NATIONAL ENGINEERING HANDBOOK - PART 652 -
NATIONAL ENGINEERING HANDBOOK - PART 652 - IRRIGATION GUIDECHAPTER 6 - IRRIGATION SYSTEM DESIGNC - CENTER-PIVOT IRRIGATION SYSTEMGENERALThe example problems in this chapter are intendedto illustrate the procedure used in the design ofcenter-pivot irrigation systems. It should beunderstood that these two examples cannot showall design situations or all alternatives to considerwhen designing a center-pivot irrigation system.Most often center-pivot irrigation systems aredesigned by the manufacturer and evaluated by theNRCS resource engineer. These examplesillustrate a design method for supplying water to thetotal root zone. A common practice is to establishand manage a moisture control zone which isusually 50% of the rooting depth.c. Equations in Steps a. and b. can be combinedas follows to relate terms into a sometimes usefulrelationship.v d. Total capacity requirements based on knownapplication depth, area, and time of application.Q 453 x A x dH x FWhere: Q Total system discharge capacityin GPMA Acreage of the design area to besprinkler irrigated in acresDESIGN CRITERIADesign criteria for center-pivot irrigation systemsare contained in the Alabama NRCS ConservationPractice Standard (CPS), Irrigation System,Sprinkler, Code 442, Field Office Technical Guide.All center-pivot systems must be designed inaccordance with applicable requirements containedin CPS 442.FORMULAS USED IN DESIGNING ANDEVALUATING CENTER-PIVOT IRRIGATIONSYSTEMSThe following formulas are used in the design andevaluation of center-pivot irrigation systems:a. Precipitation rate from center-pivot systems.I 96.3 QsD x SBWhere: I Precipitation or application rate inin/hr.Qs Discharge from individualsprinkler in gpmD Wetted diameter of sprinklernozzle in feetSB Spacing of nozzles along theboom in feetNote: 96.3 converts gpm/sqft to in/hrb. Gross application depth in inches from individualsprinklers on a center-pivot.d Qs0.62 x SB x vWhere: d Application depth in inches (gross)v Velocity of rotational speed ofindividual sprinkler along the boomin ft/min.IxD60 x dd Gross depth of application ininchesH Number of actual operating hrs/dayF Number of days allowed forcompletion of one irrigationThe following example shows the use of the aboveformulas to design a center-pivot irrigation system.EXAMPLE PROBLEM NO. 1Given:1.2.3.4.5.6.7.Location: Andalusia, Climatic Zone 7.System: Center-pivot without end gun.System efficiency: 70%Crop: CornAcreage: 160 acre field.Soil: (Evaluated on site).Slope: Between 2 and 5 percent.Solution:Step 1. Refer to chapter 3 to determine the moistureextraction rooting depth. In this example, themoisture extraction depth is 36 inches.Step 2. Determine the available water holdingcapacity (AWC). Assume a field evaluation is madeand the weighted-average AWC for the top 36 in. is0.052 in/in. The AWC for the rooting depth is0.052 in/in x 36 in 1.87 in.Step 3. Determine the net depth of application of amanagement allowed deficit of 50 percent. Netapplication depth 1.87 in x 0.50 0.94.Step 4. Determine the maximum allowed applicationrate, assuming a basic soil intake rate is 3 in/hr. Thisis also the maximum application rate.Step 5. Determine the irrigation frequency for thepeak use period. The peak consumptive use for cornin this climate zone is found to be 0.30 in/day. Theirrigation frequency is equal to the net depth of(210-vi-NEH, Part 652, Amend. AL1, April 2010)AL6-85(20)
CHAPTER 6-C - CENTRE-PIVOT IRRIGATION SYSTEMapplication divided by the peak use rate. Irrigationfrequency is 0.94 in / 0.30 in/day 3.1 days.Step 6. Determine the rate of travel at the outermostdrive wheel (generally located about 50 ft from theend of the boom).System Length 1,300 ft totalDistance to Drive Wheel 1,300 ft - 50 ft 1,250 ftLength of last drive wheel’s path in one revolution 2(Pi)r (2) (3.14) (1,250 ft) 7,854 ftVelocity, v 7,854x1 hr3.1 days x 24 hr/day60 min. 1.76 ft/minStep 7. Determine the sprinkler application rate (Qs)required to apply the required net depth ofapplication. The gross application depth is the netdepth of application divided by the system efficiency.The gross application depth is 0.94 in / 0.70 1.34 in.Using equation from Steb b. and solving for Qs, then:Qs (0.62) x (SB) x (d) x (v)Where Qs Sprinkler discharge in gpmSB Sprinkler spacing on the boom in feetd Gross application depth in inchesv Rate of travel in ft/minAssuming SB 21.5 feet, thenQs (0.62) x (21.5) x (1.34 x (1.76)Qs 31.4 gpmStep 8. Determine the minimum wetted diameter forthe sprinkler by using equation in Step a. and solvingfor D.D 96.3 x QsI x SBWhere D Wetted diameter in feetQs Sprinkler discharge in gpmI Maximum allowable application ratein/hrSB Spacing, feetD (96.3) x (31.4)(3.0) x (21.5)D 46.9 feet - this is the minimum wetteddiameter that can be selected for a nozzle to apply31.4 gpm without exceeding the allowable applicationrate.Step 9. Select a nozzle size from manufacturer’stables for use along last 100 feet of boom.Operating pressure at this point on the boom shouldbe about 60 psi. Therefore, the pressure, theminimum wetted diameter, and the requireddischarge are known. From manufacturer’s tables,select the nozzle size of 3/8 in. at 60 psi. The flowrate is 30.6 gpm with a wetted diameter of 149 feet.AL6-85(21)Step 10. Check the actual depth of application andapplication rate of this nozzle.Application Rate, I 96.3 Qs (96.3) x (30.6) 0.92 in/hrD x SB(149) x (21.5)This is the average rate applied to a point on theground surface as the moving lateral passesoverhead. A sprinkler produces a parabolicdistribution with the maximum rate directly under thesprinkler. The maximum rate is 1.5 times theaverage: 0.92 in/hr x 1.5 1.38 in/hr. This is lessthan 3.0 in/hr and is satisfactory.The actual depth,d Qs (30.6).62 x SB x v(.62)(21.5)(1.76) 1.30 inThis is satisfactory since Step 7 shows grossamount of 1.34 inches. Another way of checkingthe actual depth of application is by using equationfrom Step C. as follows:D IxD 60 v(0.92) x (149) 1.30 in.(60) x (1.76)Another way is simply to determine the time totraverse the wetted diameter and multiply it by theapplication rate.Time to traverse the wetted diameter 149 ft1.76 ft/minx1 hr60 min 1.41 hrs.Depth of Application 1.41 hrs x 0.92 in/hr 1.30 in.Step 11. Determine the total irrigation systemcapacity. In order to get equal application of water,the last 100 feet of the circular distribution mustreceive 14.8 percent of the total water. For a1,300 ft long boom, this 100 ft covers 14.8% of thetotal area of the circle (See Table AL6-C-1).Therefore, total capacity is determined as follows:Qs 30.6 gpm at 21.5 ft spacing 30.6 gpm 1.42 gpm/ft21.5 ftLast 100 ft will use 1.42 gpm x 100 ft - 142 gpmftQt x 14.8% 142 gpmQt 142.148Qt 960 gpmAn alternate procedure is to use Figure AL6-C-1.From Step 10, the depth (gross) of application is1.30 inches. The maximum irrigation period is thesame as the irrigation frequency from Step 5.d/F 1.30 / 3.1 0.42 and the system length is a¼ section(210-vi-NEH, Part 652, Amend. AL1, April 2010)
CHAPTER 6-C - CENTER-PIVOT IRRIGATION SYSTEMStep 12. The universal soil loss equation and winderosion formula are used to determine if thecropping system planned will hold soil losses to atolerable amount. These items should be checkedearly in the system planning.EXAMPLE PROBLEM NO. 2The following example illustrates a typical problem.In fact, since center-pivots are designed by themanufacturer using computers, this example wouldapply to most situations. The procedure involvesusing the “Irrigation Data Sheet - Center-PivotIrrigation system.” The NRCS technician providesthe data sheet partially completed to the irrigationcompany. The company in turn provides pertinentdesign data on the form and returns it to thetechnician. The technician then completes the datasheet.Given:1.2.3.4.5.6.7.Crop and acres: 145 acres corn.Soil: Field evaluated.Location: Foley, AL (Baldwin Co.).Well: 10 inch.No existing pump.Power unit to be diesel.Slope: 0-3 percent.b. The percent depletion allowed prior toirrigation is selected to be 50 percent.c. The net water applied per irrigation is theproduct of percent depletion allowed prior toirrigation (50%) times the available water within theroot zone (1.87 in).Net water applied 0.50 x 1.87 in 0.94.d. The water application efficiency isdetermined to be 75 percent.e. Gross water applied per irrigation is the netwater applied (0.94 in) divided by the waterapplication efficiency (0.75).Gross water applied 0.94 in / 0.75 1.25 in.f. The irrigation interval is the net waterapplied (0.94 in) divided by the crop peakconsumptive use (0.30 in/day).Irrigation interval 0.94 in / 0.30 in/day 3.1 days.g. The irrigation period to be used in theformula for determining QR is the irrigation interval3.1 days.h. The hours operating per day is 24 hours.i. The quantity of water required (in gpm) iscomputed using the formula:QR 453 xacres xin gross applicationhrs operating/day x days/irrigationSolution:QR The item numbers mentioned in the step-by-stepsolution refer to the items on the “Irrigation DataSheet - Center-Pivot Irrigation System,”Exhibit AL6-C-1.QR 1,104 gpmStep 1. Complete Items 1-4. These items providepertinent data of the site.Step 2. Complete Item 5. Make a drawing (toscale) of the field, locating trees, buildings, the well,and other features.Step 3. Complete Item 6. Refer to chapters 3 and4 to determine he moisture extraction rootingdepths and the peak consumptive use rates. Forthis example, the rooting depth is 36 in. and thepeak consumptive use rate is 0.30 in/day. (Note:This example is for illustration only and the valuesused may not be as shown in the tables.)Step 4. Complete Item 7. Obtain the soil seriesfrom a published soil survey report or from an onsite investigation. The weighted available watercapacity (AWC) is found in the field to be0.052 in/in. The basic intake rate is found to be3 in/hr; however, it has been estimated that theintake rate on this soil, under low applicationamounts, is 6 in/hr.Step 5. Complete the following parts of Item 8.a. AWC within the root zone is the product ofthe root zone moisture extraction depth (36 in.)times the AWC of the soil (0.052 in/in)AWC 36 in x 0.052 in/in 1.87 in.453 x 145 ac x 1.25 in24 hrs/day x 3.1 days/irrigationThe manufacturer used 1,100 gpm for the design ofthe system.Step 6. The manufacturer provides the data forItem 9. This must meet the criteria previouslydiscussed.a. Pivot length 1,314 ft; pivot pressure 38 psi.b. Spray nozzle.c. Gross application per revolution 1.25 in.d. Nozzle gpm and pressure along last 100 ftof span is 11.5 gpm at 20 psi on spacing of 6.5 feet.e. Nozzle wetted diameter is 45 feet.Step 7. Check the maximum application rate,Item 10.a. Time per revolution to apply grossapplication 60 hrs (from manufacturer).b. Velocity of outside tower:v, ft/hr outside circumference, fthours per revolution (2) (Pi) (1,314) 138 ft/hr60 hours.c. Time of application (i.e., time it takes thesprinkler to move past one point):d. T, hrs wetted diameter (ft)velocity of travel (ft/hr)45 ft 0.33 hrs138 ft/hr(210-vi-NEH, Part 652, Amend. AL1, April 2010)AL6-85(22)
CHAPTER 6-C - CENTRE-PIVOT IRRIGATION SYSTEMe. Average application rate, in/hr grossapplication, in / time of application, hrs1.25 in 3.8 in/hr0.33 hrsf. Maximum application rate (in/hr) is1.5 times the average application rate. Maximumapplication rate (1.5) x (3.8 in/hr) 5.7 in/hrStep 8. Complete Item 11 for sizing the mainlineand determining the total dynamic head required forthe pump.a. The mainline is 1,850 ft of 8-in diameterPVC, SDR 26, IPS pipe. The friction head loss is1.71 ft/100 ft and is taken from Appendix. The totalhead loss in the mainline is1.71 ft/100 ft x 1,850 ft 31.6 feet.b. The pressure at the pivot was given by themanufacturer and is 38 psi or 87.8 feet.c. The miscellaneous and fitting losses wereestimated to be 3.0 feet.d. The elevation difference was measured tobe 11.5 feet.e. The sum of a., b., c., and d. gives a pumpdischarge pressure required of 133.9 ft. or 58.0 psi.f. The total dynamic head (TDH) is thepumping lift plus the pump discharge pressure,133.9 ft 100.0 ft 233.9 ft 101.3 psi.Step 9. Complete the plans. The specifications,location of the pipe, check valve, air vents,pressure relief valves, etc., should be shown on theplans.CONSTRUCTION REQUIREMENTSOnce a system is designed it must be installed asplanned to function properly. The following is a listof key points that should be checked duringconstruction to assure quality installation:a. Depth of cover over the buried mainline(important for protection from vehicular traffic andfarming operation).b. Thrust block dimension and location forprevention of pipe joint separation.c. Location and size of air vents and pressurerelief valve.d. Size and proper direction of installed checkvalve.e. Riser material and dimension, as well aslocation for pivots.f. Length and quality of pipe, diameter, location,appropriate ASTM designation, size, pressurerating, and SDR as measured or found written onthe pipe.g. Determine if IPS or PIP is used; which will havean effect on the total head loss of the system.h. Verification of length of the center-pivot lateral,and if spray or impact type sprinklers.AL6-85(23)LAYOUT CONSIDERATIONSDuring planning and layout of a center-pivot systemthere are many things to be considered. Itemssuch as soil limitations, obstacles (fences, ponds,ditches, and trees), topography of the field, thefarming operation, and safety hazards such aselectrical buried gas lines.The soil limitations might affect the pivot’s ability totraverse the field and/or increase the erosionpotential from runoff if using high application rates.Obstacles, if not considered, could result in severedamage to the pivot. Bridges or culvert crossingsmay be needed for crossing wet areas or diches.Electrical lines and buried cable or gas lines mustbe located prior to burying the pipe or locating thepivot, not only to facilitate installation, but to preventa safety hazard.Topography must be considered because centerpivots are limited as to the slope on which it canfunction properly.The greater the land slope the greater the erosionpotential. Therefore, the application rate must becompatible with the slope to prevent erosion.Procedure for Determining Gross Application ofCenter-Pivot SprinklerThe objective of this procedure is to develop a tablethat relates the dial setting of the center-pivot timerto the gross water (inches) applied. The table maybe used by the irrigator to adjust the system speedto obtain a desired gross application. Theprocedure described applied to electric systemtimers which read from 0 to 100 percent. However,the procedure can be adapted to other timers.1. Determine Speed of End Tower - Select areference mark on a wheel on the end tower. Set astake by this mark. Start timing when the wheelstarts moving forward. Continue timing until thewheel has moved 20 to 30 feet, or until after thecan catch is made. Mark distance traveled byplacing a second stake by reference mark on wheeland stop timing just as the wheel starts to moveforward again. Read time and measure distancebetween the two stakes.Speed of end tower, ft/hr Distance travelled, ft x 60Time, min2. Determine Time Per Revolution - Once speedis determined, compute time of travel for onerevolution at the percent setting on the timer.Time per revolution, hrs (at % setting on timer) Distance travelled by end tower, ft.Speed of end tower, ft/hrDistance traveled by end tower, ft 2 x (Pi) x Distance from pivot to end tower, ft(210-vi-NEH, Part 652, Amend. AL1, April 2010)
CHAPTER 6-C - CENTER-PIVOT IRRIGATION SYSTEM3. Determine Hrs/Revolution for 100% Dial SettingHrs/revolution (at 100%) (Hrs/revolution) x (Dial Setting)100Note: Use the dial setting on the control panel atthe time the speed was determined andhrs/revolution corresponding to this setting.4. Determine Hrs/Revolution for Each Dial SettingHrs/revolution at X% (Hours per revolution at 100%) 100X%5. Determine Gross Application for Each DialSettingGross application, in. (Hrs/revolution for dial setting) (gpm)(453) x (Acres irrigated)Note: For acres irrigated, use designed acres. Ifnot available, use the effective wetted area.ExampleThe center-pivot timer was set on 60% and endtower traveled 87.7 ft in 19 minutes. Distance frompivot to end tower is 1,205 feet. System applies850 gpm on 130.19 acres.End tower speed (87.7) (60) 276.9 ft/hr19Time/revolution at 60% (2) (Pi) (1,205) 27.34 hrs276.9Hrs/revolution at 100% (27.34) (60%) 16.4 hrs100Hrs/revolution for each of the other dial settings:For 90% (16.4) (100) 18.22 hrs90For 80% (16.4) (100) 20.50 hrs80For 70% (16.4) (100) 23.43 hrs70For 60% (16.4) (100) 27.34 hrs60For 50% (16.4) (100) 32.80 hrs50For 40% (16.4) (100) 41.00 hrs40For 30% (16.4) (100) 54.67 hrs30For 20% (16.4) (100) 82.00 hrs20For 10% (16.4) (100) 164.00 hrs10Gross application for each dial setting:For 100% (16.4) (850)(453) (130.19)For 90% (18.27) (850)(453) (130.19)For 80% (20.50) (850)(453) (130.19)For 70% (23.43) (850)(453) (130.19)For 60% (27.34) (850)(453) (130.19)For 50% (32.80) (850)(453) (130.19)For 40% (41.00) (850)(453) (130.19)For 30% (54.67) (850)(453) (130.19)For 20% (82.00) (850)(453) (130.19)For 10% (164.00) (850)(453) (130.19) 0.24 in. 0.26 in. 0.30 in. 0.34 in. 0.39 in. 0.47 in. 0.59 in. 0.79 in. 1.18 in. 2.36 2.001.1910164.002.36(210-vi-NEH, Part 652, Amend. AL1, April 2010)AL6-85(24)
CHAPTER 6-C - CENTER-PIVOT IRRIGATION SYSTEMTABLE AL6-C-1. CENTER-PIVOT IRRIGATION SYSTEMS.% of waterapplied in1/last 100 ftTotal area ofsquare fieldtwice lengthof system(acres)Totalsystem length2/(in ft)Area covered in Acres3/With gunsprinkler usedonly in cornersWith gunsprinkler usedon entire circleWithout end .2151.612.91500206.0177.7184.7162.01/2/3/Less volume of end gun when used.Generally outside drive wheel approximately 50 ft from end.Based on 100 ft gun coverage.EXAMPLE: System is 900 ft long. Then 21% of water is applied in last 100 feet. 66.7 acres are covered withgun used in corners only.at 1,300 feet. Figure AL6-C-1 shows Qt 960 gpmIf an end gun is desired then its discharge must be added directly to the 960 gpm.AL6-85(25)(210-vi-NEH, Part 652, Amend. AL1, April 2010)
CHAPTER 6-C - CENTER-PIVOT IRRIGATION SYSTEMRules for Use of Chart1) Reading up from known water supply, readvertically to d/f line; move horizontally to area vs.radius curve; then vertically to find radius.2) Reading down from known system length, dropvertically to area vs. radius curve; move horizontallyto appropriate d/f line; then drop vertically to watersupply.Example1) Given: Water supply is 750 gpm; croprequirements are 1.29 in. with a 4-day irrigationperiod.Find: Maximum acreage that can be adequatelyirrigated, and center-pivot radius associated withthis acreage.Solution:d/f 1.29 0.32; Enter 7504gpm: Apply Rule 1; find area is about 125 acresand system radius is about 1,320 feet.2) Given: Center-pivot length is 1,320 feet; cropneeds 1.29 in. (gross) in 4-day irrigation period; d/f 0.32Find: Water supply required.Solution: Apply rule 2; find 750 gpm required.Figure AL6-C-1. Capacity Requirements for Center-Pivot Systems.(210-vi-NEH, Part 652, Amend. AL 1, April 2010)AL6-85(26)
CHAPTER 6-C - CENTER-PIVOT IRRIGATION SYSTEMU.S. Department of AgricultureNatural Resources Conservation ServiceExhibit AL6-C-1.AL6-85(27)(210-vi-NEH, Part 652, Amend. AL 1, April 2010)
CHAPTER 6-C - CENTER-PIVOT IRRIGATION SYSTEMExhibit AL6-C-1. (con’t)(210-vi-NEH, Part 652, Amend. AL 1, April 2010)AL6-85(28)
CHAPTER 6-C - CENTER-PIVOT IRRIGATION SYSTEMExhibit AL6-C-1. (con’t)AL6-85(29)(210-vi-NEH, Part 652, Amend. AL 1, April 2010)
CHAPTER 6-C - CENTER-PIVOT IRRIGATION SYSTEMExhibit AL6-C-1. (con’t)(210-vi-NEH, Part 652, Amend. AL 1, April 2010)AL6-85(30)
An alternate procedure is to use Figure AL6-C-1. From Step 10, the depth (gross) of application is 1.30 inches. The maximum irrigation period is the same as the irrigation frequency from Step 5. d/F
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