localized irrigation v2.08
TRANSCRIPT
A. Keïta, 2008-10
Localized Irrigation
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V 2.08
Introduction Irrigation : application of water supplementary to the precipitation
supply for crop production (Symposium, Lome, 1997).
Most efforts : Water resources development, few in On-Farm water use improvement
Increasing demand for higher water use efficiency, intensification and diversification of crop production
Successful experiences in many countries in pressurized irrigation
Two basic irrigation systems : open channel and pressured pipes. Focus on latter one, and more on localised irrigation.
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Objectives
• The lecture aims to qualify students and help them to become competent enough to– Collect basic irrigation data– Treat basic data to produce parameters required
for system design– Design a localized irrigation system– Produce a bill of quantity– Prepare a tender for supply
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I. Localised irrigation system layout
Traditional vs. Modern– Pressur. Piped irrig. System = network whith
pipes, fittings and other devices designed and installed to deliver water to a cropping area.
– Main differences with surface irrigation
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Designation Traditional surf. methods Pressurized piped irrigation methods
Flow regime Large stream Small stream (even 1m3/s)
Flow route Open channel and ditches along contours
Closed pipes under pressure along shortest ways
Irrigated surface Large volume per over large area
Small volume over large area
Energy Gravity only External pressure (2-3 bars)
System Layout
System Layout• From the head control unit to the hydrants :
buried piped (main, submain). Hence, protection against sunshine, agric. Machines
• Hydrants rising above the ground surface supply manifolds on which are fitted the laterals
• The laterals have water emitters : sprinklers, microprinklers, drippers etc.
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System Layout
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System Layout
• Head station :– Supply line in rigid PVC or galvanized steel. Mini height = 60 cm above ground– Air release valve, check valve, 50 mm hose outlet for connection with fertilizer injector,
a shut off valve between the 2 outlets, a fertilizer injector and a filter
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System Layout
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Centrifugal pump
Electrical motor
System Layout
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Sand filters
System Layout
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Fertilizer tank
Injection pumps
System Layout
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Fertilizer bag
System Layout
• Main pipeline :– Largest diameter in the network– Buried pipes, can be : rigid PVC, black high density poly
ethylene (HDPE), layflat hose, quick coupling galvanizsed light steel pipes 63-160 mm (2-6 in)
• Submains :– Extend from the main to the manifold – Buried pipes– Same type of pipe as the main, but smaller diameters
• Manifolds (feeder lines) : – Smaller diameter than submains– Connected to hydrants and are laid on the ground along
plot edges, supplying the laterals– Possible pipes : any kind, usually HDPE 2-3 in
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System Layout
• Laterals – Irrigating lines fitted to the manifolds and perpend. to
them– The smallest diameters– Laid along plant rows and often bearing emitters at
regular spacing
• Emitters– Device, fitted on a pipe, operated under pressure to
deliver a discharge in any form : » shooting water jets (sprinklers)» Small spray (sprayer)» Continuous drops (drippers)» Small stream (bubblers, openings on pipes, ...)
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System Layout
• Analogies between traditional and modern irrigation methods
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Traditional surf. methods Modern Pressurized piped irrig. methods
Designation Headworks (main gate) Head station
Main canal Main pipes
Submain canals Submain pipes
Canal intake gates Hydrants
Tertiary canals Manifolds
Field canals Laterals
System Layout
System classification
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3 types of classification
Pressure required for operation
Plant watering method
Dynamic of the system installation
System Layout
• Operating pressure classes (head pressure)• Low pressure : 2.0 – 3.5 bars
• Medium pressure : 3.5 – 5.0 bars
• High pressure : > 5.0 bars
• Plant watering method classes • Sprinkler irrigation (overhead or under foliage) : water delivered in
form of rain over the entire area
• Surface irrigation (furrow, basin, border...): water flows from the hydrants and spread all over the area or is side applied
• Micro-irrigation (localised irrigation) : drippers, bubblers or sprayers deliver water in low rates at the exact locations of the plants
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System Layout
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Example of low pressure system
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System Layout
• Dynamic of the system installation classes• Solid installation : fixed systems, permanent or seasonal
position of all the elements
• Semi-permanent installations : the main, submains and the manifolds are permanent ; the laterals are mobile (hand or mechanically moved)
• Portable installations : all the components of the system are portable
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System Layout
Modern vs. Traditional irrigation systems
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Designation Traditional irrigation in open channels methods
Modern Pressurized piped irrig. methods
Irrigation efficiency Water losses ≈ 40% (unlined ditches) ; 25% in lined canalsLoss locations : seepage in canals, phreatophytes, leakage in gates & spillways
Water losses ≈ 10% (local irrig.) ; 30% (overhead sprinkler and surface methods)
Economic return per unit of water
Yield increase of 10-45 % due to less water losses
Operation and maintenance (O&M)
-Requires skilled labour for operation-Expensive actions to prevent damage by roots, seepage, spread of weeds, sedimentation, clogging of outlets
-Anyone can easily operate the system- 1/10th to 1/4th of man-hour required for O&M in open channel system-Minimal maintenance during the first 7 years (except in motor pumps)-Maintenance cost ≈ 5.00% of the initial investment
Costs Reduced cost of pipes made in many countries and sizes : PVC, LDPE, HDPE, polypropylene (PP)
Initial capital investment - 2000 to 25000 US$/ha in Burkina Faso
-Function of the irrigation method and the type of installation- Solid installations for localized methods are the highest (See Table in case of Europe)
Design complexity and multiplicity
-Only apparent.-Design is simple and flexible-Mechanical difficulties only in early years. Farmer get quickly acquainted -Drastic change in irrigation management practices in the farm
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Designation Piped surface method Sprinkler classic method (hand-move)
Drip irrigation in fixed installation
Area (ha) 1 1 – 2 2 – 3 1 1 – 2 2 – 3 1 1 – 2 2 – 3
Installation cost (US$/ha) 1 700 1 600 1 400 2 800 2 700 2 100 3 950 3 300 3 000
Yearly maintenance cost (US$/ha)
85 80 70 140 135 105 200 165 150
Average 1997 prices in Europe Source : Phocaides, 2001
Components Complex installation Simple installation
Head station > 23% 13%
Mains, submains and manifolds 10% 21%
Fittings and other accessories 22% 24%
Laterals (pipes & emitters) 45% 42%
System comparative costs
System cost breakdown
System Layout
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System Layout
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System Layout
• Precise application of water to a specific area• Used where irrigating a portion of field is desired
System Layout
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System Layout
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System Layout
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System Layout
Maize
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System Layout
Cabbage
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System Layout
Tomato
30
System Layout
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System Layout
Maize
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System Layout
Olive
33
System Layout
Orange
34
System Layout
Peach
35
System design
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System design
Strawberry
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System Layout
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papaya
System Layout
Papaya
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System Layout
Strawberry
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System Layout
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System Layout
II. Preliminary system design42A. Keïta, 2iE
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• 2.1 Four stages in irrigation planning :• Stage 1
– Crop water requirements, soil type, climate implications, water quality and irrigation scheduling
– Water supply conditions, availability of electricity or energy, topography
• Stage 2 – Economic considerations– The labour, the know-how
• Stage 3 – System design : selection of emitters, laterals, manifolds, submain
and main pipelines– Head control
• Stage 4– Detailed list of equipment needed with full description, standards
and specifications for each item43
Preliminary design parameters
• 2.2 Crop Water Requirements• ETo is computed according to Penman-Monteith method• ETM = maximum evapotranspiration or loss of water from soil both by
evaporation and by transpiration from the plants growing thereon
– Method 1: Formulas using crop cover reduction coeff. Kr• In localised irrigation, the plants cover only a portion of the soil.
Hence, a ground cover reduction factor Kr is introduced
• The values of Kr suggested are presented in the table below. For design purpose, the ground cover GC is generally taken between 70 % and 100%.
Preliminary design steps
44
For sprinkler irrigation
ETM ET 0BK c
For localized irrgation
ETM loc ETMBK r ET OBK cBK r
Preliminary design parameters
Crop ground cover reduction factor Kr
Ground cover (GC) in %
Keller & Karmelli Freeman & Garzoli Decroix CTGREF
10 0.12 0.10 0.20
20 0.24 0.20 0.30
30 0.35 0.30 0.40
40 0.47 0.40 0.50
50 0.59 0.75 0.60
60 0.70 0.80 0.70
70 0.82 0.85 0.80
80 0.94 0.90 0.90
90 1.00 0.95 1.00
100 1.00 1.00 1.00
45
System design – Crop water requirements
Source : Savva & Frenken, 2001
Crop Initial Crop development
Mid-season Late harvest
Bean (green) 0.35 0.70 1.0 0.9Bean (dry) 0.35 0.75 1.1 0.5Cabbage 0.45 0.75 1.05 0.9Carrot 0.45 0.75 1.05 0.9Cotton 0.45 0.75 1.15 0.75Cucumber 0.45 0.70 0.90 0.75Eggplant 0.45 0.75 1.15 0.80Groundnut 0.45 0.75 1.0 0.75Lettuce 0.45 0.60 1.0 0.90Maize (sweet) 0.40 0.80 1.15 .09.Maize (grain) 0.40 0.75 1.15 0.70Melon 0.45 0.75 1.0 0.75Onion (green) 0.50 0.70 1.0 1.0Onion (dry) 0.50 0.75 1.05 0.85Pea (fresh) 0.45 0.80 1.15 1.05Pepper 0.35 0.75 1.05 0.90Potato 0.45 0.75 1.15 0.75Spinach 0.45 0.60 1.0 0.90Squash 0.45 0.70 0.90 0.75Sorghum 0.35 0.75 1.10 0.65Sugar beet 0.45 0.80 1.15 0.80Sugar cane 0.45 0.85 1.15 0.65Sunflower 0.35 0.75 1.15 0.55Tomato 0.45 0.75 1.15 0.80 46
Values of crop factor Kc for seasonal crops
System design – Crop water requirements
47
Values of crop factor Kc for permanent crops
Crop Young Mature
Banana 0.50 1.10
Citrus 0.30 0.65
Apple, cherry, walnut
0.45 0.85
Almond, apricot, pear, peach, pecan, plum
0.40 0.75
Grape, palm tree 0.70 0.70
Kiwi 0.90 0.90
Olive 0.55 0.55
Alfalfa 0.35 1.1
System design – Crop water requirements
– Method 2 : Formulas using the ground cover percentage GC
• Keller and Bliestner formula
• The “unit time” = day, or decade, or month, or season.
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System design – Crop water requirements
ETM loc ETM peak mm/ unit time` ab c
B 0.1 A GC %` aq
wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwD E
Where
ETM peak conventionally estimated peak ETM
GC percentage ground cover
• Exercise– Mature citrus : ETMpeak = 7.1 mm/day using modified
Penmann-Monteith method.– GC = 70%– Task : What is the value of ETMloc peak computed by the
method of : a) Keller and Karmeli, b) Freeman and Garzoli, c) Decroix CTGREF, d) Keller and Bliestner ? What conclusion do you draw while comparing the values ?
49
System design – Crop water requirements
Ans : a) 5.8 mm/day ; b) 6.0 mm/day ; c) 5.7 mm/day ; d) 5.9 mm/day.
• 2.3 Irrigation Water Requirements (IR)• Net Irrigation water requirements (IRn)
– IRn = Depth required for normal crop growing over the entire cropping area, excluding other water source contributions (FAO, 1984)
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IRn ETM loc @ R P e
b c
Where
IRn net irrigation requirement
ETM loc crop maxi evapotranspiration for localized system
R water received by crop fromsources other than irrigation groundw ater f low b c
P e effective rainfall
P e can be computed by :
P e 0.8 P where P > 75 mm/ month
P e 0.6 P where P < 75 mm/ month
System design –Irrigation requirements
• Gross irrigation requirements (IRg)– IRg = Net irrigation requirement over the field application
efficiency (taking into account the losses at field level) plus the leaching requirement
– FAO 1984 proposed for the field application efficiency :
51
System design –Irrigation requirements
IRg IRn
E a
ffffffffff LR
Where
IRg gross irrigation requirement
E a f ield application eff iciency
E a K sB EU
Where
K s Average water stored in root volume
Average water applied
ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
EU coeff icient ref lecting the uniformity of application
• Storage coefficient values Ks
• The EU of design should not be less than 0.90 for localised irrigation
Soil type Ks values
Coarse sand or light topsoil with gravel subsoil 0.87
Sand 0.91
Silt [ loose sedimentary material with rock particles usually 1/20 millimeter or less in diameter (Merriam-Webster) ]
0.95
Loam and clay 1.0
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System design –Irrigation requirements
Source : Savva & Frenken, 2001
System/method Ea %
Earth canal network surface methods
40-50
Lined canal network surface methods
50-60
Pressure piped network surface methods
65-75
Hose irrigation systems 70-80
Low-medium pressure sprinkler systems
75
Microsprinklers, micro-jets, minisprinklers
75-85
Drip irrigation 80-90
53
• Values of Ea as function of on farm irrigation methods
System design –Irrigation requirements
• Exercise– Mature citrus : ETMpeak = 7.1 mm/day. Grown on
silt soil. GC = 70% (Use Freeman and Garzoli for Kr). Neither rainfall nor groundwater contributions. Take EU=0.90.
– Task : IRn ? IRg ? (peak period)
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Ans : IRn = 6mm/d ; IRg = 7.02 + LR
System design –Irrigation requirements
• 2.4 Leaching Requirements (LR)– Soil water contains salt. So does the irrigation water– Frequent irrigation of the localised method reduces
salt accumulated in the root zone if sufficient leaching
– For salinity control :
55
2 types of Electric conductivity required Qirrigation w ater :ECw
saturated soil extract :ECe
X\
Z
System design – Leaching requirements
• Keller & Bliestner proposed the yield reduction ratio formula as first step the computation
• If ECw < minECe, then no yield reduction expected and Yr =1
• If ECw > maxECe, then the yield is zero and Yr = 0
• In between the 2 values of ECe, the Yr equation is used to find the reduction factor.
56
Y r maxECe @ECw
maxECe @minECe
fffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
Where
Y r relative yield, ratio of estimated yield to full potential
ECw dS / mor mmhos / cm` a
irrigation w ater elect conduc
maxECe dS / mor mmhos / cm` a
maximumelect conduc
of saturated soil extract that w ill reduce the yield to zero
minECe dS / mor mmhos / cm` a
minimumelect conduc below w hich
there is no yield reduction
System design – Leaching requirements
min ECe max ECe min ECe max ECeField Crops Field Crops
Cotton 7.7 27 Corn 1.7 10Sugar beet 7.0 24 Flax 1.7 10Sorghum 6.8 13 Broad bean 1.6 12
Soya bean 5.0 10 Cow pea 1.3 8.5Sugarcane 1.7 19 Bean 1.0 6.5
Fruit and nut crops Fruit and nut cropsDate palm 4.0 32 Apricot 1.6 6Fig olive 2.7 14 Grape 1.5 12
Pomegranate 2.7 14 Almond 1.5 7Grapefruit 1.8 8 Plum 1.5 7
Orange 1.7 8 Blackberry 1.5 6Lemon 1.7 8 Boysenberry 1.5 6
Apple, pear 1.7 8 Avocado 1.3 6Walnut 1.7 8 Raspberry 1.0 5.5Peach 1.7 6.5 Strawberry 1.0 4
Vegetable Crops Vegetable CropsZucchini squash 4.7 15 Sweet corn 1.7 10
Beets 4.0 15 Sweet potato 1.5 10.5Broccoli 2.8 13.5 Pepper 1.5 8.5Tomato 2.5 12.5 Lettuce 1. 9
Cucumber 2.5 10 Radish 1.2 9Cantaloupe 2.2 16 Onion 1.2 7.5
Spinach 2.0 15 Carrot 1.0 8Cabbage 1.8 12 Turnip 0.9 12Potato 1.7 10
57Rem : Max implies zero yield and min implies no yield reduction
Max and Min values of saturated soil extract electric conductivity in deci siemens/m for various crops (Source: Keller and Bliesner, 1990)
System design – Leaching requirements
• Exercise– Citrus (orange): irrigated with water of ECw = 2 dS/m.– Task : What would be the relative yield ?
– Ans :
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Yr = 0.95
System design – Leaching requirements
• Leaching is required if one wants the yield not to be worse than the 95% of the predicted in the example of the citrus.
• Keller & Bliesner proposed a formula for estimating the leaching requirement ratio :
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LR t ECw
2B maxECe
B Cfffffffffffffffffffffffffffffffffffffffffff
Where
LR t leaching requirement ratio under drip irrigation
ECw dS / mor mmhos / cm` a
irrigation w ater elect conduc
maxECe dS / mor mmhos / cm` a
maximumelect conduc
of saturated soil extract that w ill reduce the yield to zero
System design – Leaching requirements
• The leaching requirement LR is obtained by the expression :
• Exercise :– Task : What is the IRg of citrus (orange) taking into account the leaching
requirements ?• Data :IRn= 6 mm/day. Clayey soil. Consider EU = 0.90. ECw = 2 dS/m
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System design – Leaching requirements
LR LR tBIRn
E a
ffffffffff
Where
LR t leaching requirement ratio under drip irrigation
IRn net irrigation requirement
E a f ield application eff iciency
Ans :LRt=0.125; Ks=1; Ea=0.90 ; LR = 0.83 mm ; IRg = 7.5 mm
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System design
Compute
LR t
ECw
2B maxECe
B Cfffffffffffffffffffffffffffffffffffffffffff
Compute
LR LR tB
IRn
E a
ffffffffff
Read ETo, Kc
Compute ETM peak ET oBK c
Compute
ETM loc ETM peak BK r
Or ETM loc ETM peak
mm
unit time
ffffffffffffffffffffffffffff g
B 0.1 A GC %` aq
wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwD E
Read GC or Kr
Compute
IRn ETM loc @ R P e
b c
Compute
IRg IRn
E a
ffffffffff LR IRn
E a
ffffffffff 1 LR t
b c
Read Ks , EU, (Ea)
Compute E a K sB EU
Read Pe
Compute R
MeasureECw
Read maxECe, minECe
Compute Y r
maxECe @ECw
maxECe @minECe
fffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
External data
Computational sequence
Goal
Condensed
• 2.5 Irrigation application depth– Moisture content of soil is determined by taking
• An undisturbed sample of soil, measuring its volume V and mass G1. Afterwards, the sample is dried in a stove at 105°C and weighted to get the mass G2. The moisture content is determined by
• Exercise : Data V = 0.72 l; G1 = 1.63 kg ; G2 = 1.35 kg . Task : Compute Θ» Ans : Θ =
62
System design – Practical application depth
100BG 1@G 2
wB V
fffffffffffffffffffffff Where wthe density of w ater : 1 kg/ l
Or
100BG 1@G 2
G 2
fffffffffffffffffffffffB
b
w
fffffffIn case V is unknow n
b1.2 @1.4 kg/ l bulk density
X^^^^^^\
^^^^^^Z
Y^^^^^^]
^^^^^^[
– Moisture content of a soil• pF = log(h) where h = P, pressure in cm of water. If h = 10 cm, than pF =1 ; pf = 2
means h = 100 cm• Field capacity
– The soil moisture of a suction state which is in equilibrium with the gravity force and which can retain water in the pores is between 1/10 – 1/3 bar (100 and 300 cm) water head ; in other words between pF = 2 and pF = 2.5.
– The corresponding moisture is called field capacity
• Permanent Wilting Point– Starting at the plant uses the available moisture for evapotranspiration. The roots are able to
overcome soil moisture tension up to 160 m head (about 15 bars) of 16 000 cm head ; that is also pF =4.2. The corresponding soil moisture is called permanent wilting point
• The difference in moisture content over the entire root zone depth Dr at FC and WP is called Available Moisture :
63
FC
FC
WP
AM FC @WP
b cBD r
With
D r root zone depth in cmor m
moisture in % of volume
System design – Practical application depth
Crop Rooting depth Dr (m) Crop Rooting depth Dr(m)Alfalfa 1.0 -2.0 Melons 1.0-1.5Banana 0.5-0.9 Olives 1.2-1.7Barley 1.0-1.5 Onions 0.3-0.5Beans 0.5-0.7 Palm trees 0.7-1.1Beets 0.6-1.0 Peas 0.6-1.0Cabbage 0.4-0.5 Peppers 0.5-1.0Carrots 0.5-1.0 Pineapple 0.3-0.6Celery 0.3-0.5 Potatoes 0.4-0.6Citrus 1.2-1.5 Safflower 1.0-2.0Clover 0.6-0.9 Sisal 0.6-1.3Cacao/cocoa > 1.5 Sorghum 1.0-2.0Cotton 1.0-1.7 Soybeans 0.6-1.3Cucumber 0.7-1.2 Spinach 0.3-0.5Dates 1.5-2.25 Strawberries 0.2-0.3Orchards 1.0-2.0 Sugar beet 0.7-1.2Flax 1.0-1.5 Sugar cane 1.2-2.0Grains small 0.9-1.5 Sunflower 0.8-1.5Grains winter 1.5-2.0 Sweet potatoes 1.0-1.5Grapes 1.0-2.0 Tobacco early 0.5-1.0Grass 0.5-1.5 Tobacco late 0.5-1.0Groundnuts 0.5-1.0 Tomatoes 0.7-1.5Lettuce 0.3-0.5 Vegetables 0.3-0.6Maize 1.0-1.7 Wheat 1.0-1.5
64
Potential rooting depths of some crops
Source : Depeweg, 2002
- Moisture content of a soil
System design – Practical application depth
– Moisture content of a soil• The portion of AM that can be used by the plant to evapotranspire at not less than
ETm = ETrop is determined by fraction p called soil depletion factor of Management allowed deficit MAD.
• The corresponding amount of soil water is called Readily Available soil Moisture, in short RAM, and calculated by :
• The value of p is such as p = p (type of crop, ETm). The table below provides the values of p for different crop classes.
65
Dp mm` a
RAM mm` a
pB AM mm` a
With
Dp practical application depth
RAM readily available soil moisture
p depletion factor
AM available moisture
System design – Practical application depth
– Depletion factor p
66
Group Values of ETM (mm/day)2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0
1 0.500 0.425 0.350 0.300 0.250 0.225 0.200 0.200 0.1752 0.675 0.575 0.475 0.400 0.350 0.325 0.275 0.250 0.2253 0.800 0.700 0.600 0.500 0.450 0.425 0.375 0.350 0.3004 0.875 0.800 0.700 0.600 0.550 0.500 0.450 0.425 0.400Group Crops1 Onion, pepper, potato2 Banana, cabbage, grape, pea, tomato3 Alfalfa, bean, citrus, groundnut, pineapple, sunflower, water melon, wheat4 Cotton, maize, olive, safflower, sorghum, soybean, sugar beet, sugarcane, tobacco
p
SMALL for vegetables (Group 1) or if ETm high
HIGH for cereals (Group 4) of if ETm small
Vegetables are more sensitive to water stress
System design – Practical application depth
67
n.a. (1-p)AM p.AM n.a.
Legend :n.a. = not availableETA= actual (or instant) crop evapotranspirationETM = maximum crop evapotranspirationAMt = Instant available moisture
AM
Dr
FCWP soil moisture 0
1.0
0 (1-p)AM AM0AMt
If AM t > 1 @pb c
AM Then ETA ETM
Else if AM t < 1 @pb c
AM ThenETA
ETM
ffffffffffffffffAM t
1 @pb c
AM
ffffffffffffffffffffffffffffffffffff
ETA
ETM
fffffffffffffffff AM t
b c
S
AMt, 1
AMt, 2
ETA
ETM
ffffffffffffffff
System design – Practical application depth
This condition is the one used in the Design
2.6 Irrigation frequency and turn– The irrigation frequency is the time taken by the crop to deplete the
soil moisture to a certain level– Following the computation of the application depth, the irrigation
frequency is obtained by :
– The irrigation cycle (“tour d’eau” in French) is chosen on the base of farmers preference or economical considerations :
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F d` a
Dp mm
` a
IRn mm/ d` a
ffffffffffffffffffffffffffffffffffff
With
F irrigation frequency
Dp practical application depth
IRn Net irrigation requirement
System design – Irrigation frequency
T d` a
F d` a
With
T d` a
irrigation turn
F Irrigation frequency
• Actual application depth– The actual application depth Da is based upon the irrigation turn
chosen and is computed as follows :
– Next is to compute the actual depletion factor pa :
– The gross application depth will be :
69
System design – Gross application depth
Da mm` a
T d` a
B IRn mm/ d` a
pa
Da mm` a
AM mm` a
ffffffffffffffffffffffffffffffffffff
Dg mm` a
Da mm
` a
Ea
fffffffffffffffffffffffffffffffffff LR mm/ d` a
B T d` a
With :
Dg gross application depth
Da actual application depth
Ea f ield application ef f iciency
LR Leaching requirement
T irrigation turn
70
- Knowing for example the maximum working hours per day of TWmax in hr/d (based on the local experience) and the number of lateral positions or shifts Nsh done by a lateral per day (chosen in first guess in such a way that Ts would be less than 24 hours and allow the needed time for the lateral displacement), the set time in hours per set of laterals Ts is given by the relation :
- This value of Ts will be adjusted when an emitter with a know application rate is chosen. Hence, Nsh will also be adjusted
T s hr/ d` a
T Wmax hr/ d
` a
N sh
fffffffffffffffffffffffffffffffffffffff
Avec
T s number of hours to spend in a position in order
to bring the gross application depthDg
T Wmax maximumw orking hours per day
N sh number of shif ts or positionsb c
per day to be done
by a set of laterals
System design – Time per position
The specific or equipment discharge – The specific discharge qe is very practical for
• Estimating the total discharge that will be required for a localised irrigation scheme
• The comparison of different irrigation projects (eg. surface irrigation vs localised irrigation
• The computation is as it follows :
71
System design – Specific discharge
qe
l/ s/ ha` a
Dg mm
` a
T d` a
BT s hr/ d` a
BN shB 0.36
fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
Avec
qe specif ic discharge
Dg gross application depth
T irrigation turn
T s number of hours to spend in a position in order
to bring the grsoss application depthDg
N sh number of shif ts of a set of laterals per day
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AM FC @WP
b cBD r
With
D r root zone depth in cmor m
moisture in % of volume
Dp mm` a
pB AM mm` a
F d` a
Dp mm
` a
IRn mm/ d` a
ffffffffffffffffffffffffffffffffffff
With
F irrigation frequency
Dp practical application depth
IRn Net irrigation requirement
Chose T d` a
F d` a
With
T d` a
irrigation turn
F Irrigation frequency
Da mm` a
T d` a
B IRn mm/ d` a
pa
Da mm` a
AM mm` a
ffffffffffffffffffffffffffffffffffff
Dg mm` a
Da mm
` a
Ea
fffffffffffffffffffffffffffffffffff LR mm/ d` a
B T d` a
With :
Dg gross application depth
Da actual application depth
Ea f ield application ef f iciency
LR Leaching requirement
T irrigation turn
T s h/ d` a
T Wmax hr/ d
` a
N sh
ffffffffffffffffffffffffffffffffffffff
Avec
T s number of hours to spend in a position in order
to bring the grsoss application depthDg
T Wmax maximumw orking hours per day
N sh number of shif ts or positionsb c
per day to be done
by a set of laterals
qe
l/ s/ ha` a
Dg mm
` a
T d` a
BT s hr/ d` a
BN shB 0.36
fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
Avec
qe specif ic discharge
Dg gross application depth
T irrigation turn
T s number of hours to spend in a position in order
to bring the grsoss application depthDg
N sh number of shif ts of a set of laterals per day
CondensedPreliminary design parameters
73
Application 1Data : Crop = Cucumber, ETM peak = 7 mm/d at peak period, Crop spacing : 30 cm X 30 cm ; Row spacing :30 cm ;Soil : Loam and homogeneous; ΘFC = 30%, ΘWP= 15%; Ea = 95% ; GC= 70%, rainfall P=0; Groundwater contribution R=0; LR = 1 mm/d Tasks : i) Compute p, AM, Dp , IRg, ii) Determine the irrigation frequency F and turn T, the actual application depth Da, the actual depletion factor paand the gross application depth Dbiii) If the maxi working time per day is Twmax =22h and assuming that Nsh=2 shifts are to be done, compute the time per position Tsiv) What is the specific discharge qe ?
System design – Specific discharge Solution of application 1Solution i)Cocumber (ETM=7mm/d;Group 2)p 33%Dr (mm) 900ΘFC 0.3ΘWP 0.15AM (mm) 135Dp (mm) 43.9ETM peak (mm/d) 7GC(%) 70ETMLoc (mm/d) 5.9Pe (mm/d) 0R(mm/d) 0IRn(mm/d) 5.9Ea 0.95LR (mm/d) 1IRg(mm/d) 7.2Solution ii)Frequency F(d) 7.5
An irrigation turn T(d) 6
is chosen. It allows within a week of 7 days, one day for other businesses
Actual Da (mm) 35.1Actual depletion pa 26%Depht Dg(mm) 43.0Solution iii)Twmax (h/d) 22Nsh 2Ts(hr/d) 11
Solution iv)qe (l/s/ha) 0.90
74
Final design
III. Final Design
• Once obtained the preliminary design parameters, one can start the final design steps
• The required adjustments will a review of the preliminary parameters in order to adapt them to human an physical limitations, financial limitations and the localized irrigation equipment available
• The next steps are the emitters selection with their spacings
75
Final design
3.1 The emitter selection with driplines• The first factor for the emitter selection is the soil infiltration rate • The selection of the emitter should be done in such a way that its
application rate is smaller that the soil infiltration one, in order to avoid runoff
• The following catalogue extract provides the emitter as a function of several variable
• In fact :
76
Pemit in/ hr` a
Pemit soil type, crop type, emitter spacing, lateral spacingb c
With :
Pemit emitter application rate
type of crop turf ,shrub & ground cov er, R S
soil type clay ,loam ,sandR S
emitter spacing crop spacing
lateral spacing crop lines spacing
Final design
Emitter selection table (1)
NETA
FIM
-US
A T
ech
line C
V
System design
Application rate Pemit (in/hr)
+Vegetables+small trees
NETA
FIM
-US
A T
ech
line
Emitter selection table (2)Final design
Application rate Pemit (in/hr)
+Vegetables+small trees
Indication of field parameters
79
Aw Semit
D or W
Sp
Slat
2 driplines for 1 crop line
Final design
• The second (and last) factor for the emitter selection is the energy• Since energy means money, low pressure equipment would be
preferred as far as the uniformity is not compromised• The available equipments from manufacturers have different
sensitivity to pressure variation and this has an impact on uniformity
80
Final design
Conversion table Unit Unit Unit Unit Unit
Flow 1 GPH (USA) 0.000001 m3/s 0.003785 m3/h 0.06309 l/min 0.001052 l/s1 GPM (USA) 0.000063 m3/s 0.227125 m3/h 3.785412 l/min 0.06309 l/s1 m3/h 0.000278 m3/s 1 m3/h 16.666667 l/min 0.277778 l/s
Pressure 1 PSI 0.068948 bars 0.703088 m of water
Distance 1 in (or ") 0.0254 m 2.54 cm 1 Ft (or ') 0.3048 m 30.48 cm
Area 1 in² 0.000645 m² 6.4516 cm² 1 Ft² 0.092903 m² 929.0304 cm²
1 Acre 4,046.86 m²40,468,564.2
2 cm² 1 Are 100 m² 1,000,000 cm²
81
Final design
Application 2
Data : Crop = Cucumber, ETM peak = 7 mm/d at peak period, Crop spacing : 30 cm X 30 cm ; Row spacing Sr=30 cm
Soil : Loamy sand; ΘFC = 30%, ΘWP= 15%; Ea = 95% ; GC= 70%, rainfall P=0; Groundwater contribution R=0; LR = 1 mm/d
Task : Select an emitter and give its characteristics
Solution of application 2Pemit (Loamy sand,cocumber=vegetable, SeXSr=30cmX30cm)Conversion factors
1in = 2.54cm
1GPH (US) = 0.001052 l/s
1in/hr = 25.4mm/hr1PSI 0.70m of water at 4°C
Emitter spacing Se(in) 11.8Row spacing Sr(in) 11.8Emitter selected Values Units Values Units
Flow discharge qe 0.4 GPH 4.21E-04 l/sSemit 12 in 30 cm
Slat 12 in 30 cmPemit 0.39 in/hr 9.9 mm/hr
Nominal inlet pressure 35 PSI 24.6 m of water Comment : since the Sr proposed by the catalogue is a maxi row spacing, we have keptthe Slat =30 cm corresponding to the cocumber which is smaller than 18 in =46 cmFor loamy sand, the application rate must no be greater than 19 mm/hr according to the table of Keller and B liestnerTherefore, an emitter of 9.9 mm/hr is convenient
Nemit/ pl Pl can Ft²
` aB 0.75
Aw emit Ft²` a
ffffffffffffffffffffffffffffffffffffffffffffffffffffffff
Dcan Ft` ab c2
4
ffffffffffffffffffffffffffffffffffffffB 0.75
Aw emit Ft²` a
fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
Where :
Nemit/ p number of emitters per plant
Pl can plant canopy
Aw emit wetted area per emitter This value is
given byexperience or bythe manufacturer
D can diameter of the canopyprojected vertically
on the ground D can is also called canopy size
D can
Final design
3.2 Emitter selection with point source irrigation
Use for trees
83
• Exercise :– Data : loam soil. The crop is a native plant with a canopy size of 16
foot. The technique is point source drip irrigation. How many drippers of 1GPH are required per plant a)Using the graph ? , b) Using the formula ?
Ans :a) Using the black graph withDcan = 16’ (4.88 m), we obtain Nd/p= 6 drippers of 1 GPH.b) Using the formula we get : Awdrip = 24 ft²(red from the catalog) and Nd/ p
Dcan2
4
ffffffffffffffffB 0.75
Aw drip
fffffffffffffffffffffffffffffffffffffffff
3.14 B162
4
ffffffffffffffffffffffffffffffB 0.75
24fffffffffffffffffffffffffffffffffffffffffffffffffffffff6.28 6 drippers
Final design
Final design
3.3 System layout
• The final layout will permit :– fitting emitters and emitter lines spacing and the
dimensions of the cropland so as to reduce to the minimum non irrigated portions of land
– letting enough space for the roads, the drains (to be realized for the removal of excessive rainfall only), farm bunds, toilets, ...
3.4 System design for a single farmer– First, one should compute the set time Ts : it is the time during
which a set of laterals operates at the same time in order to bring the soil the gross application depth Dg, before opening a new set of laterals at the next location. The set of laterals is not moved, but is open or closed
– One should ensure that this time Ts is enough for launching the operation of the new set at new location
– The formula used is :
85
T s hr` a
Dg mm
` a
P emit mm/ hr` a
fffffffffffffffffffffffffffffffffffffff
Where
Dg gross application depth
P emit application rate of the emitter
T s application time for a set of laterals
Final design
• One must bear in mind that Ts and Nsh are linked. Once Ts is chosen then the value of Nsh is re-evaluated using the expression :
86
N sh T Wmax hr/ d
` a
T s hr/ d` a
ffffffffffffffffffffffffffffffffffffff
With
T Wmax maximumw orking hours per day
T Wmax < 24 hrs, the dif ference w ould be the equipment displacement timeb c
Ts number of hours to spend in a position in order
N sh number of shif ts or positionsb c
per day to be done
by a set of laterals to bring the grsoss application
depth Dg
Final design
Assume that Ts gives Ts=12 hrs in a first computation. In this figure, it is theoretically possible to make Nsh= 2 shifts within 24 hours since 2x12=24 hrs
In localised irrigation, the working time Tw may reach 24 hours if no displacement is required; On the other hand, irrigation turn can be T=1 day since it is possible in localized irrigation to bring permanently the soil moisture near the field capacity
87
– In the example of Ts=12 hrs with 2 shifts per day, if the shift consists of the automatic operation of a the inlet of a manifold, then no additional time is required ; otherwise an additional time is needed to the manual operation to join the new location and operate the valve. Ts could be Ts=11hrs for example, with the 1hr for the manual operation
– In the case one wants to make 2 manual shifts for example a Ts=12 hrs are is not acceptable
• What can be done if Ts is not acceptable ? There are 3 options :– Option 1 : install additional laterals and ensure a more important discharge so as
to have more laterals working simultaneously
– Option 2 : re-evaluate the soil moisture depletion pa level at which a new irrigation must be done. That means changing the irrigation turn T
– Option 3 : change the emitter selection and take one with a higher flow rate (but not exceeding the soil infiltration rate)
Final design
• The option 1 is often the less economical. The option 2 the most practical. The option 3 can be used if there is at least another sprinkler with an application rate compatible with the soil infiltration rate
• In the table below are presented some values of application rate not to exceed while selecting an emitter with a certain application rate
88
Final design
Land slopes0-5% 5-8% 8-12% 12-16%
Texture and layers Application rates (mm/hr)Gross sand till 1.8 m 50 38 25 13Gross sand on top of more compact soils 38 25 19 10Loose sand and loam til 1.8 m 25 20 15 10Loose and loamy sand over more compact soils 19 13 10 8Fine loamy sand till 1.8 m 13 10 8 5Fine loamy sand over more compact soil layers 8 6 4 2.5Heavy soils with clay and clay-loam 4 2.5 2 1.5
Maximum application rates for different types of soil and soil thickness by Keller et Bliesner (1990)
Sourc
e A
dap
ted
fro
m S
avva &
Fre
nke
n,
20
01
The emitter must be selected in such a way
that its application rate will not exceed than the above ones
89
• In case one choses option 2, the procedure for changing the value of T in order to get a new Ts is as it follows :
Final design
Ts Ok ?
Yes
No
Modify the irrigation turn T
The values of T, pa, Dg and Ts are OK
T s hr` a
Dg mm
` a
P emit mm/ hr` a
fffffffffffffffffffffffffffffffffffffff
Where
Dg gross application depth
P emit application rate of the emitter
T s application time for a set of laterals
Da mm` a
T d` a
B IRn mm/ d` a
pa
Da mm` a
AM mm` a
ffffffffffffffffffffffffffffffffffff
Dg mm` a
Da mm
` a
Ea
fffffffffffffffffffffffffffffffffff LR mm` a
With :
Dg gross application depth
Da actual application depth
Ea f ield application ef f iciency
LR Leaching requirement
Start
Because no time left for the shift
operations
90
Final design
Application 3Cocumber project continued
Data : Crop = Cucumber, ETM peak = 7 mm/d at peak period, Crop spacing : 30 cm X 30 cm ; Row spacing Sr=30 cm
Soil : Loamy sand ; the gross application depth Dg=43 mm and the first selection of an emitter yielded Pemit)9.9 mm/hr
Task : i) Compute the time per position Ts ii) the number of shifts Nsh to be done per day by a lateral if Twmax= 22hrsiii) the time assigned to the shifts operations
Solution of application 3Dg (mm) 43.0Pemit(mm/hr) 9.9Ts(hr) 4.34Twmax (hrs) 22Nsh(nbr/d) 5
Missing time to reach 22h = 0.30
There still remain 2 hours to reach 24 hours
This last result means Ts is OK
91
Bad because not parallel to any side
Bad : the laterals will be too long
Bad since will required too long laterals
Bad because the rotation of laterals is impossible
Bad : it is preferable to avoid a 90° elbow with a lateral
Good and bad layouts of the manifold according to the farm shape in case of an horizontal land
Bad since will not lead to a maximum number of laterals with equal size
The system layout
Final design
– The number emitters per lateralis calculated by :
– The discharge of a lateral isgiven by :
– The nominal discharge of the system is given by :
92
Final design
Q lat l / s` a
qemit l/ hr
` aBNemit / lat
3600
fffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
Avec :
qemit discharge of one emitter
Nemit / lat number of emitters per lateral
Q lat discharge of the lateral
Q syst l/ s` a
N lat,simBQ lat l/ s` aa
With :
N lat,sim number of laterals working
simultaneouly
Q lat discharge of a lateral
Nemit / lat L lat m
` a
Semit m` a
ffffffffffffffffffffffffff
With:
Lrp lateral length
Semit emitter spacing
Nemit / lat number of emitter
per lateral
• If Ts is acceptable, the next are described below– The number of lateral positions
on a manifold , Npos, is obtained by :
– Knowing the number Nsh of shifts per day and the irrigation turn T(d), the number of laterals working simultaneously is deduced by the expression :
93
Final design
Npos L farm m
` a
S lat m` a
ffffffffffffffffffffffffff
Npos total number of lateral positions
L farm length or width` a
of the farm
parallel to the manifold
S lat lateral spacing
N lat,sim Npos
T d` a
BN sh nbr/ d` a
fffffffffffffffffffffffffffffffffffffffffffffffffffffffff
With :
N lat,sim nombre of laterals working simultaneously
on the irrigated farm
Npos total number of lateral positions
N sh number of shifts per day for a set of laterals
T irrigation turn
94
System design
T s h` a
Dg mm
` a
P emit mm/ hr` a
fffffffffffffffffffffffffffffffffffffff
Avec
Dg gross application depth
P emit emitter application rate
T s application time for a set of laterals
Ts Ok ?
YES NODa mm
` aT d
` aBET m peak mm/ d
` aChange the turn T pa
Da mm` a
AM mm` a
ffffffffffffffffffffffffffffffffffff
Dg mm` a
Da mm
` a
Ea
ffffffffffffffffffffffffff LR mm/ d` a
B T d` a
T s h` a
Dg mm
` a
P emit mm/ hr` a
fffffffffffffffffffffffffffffffffffffff
N lat L farm m
` a
S lat m` a
ffffffffffffffffffffffffff
N lat nombre total de position de rampes
L farm length or width` a
of the farm
parallel to the manifolds
S lat lateral spacing
N lat,sim Npos
T d` a
BN sh nbr/ d` a
fffffffffffffffffffffffffffffffffffffffffffffffffffffffff
With :
N lat,sim number of laterals working simultaneously
on the farm
Npos total number of lateral positions
N sh number of shifts per day of a set of lateral
T the irrigation turn
The values of T, pa, Dg and Ts are OK
Nlat,sim Ok ?
NO
YES
Nemit / lat L lat m
` a
Semit m` a
ffffffffffffffffffffffffff
With:
Lrp lateral length
Semit emitter spacing
Nemit / lat number of emitter
per lateral
Q lat l / s` a
qemit l/ hr
` aBNemit / lat
3600
fffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
Avec :
qemit discharge of one emitter
Nemit / lat number of emitters per lateral
Q lat discharge of the lateral
Q syst l/ s` a
N lat,simBQ lat l/ s` aa
With :
N lat,sim number of laterals working
simultaneouly
Q lat discharge of a lateral
Summary of the design steps for a semi-moved system
95
Final design
105.0 m
105.0 m
105.5 m
105.5 m
106.0 m 106.0
m
106.5 m
106.5 m
107.0 m
107.0 m
107.5 m
107.5 m
Water source
60
0 m
300 m
Map of the scheme
road
108.0 m
108.0 mApplication 4
Cocumber project continued Taking into account the data of application 3, make the final design of the semi-portable system for a single farmer for the map represented on the right.
Task : Determine i) The number of laterals working simultaneously Nlat,sim, ii) the number of positions Npos and the layout on the field iii) the discharges of a lateral Qlat (l/s), a manifold Qmani(l/s) and the main Qmain(l/s)
96
Final designSolution of application 4qemit (l/s) 4.21E-04Lfarm (m) 600Wfarm (m) 300Slat (m) 0.30Semit(m) 0.30Npos1 2000.0first evaluationT(d) 6Nsh(nbr/d) 5
Nlat,sim 66.7 66
which is an even number is chosen
Npos 1980the lost area is 20*0.3m*300m =
1800 m²
Nemit/lat 1000 Qlat (l/s) 0.42 Qmani (l/s) 27.77 or 8 m3/hr We make a designer choice of dividing the lateral into 4 portions 75 m The 1980 lateral positions are clustered into sets of 66 commanded each one by a valveFour sets of 66=264 laterals; all the laterals of the 4 sets works simultaneouly Every day (from day 1), 5 (number of shifts) x 264= 1320 laterals receive waterAfter 6 days, 6*1320 = 4*1980 = 7920 laterals of 75 m lenght eachone have received water
Llat (m) 75.0 Slat (m) 0.30 Semit(m) 0.30 Nsh(nbr/d) 20 T(d) 6.0 Nlat,sim 264 Nlat/mani 66 Npos 7920 qemit (l/s) 4.21E-04 Nemit/lat 250 Qlat (l/s) 0.11 Qmani(l/s) 6.94Nmani/main 60Nmani,sim per main 2Qmain(l/s) 13.89
97
Final design
Elevation 100m
105.0 m
105.5 m
106.0 m
106.5 m
107.0 m
107.5 m
Water source
60
0 m
300 m
Day 1
108.0 m
Day 2
Day 3
Day 4
Day 5
5x264 (4x66) laterals receive water everyday. There are 5 shifts per day for 4 sets of 66 laterals
each one. After 6 days, 6x5x264 laterals have received water Each line represents a position for 66
laterals of 75 m length each one
Dire
ction
of s
ucce
ssiv
e se
ts
2 mains of 600 m
Day 6Valve
Main
Manifold
Set of 66 laterals
Roads
Qlat (l/s) = 0.11Nlat/man i = 66Qmani (l/s) = 6.94Qmain (l/s) =13.89
Middle line
Pipe connecting the tow mains
Transport pipe
i. Dimensioning the pipes
• The manifolds feed simultaneously a number of laterals (flow is distributed en route). Therefore, Christiansen’s factor is applied to compute the friction losses (which should not exceed 20% of the emitters’ operating pressure)
• The laterals, submains, mains and all hydrants are selected in such sizes that the friction losses do not exceed 15% of the total dynamic head required at the beginning of the system network
• The system flow Qsyst may be Qtot (total flow rate of the laterals) or imposed by the water resource (dam, river, spring etc.)
98
Final design
ii. Discharge-pressure relationship :• where q= emitter discharge in l/h• Kd= discharge coefficient characterising each emitter• H = emitter operating pressure in m• x = emitter discharge exponent. The less the value of X, the less the flow will
be affected by pressure changes. x=0 for fully compensated emitters. x= 0.5 for turbulent emitters like orifice; x=0.4 for vortex emitters. For tortuous path emitters, x is varies between 0.5 and 0.7 whist the value is 0.7-0.8 for long path emitters.
• After measuring (q1,H1) and (q2,H2), it is possible to compute Kd and x.
• Exercise – An equipment tester measured (q1,H1)=(2,8) and (q2,H2)=(3,20) for one
emitter. What are the values of Kd and x) ? Tip : use
99
System design
x log q
1/q
2
B C
log H 1 /H 2
B Cfffffffffffffffffffffffffffffffffffff
q q H` a
K d Hx
Ans : x=0.44; Kd=0.80
• Knowing the manufacturer average discharge under the average pressure, one can calculate the pressure required for a different discharge by the relation :
• Exercise : Data : From a manufacturer catalogue was selected a dripper with : x = 0.42, q = 4 lph at H = 10 m. Task : The suitable discharge to meet the irrigation needs is 4.32 lph. What should be the pressure ?
100
System design
H a Hq
a
q
fffffffH
J
I
K
1
x
ffff
With
qa
lphb c
average emitter discharge obtainable
under the average pressureH a
q lphb c
emitter discharge obtainable under the average pressure H
x discharge pressure relation exponent
• In practice : several emitters, various flow regimes, other factors affecting the pressure/discharge relation along the laterals in the field (local minor losses at emitters/connection, temperature fluctuations etc.)
• Hence : manufacturers should always provide charts for the optimum length of emitter laterals, based on the size of pipe, emitter spacing, operating pressure, flow rate and slope
101
System design
102
System design
103
Pressure distribution along a lateral placed on a level ground
Inlet pressure (Pi)
Average pressure (Pav)
LateralPressure at distal end (Pd)
Pre
ssure
Distance along lateral
L
∆H
3/4
∆H
1/4
∆H
0
Qlat
Qemit
L/3 L/3L/3
L
System design
• General trends– Maximum at the inlet and minimum at distal end (assuming
level lateral)– Linear variation in between? NO ! – Equations for a level lateral
Pressure variation along an horizontal lateral
104
P i m` a
Pav m` a
3
4
ffffH m` a
Pd m` a
Pav m` a
@1
4
ffffH m` a
Where :
P i inlet pressure
Pav average pressure
Pd distal pressure
H absolute value of the pressure loss
Hallow 0.20BPav for the design
System design
– The above equations assume that the third in elevation change occur upstream of the average pressure point, and the two third occur downstream of that point (even if that assumption is not exact, the equations still work pretty well)
105
Equations for a sloping lateral
!
System design
P i Pav 3
4ffffH @
1
3fff E i @E d
b c
Pd Pav @1
4ffffH
2
3fff E i @E d
b c
Where :
P i inlet pressure
Pav average pressure
Pd distal pressure
H absolute value of the pressure loss
Hallow 0.20BPav for the design
The pressure variation in between the inlet and the distal points are given by :
– Case of a level ground (slope nil) :
– Case of a lateral on a sloping land:
106
P P i m` a
@Pd m` a
Pav m` a
3
4
ffffH @Pav m` a
1
4
ffffH m` a
H m` a
With
P i inlet pressure
Pav average pressure
Pd distal pressure
H absolute value of the pressure loss
Hallow 0.20BPav for the design
P P i @Pd
Pav 3
4ffffH @
1
3fff E i @E d
b c@Pav
1
4ffffH @
2
3fff E i @E d
b c
H @ E i @E d
b c
Where :
P i inlet pressure
Pav average pressure
Pd distal pressure
H absolute value of the pressure loss
Hallow 0.20BPav for the design
System design
107
Allowable pressure variation along a lateral– The pressure variation inside an hydraulic unit must be obey to the
Christiansen criterion that ensures a high degree of uniformity within the unit
– An hydraulic unit can be a hole irrigation scheme or a sub-set of the scheme (area depending on the a sub-main, a sub-sub-main, manifold …)
The recommendation is that the difference (qmax - qmin) along any lateral must not exceed 10% of qav of the emitters (criterion of Christiansen, applicable in all pressurized systems)
Because of the square root relationship between the discharge and the pressure Q ~H0.5, this criterion means the pressure difference (Pmax - Pmin) must not exceed 20% of the average
pressure Pav of the emitter :
General rule :
H allow 20%BP av
and P P max @P min H allow
Avec:
H allow allow able pressure variation
P nom average pressure of the emitters
P max maximumpressure along the lateral can be Pi
b c
P min minimumpressure along the lateral can be Pd
b c
LLLLLLLLLLLLLLLLLLLLLLLLL
System design
• This design aims to determine the pipe diameter i) that will allow a flow with a velocity inferior to the allowable vallow, and ii) along which the head loss ∆H between any two emitters is less than allowable ∆Hallow
– The velocity in any pipe must comply with the following : • For plastic pipes V≤ 1.7 m/s• For metal pipes (steel, aluminium, cast steel) V≤ 2 m/s
– The preliminary diameter of the pipe is computed from V=Q/A, according to the process :
– The diameter obtained must be changed against a commercial available diameter, which can selected from the following abacuses or the commercial catalogues before any friction loss calculation
108
System design
Design of the laterals (distributing pipes)
D mm` a
Q m3 / h
b c
V m/ s` a
fffffffffffffffffffffffffffff
vuuuut
wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww
x 18.811For metal steel, alu,cast steel etc A
b c, Chose V 2m/ s
For plastic PVC, PE, etc Ab c
, Chose V 1.7 m/ s
X^\
^Z
Y^]
^[
• Pipes are found in the commerce in classes of nominal pressures that should not exceed any equipment :– For example, for unplasticized PVC (uPVC), are found classes 4, 6, 10
and 16 corresponding to nominal pressures of 4, 6, 10 and 16 bars
Design of the laterals– The friction head loss is
a function of the diameter of the pipe
– The friction head loss in non distributing pipe is computed using a formula or an abacus
– For example, a useful formula is the opposite one :
109
System design
Colebrook, Calm on & Lechapt :
H simple m/ m` a
aQ m3 / s
b cD EN
D m` aB CM
fffffffffffffffffffffffffffffffffffffff
a
Q m3 / hb c
3600
fffffffffffffffffffffffffffffffffffffff
H
J
I
K
N
D mm` a
B 10@3
B CM
ffffffffffffffffffffffffffffffffffffffffffffffffffffff
With :
H simple f riction head loss per meter in the pipe
a a material of the pipeb c
, a coeff icient
N N material of the pipeb c
, a coeff icient
M M material of the pipeb c
, a coeff icient
Matériaux a N M
Mortier de ciment centrifugé 1.049 .10-3 1.88 4.93
Métal neuf 1.100 .10-3 1.89 5.01
Béton centrifugé 1.160 .10-3 1.93 5.11
Fonte acier revêtement ciment 1.400 .10-3 1.96 5.19
Fonte acier non revêtu neuf 1.601 .10-3 1.975 5.25
Fonte acier non revêtu ancien 1.863 .10-3 2 5.33
PVC 1.101 .10-3 1.84 4.88
PVC diamètre D tel que 50≤D≤200 mm 0.916.10-3 1.78 4.78
PVC diamètre D tel que 250≤D≤1000 mm 0.971.10-3 1.81 4.81
110
Sou
rce :
D
eb
ois
sezo
n, 1
98
5
Coefficient values in the formula of Colebrook, Calmon & Lechapt
An alternative option consists in using an abacus among those presented in the following pages
System design
111
Example of friction head loss diagram : case of soft polyethylene
Diameters (mm)
Head loss (%)
Q(m3/hr)
Abacus 1 : Calculation of the friction loss ∆Hsimple = ∆Hsimple (Q,D,Press,v, PE pipes)
System design
Sou
rce
: S
outh
Afr
ican
Bur
eau
of
Sta
nda
rds
112
Diameters (mm)
Q(m3/hr)
Head loss
X100Pressure class :
4 →4 bars
Abacus 2 : Calculation of the friction loss ∆Hsimple = ∆Hsimple (Q,D,Press,v, uPVC pipes)
System design
113
This abacus is made considering :- Pipes of the class 16 bars from Ø 12 to 90 included.- Pipes of the class 10 bars from Ø 110 to 400 included.
How to use the abacus :Determination of the friction loss for a given diameter D and a discharge Q.Draw the vertical through Q till the intersection with the line of D.From this point, draw an horizontal line that cuts the friction loss axisat the researched value.
Example : for a discharge Q= 30 l/s, a pipe D=250 mm- The friction loss J~2,3 mm/m.- The flow velocity V is about 0,75 m/s.
Discharge in litre/second
Fric
tion
loss
in m
illim
etre
s pe
r met
er o
f pip
e
System design Abacus 3 : Calculation of the friction loss ∆Hsimple = ∆Hsimple (Q,D,Press,v, PVC pipes)
114
Source : Savva & Frenken, 2001
Abacus 4 : Calculation of the friction loss ∆Hsimple = ∆Hsimple (Q,D,Press,v, AC pipes)
System design
115
Source : Savva & Frenken, 2001
Abacus 5 : Calculation of the friction loss ∆Hsimple = ∆Hsimple (Q,D,Press,v, AC pipes)
Number of
outlets
F value Number of
outlets
F value
1 1.000 14 0.3872 0.639 16 0.3823 0.535 18 0.3794 0.486 20 0.3765 0.457 25 0.3716 0.435 30 0.3688 0.415 40 0.36410 0.402 50 0.36112 0.394 100 0.356
116
System design
For a lateral, the amount of water diminishes from upstream to downstream. The computation must determine first the friction head loss ∆Hsimple for a non distributing pipe. Afterwards, this value is corrected using a correction factor called F, which depends on the number of outlets
The values of F are provided below according to Keller & Karmelli (1975)
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 1000.300.350.400.450.500.550.600.650.700.750.800.850.900.951.001.05
Number of outlets on the lateral
Fact
or F
117
– The friction loss ∆Hlat of a lateral is computed by :
– The total pressure change (due to friction and topography difference) must be compared with the allowable pressure variation. Therefore, for a pipe diameter to be acceptable, the following condition is required :
– Otherwise, one must increase the diameter, re-select a commercial pipe and redo the pressure variation check
H lat m` a
H simple m` a
B F
With :
H simple L lat m` a
B H simple m/ m` a
F Correction factor w hich value
is a function of the number of outlets
L lat length of the lateral
P H lat m` a
@ E i @E d
b cm
` aD E
H allow m` a
System design
118
Summary of the diameter design steps for a lateral (uniform distributing pipe)
Use an abacus to determine ∆Hsimple (m/m)
YES
D, Llat and ∆Hlat are correct
NO
Increase DEND
Use Colebrook @Calm on @Lechapt :
H simple m/ m` a
a
Q m3 / hb c
3600
fffffffffffffffffffffffffffffffffffffff
H
J
I
K
N
D mm` a
B 10@3
B CM
ffffffffffffffffffffffffffffffffffffffffffffffffffffff
General rule :
H allow 20%BP av
and P P max @P min H allow
Avec:
H allow allow able pressure variation
P nom average pressure of the emitters
P max maximumpressure along the lateral can be Pi
b c
P min minimumpressure along the lateral can be Pd
b c
LLLLLLLLLLLLLLLLLLLLLLLLL
D mm` a
Q m3 / h
b c
V m/ s` a
fffffffffffffffffffffffffffff
vuuuut
wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww
x 18.811For metal steel, alu,cast steel etc A
b c, Chose V 2m/ s
For plastic PVC, PE, etc Ab c
, Chose V 1.7 m/ s
X^\
^Z
Y^]
^[
Select a commercial Diameter
H lat m` a
H simple m` a
B F
With :
H simple L lat m` a
B H simple m/ m` a
F Correction factor w hich value
is a function of the number of outlets
L lat length of the lateral
P H ramp m` a
@ E i @E d
b cm
` aD E
H allow m` a
System design
4 Constraints to bear in mind :
i) the dischargeii) the pressureiii) the velocity limit in pipes and
iv) the available pipe diameters
• Exercise : Hav = 15m, qav=0.5 GPH, Number of emitters = 650 while the lateral length Llat=100 . Polyethylene pipe. Task : a) compute ∆Hlat b) Is this value acceptable ?
119
Ans : a) We chose φ=25mm PE pipe. ∆Hsimple = 6m, ∆Hlat=2.14 m b) 20% Hav=3m > ∆Hlat. Hence 2.14 m losses are acceptable.
System design
Application 5Cucumber project continuedLet’s consider the configuration of application 4 with the main located in the mid of the scheme. The lateral in Polyethylene will have Llat=75m and 250 emitters with a nominal pressure of 35 PSI. The Qlat=0.11 l/s.
Task : Determine i) The allowable pressure variation along the lateral, ∆Hallow ii) An appropriate pipe diameter D for the lateral and its friction loss ∆Hlat
Solution of application 5Pav (m) 24.6∆Hallow (m) 4.92Plastic PE pipe, then Vlimit (m/s) 1.7Q(m3/h) 0.379 0.105l/sD initial (mm) 8.9
D(mm) 20We chose a PE commercial read
from the abacus 1Class 6Abacus ∆Hsimple 3.0%Llat (m) 75.0∆Hsimple (m) 2.25Number of outlets 250Correction factor F 0.35∆Hlat (m) 0.79Conclusion
D(mm) 20Commercial PE pipe read from the
abacus 1The laterals are parallel with the contour lines Hence Ei-Ed = 0∆P (m) 0.79 Smaller than ∆Hallow (m)
New ∆Hallow (m) 4.13Difference between ∆Hallow (m) and
∆P (m)
120
Design of the manifolds– Pipes in asbestos cement (AC) are no more recommended for domestic use due
to revealed induced health problems. – If farmers may drink irrigation water, it is preferable to avoid using AC pipes – The uPVC are sold per elements of 6 meters and service classes. The most
common classes are 4 up to 16. The internal diameters vary from 25 mm à 250 mm
– The polyethylene and uPVC pipes are often used in localized and sprinkler irrigation systems. The service pressure (allowable pressure of the pipe) must be respected (see table below)
– The selection of various commercial diameters can be performed using the abacuses (see abacus 2)
Classes Service pression de (bars)
4 4
6 6
10 10
16 16
uPVC pipes classes and service pressures
sou
rce :
Sou
rth
Afr
ican
Gu
reau
of
Sta
nd
ard
s,
19
76
System design
121
Application 6: Cucumber project continued
Tasks : i) identify the manifold with the most important constraints in term of pressure and topography, ii) determine the appropriate diameter in polyethylene pipes for the manifold that will be considered as a standard for the project
Elevation 100m
105.0 m
105.5 m
106.0 m
106.5 m
107.0 m
107.5 m
Water source
600 m
300 m
Day 1
108.0 m
Day 2
Day 3
Day 4
Day 5
5x264 (4x66) laterals receive water everyday. There are 5 shifts per day for 4 sets of 66 laterals each one. After 6 days,
6x5x264 laterals have received waterEach line represents a position for 66
laterals of 75 m length each one
Dire
ction
of s
ucce
ssiv
e se
ts
2 mains of 600 m
Day 6Valve
Main
Manifold
Set of 66 laterals
Roads
Q lat (l/ s) = 0.11Nlat/ man i = 66Qmani (l/ s) = 6.94Qmain (l/ s) =13.89
Middle line
Pipe connecting the tow mains
Transport pipe
System design
122
System design
Elevation 100m
105.0 m
105.5 m
106.0 m
106.5 m
107.0 m
107.5 m
Water source
600 m
300 m
Day 1
108.0 m
Day 2
Day 3
Day 4
Day 5
5x264 (4x66) laterals receive water everyday. There are 5 shifts per day for 4 sets of 66 laterals each one. After 6 days,
6x5x264 laterals have received waterEach line represents a position for 66
laterals of 75 m length each one
Dire
ction
of s
ucce
ssiv
e se
ts
2 mains of 600 m
Day 6Valve
Main
Manifold
Set of 66 laterals
Roads
Q lat (l/ s) = 0.11Nlat/ man i = 66Qmani (l/ s) = 6.94Qmain (l/ s) =13.89
Middle line
Pipe connecting the tow mains
Transport pipe
Solution of application 6i)The highest elevation in the system area is 108.25 m and the manifold at the top left corner is the most unfavorable in term of pressure
One should notice that the manifolds are perpendicular to the land slope ii) Slat (m) 0.30Nlat/mani 66Lmani (m) 19.5Qlat (m3/h) 0.379 0.11l/sNlat,sim 66Qmani(m3/h) 25.00 6.94l/sPav (m) 24.6∆Hallow (m) 4.13Type polyethylene pipes Vlimit (m/s) 1.7D initial (mm) 72.1
D(mm) 75commercial PE pipe, deduced from abacus N°2
a 9.160E-04N 1.780M 4.78∆Hsimple (m) Calmon Lechapt 0.613Number of outlets 66Correction factor F 0.357∆Hmani (m) 0.219Conclusion D(mm) 75The manifold are perpendicular to the contour lines Average slope 0.5%Ei (m)-Ed (m) -0.11∆P (m) 0.32smaller than ∆Hallow (m)New ∆Hallow (m) 3.81
123
Design of the main– The design process, for a main delivering water to several manifolds is the same
as for a lateral– The pipes will have here bigger diameters since carrying a more important
discharge– One must know first : the number of manifolds fed, the discharge of the manifold,
the topography of the submain– The allowable pressure variation is the rest after deduction of the variation in the
most unfavourable lateral and the related manifold– The computation continue form downstream to upstream (toward the main)
System design
Application 7: Cucumber project continued
Tasks : determine the appropriate diameter in uPVC pipes for the main that will be considered as a standard for the project. One should notice that the mains are perpendicular to the land slope.
124
System design
Solution of application 7
Spacing mani intakes (m) 20.00Nmani/main 60Nmani,sim per main 2Qmani(m3/h) 25.00Qmain(m3/h) 50.0 6.94l/sPav (m) 24.6 13.89l/s∆Hallow (m) 3.81PVC pipe Vlimit (m/s) 1.7D initial (mm) 102.0
D(mm) 200Commercial PVC class 6 pipe from abacus N°2
Abacus ∆Hsimple 0.08%Lmain (m) 600∆Hsimple (m) abacus 0.48
Number of outlets 1The most unfavorable case corresponds to the 2 mani
Correction factor F 1working at the end of the main, 600m downstream
∆Hmain (m) 0.480Conclusion D(mm) 200
The manifold are perpendicular to the contour lines Average slope 0.5%Ei (m)-Ed (m) -3.25∆P (m) 3.73smaller than ∆Hallow (m)
New ∆Hallow (m) 0.08
Upstream the inlet point of the main, the pressure variation
must be compensated by the pump. Christiansen criterion is
no more needed, though head loss should be reasonably small
Elevation 100m
105.0 m
105.5 m
106.0 m
106.5 m
107.0 m
107.5 m
Water source
600 m
300 m
Day 1
108.0 m
Day 2
Day 3
Day 4
Day 5
5x264 (4x66) laterals receive water everyday. There are 5 shifts per day for 4 sets of 66 laterals each one. After 6 days,
6x5x264 laterals have received waterEach line represents a position for 66
laterals of 75 m length each one
Dire
ction
of s
ucce
ssiv
e se
ts
2 mains of 600 m
Day 6Valve
Main
Manifold
Set of 66 laterals
Roads
Q lat (l/ s) = 0.11Nlat/ man i = 66Qmani (l/ s) = 6.94Qmain (l/ s) =13.89
Middle line
Pipe connecting the tow mains
Transport pipe
125
System design
Application 8: Cucumber project continuedIt is assumed that the pipe connecting the 2 mains has the same diameter than one (200mm, L=150 m, ∆Hinter_main= 150m*0.08%=1.2 m). From the intersection of the 2 main pipes, the PVC transport pipe runs to the water source where is located the pumping station. It carries two times the discharge of a main pipe.Task : Determine the diameter and the friction loss of the transportation pipe if its runs parallel to the contour lines and its lengts is 200 m
Elevation 100m
105.0 m
105.5 m
106.0 m
106.5 m
107.0 m
107.5 m
Water source
600 m
300 m
Day 1
108.0 m
Day 2
Day 3
Day 4
Day 5
5x264 (4x66) laterals receive water everyday. There are 5 shifts per day for 4 sets of 66 laterals each one. After 6 days,
6x5x264 laterals have received waterEach line represents a position for 66
laterals of 75 m length each one
Dire
ction
of s
ucce
ssiv
e se
ts
2 mains of 600 m
Day 6Valve
Main
Manifold
Set of 66 laterals
Roads
Q lat (l/ s) = 0.11Nlat/ man i = 66Qmani (l/ s) = 6.94Qmain (l/ s) =13.89
Middle line
Pipe connecting the tow mains
Transport pipe
Solution of application 8
Qtransport (m3/h) 100.0Twice the discharge of a main 27.77l/s
Vlimit (m/s) 1.7 D initial (mm) 144.3
D(mm) 250Commercial uPVC class 6, abacus N°2
Abacus ∆Hsimple 0.08% Ltransport (m) 200 ∆Htransport (m) 0.16
Total head or dynamic head requirements (TDH)– The TDH includes the following :
1. The suction lift at the pumping station2. Friction losses in the transport line3. Friction losses in mainline4. Friction losses in the manifold5. Friction losses in the laterals or driplines6. Geometric elevation difference the most unfavorable7. The drippers operating pressure8. Friction losses in the head control (filters, injectors etc.)9. Friction loss in fittings
126
System design
We have seen how to compute these 3 components
127
– Lifting friction loss at pumping station: • Keller & Bliestner proposed that the suction pipe diameter should be as such
suction velocity V≤ 3.3 m/s. If the pipe is not too long, the lifting suction loss can be approached as a velocity head
• The lifting friction loss can be computed using the formula :
• If one takes v=3.3m/s, then the lifting friction loss ∆Hsuction = 0.56 m
H suction m` a
V
2m/ s
` a2
2g m/s 2b c
fffffffffffffffffffffffffffffff
Avec :
H suction f riction loss in the lif ting pipe
v velocity in the pipe
g gravity acceleration
9.81 m/ s2
System design
128
– The friction loss in the transport line (from the water source to the entrance of the scheme)
• Given the length of the pipe Lpipe and its discharge, the appropriate diameter is determined using the previous process :
• Since water is note distributed en route in this pipe, Christiansen criterion does not apply. One should find a good agreement between the permissible friction loss and the diameter (that influence the cost) of the pipe
Use an abacus to determine ∆Hsimple (m/m)
Use Colebrook @Calm on @Lechapt :
H simple m/ m` a
a
Q m3 / hb c
3600
fffffffffffffffffffffffffffffffffffffff
H
J
I
K
N
D mm` a
B 10@3
B CM
ffffffffffffffffffffffffffffffffffffffffffffffffffffff
D mm` a
Q m3 / h
b c
V m/ s` a
fffffffffffffffffffffffffffff
vuuuut
wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww
x 18.811For metal steel, alu,cast steel etc A
b c, Chose V 2m/ s
For plastic PVC, PE, etc Ab c
, Chose V 1.7 m/ s
X^\
^Z
Y^]
^[
Select a commercial Diameter
System design
129
– The geometric elevation difference the most unfavorable• It corresponds to the elevation difference between the water surface at the
pumping station and the highest point in the farm where water must be delivered :
– The total head• It is computed by :
H total m` a
Xi
P average H lat H manifold H main H transport H suction
b c
H geom m` a
Z max m` a
@Z water m` a
One should notice that in the computation of Htotal, except the nominal pressure Paverage of the emitter, only friction loss terms are recorded. All the geometric elevation differences are taken into account in ∆Hgeom
System design
130
– The friction head loss in the fittings (elbow, T, valves…)• It is common to consider they amount to 10% of total of the calculated
head losses
• The head loss in filters, injectors etc.– They can be estimated to be 5 to 7 m in general
– Eventually, the Total Dynamic Head (TDH) is provided by the expression :
H fittings m` a
0.10BH total m` a
System design
TDH m` a
H total m` a
H geom H fittings m` a
H filtres,injec m` a
H filtres,injec m` a
5 to 7 m` a
131
Pump selection– The pump manufacturers usually provide the pump
characteristic curves– Knowing the total discharge Q and the TDH, one can select an
appropriate pump so as to ensure a high degree of efficiency and an engine to drive the pump (unless it is a diesel motor pump)
– In the selection process , the NPSH plays also an important role (see a pumping station lecture)
– The pump must be selected so as that the NPSHA (Net Positive Suction Head Available) is greater than the NPSHR (NPSH Required )
System design
– The basic formula for the required power of the engine to drive the pump is the brake horsepower formula :
Pump efficiency: e1 = 0.5-0.8 Electric motor efficiency: e2 = 0.7-0.9 ; Diesel engine efficiency: e2 = 0.5-
0.75 Hence overall pumping efficiency = 0.35 in diesel engine driven units to
0.50 in electric driven pumps. Higher efficiencies are not realistic.
132
P kW` a
Q m3 / h
b cB TDH m
` a
360Be1Be2
ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
P HP` a
Q m3 / h
b cB TDH m
` a
273Be1Be2
ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
Where
Q total discharge
TDH Total dynamic head
e1 the pump eff iciency fraction` a
;
e2 the driving eff iciency fraction` a
360 or 273 a constant for metric units
133
System design
Application 9: Cucumber project continuedConsidering the left main pipe (this one will impose the minimum pressure condition at the intersection of the two mains) and the top left edge farm (the highest) ,
i) compute the total dynamic head TDH assuming 7m of head loss in the filters and injectors, and ii) select the electric engine for the pump
Elevation 100m
105.0 m
105.5 m
106.0 m
106.5 m
107.0 m
107.5 m
Water source
600 m
300 m
Day 1
108.0 m
Day 2
Day 3
Day 4
Day 5
5x264 (4x66) laterals receive water everyday. There are 5 shifts per day for 4 sets of 66 laterals each one. After 6 days,
6x5x264 laterals have received waterEach line represents a position for 66
laterals of 75 m length each one
Dire
ction
of s
ucce
ssiv
e se
ts
2 mains of 600 m
Day 6Valve
Main
Manifold
Set of 66 laterals
Roads
Q lat (l/ s) = 0.11Nlat/ man i = 66Qmani (l/ s) = 6.94Qmain (l/ s) =13.89
Middle line
Pipe connecting the tow mains
Transport pipe
Solution i) Paverage (m) 24.6
∆Hlat (m) 0.79 ∆Hmani (m) 0.219 ∆Hmain (m) 0.480 ∆Hinter_main(m) 1.2 ∆Hsuction (m) 0.56
∆Htotal (m) 27.9
∆Hfittings (m) 2.8 Zmax(m) 108.5 Zwater(m) 98 ∆Hgeom (m) 10.5 ∆Hfilters,inject (m) 7 TDH (m) 48.1 Solution ii) Qtot (m3/h) 100.0 TDH(m) 48.1 e1xe2 0.5 P(KW) 27 Electric engine
Bill of quantities• After completion of the design computations,
make :– relevant drawings– the bill of quantities : a detailed list of all
equipment needed• In addition to the quantities specify :
– size and name (2-in ball valve, 50-mm pipe, etc.);– kind of material (brass, uPVC, etc.);– pressure rating (PN 16 bars, 6 bars, etc.);– type of joints (screw, solvent welded, etc.);– standards complied with (ISO 161, 3606, BS 21, ISO 7, etc.).
134
System design
Table of contents
The 3 required lists, Equipment working pressure, Main, Submains, Manifolds and Hydrants, Laterals and fittings, Head control, Pumping unit, Standards
Tenders
136
Required lists• Three different lists to be prepared :
– List 1 : for main, submains and manifolds with hydrants
– List 2 : laterals with emitters– List 3 : head control
137
Equipment
Equipment working pressure• Working pressure of the pipes
– Should always be higher than the system operating pressure
– Example :
138
Equipment
Operating pressure of a mini sprinkler system (bar)
Possible equipment
pressure (bar)
Advised pipe pressure (bar)
Laterals 2.3 – 2.5 4.0 4.0
Manifolds 2.5 – 2.7 4.0 4.0
Main 2.7 – 3.0 4.0 6.0(*)
(*) Desire to prevent water hammer problems
Main, submain, manifolds pipes and hydrants
• Most frequent material used : rigid PVC, HDPE, LDPE, quick coupling aluminium or light steel
139
Equipment
To be determined Description Quantity
Pipes and related pieces Total length x 1.05
Pipe connector fittings with 2 identical types of connections
bends, tees, end plugs, reducers, etc.To be used with the above pipes
Total required + 5%
Fittings with 2 different types of connections
Bends, tees, reduces, etc.e.g. bend 110 mm x 3 in (flanged)
Total required + 5%
Adaptors (starters) Fittings with one end threaded or flanged and the other end arranged in the same type of connection as the pipes. Used at the starting point of the pipelines and at any other point where valves are fitted
Total required
Shut-off and air valve The air valves are fitted on riser pipes connected with clamp saddleson the mains
Total required
Riser pipes for hydrants (case where the mains are buried)Shut –off valves, special hydrant valves
If the mains are not buried, the clamp saddles/shut-off valve fittings must be determined (same number as hydrants)
Total required
• Generally quick coupling LDPE pipes are used as surface laterals
140
Equipment
Laterals
To be determined Description QuantityPipes with related pieces
Total length x 1.05
Fittings & filters Adaptors, tees, bends, end plugs and line filters
Total required + 5%
Emitters and their connector fittings
e.g. Mini sprinkler complete set Total required + 5%
Shut-off and air valve
The air valves are fitted on riser pipes connected with clamp saddleson the mains
Total required
141
Equipment
To be determined Description Quantity
shut-off valves Total required
check valve Total required
fertilizer injector Total required
air valve Total required
filters, Total required
pressure regulators Total required
Pressure gauge etc. Total required
Head control
Example of simplified head control
142
Equipment
Source : NETAFIM FDS
143
Equipment
To be determined Description Quantity
Average P (kW)
Type of pump - e.g. centrifugal single or multi-stage, turbine, electro, submersible …- inlet and outlet diameter- the type and number of stages
Total required
Capacity and output of the pumping unit
the water deliveryversus the dynamic head.
Pumping unit
• Equipments are manufactured according to various standards• Standards, though equivalent, differ in terms of dimensioning,
class rating, safety factor, nomenclature • The International Standard Organisation (ISO) has devoted
great technical engineering effort to make standards , for regional and national conformity , reducing confusion
• But there are still a lot of national standards and small farmers are often confused
• Example : 4-in rigid PVC pipe, 6.0 bars, in two different national standards
144
Equipment
Standards
Source : Phocaides, 2001
• Produce a simple and clear description of the required equipment
• Example : black LDPE, PN 4.0 bars, to DIN 8072 or equivalent standards in compliance to ISO, supplied in coils of 200 m :– a) 32 mm DN, 1800 m– b) 25 mm DN, 3200 m
• If the equipment does not comply any standard, a full and clear technical description must be provided in term of material it is made of, working pressure and use
• Most of the irrigation equipment should meet the standard requirements in the table below
145
StandardsStandards
146
Standards
Source : Phocaides, 2001
147
Standards
Source : Phocaides, 2001
Tenders• For equipment & services up to €500, purchase possible
through « quotations » with 2-3 suppliers• When the value of purchase > €600, it is effected through a
tender. This is done in accordance with the “store of regulations” of the project or the country
• Wide advertisement should be given to any tender
148
Tenders
Tender advertisement
Name of the buyer
Description of the items
Address for equipment delivery
Closing date and time for the tenders
Indication of where bidders can have
further information
Statement that buyer is not bound to lower offer, etc.
149
Tender document
Make available for propect. bidders
General conditions
Technical specifications of equipment
Time & method of delivery (FOB, CIF, ex-stock, method
of payment, ...)
Tenders of more than €3000, bidders should
provide a bank guarantee or check = 10% of tender
value price
Tenders
• Include, in case of international bids, in the contract the following documents (source : Phocaïdes 2001) :– invitation for bids (as described above);– instructions to bidders (source of funds, eligible bidders, goods and
services, cost, content of bidding documents, preparation and submission of bids, opening and evaluation, award of contract, etc.);
– general conditions of contract (definitions, country of origin and standards, performance, security, inspection and tests, insurance, transportation, warranty, payment, amendments, delays, force majeure, etc.);
– special conditions;– technical specifications (general, materials and workmanship,
schedules of requirements/bill of quantities, and particular technical requirements/specifications);
– bid form and price schedules;– contract form, bid security and performance security.
150
Tenders
• Example (source : Phocaides, 2001)
151
Tenders
• Example (source : Phocaïdes, 2001)
152
Tenders
153
• Example (source : Phocaïdes, 2001)
Tenders
Table of contents
Review NETAFIM Landscape and Turf Products Catalogue
Review JAIN Catalogue for Micro irrigation
SENMINGER catalogue
155
Table of contents
Clogging and water treatment
Filtration : settling basin, vortex sand separator, screen mesh filter, sand media filter, disc filter
Fertigation : closed tank, venture, piston pump
157
• Narrowness of flow paths makes emitters sensitive to clogging• Water analysis is required to determine the appropriate treatment
• Water analysis required : 1) Total suspended solids, 2) Comple cathion-anions analysis, 3) Hardness and PH, 4) Total dissolved solids, 5) Iron (both ferrous and ferric) and hydrogen sulfide, 6) Bacteria population and possibly iron bacteria 158
Clogging
Causes of clogging
Physical Chemical Biological
Sand, silt, plastic chips,
metallic flakes
Precipitation of iron and salts
(calcium carbonate), of
fertilisers
Growth of algae/bacteria
in water source, in drippers or
sprinklers,
Clogging & Water treatment
159
Clogging & Water treatment
Confront the results to table below to determine water treatment required
– Install the WT plant (filtration and chemigation) at the pumping area– Since debris can enter the system after any pipe breakage, additional
protection should be provided as small screens, disc filters at the header of laterals or manifolds (cost optimisation)
– Small screens or disc filters required for multi-users system, after the fertigation unit of each user
160
Simple filtration (screen filter, disc filter) often
enough to remove sand
Water type
Groundwater
Open sources
Precipitations may require treatment
of chemicals at times
• Pretreatment with settling basins or vortex extractor
• Sand filter• Screen filter• Chemical treatment
Water treatment
Clogging & Water treatment
• Generally particles of larger than 0.075 mm (0.150 mm for some manufacturers) must be removed
• Consider that fine sand and very find sand may settle in the system areas (different from emitters) where flow is small (e.g. the end of laterals). Hence a 200 mesh screens may be required even if the diameter of emitters is close to 1mm
• The 200 mesh screen is the most common used screen filter used. It cannot arrest fine sand and silts as shown on the table below.
• The following table provides the appropriate mesh screen for the particle size found.
161
Filtration
Clogging & Water treatment
• Settling basin• Vortex sand separator• Screen mesh filter• Sand media filter• Disc filter
162
Type of filtration
Clogging & Water treatment
163
Clogging & Water treatment • Settling basin– Design in such a way that
it takes 15 mn to water particle to reach the pump (Keller & Bliestner). Most inorganic particles greater than 0.080 mm will settle
– Equivalent to 200 mesh screen filtration
– Used together with other filtration methods
Clogging & Water treatment
• Vortex sand separator– Suitable if the sand load
is high– Modern one can remove
up to 98 % of the sand that otherwise will be trapped in the 200 mm mesh screen
164
Clogging & Water treatment
Picture : Phocaides 2001
• Screen mesh filter– The simplest of all filters– Generally cannot be back flushed (some manufacturers proposed
this possibility when using a pair of filters)– Not recommended in removal of algae or organic material– To be cleaned manually
165
Clogging & Water treatment
Picture : Phocaides 2001
166
Screen mesh filter
Clogging & Water treatment
167
Clogging & Water treatment
• Sand media filter– Developed to arrest
particle that other filters cannot
– Effective in filtering particles from 25-200 microns
– Recommended for filtering out algae, when mesh screen or disc filter require too frequent cleaning...
– Cleaned through back flushing : 2 filters at least are required (if irrigation must be continue during the process)
168
Clogging & Water treatment
Picture : Phocaides 2001
169
Clogging & Water treatment
• Disc filter– Composed of many plastic discs
with grooves (long narrow channels) which are very tightly spaced
– Water can flow through the disc in both directions. Hence, back flushing is possible (great advantage compare to mesh screen filters)
– Has become more and more popular because of back flushing and cost less than mesh screens
– When used together with sand filter, make provision for a hose with clean water for cleaning the filter 170
Clogging & Water treatment
Picture : Phocaides 2001
• There are 3 main types of fertilizer injectors :– Closed tank– Venturi– Piston pump
171
Fertigation
Fertigation
• Closed tank– Part of the flow is diverted to the tank entering by the
bottom, mixed up with the fertilizer and is reinjected into the system
172
Fertigation
Picture : Phocaides 2001
• Venturi– Based on the principle of Venturi tube : a pressure difference is
needed between inlet and outlet– Relatively cheap
173
Fertigation
Picture : Phocaides 2001
• Piston pump– Powered by the water pressure– Can be installed directly on the supply line– System flow activates the pistons and operates the injector
174Picture : Phocaides 2001
175Thank you for your attention
References• Ankum, P. 2004. Flow control in Irrigation systems. UNESCO-IHE Lecture Notes.
Delft,The Netherlands.• de Laat, P.J.M. 2006. Soil Water Plant Relations. UNESCO-IHE Lecture Notes. Delft, The
Netherlands.
• PHOCAIDES, A. (2001) Handbook on Pressurized irrigation techniques, Rome, Italty, FAO.
• Savva P. A., Frenken K, 2001. Irrigation manual module 8 : sprinkler irrigation systems, planning, design, operation and maintenance. FAO Subregional office for East and southern Africa
• SAVVA, A. P. & FRENKEN, K. (2002) Localized irrigation systems. Planning, design, operation and maintenance. Harare, Zimbabwe, FAO.
• VERMEIREN L. & JOBLING, G. A. (1980) Localized irrigation - Design, installation, operation, evaluation Rome, Italy, FAO.
176
177
Appendix I
Point-source irrigation of 44 ha of orange
Denomination Values UnitsData :• A mature orange orchard will be irrigated using point-source technology• Use a single lateral per row of trees.• Fine loamy sand soil• Irrigated area is 44 ha with A 44ha length L (m) = 1000 L (m) 1000 width W (m) = 440 W (m) 440• The laterals are parallel to the width• The manifold is perpendicular to the main slope• The main slope of the farm ascending from the water ressource is i = 0.20%• Trees spacing is 5m x 5m and similar to emitter spacing and lateral spacing so that Sl(m) = 5 Slat 5m Stree(m) =5 Stree 5m•Wetted area per 2GPH emitter AW (Feet²) = 28 Aw/emit(Feet²) 28• Canopy size Dcan(m) = 2.03• Peak daily ETMpeak (mm/d) = 7.0 ETMpeak 7mm/d• Effective rain, peak-use period: assume zero. Pe 0• Residual R soil water : assume zero. R 0• Use a moisture content (ΘFC-ΘWP) of 25% (ΘFC-ΘWP) 25%• The depletion factor p is to be found for orange (citrus) p• Water source: widely available• Irrigation water quality ECw (dS/m) 1.2 Ecw (dS/m) 1.2• Field application efficiency Ea= 0.9 Ea 90%• Root zone depth Dr (citrus) : take the minimum value in the litterature Dr ???• Shaded area is GC= 75%. GC 0.75• Nominal emitter flow rate qemit (GPH)= 2 qa 2GPH• Emitter average or working pressure : 45 PSI Pav 45PSI
178
Appendix I - continued
Tasks Solutions1. Determine the number of 2 GPH emitters per tree, Nemit/pl Nemit/tree= 12. What is the value of the discharge for one orange qtree (l/s) ? qtree(l/s)= 0.0023. What is the discharge of one point source Qsource (l/s) ? Qsource(l/s) 0.0024. What is the application rate for a 2 GPH emitter Pe(mm/h) ? Pemit(mm/h) 2.95. What is the allowable infiltration rate for fine loamy sand soil with a slope of 0.20%, Inf (mm/h )? Inf (mm/h) 136. What is the value of the ETMLoc (mm/d) ETMLoc (mm/d) 6.17. How much is required for leaching LR (mm/d) LR (mm/d) 0.518. What is the gross irrigation requirement IRg (mm/d) IRg(mm/d) 7.249. To what amounts the readily available moiture RAM(mm) RAM=Dp (mm) 127.510. What is the irrigation frequency F(d) ? F(d) 21.011. To what amounts the irrigation Turn if it is 17 days lower than F, T(d) ? T(d) 4.012. What is the value of the gross application Depth, Dg (mm) ? Dg(mm) 29.013. The number Ts(h) of hours in a position is... Ts(h) 10.014. If 4 hours are to be left for activities other than irrigation, how many shifts Nsh are possible per day ? Nsh 2.015. Assuming a full lateral is 440 m long, how many lateral positions do we have, Npos Npos 20016. Now assuming the main passes in de middle of the rectangular farm, how many half-laterals work simultaneously Nlat,sim Nlat,sim 5017. How many point-source sets of 2GPH emitters do we have along a half-lateral, Nsource/lat ? Nsource/lat 4418. What is the allowable pressure variation for this farm ∆Hallow (m) ∆Hallow (m) 6.3319. What is the appropriate class 6 PVC pipe diameter for the half-lateral, Dlat (mm) ? Dlat (mm) 1620. What is the remaining pressure variation after removing that (Calmon-Lechapt) of the remotest and highest half-lateral, New ∆Hallow ? New ∆Hallow 4.4821. What is the discharge of the Manifold giving water to a number of simultaneous half-laterals , Qmani (m3/h) ? Qmani (m3/h) 8.3222. What is the appropriate (Calmon-Lechapt) PVC diameter for the manifold which is perpendicular to the slope, Dmani (mm) ? Dmani (mm) 5023. How mani manifolds are fed simultaneously Nmani,sim ? Nmani,sim 224. What is the discharge of the main, Qmain (m3/h) ? Qmain(m3/h) 16.6325. What is the appropriate (Calmon-Lechapt) PVC diameter for the main which is perpendicular to the slope, Dmain (mm) ? Dmain (mm) 63
179
Appendix I - end
Day 12x2 shifts per day
Day 2
Day 3
Day 4
25 lat φ16
25 latTotal number of laterals : (2x25*4)x2 = 400Qh-lat = 0.093 l/sQmani = 2.31 l/sQmain = 4.63 l/sqe(l/s/ha) = 0.11
Manifold φ50
1000
m
440 m
Main φ63
System layout