This document provides an internship report on farm water management in Jhang, Pakistan. It discusses the benefits of high efficiency irrigation systems over conventional flooding methods. Conventional methods have low irrigation efficiencies below 50% and cause water and fertilizer losses. The report outlines components of a drip irrigation system design for a 6.5 acre cotton field, including filters, pipes, emitters, and calculations of crop water requirements using reference evapotranspiration data and crop coefficients. Tables provide soil properties, crop information, and climate data used in the design process.

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ON FARM WATER MANAGEMENT
(OFWM) JHANG
INTERNSHIP REPORT
Prepared by the Internees:
ASAD SHAFIQ 2009-ag-3200
of final year B.Sc. (Agri. Eng.) from University of Agriculture, Faisalabad.
Under the Co-Supervision of:
Engr. Sheikh Arshad Nawaz WMO, OFWM
Engr. Muhammad Sadiq Anjum WMO, OFWM

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Chapter .1
WHY HIGH EFFICIENCY IRRIGATION
SYSTEMS (HEIS)?
Fig. 1.1 Punjab Water Budget
Flooding is the most common irrigation method practiced by the farmers and its
efficiency is not more than 50 percent. Such low irrigation efficiencies at farm level are major
constraint in attaining potential production from otherwise highly productive agricultural
lands. In addition, more than 40 percent of canal water is lost between mogha outlet and
farmers’ fields due to poor condition of tertiary conveyance system (water courses). The crop
water requirements are not met timely because of supply based irrigation water delivery,
which negatively affects the overall agricultural production.

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Conventional Irrigation System is Great Barrier in Economic Betterment of Farmer/Country
1.1 DEMARITS OF CONVENTIAL IRRIGATION
METHODS
a) Water Losses:
 Evapotranspiration
 Leaching
 Used by weeds & other plants
b) Fertilizer Losses:
 Excess uses
 Volatilization
 Deep peculation below root zone
 Used by weeds
 Fertilizers must be applied around the root zone for its more effectiveness
and for decreasing their losses due to mobility of nutrients.
 Increase soil pH
Fig. 1.2(a) Maximum distance from root for nutrient absorption (mm)
0
4
8
12
16
20
N K Mg Ca P
20
7.5
5
5
1
Maximumdistancefrom
rootfor
nutrientabsorption
(mm)

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Fig. 1.2(b)
c) Low Yield
d) Low Quality Yields:
 Plant remain in stress.
 Uneven distribution of requirements
e) Soil Erosion
f) Land Leveling Required
g) More Labor Cost
Sr. No. Country Area under HEIS (%)
1. Germany 100
2. Israel 100
3 France 90
4 Greece (Unnan) 65
5 Urdan 62
6 South Africa 37
7 USA 21
8 Italy 16
9 China 3
10 India 2
Reporting Year: 2007-08
Table. 1.1 Area of different countries under HEIS
Table. 1.2

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Chapter .2
Soil – Plant - Water Relationship
2.1 Soil:
2.1.1 Role of Soil in Plant growth:
 Act as water reservoir between rains and/or irrigation
 Act as nutrient/O.M. reservoir and
 Mechanically support and stabilize plants
 Home for natural Habitat
 Temperate regime for plant growth
2.1.2 Physical Characteristics:
a) Soil Composition
 Mineral matter (47 % of the soil volume)
 Organic matter (3 % of the soil volume only in upper 0.3m of soil)
 Air (25 % of the soil volume)
 Water (25 % of the soil volume)
Fig. 2.1
b) Soil Texture
 Sand (2mm to 0.05mm)
 Silt (0.05 to 0.002mm)
Mineral
47%
O.M
3%
Air
25%
Water
25%

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 Clay (less than 0.002)
Fig. 2.2 Soil texture Diagram
2.2 Water
2.2.1 Field capacity:
Field capacity is the amount of soil moisture retained in soil under
gravitational force.
2.2.2 Permanent wilting point:
Wilting point (PWP) is defined as the point at which plant is unable to uptake
water.
2.2.3 Available water content:
Soil texture is
determined by
the mass ratio,
or percent by
weight of the
above three
soil factions
(there are 12
major texture
classes)

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The amount of water actually available to the plant.
Fig. 2.3 Available water
Fig. 2.4(a) Water Movement in Soil
Fig. 2.4(b) Water Movement in Soil

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Soil Texture FC PWP Available Water
Coarse Sand 0.06 0.02 0.04
Fine Sand 0.10 0.04 0.06
Loamy Sand 0.14 0.06 0.08
Sandy Loam 0.20 0.08 0.12
Light sandy loam 0.23 0.10 0.13
Loam 0.27 0.12 0.15
Sandy clay loam 0.28 0.13 0.15
Clay loam 0.32 0.14 0.18
Clay 0.40 0.25 0.15
Table. 2.1 Field Capacity and Wilting Point (m3
/m3
)
2.3 Soil Water Measurement
2.3.1 Feel and Appearance Method
 Take field samples and feel them by hand
 Advantages: low cost; Multiple locations
 Disadvantages: experience required; Not highly accurate
2.3.2 Gravimetric Method
 Take field samples  weigh  oven dry  weigh
 Advantages: accurate; Multiple locations
 Disadvantages: labor; Time delay(24 hours @ 105o
C)
sevaporatedwaterofMass
soildryofMass
contentwaterMass 
2.3.3 Tensiometer
 Measure soil water potential (tension)
 A pipe, comprises of a ceramic cup on a lower end, & pressure gauge on other
end
 Fits up to the root depth
 Fills water in the tube
 Under water deficiency, soil extracts water from it
 Which creates –ive pressure on gauge
Vice versa

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Fig. 2.4 Tensiometer
INTHEFIELD

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Chapter .3
Drip Irrigation System Components
Fig. 3.1(a) Components
Fig. 3.1(c) Components Fig. 3.1(b) Components

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3.1 Head Control Unit:
3.1.1 Filters:
Prevent entry of impurities that can cause system clogging.
1. Primary Filters
 Hydro-cyclone Filter
 Gravel Filter
2. Secondary Filters
 Disc Filter
 Screen Filter
a) Hydro-cyclone Filter:
Tangential flow of water into the filter creates centrifugal action that separates
sand from incoming water. Below there is a sedimentation tank.
If ¾ of the sedimentation tank is filled with sand then remove sand from tank.
Fig. 3.2
b) Gravel Filter:
Prevent entry of organic impurities (e.g. algae, silt, and clay)
Used when water source is exposed to sunlight (e.g. canal water stored in a pond)

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Fig. 3.3
c) Disc Filter:
Micron precise filtration of solids, No filtration failure &higher filtration
efficiency.
 Filtration Media, Sharp edged silica sand
Fig. 3.4
d) Screen Filter:
Removes small size suspended materials (e.g. fine sand) or a little amount of
algae
 Filtration Media, Stainless steel/nylon/polyester mesh (16-400 mesh)
Fig. 3.5

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3.1.2 Fertigation Equipment:
a) Fertilizer Tank:
No external energy source required, Low
fertilizer application uniformity, each irrigation
turn requires tank cleaning & Low cost
Fig. 3.6
b) Venturi Injector:
No external energy source required, Better
fertilizer application uniformity, Limited operating
range & high pressure loss.
Fig. 3.7
c) Injection Pump:
High fertilizer application uniformity, no pressure loss
in the system, required external source of power &
limited operating range.
Fig. 3.8
d) Foot valve
Foot valve is resides in the pond at the end of suction pipe to filter pumping
water initially.
3.2 Pipes & Fittings:
3.2.1 Mainline:
 PVC made
 Placed underground
 Connect head control unit & sub-mains
Fig. 3.9

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3.2.2 Sub-Main:
 PVC / PE made
 Placed on surface / subsurface
 Connect mainline & laterals Fig. 3.10
3.3 Emitting Devices
3.3.1 Plain/Online Laterals:
 Emitters can be fixed at desired spacing
 Sizes varies from 12 to 20 mm
 Wall thickness 0.5-1.5 mm
 Suitable for orchards
 PE made
 Placed on surface /subsurface
Fig. 3.11
3.3.2 Integrated Drip-lines:
 Emitters are inbuilt at different spacing
 Suitable for row crops
 Sizes12 to 16 mm
 Wall thickness 0.5 to 1.5 mm
 Long Life
 High initial cost
Fig. 3.12
3.3.3 Drip Tape:
 Inbuilt hole at different spacing
 Suitable for row crops

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 Suitable for plain terrain
 Low initial cost
 Recyclable
 Short life
Fig. 3.13
3.3.4 Emitters/Drippers:
 Made of high quality plastics
 Mount on soft PE pipes (at frequent spaces)
 1.0 bar pressure
 Droplets discharge rate 1.0-24 liters/h.
 Drippers are also characterized by on-line &
in-line Fig. 3.13
Fig. 3.14 Schematic Diagram Of Drip Irrigation System
Fig. 3.14 Drip Irrigation System

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Chapter .4
Drip Irrigation System Design
To design a drip irrigation system for 6.5 acres of cotton crop. Soil is sandy loam.
System should be farmer friendly.
Canal water is available for 2 hours and discharge is 28 lps. Farmer is interested to use canal
water only but he can manage, ground water also.
Fig. 4.1
Crop Information
Parameters Crop Area
(acres)
Plant
spacing
(m)
Row
spacing
(m)
Age of
crop
(years)
Sowing
Date
Harvest
ing
Date
Block-
I
Row
crops
Proposed Cotton 6.5 0.3 1.52 new 15 May 15 Nov
Rotation
Total / Design Value 6.5 0.3 1.52 15 May 15 Nov
Table. 4.1. Crop Information
Values of ETo for Major Agro-ecological Regions of Pakistan
Sr.
No.
Districts
Average Daily Reference Crop Evapotranspiration (mm per day)
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
1
Gilgit, Chitral,
Swat
0.8 1.1 2.0 3.1 4.2 5.1 5.1 4.7 3.8 2.5 1.4 0.8
2 Peshawar 1.4 2.0 3.1 4.4 5.7 6.3 5.2 4.9 4.2 2.8 1.7 1.2
3 Kohat 1.6 2.3 3.5 4.7 6.3 7.3 7.2 6.5 5.5 4.1 2.6 1.7
4 D I Khan 1.6 2.4 3.8 5.0 6.4 6.9 6.0 5.5 4.8 3.3 2.0 1.5
5 Rawalpindi 1.7 1.7 3.0 4.9 6.3 7.1 6.7 4.9 5.4 4.0 2.7 1.7
6 Jehlum 1.4 2.1 3.4 4.8 5.9 6.2 4.7 4.0 4.5 4.0 2.9 1.8
7 Sialkot 1.2 1.9 3.1 4.4 5.4 5.8 4.4 4.3 3.8 2.7 1.6 1.1
8 Mianwali 1.4 2.0 3.3 4.6 6.2 6.5 5.6 5.0 4.6 3.1 1.8 1.2
9 Sargodha 1.7 2.5 4.0 5.6 7.2 7.6 6.1 5.3 5.1 3.8 2.3 1.7
10 Lahore 1.4 2.2 3.6 5.0 6.3 6.5 5.2 4.7 4.5 3.2 1.8 1.3

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Table. 4.2. ETo
Values of the crop factor kc for various crops (kc)
Crop Initial stage Crop development
stage
Mid- season
stage
Late season
stage
Wheat
Bean-green
Bean-dry
Cabbage/Carrot
Cotton
Cucumber/Squash
Eggplant/Tomato
Lentil/Pluses
Lettuce/Spinach
Maize, sweet
Maize, grain
Melon
Millet
Onion, green
Onion, dry
Groundnut
Pea, fresh
Pepper, fresh
Potato
Radish
Sorghum
Soybean
Sunflower
Tobacco
0.35
0.35
0.35
0.45
0.45
0.45
0.45
0.45
0.45
0.40
0.40
0.45
0.35
0.50
0.50
0.45
0.45
0.35
0.45
0.45
0.35
0.35
0.35
0.35
0.75
0.70
0.70
0.75
0.75
0.70
0.75
0.75
0.60
0.80
0.80
0.75
0.70
0.70
0.75
0.75
0.80
0.70
0.75
0.60
0.75
0.75
0.75
0.75
1.15
1.10
1.10
1.05
1.15
0.90
1.15
1.10
1.00
1.15
1.15
1.00
1.10
1.00
1.05
1.05
1.15
1.05
1.15
0.90
1.10
1.10
1.15
1.10
0.45
0.90
0.30
0.90
0.75
0.75
0.80
0.50
0.90
1.00
0.70
0.75
0.65
1.00
0.85
0.70
1.05
0.90
0.85
0.90
0.65
0.60
0.55
0.90
Table. 4.3. Kc
11 Faisalabad 1.4 2.2 3.5 4.8 6.2 6.5 5.4 4.9 4.7 3.3 1.9 1.3
12 Jhang 1.6 2.5 3.9 5.5 7.2 7.8 6.5 5.8 5.0 3.6 2.2 1.5
13 Multan 1.5 2.5 4.1 5.2 6.4 7.6 6.2 5.9 5.2 3.4 2.0 1.5
14 D G Khan 1.7 2.4 3.7 5.2 6.8 6.9 6.3 6.0 5.1 3.8 2.3 1.6
15 Bahawalpur 1.8 2.8 4.1 5.5 6.8 7.7 6.2 6.0 5.3 3.9 2.5 1.7
16 Bahawalnagar 2.0 3.0 4.5 6.5 8.3 9.0 6.6 6.3 5.7 4.4 2.8 1.9
17
Rahim yar
khan
2.0 3.0 4.5 6.0 7.2 8.0 6.6 6.2 5.2 4.0 2.6 1.9
18 Rohri 2.6 3.3 4.5 6.0 7.3 7.9 6.4 6.7 5.9 5.8 4.3 2.7
19 Nawabshah 2.8 3.7 5.1 6.7 8.7 9.5 7.2 6.2 6.1 4.9 3.3 2.7
20 Hyderabad 2.8 3.7 5.1 6.9 8.6 7.9 6.2 5.6 6.1 4.9 3.3 2.8
21 Larkana 2.4 3.4 5.3 7.0 9.0 9.8 7.4 6.4 6.1 4.4 3.1 2.3
22 Jacobabad 2.6 3.5 5.6 7.5 9.3 10.2 7.9 6.9 6.0 4.7 3.0 2.2
23 Karachi 2.5 3.2 4.4 5.8 7.2 7.8 7.5 6.9 5.8 4.6 3.3 2.5
24 Quetta 1.0 1.2 3.2 4.6 5.9 7.2 7.4 5.7 4.4 3.7 2.3 1.0
25 Loralai, Kalat 1.6 2.3 3.5 4.7 6.3 7.3 7.2 6.5 5.5 4.1 2.6 1.7
26 Sibi 3.2 4.3 6.3 8.5 11 11.1 9.2 8.1 7.6 6.2 4.1 3.2
Sources- FAO, IWMI & Agromet. Pakistan

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Approximate duration of growth stages for various field crops
Crop Duration of growth stages – days
Total Initial stage Crop development
stage
Mid-season
stage
Late-season
stage
Barley 140 20 25 65 30
Cauliflower (after
transplanting)
80 20 30 20 10
Chilies (after
transplanting)
150 30 40 55 25
Cotton 180 30 50 55 45
Cucumber 75 20 30 15 10
Cumin 80 10 20 35 15
Eggplant (after
transplanting)
170 30 45 65 30
Garlic 210 20 35 110 45
Gourds 90 15 25 40 10
Groundnuts 165 35 45 60 25
Ladyfinger 75 10 20 35 10
Lentils 85 15 25 35 10
Lettuce (after
transplanting)
75 20 30 15 10
Maize 105 15 25 40 25
Mug and mash 70 15 20 25 10
Musk melon 100 20 30 30 20
Onions 120 15 25 40 40
Peas 90 15 25 35 15
Potatoes 110 20 30 35 25
Radish 60 10 15 25 10
Rape and Mustard 155 15 35 80 25
Sesame 100 10 25 50 15
Sorghum 120 10 35 45 30
Soya bean-Kharif 95 15 20 40 20
Soya bean-Spring 125 15 25 55 30
Spinach 180 20 30 90 40
Sugar beet 205 30 40 95 40
Sunflower 95 15 25 35 20
Tomato 130 20 30 50 30
Turnips 60 15 25 15 5
Water melon 100 15 20 50 15
Wheat 140 20 30 60 30
Table. 4.4. Growth Stages

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CALCULATIONS:
4.1 Crop Water Requirement
Crop water requirement, CWR = Eto * Kc
Month J F M A M J J A S O N D
Eto (mm/d) 1.8 2.7 4.0 6.4 8.6 9.2 7.6 6.5 5.8 4.5 2.7 1.9
Kc (row crop) 0.4 0.6 0.8 1.2 1.1 0.4 0.4
CWR (row crop) 3.4 5.5 6.1 7.8 6.4 2.1 1.1
Table. 4.5
4.2 Measuring Canopy Factor (cf)
Canopy Area = π r2
= 3.14 x 52
= 76 ft.
Canopy Factor = Canopy area /plant area
=76 / (10x10)
=76 %
 Canopy Factor has direct relationship with Operation Time
 Canopy of the fully grown crop must be used for designing
purpose
Fig. 4.1
Further Calculations:
S.N. Parameters Unit Calculations
1 Total area under HEIS Acres 6.5
2 Design reference evapotranspiration, ETo mm/day 6.5
3 Design Crop factor at maturity, Kc - 1.2
4 Plant spacing m 0.3
5 Lateral spacing m 1.52
6 Shaded area % 100
7 Irrigation system efficiency % 90
8 Emitter flow rate LPH 2
9 Emitter spacing m 2
10 No. of crop rows per drip line Nos. 1
11 Irrigation cycle (assume one day) Days 1
12 Peak daily consumptive use per day mm = (ETo * Kc * shaded area) / (efficiency)

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Do We Have Sufficient Water?
Peak Water Requirement = 8.7 mm/day (calculated previously)
Volume of water required (m3
/d) = PWR x Area (m2
)
= 8.7/1000 * 6.5 * 4047
= 229 m3
/d
Water Availability
a) For canal water
Outlet discharge = 28 lps
Warabandi period = 2 hrs
Total water availability = 28/1000*3600 * 2 = 202 cubic meters for seven days
Water is not sufficient to design a drip irrigation system for 6.5 acres.
Options:
a) Reduce cropping area
Or
b) Arrange alternate water source
b) As farmer has arranged water source i.e., tubewell water
Proceed further
=6.5 * 1.2 * 1 /0.9 = 8.7 mm
13 Total no. of plants Nos = Area x 4047 /(R-R * P-P)
= 6.5*4047/(1.52*0.3) = 57,688
14 Total drip line length M = Area*4047/L-L
= 6.5*4047/(1.52) = 17,306
15 Total no. of emitters Nos = Total drip length/ emitter spacing
= 17306/0.4 = 43,266
16 Average emitter spacing M 0.4
17 Total flow rate Lph = total no. of emitters x emitter flow rate
= 43266 * 2 = 86,531
18 Application rate mm/hr Total flow rate / area
= (86,531/1000)/(6.5*4047) *1000 = 3.29
19 Operation time hrs = Peak daily consumptive use/ Application
rate
= 8.7/3.29 = 2.64 hrs

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4.3 Option-I (Irrigate area into Zones: Say 2
Zones)
Fig. 4.2
4.3.1 Equations for Calculating frictional Head Loss
of pipe lines.
4.3.1.2 Darcy- Weisbach and Blasius Equation
Darcy- Weisbach and Blasius Equation can be combined to give accurate prediction
of friction head loss for smooth plastic pipes of less than 125 mm (5 in.) diameter.
J = 100 hf/L = k * [(Q1.75)/ (D4.75)]*F
J = Head loss gradient, m/100 m
Hf= pipe friction head loss, m
K= Constant, 7.89 * 10 ^7 for metric units
Q= Flow Rate, LPS
L= Pipe length, m
D= Inside Diameter of Pipe, mm
F= opening factor
&
4.3.1.3 Keller Equation
Keller Equation can give accurate prediction of friction head loss for smooth plastic
pipes of greater than 125 mm (5 in.) diameter.
J = 100 hf/L = k * [(Q1.83)/ (D4.83)]*F
J = Head loss gradient, m/100 m
Hf= pipe friction head loss, m
K= Constant, 9.58 *10 ^7 for metric units

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Q= Flow Rate, LPS
L= Pipe length, m
D= Inside Diameter of Pipe, mm
F= opening factor
a) Lateral Design
Lateral length = 60 m
No. of emitters = 60/0.4= 150
Lateral flow rate = 2*150=300 lph
Assume 13 mm lateral line
Inside diameter = 10 mm
J = 7.89*10^7 * [(300/3600) ^1.75)/ (10^4.75)] * 0.36
J= 6.53 m/100 m
For 60 m length
Hf = 3.92m Hf must be < 2
Table. 4.6
So again assume 16 mm lateral line
Inside diameter = 13 mm
J = 7.89*10^7 * [(300/3600) ^1.75)/ (13^4.75)] * 0.36
J= 1.9 m/100 m
For 60 m length
Hf = 1.1m
b) Sub-Mainline Design:
Length = 110 m
No. of laterals = 110/1.52= 72
Sub-main flow rate = 86531/2
=43265lph
Assume 70(2.5”) mm submain
Inside diameter = 63 mm
J = 7.89*10^7 * [(43265/3600) ^1.75)/ (63^4.75)] * 0.36
J= 6.3 m/100 m
For 110 m length
Hf = 6.93 m Hf must be < 2
Again assume 110 mm Sub-Mainline
Inside diameter = 104 mm
J = 7.89*10^7 * [(43265/3600) ^1.75)/ (104^4.75)] * 0.36
J= 0.6 m/100 m
For 110 m length
Hf = 0.66 m
c) Mainline Design:
Velocity of Mainline must be < 1.5 m/s
No. of
Outlets
F
1 1.00
2 0.52
3 0.44
4 0.41
5 0.40
6 0.39
7 0.38
8 0.38
9 0.37
10 -11 0.37
12 -15 0.37
16 – 20 0.36
21 – 30 0.36
≥ 31 0.36

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v = 1.274 q / d2
Where
v = velocity (m/s)
q = volume flow (m3/s)
d = pipe inside diameter (m)
d =SQRT (1.274*((86531/ (1000*3600))/1.5)) = 0.142 m =5.59 inch = 142 mm
Length = 60 m
Main flow rate = 86531 LPH
Mainline Inside diameter = 142mm
J = 9.58*10^7 * [(86531/3600) ^1.83)/ (142^4.83)]
J= 1.2 m/100 m
For 60 m length
Hf = 0.8 m
4.3.2 Filters Head Loss:
Fig. 4.3. Disc filter
Fig. 4.4. Media Filter

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4.3.3 Total Dynamic Head
Emitter operating pressure = 10 m = 1 bar
Head loss in lateral = 1.1 m
Lateral Elevation = 0 m
Head loss in sub mains = 0.7 m
Head loss in Valve (assume 2 m) = 2.0 m
Field fitting head loss = 2.0 m
Head Loss in Main line = 0.8 m
Primary filter head loss (Gravel 2 x40 M3
) 0.2x2 = 4 m
Secondary filter head loss (Disc 2 x 40 M3
)0.2x2 = 4 m
Fertigation equipment head loss (5 m for ventury) = 5 m
Suction head = 1 m
Delivery head = 1 m
Safety equipment head loss = 2 m
Other losses = m
Total Head Required = 33.6 m
Say 35 m
4.3.4 Pump Requirement:
Total Flow required = 85.531m3/hr = 23.8 lps
Pump H.P = Q (LPS) * H (m)
75 x pump efficiency x motor efficiency
= 23.8 * 35 / (75 * 0.65 * 0.9)
= 19
Say 20 H.P

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4.4 Option-II (Irrigate area into Zones: Say 4
Zones)
Fig. 4.5
a) Lateral Design
Lateral length = 60 m
No. of emitters = 60/0.4= 150
Lateral flow rate = 2*150=300 lph
Assume 13 mm lateral line
Inside diameter = 10 mm
J = 7.89*10^7 * [(300/3600) ^1.75)/ (10^4.75)] * 0.36
J= 6.53 m/100 m
For 60 m length
Hf = 3.92m Hf must be < 2
So again assume 16 mm lateral line
Inside diameter = 13 mm
J = 7.89*10^7 * [(300/3600) ^1.75)/ (13^4.75)] * 0.36
J= 1.9 m/100 m
For 60 m length
Parameters Unit HZ-1 HZ-2 HZ-3 HZ-4 Total
Area Acres 1.625 1.625 1.625 1.625 6.5
Peak daily consumptive
use per day
Mm 8.7 8.7 8.7 8.7 8.7
Total no. of plants Nos 14,422 14,422 14,422 14,422 57,688
Total drip line length M 4,327 4,327 4,327 4,327 17,306
Total no. of emitters Nos 10,816 10,816 10,816 10,816 43,266
Average emitter spacing M 0.4 0.4 0.4 0.4 0.4
Total flow rate Lph 21,633 21,633 21,633 21,633 86,532
Application rate Mm/hr 3.3 3.3 3.3 3.3 3.3
Operation time* hrs 2.6 2.6 2.6 2.6 10.5

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Hf = 1.1m
b) Sub-Mainline Design:
Length = 110 m
No. of laterals = 110/1.52= 72
Sub-main flow rate = 86531/4
=21,633 lph
Assume 70 mm (2.5”) submain line
Inside diameter 63 mm
J = 7.89*10^7 * [(21,633/3600) ^1.75)/ (63^4.75)] * 0.36
J= 1.9 m/100 m
For 110 m length
Hf = 2.1 m
c) Mainline Design:
Velocity of Mainline must be < 1.5 m/s
v = 1.274 q / d2
Where
v = velocity (m/s)
q = volume flow (m3/s)
d = pipe inside diameter (m)
d = SQRT {[1.274*((21633/ (1000*3600))/1.5)]} = 0.071441
=2.81 inch = 71.44 mm
Length = 60 m
Main flow rate = 21633 LPH
Mainline Inside diameter = 71.44 mm
J = 7.89*10^7 * [(21633/3600) ^1.75)/ (87^4.75)]
J = 1.115027 m/100 m
For 60 m length
Hf = 0.7 m
4.4.1 Filters Head Loss:
Fig. 4.6. Disc filter

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Fig. 4.7. Media Filter
4.4.2 Total Dynamic Head
Emitter operating pressure = 10 m = 1 bar
Head loss in lateral = 1.1 m
Lateral Elevation = 0 m
Head loss in submains = 2.2 m
Head loss in Valve (assume 2 m) = 2.0 m
Field fitting head loss = 2.0 m
Head Loss in Main line = 0.7 m
Primary filter head loss (Gravel 20 m3) = 1.5 m
Secondary filter head loss (Disc 20 m3) = 1.3 m
Fertigation equipment head loss (5 m for ventury) = 5 m
Suction head = 1 m
Delivery head = 1 m
Safety equipment head loss = 2 m
Other losses = m
Total Head Required = 29.8 m
Say 30 m
4.4.3 Pump Requirement
Total Flow required = 21.633 m3/hr = 6 lps
Pump H.P = Q (LPS) * H (m)
75 x pump efficiency x motor efficiency
= 6 * 30 / (75 * 0.65 * 0.9)
= 4.1
Say 5 H.P

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Chapter .5
Water storage pond Designing
It is designed on the following parameters:
 If Available water is greater than the crop requirement then the pond is designed on
requirement of the crop.
 If Available water is less than the crop requirement then the pond is designed on
Available water.
Pond can be of two types:
 Rectangular pond
 Trapezoidal pond
Example:
To design a water storage tank of sprinkler irrigation system for 15 acres of maize
crop in Jhang. Canal water is available for 20 min/acre and discharge is 40 lps and warabandi
turn comes after 7 days. Crop Peak daily consumptive use is 4.68 mm/day. His required depth
of pond is 3 m.
Volume of Available water:
Available water = Q*t*A*0.06
Where
Q = sanctioned discharge, lps
t = time, min/acre
A = total area under HEIS (acre)
So
Available water = Q*t*A*0.06
= 40 * 20 *15 * 0.06
= 720 m3
Volume of Water on Crop Requirement:
Crop water requirement for one day = 4.04 * PDCU * A
Here
PDCU = Peak Daily consumptive use, mm/day.
A = total area under HEIS, acre
So
Crop water requirement for one day = 4.04 * PDCU * A
= 4.04 * 4.68 * 15
= 283.608

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As we have to store water for 6 days
Then,
Volume of water to store = 283.608 * 6 = 1701.648 m3
Its Available water is less than the crop requirement so the pond will be design on
Available water.
5.1 Rectangular Pond
Length * width * depth = Available water
Length * width = Available water / Depth
Length * width = 720 / 3
= 240 m2
Assuming
Length = 20 m
Then
Width = 12 m
So we shall design the pond on the following dimensions:
Top length = 20 m
Top width = 12 m
Depth = 3 m
Normally trapezoidal pond is designed for the storage of water.
Because when the pond is empty, the lateral pressure of the soil against walls of the
pond cause cracks/damage.
Where as in case of trapezoidal pond this pressure is compensated by its walls due to
some slope.
Fig. 5,1. Rectangular Pond
5.2 Trapezoidal Pond
Avg. area * Depth = Available water
Avg. area = (Top length * Top width) + (Bottom length * Bottom width)
2
Z = side slope

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Factor = 2*(Depth/ Side slope)
If
z = 1
Then
Factor = 2*(3/1) = 6
Let us suppose
Top length = 22
Top width = 15
Depth = 3
Then
Bottom length= Top length – factor
= 22 – 6
= 16
Bottom width= Top width – factor
= 15 – 6
= 9
SO
Avg. area = (Top length * Top width) + (Bottom length * Bottom width)
2
Avg. area = (22 * 15) + (16 * 9) =237
2
We know that
Avg. area * Depth = Available water
= 720 m3
Avg. area * Depth = 237 * 3 = 711 m3
It means there is deficiency of 9 m3
We shall suppose again
If
Top length=24
Top width= 14
Then
Bottom length= 24 – 6 = 18
Bottom width= 14 – 6 = 8
SO
Avg. area = (24 * 14) + (18 * 8)
2
=240
We know that,
Avg. area * Depth = Available water
Avg. area * Depth = 720 m3 ∵ Available water = 720 m3
Avg. area * Depth = 240 * 3 = 720 m3
Ok

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So we shall design the pond on the following dimensions:
Top length=24 m
Top width= 14 m
Bottom length= 18 m
Bottom width= 8 m
Depth = 3 m
Fig. 5,1. Trapezoidal Pond

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Chapter .6
Irrigation Scheduling
Table. 6.1 Excel sheet

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Chapter .7
Fertigation Scheduling
Table. 7.1 Excel sheet

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Chapter .8
Chemical Treatment
It is recommended to perform chemical treatment to prevent and remove any salt
Deposition or biological impurities settled inside the Drip Irrigation System
Chemical Treatments for System
a) Acid Treatment:
Acid treatment is given to prevent and also to remove the precipitated salts in
laterals and drippers.
b) Chlorination:
 Preventing the clogging and sedimentation of organic substances
 Organic sedimentation like bacteria, slimes and algae
 Destroying and decomposing Sulphur and Iron bacteria, as well as
accumulated bacterial slime in the system
8.1 Acid Treatment:
Most ground water and surface water are high in Calcium and Magnesium salts and
have alkaline pH (more than 7). Under such conditions, the Calcium and Magnesium salts
settle down and coat the inside of the pipes and block emitters. This can effectively be
removed by introducing acidic water that can dissolve the precipitates.
8.1.1 Acids Available for Treatment:
 Sulfuric acid (H2SO4, Conc. 98%)
 Hydrochloric acid (HCl, Conc. 35%)
 Nitric acid (HNO3, Conc. 60%)
 Phosphoric acid (H3PO4, Conc. 85%)
Site visit: A site of Abdul Raoof on Mehmood Kot in A.P Tehsil had
clogged/choked emitters due to salt accumulation was treated as:
8.1.2 Acid Treatment Procedure:
 Take 1 liter of water from water source used for micro irrigation in the
cylinder
 Use syringe for injection of acid

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 Mix thoroughly after adding acid
 Use pH paper to determine when desired level is reached (3 pH)
Fig. 8.1
8.1.3 Calculation:
For 1 liter irrigation water
Volume of acid required to bring pH up to 3 = 0.4 ml/L
Q of System = 100L/8sec = 12.5 L/s
Time of treatment = 2 minutes/ section
Vol. of water for 2 min = 1500 L
Acid Volume Required for Treatment = 1500 L * 0.4 ml/L = 600 ml

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Shut the system for 24 hr.
Flush all sub mains and laterals.
8.1.4 Precautions:
 Acids are dangerous, they must be handled with care.
 NEVER add water to acid, always add acid to water.
 Ensure that equipment used to handle the acid is resistant to acid
attack.
Always use ventury
8.2 Chlorination:
Chlorine treatment is given to remove & prevent the growth of algae in the Drip
Irrigation System and other organic slimes.
Fig. 8.2
8.2.1 Products Available for Treatment:
 Bleaching Powder (CaOCL2)
 35% Free Chlorine
 Sodium Hypo chloride (NaOCL)
 19.67% Free Chlorine
8.2.2 Chlorination Procedure:
 Mark 5 – 10 drippers with problem
 Start injection immediately after making solution of Bleaching Powder
 Adjust the injection rate if needed
 After injecting for 10 minutes, Stop injection
 Shut the system for 6 – 7 hours
 Flush all sub mains and laterals.
8.2.3 Calculation for Free Chlorine:
Required ppm =10
LPS of System = 12.5
Time of treatment = 2 minutes/section
Volume of water to be treated = LPS X Treatment Time
=12.5 X (2 X 60)
=1500 Litters
Free Chlorine Required for Treatment= 1500 X 10

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= 15000 milligram
= 15 gm.
Bleaching Powder required = 15/0.35
= 43 gm

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Chapter .9
2011-13
OFWM HEIS Progress Report
BRIEF SUMMARY OF HEIS SITES FOR THE YEAR 2011-12 DISTRICT JHANG
BRIEF SUMMARY OF HEIS SITES FOR THE YEAR 2012-13 DISTRICT JHANG

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Summary:
Drip and sprinkler irrigation enables timely application of water and other inputs
(fertilizers, nutrients etc.) as per plant requirements at various stages of its growth. These
systems enable irrigating variety of soil Conditions e.g. uneven topography, odd field
configurations, rolling sandy areas, long lengths of run etc.
The Drip is best suited for orchards and high value row crops such as vegetables,
cotton, maize, sugarcane etc.
Whereas sprinkler systems are more suitable for field crops e.g. wheat, fodder, gram
etc.
Drip and sprinkle irrigation technologies enhance agricultural production manifolds as
well as improve quality of produce by enabling precise and timely application of water and
all other nutrients. These systems are Versatile in their applicability and provide complete
control in irrigation operations. Very little labor is required to operate drip and sprinkler
equipment, which almost alleviates drudgery of irrigation functions. These Systems can
rather be automated to further minimize labor requirements besides achieving more precision
in application of water and other inputs to the crops
Conclusion:
In highly dry areas it is possible to cultivate vegetables with drip irrigation method.
The use of drip irrigation allows changing the cropping patterns that encourages land
cultivation.
Drip irrigation is very adaptable to the soil conditions and local sources of fresh
water. System of drip irrigation can be successfully located and used as at the small holdings
so at farms of different type of ownership.
Drip irrigation is more conservative in water use and increases water availability for
households.
Initial investments to install drip irrigation systems are the main bottleneck.
Recommendations:
Irrigation water (not land) is the limiting factor in improving agricultural production
in Pakistan.
The adoption of high efficiency irrigation systems (HEIS) can give boom to Pakistan's
agriculture as current water application methods not only consume huge amount of water but
also limit effectiveness of other inputs. As such, use of HEIS can not only shift existing
conventional inexact crop water application practices to exact and precise irrigation but will
also increase productivity of fertilizers, energy, Labor, pesticides, & weedicides etc...
Literature sited:
From On farm water management.