Localized Irrigation v2.08

A. Keïta, 2008-10

Localized Irrigation

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NETA

FIM

-USA

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.

2

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

3

I. Localised irrigation system layout

http://www.2ie-edu.org/

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

5

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, ...)

14

System Layout

• Analogies between traditional and modern irrigation methods

15

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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08

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

25

System Layout

26

System Layout

27

System Layout

Maize

28

System Layout

Cabbage

29

System Layout

Tomato

30

System Layout

31

System Layout

Maize

32

System Layout

Olive

33

System Layout

Orange

34

System Layout

Peach

35

System design

36

System design

Strawberry

37

System Layout

38

papaya

System Layout

Papaya

39

System Layout

Strawberry

40

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FIM

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System Layout

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System Layout

II. Preliminary system design42A. Keïta, 2iE

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2008


• 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 irrigationETM ET 0BK c

For localized irrgationETM 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


47

Values of crop factor Kc for permanent crops

Crop Young MatureBanana 0.50 1.10Citrus 0.30 0.65Apple, cherry, walnut

0.45 0.85

Almond, apricot, pear, peach, pecan, plum

0.40 0.75

Grape, palm tree 0.70 0.70Kiwi 0.90 0.90Olive 0.55 0.55Alfalfa 0.35 1.1


– Method 2 : Formulas using the ground cover percentage GC

• Keller and Bliestner formula

• The “unit time” = day, or decade, or month, or season. 48


ETM loc ETM peak mm/ unit time` ab c

B 0.1 A GC %` aq

wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwD E

WhereETM 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


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)

50

IRn ETM loc @ R P e

b c

WhereIRn 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/ monthP 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


IRg IRn

E a

ffffffffff LR

WhereIRg gross irrigation requirementE a f ield application eff iciency

E a K sB EUWhere

K s Average water stored in root volume

Average water appliedffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

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/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 70-80Low-medium pressure sprinkler systems

75

Microsprinklers, micro-jets, minisprinklers

75-85

Drip irrigation 80-90

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• Values of Ea as function of on farm irrigation methods


• 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)

54

Ans : IRn = 6mm/d ; IRg = 7.02 + LR


• 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 Q irrigation 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

WhereY r relative yield, ratio of estimated yield to full potentialECw dS / mor mmhos / cm

` airrigation w ater elect conduc

maxECe dS / mor mmhos / cm` a

maximumelect conducof saturated soil extract that w ill reduce the yield to zero

minECe dS / mor mmhos / cm` a

minimumelect conduc below w hichthere is no yield reduction


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)


• Exercise– Citrus (orange): irrigated with water of ECw = 2 dS/m.– Task : What would be the relative yield ?

– Ans :

58

Yr = 0.95


• 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 :

59

LR t ECw

2B maxECe

B Cfffffffffffffffffffffffffffffffffffffffffff

WhereLR t leaching requirement ratio under drip irrigationECw dS / mor mmhos / cm

` airrigation w ater elect conduc

maxECe dS / mor mmhos / cm` a

maximumelect conducof saturated soil extract that w ill reduce the yield to zero


• 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

60


LR LR tBIRn

E a

ffffffffff

WhereLR t leaching requirement ratio under drip irrigationIRn net irrigation requirementE a f ield application eff iciency

Ans :LRt=0.125; Ks=1; Ea=0.90 ; LR = 0.83 mm ; IRg = 7.5 mm

61

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

mmunit timefffffffffffffffffffffffffff

f gB 0.1 A GC %

` aqwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwD E

Read GC or Kr

ComputeIRn 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 Vfffffffffffffffffffffff Where w the density of w ater : 1 kg/ l

Or

100BG 1@G 2

G 2

fffffffffffffffffffffffB b

w

fffffffIn case V is unknow n

b 1.2 @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

WithD r root zone depth in cmor m moisture in % of volume


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


– 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

WithDp practical application depth

RAM readily available soil moisturep depletion factor

AM available moisture


– 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


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 AM0 AMt

If AM t > 1 @pb c

AM Then ETA ETM

Else if AM t < 1 @pb c

AM ThenETAETMffffffffffffffff

AM t

1 @pb c

AMffffffffffffffffffffffffffffffffffff

ETAETMfffffffffffffffff AM t

b c

S

AMt, 1

AMt, 2

ETAETMffffffffffffffff


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 : 68

F d` a

Dp mm

` a

IRn mm/ d` a

ffffffffffffffffffffffffffffffffffff

WithF irrigation frequency

Dp practical application depthIRn Net irrigation requirement

System design – Irrigation frequency

T d` a

F d` a

WithT d

` airrigation 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 :

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System design – Gross application depth

Da mm` a

T d` aB IRn mm/ d

` a

pa Da mm

` a

AM mm` a


Dg mm` a

Da mm

` a

Eafffffffffffffffffffffffffffffffffff LR mm/ d

` aB T d` a

With :Dg gross application depthDa actual application depthEa f ield application ef f iciencyLR 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

AvecT 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

q e l/ s/ ha` a

Dg mm

` a

T d` aBT s hr/ d

` aBN shB 0.36fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

Avecqe specif ic dischargeDg gross application depthT irrigation turnT 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

72

AM FC @WP

b cBD r

WithD 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


WithF irrigation frequency

Dp practical application depthIRn Net irrigation requirement

Chose T d` a

F d` a

WithT d

` airrigation turn

F Irrigation frequency

Da mm` a

T d` aB IRn mm/ d

` apa D a mm

` a

AM mm` a


Dg mm` a

Da mm

` a

Eafffffffffffffffffffffffffffffffffff LR mm/ d

` aB T d` a


T irrigation turn

T s h/ d` a

T Wmax hr/ d

` a

N sh

ffffffffffffffffffffffffffffffffffffff

AvecT s number of hours to spend in a position in order


T Wmax maximumw orking hours per day


per day to be done

by a set of laterals

q e l/ s/ ha` a

Dg mm

` a

T d` aBT s hr/ d

` aBN shB 0.36fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

Avecqe specif ic dischargeDg gross application depthT irrigation turnT s number of hours to spend in a position in order


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 spacinglateral spacing crop lines spacing

Final design

Emitter selection table (1)

NETA

FIM

-USA

Tech

line

CV

System design

Application rate Pemit (in/hr)

+Vegetables+small trees

NETA

FIM

-USA

Tech

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

SpSlat

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

4ffffffffffffffffffffffffffffffffffffffB 0.75

Aw emit Ft²` a

fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

Where :Nemit/ p number of emitters per plant

Pl can plant canopyAw emit wetted area per emitter This value is

given byexperience or bythe manufacturerD 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 irrigationUse 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

4ffffffffffffffffB 0.75

Aw drip

fffffffffffffffffffffffffffffffffffffffff

3.14 B162

4ffffffffffffffffffffffffffffffB 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


WhereD g gross application depthP emit application rate of the emitterT 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

WithT 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


per day to be done

by a set of laterals to bring the grsoss applicationdepth 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)

Sour

ce A

dapt

ed fr

om S

avva

& Fr

enke

n,

2001

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


WhereDg gross application depthP emit application rate of the emitterT s application time for a set of laterals

Da mm` a

T d` aB IRn mm/ d

` a

pa Da mm

` a

AM mm` a


Dg mm` a

D a mm

` a

Eafffffffffffffffffffffffffffffffffff LR mm

` a


Start

Because no time left for the shift

operations

90

Final design

Application 3Cocumber project continuedData : 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/hrTask : 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

3600fffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

Avec :qemit discharge of one emitterNemit / lat number of emitters per lateralQ 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

simultaneoulyQ lat discharge of a lateral

Nemit / lat L lat m

` a

Semit m` affffffffffffffffffffffffff

With:Lrp lateral lengthSemit emitter spacingNemit / 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 positionsLfarm length or width

` aof the farm

parallel to the manifoldS lat lateral spacing

N lat,s im Npos

T d` aBN sh nbr/ d

` afffffffffffffffffffffffffffffffffffffffffffffffffffffffff

With :N lat,s im nombre of laterals working simultaneously

on the irrigated farmNpos total number of lateral positionsN sh number of shifts per day for a set of lateralsT irrigation turn

94

System design

T s h` a

Dg mm

` a

P emit mm/ hr` a


AvecDg gross application depthP emit emitter application rateT s application time for a set of laterals

Ts Ok ?

YES NODa mm

` aT d

` aBET m peak mm/ d` a

Change the turn T pa Da mm

` a

AM mm` a


Dg mm` a

Da mm

` a

Eaffffffffffffffffffffffffff LR mm/ d

` aB T d` a

T s h` a

Dg mm

` a

P emit mm/ hr` a


N lat Lfarm m

` a

S lat m` a


N lat nombre total de position de rampesL farm length or width

` aof the farm

parallel to the manifoldsS lat lateral spacing

N lat,s im Npos

T d` aBNsh nbr/ d

` afffffffffffffffffffffffffffffffffffffffffffffffffffffffff

With :N lat,s im number of laterals working simultaneously

on the farmNpos total number of lateral positionsN sh number of shifts per day of a set of lateralT the irrigation turn

The values of T, pa, Dg and Ts are OK

Nlat,sim Ok ?

NO

YES

N emit / lat L lat m

` a

Semit m` a


With:L rp lateral lengthSemit emitter spacingN emit / lat number of emitter

per lateral

Q lat l / s` a

qemit l/ hr

` aBNemit / lat

3600fffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

Avec :qemit discharge of one emitterNemit / lat number of emitters per lateralQ 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

simultaneoulyQ 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

600

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

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 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 H x

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

qfffffff

HJ

IK

1

xffff

Withq a lph

b caverage 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)

Pres

sure

Distance along lateralL

∆H

3/4∆

H1/

4∆H

0Qlat

Qemit

L/3 L/3L/3L

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

34ffffH m

` a

Pd m` a

Pav m` a@1

4ffffH m

` a

Where :P i inlet pressurePav average pressurePd distal pressureH 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 34ffffH @1

3fff E i @E d

b c

Pd Pav @14ffffH

23fff E i @E d

b c



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

34ffffH @Pav m

` a

14ffffH m

` a

H m` a

WithP i inlet pressurePav average pressurePd distal pressureH absolute value of the pressure loss


P P i @Pd

Pav 34ffffH @1

3fff E i @E d

b c@Pav

14ffffH @2

3fff E i @E d

b c

H @ E i @E d

b c



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 variationP 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

fffffffffffffffffffffffffffffvuuuut

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


a

Q m3 / hb c

3600fffffffffffffffffffffffffffffffffffffff

HJ

IK

N

D mm` aB 10@3B CM

ffffffffffffffffffffffffffffffffffffffffffffffffffffff

With :H simple f riction head loss per meter in the pipe

a a material of the pipeb c

, a coef f 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

Sour

ce :

Debo

issez

on, 1

985

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

: Sou

th A

frica

n B

urea

u of

S

tand

ards

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 per

met

er o

f pip

e

System design Abacus 3 : Calculation of the friction loss ∆Hsimple = ∆Hsimple (Q,D,Press,v, PVC pipes)

114


Abacus 4 : Calculation of the friction loss ∆Hsimple = ∆Hsimple (Q,D,Press,v, AC pipes)

System design

115


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` aB F

With :H simple L lat m

` aB H simple m/ m` a

F Correction factor w hich valueis 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 EH 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


HJ

IK

N

D mm` aB 10@3B CM


General rule :

H allow 20%BP av

and P P max @P min H allow

Avec:H allow allow able pressure variationP 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




b c, Chose V 2m/ s


, Chose V 1.7 m/ s

X^\^Z

Y^]^[

Select a commercial Diameter

H lat m` a

H simple m` aB F

With :H simple L lat m

` aB H simple m/ m` a

F Correction factor w hich valueis 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 EH 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

sour

ce :

Sour

th A

frica

n Gu

reau

of S

tand

ards

, 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


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


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




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


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




Dire

ction

of s

ucce

ssiv

e se

ts

2 mains of 600 m

Day 6Valve

Main

Manifold

Set of 66 laterals

Roads


Middle line


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




Dire

ction

of s

ucce

ssiv

e se

ts

2 mains of 600 m

Day 6Valve

Main

Manifold

Set of 66 laterals

Roads


Middle line


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 2 m/ s

` a2

2g m/s 2b cfffffffffffffffffffffffffffffff

Avec :H suction f riction loss in the lif ting pipev velocity in the pipeg 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


HJ

IK

N

D mm` aB 10@3B CM


D mm` a

Q m3 / h

b c

V m/ s` a




b c, Chose V 2m/ s


, 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

` aX

i

P average H lat H manifold H main H transport H suction

b c

H geom m` a

Zmax m` a@Zwater 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

WhereQ total discharge

TDH Total dynamic heade1 the pump eff iciency fraction

` a;

e2 the driving eff iciency fraction` a

360 or 273 a constant for metric units

133

System designApplication 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




Dire

ction

of s

ucce

ssiv

e se

ts

2 mains of 600 m

Day 6Valve

Main

Manifold

Set of 66 laterals

Roads


Middle line


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

IV.Equipment standards and

tenders for supply

135A. Keïta, 2iE


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 Quantityshut-off valves Total required check valve Total required fertilizer injector Total required air valve Total requiredfilters, 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 QuantityAverage 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


147

Standards


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

VEquipment and fittings

(see catalogues)

154A. Keïta, Sept-Nov 2008


Table of contents

Review NETAFIM Landscape and Turf Products Catalogue

Review JAIN Catalogue for Micro irrigation

SENMINGER catalogue

155

VIMaintenance & Fertigation

156A. Keïta, 2iE


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


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


• 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


• Settling basin• Vortex sand separator• Screen mesh filter• Sand media filter• Disc filter

162

Type of filtration


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


• 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


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



166

Screen mesh filter


167


• 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



169


• 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



• 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


• Venturi– Based on the principle of Venturi tube : a pressure difference is

needed between inlet and outlet– Relatively cheap

173

Fertigation


• 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

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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

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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

Localized Irrigation v2.08

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