Center Pivot Irrigation Design 1x6w2k

Center Pivot Irrigation System Dr. Mostafa Ahmad Ghaith

Applying Irrigation Water in Circles (vs. squares)

Why (briefly) 1) Economical 2) Low O & M 3) High Reliability

4) Central Delivery Point

Applying Irrigation Water in Circles (vs. squares) Why it’s a little trickier? In a rectangular system each sprinkler applies water to an Identically sized Area (A)

In a circular system the area increases as the radius increases Hence, each sprinkler applies water to a differently sized Area (A)

1 2

1

2

3

A1 = A2 = A3 = A4

3 4

4

A1 < A2 < A3 < A4

How Does this Weigh up on a Typical System? (System Capacity = 6 gpm / Fed = 22.74 lpm/Fed) Circle Area Computations Area = π R2 Radius

m

Area

Ring Area

Flow

Feddans Feddans

lpm

50

2

2

46

100

8

6

182

150

17

9

387

200

30

13

683

250

47

17

1069

300

68

21

1547

350

92

24

2093

400

120

28

2729

450

152

32

3457

500

187

35

4253

Sprinklers are sized appropriately along length of pivot to maintain uniform applications along linear length of the center pivot machine

How Does this Weigh up on a Typical System?

High Pressure

How Does this Weigh up on a Typical System?

Medium Pressure

How Does this Weigh up on a Typical System?

Low Pressure

Soil / Water Intake Curves 100

Intake Rate (mm / hr)

75 1.0 Family

50 0.5 Family 0.3 Family

25

0.0 0.0

0.1

0.2

0.3 Time (hrs)

0.4

0.5

Sprinkler Pressure vs. Intake Characteristics

• Center pivots are used on about half of the sprinkler-irrigated land in the USA • Center pivots are also found in many other countries • Typical lateral length is 1,320 ft (400 m), or 1/4 mile • The lateral is often about 3 m above the ground • Typically, 120 ft (40 m) pipe span per tower (range: 30 to 85 m), often with onehorsepower electric motors (geared down) • At 40 m per tower, a 400 m lateral has about 10 towers; with 1-HP motors, that comes to about 10 HP just for moving the pivot around in a circle

• The cost for a 1/4-mile (400 m ) center pivot is typically about $55,000 (about $435/ac 3000 LE. /fed or $1,100/ha), plus about $20,000 (or more) for a corner system • For a 1/2-mile lateral, the cost may be about $75,000 (w/o corner system) • In the state of Nebraska there are said to be 43,000 installed center pivots, about 15% of which have corner systems • Center pivots are easily (and commonly) automated, and can have much lower labor costs than periodic-move sprinkler systems

• The end tower sets the rotation speed; micro switches & cables keep other towers aligned • Corner systems are expensive; can operate using buried cable; corner systems don't necessarily irrigate the whole corner

• IPS 6" lateral pipe is common (about 6-5/8 inches OD); lateral pipe is generally 6 to 8 inches, but can be up to 10 inches for 880 m laterals Long pivot laterals will usually have two different pipe sizes

• Typical lateral inflow rates are 45 - 65 Ips (700 to 1,000 gpm) • At 55 Ips with a 6-inch pipe, the entrance velocity is a bit high at 3 m/s Typical lateral operating pressures are 140 - 500 kPa (20 to 70 psi)

• Center pivot maintenance costs can be high because it is a large and fairly complex machine, operating under "field" conditions • The typical maximum complete rotation is 20 hrs or so, but some (120-acre pivots) can go around in only about 6 hrs

• Without a corner system or end gun, 79% of the square area is irrigated

• For a 400 m lateral (without an end gun), the irrigated area is 125.66 Fed. • For design purposes, usually ignore soil WHC (WaZ); but, refill root zone at each irrigation (even if daily) • Center pivots can operate on very undulating topography • Some center pivots can be moved from field to field

Cairo Ismailia Desert Rd

Dina Farms Cairo Alex Desert Rd

Near Cairo Alex Desert Rd

Cairo Alex Desert Rd

Toshka Region

Toshka Region

Control of Large Farm

Electrical Center Pivot Operation

Electrical Center Pivot Operation

Last Tower Controlled By Percent Timer

Electrical Center Pivot Operation

Next Tower Follows When Micro-switch Triggers

Electrical Center Pivot Operation

All Other Towers Follow Similarly

Center Pivot 10 Conductor Span Cable

Timer End Gun

Forward

Reverse

Neutral

Safety

Ground Power

Power Power

Hydraulics of Center Pivot

Q= Q=

Ii T

T

R= Lateral Length Da= Irrigation depth applied per Irrigation T = Rotation time (usually 18-22 hrs during peak conditions) Da = Id / Ii

or = ETmax / Ea

Ea = irrigation Efficiency (80 -90 %)

Q   R2 Qr  ( R2 -  r2) Qr ( R2 -  r2) = Q  R2 Qr = Q

1- (r / R)2

Q

r

Qr

R

zero

Center Pivot with Booms

Center Pivot with Uniform Sprinkler Spacing and increasing Sprinkler discharge

Center Pivot with Uniform Sprinkler size and decreasing Sprinkler Spacing

Hf

H

Hr HR = Ho r R

Hf = 8/15 ( hf similar supply Pipe)

In usual lateral for n =  F= 1/3 = 0.333 In Center Pivot F = 8/15 = 0.533

Why ?

Irrigation System Design (Some Basic Concepts) Don’t Over - Complicate We Want To Get This

FIELD

WATER

Up Here

Irrigation System Design (Some Basic Concepts) 2 Important Parameters

1)Flow (most commonly given in lpm)

Bucket–Fulls Per Unit Time

2)Pressure or Head (given in m. of water) Squirting Distance

FLOW DETERMINATION 1) Crop / Soil Requirements a) effective root zone b) soil texture 2) Field Size 3) Water Source Limitations a) physical b) by permit c) other

Crop Requirements (lpm / Feddan) Table 1. System Capacity in gallons per minute per acre (gpm/acre) for different soil textures needed to supply sufficient water for each crop in 9 out of 10 years. An application efficiency of 80% and a 50% depletion of available soil water were used for the calculations. -----------------------------------------------------------Root Coarse Loam Zone Sand Fine and Depth and Loamy Sandy Sandy Silt Crop (ft) Gravel Sand Sand Loam Loam Loam -----------------------------------------------------------POTATOES* 2.0 8.2 7.5 7.0 6.4 6.1 5.7 DRY BEANS 2.0 7.9 7.1 6.4 6.1 5.7 5.4 SOYBEANS 2.0 7.9 7.1 6.4 6.1 5.7 5.4 CORN 3.0 7.3 6.6 5.9 5.5 5.3 4.9 SUGARBEETS 3.0 7.3 6.6 5.9 5.5 5.3 4.9 SMALL GRAINS 3.0 7.3 6.6 5.9 5.5 5.3 4.9 ALFALFA 4.0 6.8 5.9 5.6 5.1 5.0 4.5 -----------------------------------------------------------*Adjusted for 40% depletion of available water

General Rule = 6 gpm / feddan

(Crop Requirement) x (Field Size) = Flow Requirement EXAMPLE (6 gpm / feddan) x (125 Feddan) = 750 gpm

(Not Written in Stone but good guidelines to follow) May also be physical or permit demanded constraints on pumping rate which dictate

PRESSURE or HEAD 4 Main Considerations 1) To offset Elevation difference between source and delivery point 2) To compensate for Friction losses in the mainline delivery system 3) System Operational Requirements

4) Other Minor losses

Elevation Difference between water source and point of distribution Vertical distance between pumping water surface and the field delivery point (for center pivots use the highest point in the irrigated field for conservative calculations)

Example 15 m

Surface Water

Ground Water

Friction Losses Most friction losses in irrigation systems are developed in the system mainline (transmission pipeline) (Significant friction loss also occurs in the pivot itself but Is usually calculated and included as part of the operational pressure requirements) Transmission Pipeline Most often PVC but may also be aluminum, steel or PE

Friction Losses Important factors in the calculation pipe friction loss are:

• Pipe Inside Diameter (id) • Pipe Material • Pipe Length • Fluid Velocity or Flow Rate

Friction loss is typically calculated using one of several common equations: (Hazen Williams equation or Darcy equation)

Friction Losses Hazen Williams Equation H

= 10.65 Q1.852 L C1.852 d4.87

Where:

• H = head loss from friction (m.) • L = length of pipe (m.) • Q = flow (m3/sec) • C = friction factor (140 – 150 for PVC pipe higher number means smoother pipe) • d = inside diameter of pipe (m.)

Friction Losses Hazen Williams Equation H

= 10.44LQ1.85 C1.85d4.87

Example If 47.25 LPS is flowing through 500 m of new 8 inch (20 cm) ID PVC pipe the friction loss will be

3.75 m

Operational Pressure Requirements At the Center Pivot Consist of: 1) Pressure necessary to operate sprinklers and regulators satisfactorily (5 psi or greater above rated pressure of regulator) 2) Friction losses incurred in span pipe Calculation is usually combined together with sprinkler package spreadsheet Requirements are commonly given at pivot point location

Elevation differences along pivot may also be included

Example pivot point requirement: 3.1 bar @ 47.25 l/sec

Minor Losses The majority of minor losses which will increase the overall head requirement can be caused by: 1) Small friction losses which occur due to fittings and deviations in pipeline alignment 2) Extra losses through pump and suction pipe 3) Friction loss incurred in well tubing 4) Other

In large pipeline networks minor losses can be a substantial portion of the total head requirement Typically in irrigation systems minor losses are not a large part of the total head requirement – Often times it is good enough to simply add 1.5 to 3 m to the final head calculation as an adjustment for any minor losses which may occur in the system

Example Pressure Totals 1) Elevation Head = 15 m. 2) Friction losses in the mainline delivery system = 3.75 m

3) System Operational Requirements = 3.1 bar or 31 m 4)Minor losses estimate = 3.25 m Total Dynamic Head = 53 m

PUMP SELECTION

Total Dynamic Head (m)

70 Full Impellor 10% Trim 85%

20% Trim 53

82%

30% Trim

79%

0

47.25

Flow (l/sec)

80

POWER REQUIREMENTS

Horsepower Required =HxQ 75 x  Where  = (pump efficiency good first guess is .75)

EXAMPLE {(53 m. ) x (47.25 l/sec)} / {75 x .75} = 44.5 hp

Potential Run off Calculations

AR & IR mm/hr

AR Pattern at Downstream end Potential Run off AR Pattern at Upstream end Soil Intake Rate

Time

Vr =  r

Tor = Dw / Vr

 = 2/T

ToR = Dw / VR ToR < Tor

=

Vr 2 r /T VR = 2 R /T VR > Vr

 r Vr

VR 

R

ARMax

ARavg = Da / Tor ARavg

Area of half Ellipse = ½  a b Area of half Ellipse = ½  Tor /2 ARMax = Aravg Tor ARMax = 4/ ARavg

Tor

I=atb +c Di = A T

B

By integration from t=0 to t=T

+cT

A = a/b+1 B = b+1

Dr = Da - Di Dr : Potential runoff depth Da : Irrigation depth per rotation Di : Infiltrated depth