IRRGTN & DRAINAGE ENGR
IRRGTN & DRAINAGE ENGR BAEN 464
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Date Created: 10/21/15
Walsr lost by Iranxphauon by m haIrs KNDTHEAVAILABILIIY OF T HE WATER TO PLANTS LEADS TVOTDEFINTIPNSYOFi rwnewqmmir SO39RWATE IDIENT39A AITEN QN COMPOSITION OF AN UNSATURATED SOIL SAMPLE SOIL PARTIOL FORE SPACE SOLID MATERIAL SOIL PHYSICAL PROPERTIES Volumes Vv Volume Voids V Volume Air M I Mass of alr zero Vw I Volume Water M I Mass of sand VI Volume Solids Mw Mass of water vv Vl Vw SOIL PHYSICAL PROPERTIES Volume V V V Porosity m I vvlvb Void Ratio 9 vW Vb V Vw v Saturation S I VWNV 0 S S S 10 SOIL PHYSICAL PROPERTIES Mass and Density Soil 9 Speci c Gravity of Solids 1 Mineral 235 p I MIVI 1 Clay 270 Organic 260 SOIL PHYSICAL PROPERTIES V39 Mass and Denslg Bulk Density ph Apparent Speci c Gravity A P 39 VinVa A 39 Palm 39 MIVu 9w ONE CAN SHOW THAT D I 1 AIp RELATIVESIZES OF SOIL PARTICLES SAND 005 20 mm SILT 0002 005 mm SOIL TEXTURAL TRIANGLE r CLAY 39 lt 0002 mm 330 3 MASS WATER CONTENT DRY SOIL SAMPLE WATER WET WEIGHT DRY WEIGHT WEIGHT OF OF SOIL 0F SOIL WATER VOLUMETRIO WATER CONTENT WEI39 SOIL DRY SOIL SAMPLE SAMPLE WATER BULK VOLUME VOLUME OF SOIL 0F SOIL 0F Water DEPTH OF WATER I IN A SOIL LAYER AIR d 9 L WAYER SOIL r 3qu Walnut w Wahr Valum w vnr2L vrnIu VOLUMEI39RIC WATER CONTENT VOLUMEI39RIC WATER CONTENT 3qu v m 0v 39 VJV pw WWW 0 wpw I Wu on p I pw Pb 39 MJV 0v 39 VJWu 39 quotJIMquotJpn 0 39 WPwMlpn399m Pulp evemphpw emA EXAMPLE OF SOIL WATER PROPERTIES A field soil sample prior to being disturbed has a volume of 80 cm3 The sample weighed 120 grams After drying at 105 C the dry soil weighs 100 grams What is the water content by weight What is volumetric water content What depth of water must be added to increase the volumetric water content of the top onefoot of soil to 030 Given Soil sample volume Vb 80 cm3 Dry weight of soil sample M5 100 g Wet weight of soil sample MS M 120 g EXAMPLE OF SOIL WATER PROPERTIES Find 9quot water content on a dry weight basis 0 water content on a volume basis d depth of water EXAMPLE OF SOIL WATER PROPERTIES Solution em Mquot I M Mw 120 100 20 9 em Illw I M 209l1009 020 g of waterlg of soil EXAMPLE OF SOIL WATER PROPERTIES Solution 0 I VwVh and 9v phi pw 9 pb MNh 1009I80 cm39 125 glama e pbl pw 9 1251100 020 025 cm of water per cm ofsoil lt EXAMPLE OF SOIL WATER PROPERTIES Solution Current depth of water in one foot of soil d 0 L d 0 L 025 12 in 3 inches ofwater The depth of water in the soil when 0 030 is d 0 L 030 12 in 360 inches of water Thus the depth of water to be added is 36 minus 30 or 06 inches GRAVITATIONAL POTE NTIAL POTENTIAL O DATUM CAPILLARY FORCES 23 FORCE amuse Tm Ill 0 A WWW roman rmwmm I F1 2 n r a cosm uninan h huhth um FUN quE 1 F2pg nr2h F1F1 iznriccosmpg 111th h 2 039 cosiNpgr h coMsTAMT r CAPILLARV FORCES sun u may m r H nunnuan PREIIUIE In between murmur opIhyn h PIEBSUHE n CAPILLARY FORCES I I k l rldlulv39hlhl Capillary pressure pw pa klr NEARLY SATURATED SOIL LARGE RADILI s m SMALL MATRIC POTENTIAL Le ttle capillary action DRY SOIL SOIL PARTICLE SMALL RADILI s m LARGE MATRIC POTENTIAL Le large capillary action DEVELOPING MOISTURE RELEASE CURVES USING A HANGING WATER COLUMN mew Ponws mm usuwm HANGING WATER COLUMN FOR UNSATURATED CONDITIONS BDRQUS39ELATE REMAINS TURATED D THE39SDIL WATER IS UNDER A vTENsmw 39 wHE FLQW TOIgtSTHE T ION ml ENS IL HANGING WATER COLUMN FOR UNSATURATED CONDITIONS WHEN NO FLOW w 0 HAT Is THE W MATRIC POTENTIAL w w w M 0 tug 100 cm wm 100 cm DATUM 420 HANGING WATER COLUMN FOR MOISTURE RELEASE CURVES mmnmu Iu I DETERMINETHE QLIJMETRIC WATER WE39I39I WVEIGIIT shi g BULK VOLUME 300 cmi s bibdwoo VOLUMETRIC WATER CONTENT MOISTURE RELEASE CURVE SOIL WATER TENSION cm of water SOIL WATER POTENTIAL MEASURE OF WATER AVAILABILITY COMMON UNITS OF PRESSURE AND HEAD AND THEIR EQUIVALENTS MOISTURE RELEASE CURVES SILT LOAM AVAILABLE WATE R UNAVAILABLE WATE R VOLUMETRIG WATER CONTENT 39 1n Inn 1 1 m5 SOIL WATER TENSION cm FIELD CAPACITY A SATURATION RESERVOIRANALOGY SATURATION FIELD CAPACITY DRAIN TUBE TI 3 39 wlI 739 EH J Q 1 AVAILABLE WATER RESERVOIR ANALOGY PUMP SATU RATIOquot FI ELD CAPACITY AVAILABLE WATER RESERVOIR ANALOGY PUMP SATURATION FIELD CAPACITY DRAIN TUBE UNAVAILABLE WATER RESERVOIR ANALOGY PUMP DRAIN TUBE SATURATION FIELD CAPACITY U NAVAILABLE WATER WATER STATUS RESERVOIR ANALOGY PUMP SATURATION FIELD CAPACITY DRAIN TUBE AVAILABLE WATER UNAVAILABLE WATER MOISTURE RELEASE CURVES SATURATION FIELD WILTING POINT VOLUM ETRIC WATER CONTENT 10 100 1000 10000 105 SOIL WATER TENSION cm 10 AVAILABLE WATER amp SOIL WATER RESERVOIR The water held between field capacity and permanent wi 39n int is called the available water or the available water capacity AWC and is sometimes called the availa le water holding aci The AWC is calculated by AWC IE 0MP AWC is primarily a function of soil texture Table 23 Exmnle mlues n1 snll water chzrzumsmsmvmnus snll textures Considerable mum exits rmm thesewhlzs withmeachsod mm VARIATION IN SOIL WATER EV SOIL TEXTURE a IGRMITATIOIALWATER nmmLmL STRESS E WATER in ONE FOOT or SOlL lnches Sllyuzyanm Sand Loam The capacity of the available Ioil water reservoir T AW DependI on both the AWC and the depth thatthe plant mete have penetrated This rslltionllip is given by TAW AWC N where TAW the total available wetr capacity in the plant roalzone and RI depth lith plant motmne mmmwumungmmlu Plant can rams only a portion of the available soil wotor morvoir 139 AW adore growth and ultimotlly yiold arc oftcud This portion in ll39lll d rendin waillhls wlhr MM and for molt cropl it rang bolwoon 40 and 65 pm oftho ova able water in the cop rod mono RAW AWC MAD whore MAD ilthn monogomsnt allow do oioncy dodmol which can b runavod EXAMPLE PROBLEM WM mm m MM WM 1 C n crop WIth root zone of 3 feet FIND READILY AVAILABLE WATER RAW TAW I AWC Rd From TABLE 141 AWC 14 inohealfoot Rd 3feet TAW I 14 3 feet I 42 Inches RAW Awe MAD From TABLE 142 MAD 06 RAW I 42 06 I 25 inches AVAILABLE WATER amp SOIL WATER RESERVOIR De ne depleted and remaining water as fraction of available water depleted or fraction of available water remaining Fractlon ofavallable water depleted I f g I etc 39 vleio 39 evop 9 Fraction of available water remaining f t 0v 0l 9 10 Also I 1 I 10A AVAILABLE WATER amp SOIL WATER RESERVOIR It Is very useful In irrigation management to know the depth of water requlred to fill a layer of soil to eld capacity This depth is equal to 8WD 8059 G AWCL 11 by through a bit of algebra you will nd the equivalent to 8WD39 9m 9v I 1 2 quotI I I I A SAMPLE OF A SILT LOAM SOIL HAS A VOLUMETRIC WATER CONTENT OF 026 CALCULATE fd f AWC AND SWD ASSUME THE SOIL IS 36 INCHES DEEP GIven 9 I 026 c 034 a 016 Find fd fr AWC Available Water Holding Capacity SWD Depth of Soll Water Depleted 14 EXAMPLE OF SOIL WATER PROPER39H ES Solution Q 9n Mian 0quot Equation 10 034 026 I 034 016 044 I 1 Equation10A 10 044 056 AWE 0 WP 034 016 018 inIin 216inl 8WD I ch L Equation 11 I 044 018 inlin 36 in I 235 in 15 SOLID SET SPRINKLER SYSTEM The sprinkler SPREADER head is the heart NOZZLE of a sprinkler l irrigation system mum in mzzta sum1a nnzzh shown with vane NOZZLE TYP mnge nozzb Ksyhub quotnab The small diameter nozzle causes pressure to build MMWM up in the pipe and WWW sprinkler body The discharge ow rate of pressure and the size of the nozzle mnge nozzb Q MANUFACTURER S DATA Brass Simiullt Bore nozzle sane Wllll P lslmlnlleu 30quot 34 ull Cllcle Brass Impch Smlnklm Bearing 34quot Male lll l Brass 39llrllaclary Angle 2 an nlll clrcle Brass lmlml 5Wquotle mm W lllmu llPl Emss 11mm nnula 77 upemllllg Range A m m aw 3 3l DPM Radllls MANUFACTURER S DATA Bmss Sl li lll me Home willl Valle sun3U willl Smomtur LAM1 lslmnl Helqlll lull uozllEslx e pan circle sprinkler has a special mechanism so that a latch engages when the sprinkle s rotated to the desired angle It then rotates the opposite direction to the original position 11239 Full or Pan allele BIAS llllllml Sprinklers Begum lrz MilleNFI mm mummy Angle 25 uniman Range 3U 5U ml Flm l Rule 1 H 0 cm Rnnlus 35 ll ll MANUFACTURER S DATA Straight Bore Nozzle SHIN Stream Helgm T rle llnmE SIZE n 314 Full or Part cum Brass 1mm smnxlus Br A ass Sllddll Bore INN3 or DAN3 Mauls yjstramn quotnight on 1 o Sprlnkler Head Lowangle sprinklers were develop d for systems where the sprlnklers are carried above the crop e low angle reduces drift and evaporation losses 1239 Full Circle Brass impact Sprinkle Baalinu It Mala NPT Bias Txnjactory Am le 10 npaming Range 23 3 psi Haw Ru 2 I 734 15 Gm 7 n Radius distances us little pressur possible e as g nger LinuxquotIn ll Ihmwlhvm Mum Rolan Small plastic inserts called Obviously there must be trade offs as some strabghteTng T are solmft39meds of these criteria are mutually exclusive 5 lups mans e n01 ed a uce Many types of nozzles have been developed rthquot ence an Increase e 395 ance to accommodate these objectives 0 row When straight bore nozzles are operated at low pressure the jet does not break up into droplets ve well leading to poor uniformity and soil compactio As a result many new lowpressure nozzles were developed munmnan Lawmnnnzzh A smaller qlameter plpecailed a rlser Is usedlo V conduc iwa er quot9m hev Herd 0 he SPrlnkler Rlsers are also used ta rduce the turbulenceof deVIce The SP39M39 should beiloca ed above the Water stream Las Itreachesihe s39p rlnkler This Interferan with thejet from the nozzle 1 39quot39P39 quot s he performance the39sprmkler IMPACT r 2 M39SPRINKLER VRISERI SPRINKLER TRIANGULAR SPACINGS Sprinkler head Lateral 086xS SPRINKLER RECTANGULAR SPACINGS Lateral J Sprinkler Performance The discharge or volume flow rate of water leaving the nozzle can be described by qs 2982 Cd NJ where q5 discharge through the nozzle gpm Cd discharge coef cient for the sprinkler head 096 D inside diameter of the nozzle orifice in P pressure of the water at the nozzle ps0 and 2982 is a unit conversion constant 1quot 111 DISCHARGE gpm Nozzle Size PRESSURE In 13425 so 35 4o 45 3I32 6 127 139 150 161 171 160 7I64 7 173 190 205 219 232 245 1I6 6 226 246 266 266 303 320 964 9 266 313 336 362 364 405 5I32 10 35 39 42 45 47 50 11I64 11 43 47 51 54 57 60 316 12 51 56 60 64 66 72 Sprinkler Performance The discharge from a speclllc nozzle and sprinkler may vary slightly from the data presente n e 1 owever this Information Is adequate for most needs WHEN SPECIFYING SPRINKLER TYPES AND NOZZLES ALWAYS FOLLOW IIMNUFACTURE S CATALOG DATA Brn 1mlulrl Bore quotmus mm mm snmmut r5 Id Full circle Brass Imwrclsprlnkler J wmmumum n no Am 9139 Onemlm Ram 75 m rsw HHWHME 29 H mm am an m u 3 Sprinkler Performance quotquot quot 39 39 luau with two nozzle outlas is simply the sum of the discharge from each nozzle separately for that pressure The discharge for pressures between those listed in Table 1 can be determined by Interpolation Sprinkler Performance The second Important characteristic of A f llli er coverage The distance where the water application rate exceeds 001 Inches per houn Table 112 Nozzle size DIAMETER 0F COVERAGE feet NOZZLE PRESSURE psi inches 6 25 1o 15 4o 45 1132 6 64 66 68 69 7o 71 7164 7 65 67 69 7o 71 72 1I8 a 78 79 80 81 82 61 9164 9 80 81 82 61 64 85 5112 1o 82 85 87 88 89 9o 11164 11 61 88 9o 92 91 95 1116 12 85 91 94 96 98 1oo 11164 11 91 97 1oo 1o1 1o5 1o7 7112 14 92 99 1o2 1o5 108 11o 15164 15 91 100 104 107 110 112 mama 39 mm 0 114 16 94 1o2 1o5 1o9 112 115 17164 17 95 1o1 1o7 11o 114 117 9112 18 96 1o4 108 112 116 119 1 N0 WIND 2 IMPACT with STRA39I BORE NOZZLES 3 23 SPRINKLER Example 1 leen A stralght bore nozzle Is used In a sprinkler The discharge is 10 gpm m Would a 20 increase In nozzle dlameter produce more ow than a 20 Increase In pressure Example 1 Solution Let qs1 10 gpm the initial flow rate Using Equation 1 one can develop a term called the discharge ratio where 2 denotes the new condition and 1 the original condition 9 D22SQRTP2 1151 D SQRTP1 Example 1 For a 20 increase in Diameter D2 12 D1 and P2 P1 Solution Let 03951 10 gpm the inltial ow rate qsz tag 12 D1 I D1 2 144 qs1 144 9pm Thus a 20 increase in diameter provides a 44 increase in flow rate Example 1 For a 20 percent Increase in pressure P2 12 P1 and D2 D1 9 D 2S RT P qs1 D SQRT P1 qsz qs1l12P1IP1m 1391qs1 11 9pm Thus a 20 increase in pressure only changes the flow rate by 10 percent Sprinkler Performance The patterns shown in Figure following are typical of distributions where the operating pressure is inappropriate When the pressure is within the proper range the pattern in nearly elliptical with distance from the sprinkler Careful wl h pressure and operate within guldellnes When the pressure Is too hlgh the water jet breaks up Into a hlgh percentme of small drops giving reduced diameter of coverwe PRESSURE GUIDELINES Ilmss Slmiulrl quotME quot01119 San3 will Pluu mmmuaum u 4 Full Bilcle mss ImpmlSprlnkler wing 3Mquot Male NPY Brass niacin An I 39 When the pressure is too diameter of coverwe Overlap requirements have been developed for sprinkler systems mm a The narrowlng ol39 the Note that the m Wetted 3a me m mm wetted pattern is 39p m permquot 398quot ar e p m not only displaced 7 d39rem39m or quot39nd m I travel has an downwmdv 39t 3 5 important impact on narrower In th waw mlnm spnnkler spacings WWIrm direction VIIquotMen an nthe dlrectl n m quot perpendicular to I T ofthe lateral relatlve T the dlrectlon of mm tqtzeg domnam M 39nd rave quotm w39quot quotquot9 e quotW W I 4 irrigation season am EIl wrw mrw n wwmm With no wind the When the laterals are oriented individual parallel to the sprinkler pattern mr d39red39on of W39quotd lama1 ls circular and the m travel The wmd laterals appear to 4 causes the wetted gt have adequate pattern from a overla sprinkler to narrow If p Wm into a tlghter m pattern along the lateral A dry zone may m I Where the laterals um result between the rm f quot are laid out m fquot laterals because of perpendicular to insufficient overlap 23232 mum m mum T quot 39 quot m prOblem more gt is still narrower laterals WOUId be due to the wind needed with a however now the smaller spacing mm spacing of mm between laterals sPI inklers along the lateral is This leads to a more expensive system lll spacing between laterals smaller than the SPRINKLER SYSTEMS Maximum Spacing ol39 Spacing beMeen Sprinklers Laterals along the wind speed on Lateral mainline m h it ill of ill 07 40 of dla 05 of dla 710 40 of dla 00 of dla gt10 30 of dla 50 of dla Sprinkler Performance Example 2 Example 2 Given Impact sprinkler heads 23 exlt angle with two nozzles per head are spaced 40 feet along a lateral Laterals are spaced at intervals of 60 feet along the mainline The nozzle sizesare 1164 x 3I32 and the operating pressure Is 50 psl Wlnd in the area usually averages 5 mph Fi d Will thlslayout provide acceptable uniformity Solution The diameter of coverage for the range nozzle ie the larger 1164 inch nozzle Is 95 feet from Table 2 At this wind speed the maxlmum spacing between sprinklers on the lateral is 45 of the diameter of coverage The maxim spacing between laterals Is 60 of the diameter of coverme Exmploz Solution Maximum spacing along numeral 06 x W 43 1 Maximum Spacln bemmn numeral 060 X 5 57 R so mus Iayomjusnails the wind speed criteria and unifome may be only a blem gm app cmlnnlrnmzxw u rcnnsld r I mth ma u Ilrlnkluirllnmnn Systuls the mnIcmnn mum snnn lers nnslllnned in z mmnuulzrsnzcmmsm n k we try AIM QJSyxsm 5 12 A thE III M Walef Zrmllczllon Inhm q the s quotm Immune me mnmh 5 mun o sumklers unnu the lateral NV and m mm m lmmls 2mm me Immune an 963 s m unn cnmersnn Sprinkler Performance The application rate of the sprinkler system is important for two reasons First the depth of water applied for a given set time is proportional to the application rate d9 Ar To Equation 4 where d9 the gross depth of water applied per irrigation in Ar the application rate inhr To the actual time of operation hr sails and ne mas sandy lnams ne Slnpe and lmmy ne may lnzms and sin lnzm sails 030 r 045 006 r 015 020 r 030 010 r 020 004 r 010 010 r 020 005 r 010 002 r 005 water will Example 3 Given A sprinkler irrigation system is used to irrigate ayoung row crop with an unprotected soil surface Sprinkler spacing is 40 feet between sprinklers and 60 feet between the lateral positions along the mainline A pressure of 50 psi is available at the design location along the lateral Wind in the region averages 5 mph during the irrigation season The soil has a silt loam texture Example 3 Find Determine the smallest nozzle size that is acceptable for this system What would the application rate be for this system Is this system acceptable for a silt loam soil with a slope of 2 Example 3 Solution Given a wind speed of 5 mph the maximum spacing of the sprinkler along the lateral is 45 of the diameter of coverage and the maximum spacing of laterals is 60 of the diameter of coverage Since we already know the actual spacing of the sprlnklers and the lateral spaclng we need to determlne the dlameter of coverage dc needed for this system Example 3 d3 needed maximum of S I 045 and Sm I 060 or do needed maximum of 40l045 89 ft and 60I060 100 ft Based on the wlnd speed crlterla a diameter of coverage of 100 feet is needed to meet the wlnd requlrements of thls system Example 3 From Table 2 a nozzle size of 316 inch will provide a diameter of coverage of 100 feet when operated at a pressure of 50 psi From Table 1 a 316 inch nozzle operated at 50 psi will produce a discharge of 72 gpm Using Equation 2 the application rate would be SPRINKLER SYSTEMS DISCHARGE 9pm Nozzle Size NOZZLE PRESSURE psi inches I64quot 25 30 35 40 45 50 832 6 127 189 150 161 171 180 7I64 7 173 190 205 219 232 45 8 9 2 13 226 243 268 286 303 320 964 286 313 339 362 334 405 32 10 39 4 47 1164 11 43 47 51 54 51 so 3116 12 51 56 60 64 68 12 Table 1 14 Example 3 Ar 963 Cls I S Sm Cmsmmgs Medium tlextured 333ml 5 a quotMM 5 sands73quot ne 3 T sandy loam and Ar 963 X 72 gpm I 40 ft X 60 ft siltloam soils 029 quot1quotquot n Snilsurfacenmprmeclednnrr The maximum recommended application quot5quot 5 mm rate for these soil conditions is between 0585 33752 01quot jug5 33 025 and 05 inches per hour Table 4 so Hz mm mm mm 13 7 Zn 1 r J 1 r J mu 7 5 thIs desIgn probably Is acceptable gt2n mm rrTuri gxassnrhezvymsiduecnverrquotwquot 5 n85 r 13 5 r H575 His 7 H35 678 7 rl J r JS l r ls 20 The friction loss for a lateral Is abdut 35 of the less Pmun In quotmLataral Design Ava iii1MP Ln Funny I and mi Fa L L 395 Dim um Mull The average ressure uccun 14 to 13 of the inlet of the lateral Sprinkler E systems are a usually designed to select the nozzle size for the average pressure alang the lateral dietl and Pa L z Dlltxmx Ilnnu bud Lateral Design The diagram in Figure 119 shows that the average pressure is closer to the pressure at the distal end of the lateral than to the pressure at the inlet ofthe lateral For practical purposes the pressure at each end of a lateral on level ground can be computed by pl p I Equation 115 pd pa y PI Equation 116 PI PI 4P Equation 115 Pd P 14 P where P pressure at the inlet into the lateral psi Pa average pressure along the lateral psi Pd pressure at the distal end of the lateral psi and P pressure loss along the lateral Psi Lateral Design From a practical perspective the maximum acceptable pressure loss along a lateral placed on level ground equals 234 of the average or design pressure of the lateral In other words Maximum Pl lt 0234 Pa Eq 116 Lateral Design When a lateral runs up or down hill the change in elevation causes changes in pressure elevation change of 10 feet is equal to a pressure change of 43 psi Thus when laterals run downhill there is less pressure variation from the inlet to the distal end than for laterals on level ground because the slope provides some pressure increase When laterals run uphill the pressure in the lateral drops because of friction and because of the change in elevation 22 Lateral Design EQUATION 117 H 2pi j 4 2 23 1 P 2 I 1 P 1 j 4 2 23 1 where E the elevation of the inlet to the lateral and Ed the elevation of the distal end of the lateral Lateral Design Example 114 Given A sprinkler lateral was designed for an average pressure of 50 psi and sprinkler heads with one 532 inch nozzle in each sprinkler head The sprinkler lateral is made of 4 inch diameter aluminum pipe with joints 30 feet long There is one sprinkler outlet at the end of each joint of pipe The lateral is 1320 feet long Example 114 m a The pressure at the inlet and distal ends of the lateral if the lateral is on level ground b The pressure at each end of the lateral if the lateral runs down a uniform 2 grade c The pressure at each end of the lateral if the lateral runs up a uniform 2 grade Which of these systems meet the ASAE criteria for pressure variation in laterals Example 114 Solutlon There are 44 sprlnklers on the lateral ie 1320 feet wlth 30 feet between sprlnklers Wlth 5R2 Inch nozzles the average ow is 5 gpm per sprinkler Table 111 and the total flow for the lateral is 220 gpm 5 x 44 sprinklers Aluminum pipe with couplers has a C value of 120 in the HazenWilliams equation so the friction loss for a mainline with a ow rate of 220 gpm through a 4 inch aluminum pipe is given by 23 Example 114 P 456 QCJ 52 1d 486 1 Eq 88b P 456 220120 52 14o 4 J 215 psi where Pm the pressure loss in a mainline of constant diameter and ow Example 114 The multiple outlet friction factor F for a lateral with 44 sprinklers is about 036 Table 83 so the friction loss for the lateral is P F Pm 036 x 215 psi 77 psi Example 114 The pressure at the inlet to the lateral for level ground is PI Pa 34 PI 50 075 x 77 56 psi The pressure at the distal end of the lateral for level ground is PdPa1I4P50025x7748psi Example 114 The pressure variation along the lateral is 77 psi compared to the average pressure of 50 psi The variation is 154 of the average pressure and is less than the maximum permissible pressure variation so the lateral meets the ASAE standard 24 Example 114 When the lateral runs down a 2 grade the elevation change along the lateral is Ei Ed 002 x 1320 ft 264 ft So the Inlet is about 26 feet above the distal end Example 114 The pressures at the inlet and distal ends are then Pi Pa 34 PI 05 EI EdI231 Pi Eq 117 50 015 X 17 05 x 264231 501 psi Pd Pa 1I4 PI 05 Ei EdI231 Eq 117 50 025 X 77 05 X 264I231 538 psi Example 114 Here the pressure variation is only 37 psi well within the allowable variation Note that the highest pressure occurs at the distal end of the lateral for this case When the lateral runs uphill the elevation of the inlet is now below the distal end so the value of Ei E 264 feet Using thls value and the method in Part B the pressures at the ends of the lateral are Example 114 PI 615 psi and Pd 424 psi Now the pressure variation is about 19 psi or 38 ofthe average pressure which is unacceptable according to the standard 25 115 Minimum Lamrd Inflow SKIP