zone. If the groundwater
table is too high, there will be a continuous capillary rise
into the root zone and if the groundwater is salty, a high
concentration of salts will accumulate in the root-zone.
Figure 101 demonstrates this phenomenon.
Leaching is the procedure whereby salt is flushed away from
the root-zone by applying excess water, sufficient in
quantity to reduce the salt concentration in the soil to a
desired level. Generally, about 10-30% more irrigation
water than is needed by the crops should be applied to the
soil for this purpose. This excess water has to be drained
away by the subsurface drainage system.
131
Irrigation manual
Figure 101
Salt accumulation in the root zone and the accompanying capillary rise
132
Chapter 9
Bill of quantities
During the design stage, detailed technical drawings have to
be made. These drawings are not only needed during the
implementation stage, but they are also needed for the
calculation of the bill of quantities and costs. An
implementation programme or time schedule should be
prepared as well, to give an estimate of the labour and
equipment requirements. Details on the preparation of the
implementation schedule are shown in Module 13. This
chapter provides examples of the calculation of bill of
quantities for a concrete-lined canal, a saddle bridge and a
diversion structure at Nabusenga irrigation scheme and for
a piped system at Mangui irrigation scheme. At the end, the
overall bill of quantities for each scheme is given (excluding
the headworks, conveyance system and night storage
reservoirs).
two slanting sides and the two lips at the top. The
dimensions can be measured from a design drawing of the
canal cross-section. In our example of the 350 mm bottom
width canal section, the volumes of the different sections
are calculated as shown below:
Y
sin 60° = (0.05 + 0.30 + 0.05)/L
o L 0.40/0.866 = 0.46 m
(length x thickness) = 0.46 x 0.05 = 0.023 m
o for two sides, the area is 2 x 0.023 = 0.046 m2
Y
The width of the bottom is 0.35 m. However, this
includes part of the slanting side (about 0.015 m at
each side), when drawing the slanting sides down
diagonally till the lower side of the bottom. Therefore,
the length of the bottom to be used in the calculation
is (0.35 - 0.03) = 0.32 m, giving a concrete area of
(0.32 x 0.05) = 0.016 m2
Y
The length of the lip is 0.15 m. However, this also
includes part (about 0.05 m) of the slanting side.
Therefore, the length of the lip to be used in the
calculation is (0.15 - 0.05) = 0.10 m, giving a concrete
area of 0.10 x 0.05 = 0.005 m2 for one lip or 2 x
0.005 = 0.010 m2 for both lips.
9.1. Bill of quantities for Nabusenga
irrigation scheme
9.1.1. The construction of a concrete-lined canal
From the Nabusenga design prepared, it can be seen that a
total of 980 m of concrete-lined trapezoidal canal with a
bed width of 350 mm has to be constructed. The crosssection for this canal is given in Figure 102.
Materials for the preparation of concrete
The volume of concrete required per metre of canal length
is the sum of the volumes represented by the bottom, the
The area of the concrete for a side of the canal is
(length L x thickness) and is calculated as follows:
Thus, the concrete volume required for one metre length of
canal is 0.016 + 0.046 + 0.010 = 0.072 m3. Since the
canal length is 980 m, the concrete volume required is
Figure 102
Cross-section of a concrete-lined canal at Nabusenga
133
Irrigation manual
(980 x 0.072) = 70.56 m3. It is advisable to add 10% to
the volume to cater for waste and uneven concrete
thickness in excess of the 5 cm, thus the volume of concrete
will be 1.10 x 70.56 = 77.6 m3. This is the figure that will
appear in the bill of quantities.
Table 40 gives the volume of concrete required for a
number of trapezoidal cross-sections, similar to the one in
Figure 102, whereby only the bed width changes.
Table 40
Concrete volume for different trapezoidal canal crosssections
Bed width
(mm)
Concrete volume
(m3 per100 m)
250
6.70
300
6.95
350
7.20
400
7.45
450
7.70
500
7.95
Different structures require different types of concrete
grades, as discussed in Module 13. For concrete canals, a
good concrete mix is 1:2:3, by volume batching. The
materials required for such a mix per m3 of concrete are
calculated as follows:
Measuring by volume is based on loose volume. It can be
assumed that a 50 kg bag of cement is equivalent to 40 litres
of loose volume and that the yield of the mix is 60% of the
loose volume of cement, fine aggregate (sand) and coarse
aggregate (stone). This means that about 1.68 m3 or 1 680
litres of cement, sand and stone required for the
preparation of 1 m3 of concrete. For a mixture of 1:2:3,
this means that the loose volume is: 40 x 1 (cement)
+ 40 x 2 (sand) + 40 x 3 (stone) = 240 litres. Thus the
yield is 0.6 x 240 = 144 litres. This gives the following
quantities:
Cement: 1000/144 = 6.94 = 7 bags of 50 kg each
Sand:
(7 x 40 x 2) = 560 litres or 0.56 m3
Stones: (7 x 40 x 3) = 840 litres or 0.84 m3
Thus for 980 m of canal, requiring 77.6 m3 of concrete, the
material requirements are:
Cement: 77.6 x 7 = 543 bags
Sand:
77.6 x 0.56 = 44 m3
Stones: 77.6 x 0.84 = 65 m3
134
Transport of materials
The materials for the construction of concrete (cement,
sand and stone) are bulky and are therefore very expensive
to transport to construction sites. To save on costs, cheap
forms of transport should be sought. For example, if the site
is close to a railway line, it is advisable to use this kind of
transport, as in most countries it is cheaper than transport
by road. One can also decide to combine the two modes of
transport: rail can take the materials to some point and then
the remaining distance can be covered by road. Where
transport by road is used, it may be wise to go for big
tonnage trucks, if possible, as these tend to be cheaper than
smaller trucks because of reduced number of truck loads
needed to deliver a given quantity of construction materials.
For the Nabusenga scheme, cement and coarse aggregate
have to be transported from factories to the project site,
while good quality sand is available from local rivers. In this
example, let us assume that the cement (packed in 50 kg
bags) and coarse aggregate would be transported by rail
from the factory to nearest railway siding in the project
vicinity, a distance of 240 km. The transport from the
factory to that point is charged per ton. The weight of 1 m3
of coarse aggregate (crushed stone) is approximately
1 600 kg. Thus in our example the total tonnage for the
cement and coarse aggregate is:
(543 bags of cement x 50 kg per bag) +
(65 m3 of stones x 1.600 kg/m3)
= 131 150 kg | 131 tons.
From the siding, the materials are transported by road to
Nabusenga over a distance of 240 km. A 15 or a 30 ton
truck can be hired for this purpose. The hire price can be
charged either per ton per loaded km or include a charge
for the empty return trip. In our case, the charge will be per
loaded km.
Assuming the use of a 30 ton truck, the number of trips
required for the transport of cement and coarse aggregate
would be (131 tons/30 tons per trip) = 4.4. If this is to be
the total load to be transported for the scheme, 5 trips have
to be made. However, as cement and coarse aggregate are
also needed for other works, such as structures, a noninteger figure can be used for this particular item in the bill
of quantities and cost estimates.
Fine aggregate is usually collected from nearby rivers and
sometimes at no cost, except for transport costs. This
depends on the area or country in question. In this example,
let us assume that large deposits of river sand are found
within a distance of 20 km from Nabusenga. Due to the
rough terrain conditions, a small 7 ton lorry would
preferably be used, as very large lorries would have problems
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Table 41
Summary of the bill of quantities for the construction of the 980 m long lined canal at Nabusenga
Item
Quantity
Unit
Material:
– Cement
– Coarse aggregate (stones)
– Fine aggregate (sand)
543
65
44
bag
m3
m3
Transport:
– Rail (cement and stones)
– Road (cement and stones)
– Road (sand)
131
4.4 x 240
11 x 20
ton
trips x km
trips x km
80
640
person-day
person-day
16 x 1
16 x 20
16 x 5
16 x 50
days x no.
days x km/day
days x hr/day
days x km/day
Labour:
– Skilled
– Unskilled
Equipment:
– Concrete mixer
– Motorized bowser
– Tractor + trailer
– 7 ton lorry
Unit cost
Total cost
SUB-TOTAL (including 10% contingencies)
negotiating the bad roads. For this item, one has to know the
running cost for a 7 ton lorry per loaded km or the hire
price. Assuming damp sand will be collected from the river,
the weight per m3 will be approximately 1 700 kg (dry, loose
sand weighs approximately 1 400 kg/m3 and wet sand
approximately 1 800 kg/m3). Thus, 44m3 of sand weighs an
estimated (44 x 1700)/1 000 = 75 tons. This would require
approximately (75/7) = 11 trips, using a 7 ton lorry.
Labour
A time schedule for the construction has to be drawn up,
as discussed in Module 13. In general terms, a gang of 5
skilled and 40 unskilled workers should be able to complete
70 m of 350 mm bed-width canal per day. Thus, the 980
m length of canal could be completed in (980/70) =
14 days. It again is advisable to add 10% to the days for
unforeseen circumstances to the labour requirements,
which means that the work could be completed in
(1.10 x 14) = approximately 16 days. The total labour
requirement becomes:
Skilled: 5 persons x 16 days = 80 person-days
Unskilled:40 persons x 16 days = 640 person-days
For the calculation of the cost, one has to know the rates
for skilled band and unskilled labour per person day, which
differs from one country to the other.
Equipment
The following equipment will be required during the
construction period of 16 days, the rates of which should be
obtained from those who hire out construction equipment:
Y
Y
1 motorized water bowser (at a cost per km, assuming
the water is 10 km (one way) from site)
Y
1 tractor (at a cost per hour) + 1 trailer (at a cost per
day), assuming that the running hours per day are five
Y
1 lorry of 7 tons (at a cost per km, assuming that the
lorry runs 50 km per day for jobs like collecting
materials, diesel, etc.)
The summary of the bill of quantities for lining the 980 m
long canal are summarized in Table 42.
From Table 41, the cost per metre of the construction of a
350 mm canal in Nabusenga can be determined.
The bill of quantities and cost estimates for the other canal
cross-section sizes can be determined in a similar way. The
same gang of 5 skilled and 40 unskilled workers would
construct about 100 m of 250 mm bed width canal and 50
m of 500 mm bed width canal per day.
9.1.2. The construction of a saddle bridge
Figure 103 shows a typical design of a saddle bridge or
drain-road crossing.
Materials for the preparation of concrete
The volume of concrete required for the structure is the sum
of the volume of the slab and the toe around the structure.
The dimensions can be measured from the design drawing.
The slab volume (minus the area covered by the toe) is
(length x width x thickness) = (4 x 2.5 x 0.10) = 1.0 m3.
The toe volume is (length x height x thickness) =
[(2.5 x 2) + (4.5 x 2)] x 0.60 x 0.25) = 2.1 m3.
1 concrete mixer (at a cost per day or per month)
135
136
NOTES: All dimensions in metres unless otherwise stated.
Earthworks not shown.
Concrete mix is 1 : 2 : 3.
Reinforced steel bars of ‡10 mm to be placed in 0.15 m grid
Compacted sand
Reinforced concrete
Drawing: NABU/12
Scale as shown
Irrigation manual
Figure 103
Saddle bridge for Nabusenga
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Transport of cement and coarse aggregate by road:
5.8 tons/30 tons per trip = 0.2 trips
Thus the total concrete volume, inclusive of 10%
contingencies, is 1.1 x (1.0 + 2.1) = 3.4 m3.
The concrete mix will again be 1:2:3, thus the material
requirements are:
Transport of fine aggregate from river:
1.91 m3 x 1.700 kg/m3 = 3 247 kg
= 3.25 tons/7 tons per trip = 0.47 trips.
Cement: 3.4 x 7 = 24 bags
Sand:
3.4 x 0.56 m3 = 1.91 m3
Stones: 3.4 x 0.84 = 2.86 m3
Labour
It can be assumed that the saddle bridge can be completed
in 2 days with a gang of 2 skilled workers and 4 unskilled
workers. Thus the labour requirements are:
Reinforcement steel
Plain steel bars of 10 mm diameter will be placed in the
floor at a grid spacing of 15 cm. At the ends there should
be a concrete cover of approximately 7.5 cm. In the
direction of the width of the structure there should be
(3.0/0.15) = 20 steel bars of 4.35 m each. In the direction
of the length of the structure there should be (4.50/0.15)
= 30 steel bars of 2.85 m each. Thus the total length of
steel required, inclusive of 10% contingencies, is: {(20 x
4.35) + (30 x 2.85)} x 1.10 = 190 m
Skilled:
Unskilled:
2 persons x 2 days = 4 person-days
4 persons x 2 days = 8 person-days
The wages would be similar to those applicable for the
construction of the canal.
Equipment
The equipment required for the construction is:
Transport of materials
Y
1 concrete mixer
The same procedure, as followed for the transportation of
concrete canal lining materials, will apply for the saddle
bridge:
Y
1 motorized water bowser
Y
1 tractor and trailer
This equipment will be required during the entire two days
of construction. The charges would be the same as those
for canal construction.
Transport of cement and coarse aggregate by rail:
(24 bags x 50 kg per bag) + (2.86 m3 x 1.600 kg/m3)
= 5 776 kg » 5.8 tons
Table 42 is a bill of quantities for the saddle bridge.
Table 42
Summary of the bill of quantities for the construction of a saddle bridge
Item
Material:
–
–
–
–
Cement
Coarse aggregate (stones)
Fine aggregate (sand)
Reinforcement steel bars
Transport:
– Rail (cement and stones)
– Road (cement and stones)
– Road (sand)
Labour:
– Skilled
– Unskilled
Equipment:
– Concrete mixer
– Motorized bowser
– Tractor + trailer
Quantity
Unit
24
2.86
1.91
190
bag
m3
m3
m
5.8
0.2 x 240
0.47 x 20
ton
trips x km
trips x km
4
8
person-day
person-day
2x1
2 x 20
2x2
days x no.
days x km/day
days x hr/day
Unit cost
Total cost
SUB-TOTAL (including 10% contingencies)
137
Irrigation manual
9.1.3. The construction of a diversion structure
Transport of materials
A standard diversion structure could be constructed in one
day. The calculation of the bill of quantities is similar to the
one for the saddle bridge.
Following the same procedure as in the previous two
examples, the transport requirements are as follows:
Transport of cement and coarse aggregate by rail:
(9 bags x 50 kg/bag) +(0.30 m3 x 1 600 kg/m3)
= 930 kg = 0.93 tons
Materials for the preparation of concrete
The floor is made of reinforced concrete. The concrete mix
is again 1:2:3. The concrete volume, including 10%
contingencies, is (length x width x thickness) = (2.25 x
1.45 x 0.10) x 1.10 = 0.36 m3
Transport of cement and coarse aggregate by road:
0.93 tons/30 tons per trip = 0.03 trips
Transport of fine aggregate from river:
1.11m3 x 1 700 kg/m3 = 1 887 kg
= 1.89 tons/7 tons per trip = 0.27 trips.
The walls are built up with concrete blocks. Assuming a
mortar mix of 1:4, the material requirements per m3 of
mortar are 8 bags of cement and 1.28 m3 of fine aggregate.
The volume of mortar for the walls is (height x thickness x
length). The openings for the canal and sluice gates should
be excluded. Thus the volume, including 10%
contingencies, is (0.25 x 0.50 x 5.15) x 1.10 = 0.71 m3
Labour
As indicated earlier, a gang of 2 skilled workers and 4
unskilled workers could complete a diversion structure in
one day. Thus the labour requirements are:
The material requirements for the floor and the walls are:
Skilled: 2 persons x 1 day = 2 person-days
Unskilled: 4 persons x 1 day = 4 person-days
Cement: (0.36 x 7) + (0.71 x 8) = 9 bags
Sand:
(0.36 x 0.56) + (0.71 x 1.28) = 1.11 m3
Stones: (0.36 x 0.84) = 0.30 m3
Equipment
The same equipment as required for the construction of
the saddle bridge is also required for the construction of the
diversion structure. Therefore, one concrete mixer, one
water bowser and one tractor and trailer are needed for the
construction of the diversion structure.
Reinforcement steel and gates
The grid of steel bars is again 15 cm, thus the length of steel
bars (assuming 7.5 cm concrete cover), including 10%
contingencies, will be [(1.50/ 0.15) x 2.10 + (2.25/0.15)
x 1.30] x 1.10 = 45 m.
Table 43 is a bill of quantities for the diversion structure.
The structure has two sliding gates to control the water
distribution.
Table 43
Summary of the bill of quantities for the construction of a diversion structure
Item
Material:
–
–
–
–
–
Cement
Coarse aggregate (stones)
Fine aggregate (sand)
Reinforcement steel bars
Sliding bar
Transport:
– Rail (cement and stones)
– Road (cement and stones)
– Road (sand)
Labour:
– Skilled
– Unskilled
Equipment:
– Concrete mixer
– Motorized bowser
– Tractor + trailer
SUB-TOTAL (including 10% contingencies)
138
Quantity
Unit
9
0.30
1.11
45
2
bag
m3
m3
m
no.
0.93
0.03 x 240
0.27 x 20
ton
trips x km
trips x km
2
4
person-day
person-day
1x1
1 x 20
1x1
days x no.
days x km/day
days x hr/day
Unit cost
Total cost
Module 7: Surface irrigation systems: planning, design, operation and maintenance
9.1.4. The overall bill of quantities for Nabusenga
irrigation scheme
specific job. Table 44 shows the bill of quantities for the
construction of Nabusenga, downstream of the night
storage reservoir. The material requirements could be
summarized in a separate table to facilitate procurement
(Table 45).
The bill of quantities and costs are usually summarized in a
table, that shows the material, labour, transport and
equipment requirements, as well as the costs for the
Table 44
Bill of quantities for Nabusenga scheme, downstream of the night storage reservoir
Item
1.
2.
3.
Quantity
Unit
3
set
3
set
250 mm bottom width canal section (1 325 m)
2.1.
Cement
2.2.
Coarse aggregate
2.3.
Fine aggregate
2.4.
Labour skilled
2.5.
Labour unskilled
2.6.
Equipment
2.7.
Transport
683
82
55
75
600
–
–
bag
m3
m3
person-day
person-day
lump
lump
350 mm bottom width canal section (980 m)
3.1.
Cement
3.2.
Coarse aggregate
3.3.
Fine aggregate
3.4.
Labour skilled
3.5.
Labour unskilled
3.6.
Equipment
3.7.
Transport
543
65
44
80
640
–
–
bag
m3
m3
person-day
person-day
lump
lump
Templates and formers
1.1.
1 former and 3 screeding frames for
250 mm width canal section
1.2.
1 former and 3 screeding frames for
350 mm width canal section
4.
Drainage channel
1 400
m
5.
Road
5.1.
Perimeter road, 5 m wide
5.2.
Field road, 2.5 m wide
1 600
650
m
m
6.
Land levelling
15
ha
7.
Measuring device (2 pieces)
7.1.
Steel bar 10 mm
7.2.
Cement
7.3.
Coarse aggregate
7.4.
Fine aggregate
7.5.
Labour skilled
7.6.
Labour unskilled
7.7.
Equipment
7.8.
Transport
40
16
1.6
1.0
4
8
–
–
m
bag
m3
m3
person-day
person-day
lump
lump
Diversion structure (5 pieces)
8.1.
Steel bar 10 mm
8.2.
Cement
8.3.
Coarse aggregate
8.4.
Fine aggregate
8.5.
Labour skilled
8.6.
Labour unskilled
8.7.
Equipment
8.8.
Sliding gate
8.9.
Transport
225
45
1.5
5.6
10
20
–
10
–
m
bag
m3
m3
person-day
person-day
lump
each
lump
Canal-road crossing (1 piece)
9.1.
Steel bar 10 mm
9.2.
Cement
9.3.
Coarse aggregate
9.4.
Fine aggregate
9.5.
Labour skilled
9.6.
Labour unskilled
9.7.
Equipment
9.8.
Transport
344
30
3.2
2.1
6
18
–
–
m
bag
m3
m3
person-day
person-day
lump
lump
8.
9.
Unit cost
Total cost
139
Irrigation manual
Item
10.
11.
Quantity
Unit
Drain-road crossing/saddle bridge (3 pieces)
10.1. Steel bar 10 mm
10.2. Cement
10.3. Coarse aggregate
10.4. Fine aggregate
10.5. Labour skilled
10.6. Labour unskilled
10.7. Equipment
10.8. Transport
570
72
8.6
5.7
12
24
–
–
m
bag
m3
m3
person-day
person-day
lump
lump
Tail-end structure (5 pieces)
11.1. Steel bar 10 mm
11.2. Cement
11.3. Coarse aggregate
11.4. Fine aggregate
11.5. Labour skilled
11.6. Labour unskilled
11.7. Equipment
11.8. Transport
100
30
0.8
3.5
10
20
–
–
m
bag
m3
m3
person-day
person-day
lump
lump
12.
Drop structures
–
lump
13.
Check plates
20
each
14.
Siphons
250
m
15.
Fencing
15.1. Anchor
15.2. Barbed wire, 4 lines
15.3. Corner post
15.4. Dropper
15.5. Gate, large 4.25 m
15.6. Labour skilled
15.7. Labour unskilled
15.8. Pignetting (4 ft, 3 inch)
15.9. Standard
15.10. Straining post
15.11. Transport (7 ton lorry 510 km)
15.12. Tying wire
16.
Miscellaneous
16.1. Grain bags
16.2. Labour skilled1
16.3. Labour unskilled2
16.4. Materials and equipment (wheelbarrow,
trowels, shovels, clothing)
16.5. Preparatory work (site establishment)3
Unit cost
Total cost
48
2 500
23
340
3
20
200
2 500
170
1
–
3
200
945
270
each
person-day
person-day
–
–
lump
lump
TOTAL (including 10% contingencies)
Notes:
1. It is assumed that 15 extra skilled workers are on site for 3 months. These include drivers, surveyors and a storekeeper.
2. Unskilled labour is required for setting out the irrigation works and finishing/cleaning up after construction is finished.
3. Site establishment on this project mainly consists of setting up tents. The water supply and other site requirements already exist at the project site.
140
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Table 45
Summary of material requirements for Nabusenga (including 10% contingencies)
Description
Quantity
Unit
Steel bar 10 mm
1 175
m
Cement
1 088
bag
20
each
125.3
m3
Fine aggregate
90.3
m3
Sliding gate
10
each
Siphon (38 mm diameter)
250
m
48
13
23
340
3
50
170
1
3
each
roll
each
each
each
roll
each
each
roll
3
3
set
set
200
each
Check plate
Coarse aggregate
Fencing:
–
–
–
–
–
–
–
–
–
Anchor
Barbed wire
Corner post
Dropper
Gate
Pignetting
Standard
Straining post
Tying wire
Former and screeding frames:
– 250 mm width
– 350 mm width
Grain bag
9.2. Bill of quantities for Mangui irrigation
scheme
In Table 46 only the bill of quantities for the pipes and
fittings and pumping plant at Mangui scheme are given. All
the other requirements (labour, transport, fencing, roads,
structures, equipment, etc.) are calculated in a similar way
as was done for Nabsenga scheme.
141
Irrigation manual
Table 46
Bill of quantities for pipes and fittings and pumping plant at Mangui scheme
Item
Quantity
1.
Piping
1.1.
PVC pipe, 160 mm, class 4
198
1.2.
PVC pipe, 140 mm, class 4
90
1.3.
PVC pipe, 110 mm, class 4
36
1.3.
PVC pipe, 90 mm, class 4
36
2.
Fittings on pipelines
2.1.
BP 160 mm 90°
1
2.2.
RBP 160 mm to 140 mm
1
2.3.
RBP 140 mm to 110 mm
1
2.4.
RBP 110 mm to 90 mm
1
2.5.
TCP plus TRBP 90 mm
1
2.6.
Cast iron gate valve, 6 inch
1
2.7.
Cast iron gate valve, 4 inch
1
2.8.
TCP with TRBP 160 mm
2
2.9.
Bolts and nuts to secure CI gate valves
lump
3.
Hydrant assemblies
3.1.
Saddle 160 mm with 3 inch BSP socket
4
3.2.
Saddle 140 mm with 3 inch BSP socket
3
3.3.
Saddle 110 mm with 3 inch BSP socket
1
3.4.
Saddle 90 mm with 3 inch BSP socket
1
3.5.
GI pipe 3 inch x 1.5 m long, male threaded on both ends
9
3.6.
GI 3 inch equal Tee, female threaded on three ends
9
3.7.
3 inch x 2 inch reducing bush, male threaded
18
3.8.
Brass gate valve 2 inch
18
3.9.
Reinforced plastic hose, 32 mm x 20 m long, 4 bar pressure 18
3.10. Hose clips, 32 mm
18
3.11. Hose adapters, 32 mm
18
4.
Pumping plant
Pumping (unit) plant capable of delivering 34.56 m3/hr against a
head of 11.5 m, with the highest possible efficiency. Pump to be
directly coupled to a diesel engine of appropriate horse power
rating or electric motor of acceptable kilowatt power rating.
Pumping unit to be complete with suction and delivery pipes,
valves, strainer, non-return and air release valves, pressure gauge.
SUB-TOTAL
Contingencies 10%
TOTAL
142
Unit
m
m
m
m
no.
no.
no.
no.
no.
no.
no.
no.
lump
no.
no.
no.
no.
no.
no.
no.
no.
no.
no.
no.
Unit cost
Total cost
Chapter 10
Operation and maintenance of surface irrigation systems
10.1. Operation of the irrigation system
10.1.1. Water delivery to the canals
There are three methods for delivering water to canals:
Y
Continuous delivery
Y
Rotational delivery
Y
Delivery on demand
Continuous water delivery
Each field canal or pipeline receives its calculated share of the
total water supply as an uninterrupted flow. The share is
based on the irrigated area covered by each canal or pipeline.
Water is always available, although it may not always be
necessary to use it. This method is easy and convenient to
operate, but has a disadvantage in its tendency to waste water.
The method is rarely used in small irrigation schemes.
Rotational water delivery
Water is moved from one field canal or pipeline or from a
group of field canals or pipelines to the next. Each user
receives a fixed volume of water at defined intervals of time.
This is a quite common method of water delivery.
Water delivery on demand
The required quantity of water is delivered to the field when requested by the user. This on-demand method requires complex
irrigation infrastructure and organization, especially when it has
to be applied to small farmer-operated schemes where the
number of irrigators is large and plot holdings are small.
10.1.2. Water delivery to the fields
The water, delivered in an open canal or pipeline, can be
supplied onto the fields in different ways, which are briefly
explained below.
Bank breaching
Bank breaching involves opening a cut in the bank of a field
canal to discharge water onto the field. Although this
method is practiced widely, it is not recommended, as the
canal banks become weak because of frequent destruction
and refill. It also becomes difficult to control the flow
properly. Figure 104 shows how bank breaching is done.
Permanent outlet structures
Small structures, installed in the bank of a field canal are
used to release water from the field canal onto the fields.
Figure 104
Field canal bank breaching in order to allow the water to flow from the canal onto the field
143
Irrigation manual
Figure 105 shows a permanent outlet structure.
The structures can be made of timber with wooden stop
logs or of concrete with steel gates. This method is
especially used for borderstrip and basin irrigation. It
usually gives good water control to the fields. The
disadvantage is that the structures are fixed, thereby
reducing the flexibility of water distribution. Table 48 gives
approximate discharges of small wooden field outlets like
those shown in Figure 105 (FAO, 1975a).
Table 48
Discharge of permanent wooden field outlet structures
Depth of water over
the sill at the intake
(cm)
Discharge per 10cm
width of the sill
(l/sec)
10
6
15
11
20
17
25
22
Spiles
Spiles are short lengths of pipes made from rigid plastic,
concrete, steel, bamboo or other material and buried in the
canal bank as shown in Figure 106.
The discharge depends on the pipe diameter and the head
of water available. A plug is used to close the spile on the
inlet side. Since spiles are permanently installed, they have
Figure 105
Permanent outlet structure used to supply water from the canal onto the field (Source: FAO, 1975a)
144
Module 7: Surface irrigation systems: planning, design, operation and maintenance
the same disadvantage as the permanent outlet structures.
The approximate discharge can be calculated using
Equation 34 (see Section 6.1.3):
Q = C x A x —(2gh)
Where:
Q
= discharge through the spile (m3/sec)
C
= discharge coefficient
A
= cross sectional area of outlet (m2)
g
= gravitational force (9.81 m/sec2)
h
= head of water, measured from the centre of
the spile (m)
Table 49 gives approximate flows through small spiles.
Table 49
Rates of discharge through spiles (l/sec)
Diameter
of pipe
Pressure head
(cm)
(cm)
5
10
25
20
25
20
18.7
26.4
32.3
37.3
41.7
25
29.2
41.3
50.5
58.3
65.2
30
42.0
59.4
72.8
84.0
93.9
35
57.2
80.9
99.0
114.4
127.8
For piped systems, the openings at hydrants act in the same
manner as spiles and the discharge at the hydrant opening
is calculated using Equation 34 (see Section 6.1.3).
Siphons
Siphons are short lengths of pipe usually made of plastic,
rubber hose, or aluminium and are used to convey water
from open channels to the field. They are portable and easy
to install and to remove without disturbing the canal bank.
The discharge of water onto the irrigated area varies
according to the number of siphons in the furrow, border
strip or basin.
In order to use a siphon, it is put with one end in the water
and then filled with water (through suction by hand) to take
out the air. It is then laid over the canal bank while a hand
placed over the end of the pipe prevents air re-entering the
pipe. This process is called priming.
The discharge through the siphon depends on its diameter,
its length and the difference in level, h, between the water
level in the canal and the water level on the adjacent field
(or the centre of the pipe outlet if the pipe is not
submerged in water (see also Section 1.3.3). Figure 107
shows a siphon in operation. Since the pipe is usually short,
the influence of its frictional losses on the flow is negligible.
The water level in the canal should always be above the level
of the siphon outlet. A proper siphon command (h) should
be between 10 cm and 30 cm. The discharge through the
siphon can be calculated using Equation 34. The C value is
approximately 0.55.
Figure 106
An example of a spile used to supply water from the canal onto the field (Source: FAO, 1975a)
145
Irrigation manual
Figure 107
A siphon supplying water from a canal onto the field
Example 42
The flow through siphon is Q = 4.55 l/sec. The head
of water h = 0.18 m. What should be the diameter of
siphon?
Substituting the above data in the Equation 34 gives:
0.00455 = 0.55 x
1
4
x S x d2 x (2 x 9.81 x 0.18)1/2
Solving this equation results in a required siphon
diameter of 7.5 cm
Example 43
A field canal carries a flow of 78.3 l/sec to irrigate a
field using furrow irrigation. Each furrow requires a
flow of approximately 3.31 l/sec. What should be the
number of siphons that can be used to irrigate a
furrow and what is the total number of siphons for the
discharge of 78.3 l/sec?
From Table 50 it follows that for each furrow for
example two siphons with a diameter of 6 cm each can
be used, if the available head is 5 cm, or one siphon
with a diameter of 6 cm can be used, if the command
is 20 cm. In order to utilize the total discharge of 78.3
l/sec, 24 (=78.3/3.31) furrows can be irrigated at the
same time. In case the command is 5 cm only, this
means that the total number of siphons for the 24
furrows is equal to 24 x 2 = 48 siphons.
Table 50 gives rates of discharge of 2 m to 3 m long siphons
for different diameter, d, and head, h (see also Section 1.3.3
for smaller sizes). From the table it can be concluded that
the discharge changes when the head changes. It is
therefore important to maintain a constant head in the
canal.
Using more than one siphon gives the opportunity to
remove one (cut back the flow), once the water reaches the
end of the furrow (see Section 4.3). According to the
quarter contact time rule, water should reach the end of the
furrow in about 1/4 of the contact time. In order to reduce
runoff losses after the water has reached the end of the
furrow, the flow should be reduced, ideally such that the
inflow equals the actual infiltration. This reduction is easier
when there is initially more than one siphon.
10.1.3. Operational success determinants
Proper operation of irrigation schemes requires attention
to the following points:
Y
The water distribution should be in line with the design
and crop water requirements
Y
There should be equitable water distribution among
farmers
Y
Advice on proper water management in order to
minimize water losses should be given
Table 50
Discharge for siphons for different head and pipe diameter (l/sec)
Pipe I (cm)
4
5
6
7
8
9
10
146
Head (cm)
5
7.5
10
12.5
15
17.5
20
0.75
1.17
1.68
2.29
2.99
3.78
4.67
0.91
1.43
20.6
2.80
3.66
4.63
5.72
1.06
1.65
2.38
3.24
4.23
5.35
6.60
1.18
1.85
2.66
3.62
4.72
5.98
7.38
1.29
2.02
2.91
3.96
5.18
6.55
8.09
1.40
2.18
3.14
4.28
5.59
7.07
8.73
1.49
2.33
3.36
4.58
5.98
7.56
9.34
Module 7: Surface irrigation systems: planning, design, operation and maintenance
Water distribution and application
10.2. Maintenance of the irrigation system
As discussed in the previous sections, there are three
methods of distributing water: continuous flow, rotational
water supply and on-demand water delivery. The best
method to adopt depends entirely on the situation at hand.
There are three main types of maintenance namely:
As a rule, rotational water supply is used for smallholders
because of its simplicity. However, fixed rotation does not
correspond to the different water requirements of the
crops at different stages of growth. Thus, farmers are
obliged to apply the same frequency and to some extent the
amount of water, irrespective of water demand by the
crops. This results in reduced yields and water wastage. To
improve the rotational distribution, blocking has been
introduced in Southern Africa. The total scheme is divided
into four blocks, one for each major crop. Each farmer
would then be allocated a plot within each block and a
rotation of water supply is used among the four blocks. This
improves the potential for applying a rotational irrigation
schedule and improves the equitable distribution of water
among users.
3. Routine or normal maintenance
Equitable water distribution among farmers
Ideally, irrigators should get their fair share of irrigation
water. However, this is often not the case. The most
common problems are unauthorized water abstraction and
lack of sufficient water for tail-end users. In the latter case,
farmers at the head of the irrigation system receive and
tend to use more water than they need, while those at the
tail-end receive less than they need. In order to solve these
problems, good cooperation and trust among the irrigators
is important. If all the water can be diverted into one or a
few canals at a time, there is less chance of illegal water
abstraction. Once the water is diverted into a few canals,
tail-end problems could be further reduced by allowing
farmers to irrigate in groups, starting from the bottom end
of the canals, going upwards. The incorporation of stiffer
penalties in the farmers’ bylaws and their enforcement also
helps to reduce the problems.
Advice to farmers on proper water management
In many new irrigation projects, the farmers involved do
not have experience with irrigation. They need agronomic
advice as well as assistance in water management. With
regards to water management, the farmers should be
assisted in determining parameters like contact time,
advance and recession and the number of siphons to use in
each furrow, border or basin. Similarly, they should be
trained in operating structures such as measuring devices
and night storage reservoirs.
1. Special maintenance
2. Deferred maintenance
10.2.1. Special maintenance
Special maintenance includes work that is done to repair
the irrigation system in response to unforeseen damages,
such as those caused by floods or earthquakes. In this case
no specific preventative measures would have been taken to
circumvent the damage.
10.2.2. Deferred maintenance
Deferred maintenance or rehabilitation includes any work
that is done on the irrigation infrastructure in order to
restore the capacity of the system. In this case, the system
is allowed to deteriorate to a certain level, beyond which it
would not operate well, before it is restored to its design
operational level. Sometimes, deferred maintenance and
rehabilitation are differentiated on the basis of the source of
funds. The funds for deferred maintenance come from the
operation and maintenance budget, while that of
rehabilitation comes as an investment funded by loans or
national development budgets.
10.2.3. Routine maintenance
This includes all the work that is done in order to keep the
irrigation system operating satisfactorily. It is normally done
annually.
During the construction of the irrigation scheme, the future
irrigators should provide labour for construction activities.
Besides the advantage of promoting scheme ownership by
farmers, farmer involvement in construction work teaches
them several aspects of repair and maintenance.
Once the scheme is operational, the irrigation committee
should mobilize the farmers for repair and maintenance
activities. The works to be included in a maintenance
programme are discussed below.
Headworks
The main problems with the headworks are leakages.
Regular desilting is also necessary.
Night storage reservoirs
Night storage reservoirs should not stay dry for a long time
as this allows cracks in the clay in the core and bed to
147
Irrigation manual
develop. It is necessary, however, to empty the reservoirs
from time to time in order to clear them of weeds. Weeds,
besides harbouring snails, tend to reduce the capacity of
night storage reservoirs. It is also recommended to allow
the water level in the reservoirs to fluctuate to control
snails.
Canal system
The main problems are with unlined canals siltation, weed
growth, bank breaching, erosion caused by rainfall or
burrowing by animals. Lined canals have problems of
damaged joints, siltation, cracked sections or erosion of canal
banks. Weed growth can also be a problem in lined canals,
especially if silt is allowed to accumulate. As soon as these
problems are noticed, they should be rectified. Regular
desilting and weed removal is required. Both can be done by
hand. Table 51 gives a simplified typical weed management
programme for some schemes in some hot areas of
Zimbabwe. This should be used as a guide only, since
management depends on the climate of a particular area.
Drains
The most common problem with drains is weed growth.
Weeds should be frequently removed so as to maintain the
design capacity of the drains. Table 51 gives the guidelines.
Roads
Roads need refilling of potholes and gullies that may develop.
Embankments
The common problems of embankments are erosion,
leakages and weed growth. Refill and soil compaction should
be done when repairing embankments. Weeds should be
slashed.
Land levelling
After the initial land levelling during project construction,
it is necessary to periodically level the fields in order to
maintain the desired field slope. This can be done by
machinery or manually. If levelling is done manually, it is
still recommended that after every two to four seasons
farmers use machinery, such as a land plane.
Structures
The common maintenance problems are with structures
siltation, leakages caused by cracking and weed growth. They
should be maintained accordingly.
Gates
Gates can have problems of rusting or sticking over time
and leaking. They should be painted to prevent rusting. Any
movable parts should be greased or oiled to prevent
sticking. Replacing warn-out water seals, if there are any,
can minimize leaking.
10.3. Operation and maintenance responsibilities
The operation and maintenance of smallholder irrigation
schemes can be the responsibility of either the government,
the irrigation agency, individual farmers or groups of
farmers. It can also be a joint responsibility between groups
of farmers and the government, depending on the size of
the scheme. In large schemes or government-run schemes,
the irrigation agency and the farmers often share the
responsibility of operating and maintaining the irrigation
infrastructure. In such cases, the operation and
maintenance of the water delivery and storage system is
normally the responsibility of the agency, while the farmers
are responsible for maintaining field level infrastructure
such as canals and small hydraulic structures. The dividing
line, however, is not very clear. Therefore, the agency and
the farmers need to agree on their responsibilities and write
them down in bylaws. Where irrigation projects are
operated and maintained by farmers, as is the case for small
community schemes, the farmers themselves bear all
responsibilities for operation and maintenance. But even in
this case, rules and regulation should be written down in
bylaws.
Table 51
Weed management and effectiveness
Canal/drain
Maintenance
Effectiveness
Concrete-lined field canal
Hoeing within canal
Slashing/hoeing sides 2-3 times per year
up to 4 weeks
slashing 4 weeks; hoeing 6-8 weeks
Concrete-lined main canal
Slashing canal shoulders 3 times per year
up to 4 weeks
Night storage reservoirs
Desilting every 5 years
every 5 years
Infield drains
Slashing within drain
up to 4 weeks in wet season
Main drains
Slashing 2 times per year
hoeing and reprofiling once per year
up to 3-4 months in dry season
up to 6 months; up to one year
148
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