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END PERFORATION of A PERFORATED LATERAL <br /> PRESSURE DISTRIBUTION SYSTEM G,M, G. <br /> 1. Select number of perforated laterals 3 - i. <br /> • 1144.44.9$04. Fabric 1l-' <br /> • Lb.••y Simi Layer LA, .l 1144.44.9$01144.44.9$04.1144.44.9$04.Inc'NW*.t her«V.v.to <br /> 1•01, td r.eln now) <br /> 2. Select perforation spacing = 3 feet . ,warn , ...�, P.tl«:Irn Gra led Nerlt.nl..,- <br /> /� :Veer <br /> o u.« �.► <br /> .'/.'Pius �, ♦I Lc. 17' o. <br /> eet; r. EdUwr <br /> 3. Since perforation should not be placed closer, than 1 ft. to . «n us Ree ill <br /> • •_P.,l«.Ibn.Laae;.. ., <br /> the edge of the rock layer (see diagram), subtract 2 ft. from ueon Send L. .. Wien. N L.l.r.l <br /> the rock layer length. lL a - ..loin.; sell Prep' <br /> S erNhd <br /> L) ) Selo* Placing Send Low <br /> Rock layer length- 2 ft. = 3? feet <br /> TABLE Or PERFORATION DISCHARCL•S 1N CPM <br /> 4. Determine the number of spaces between perforations. Head Perforation diameter(inches) <br /> Divide the length above by perforation spacing and round <br /> down to nearest whole number. "s' li' <br /> 1.0a 0.56 0.74 <br /> Length perf. spacing = 3 9 ft. + 3 ft. = 13 s aces 13 0.69 0.90 <br /> p 2.06 0.80 1.04 <br /> ( 13) 02) 2.5 0.89 1.17 <br /> 5. Number ofperforations is equal to oneplus the number of1.13 <br /> 3.0 0.98128 <br /> �] 4.0 1 1.477 <br /> perforation spaces . 5.0 1.26 1.65 <br /> aUse 1.0 foot of head for residential systema. <br /> 13 .spaces + 1 = I y' perforations per lateral bUsc 2.0 feet of head for other establishments <br /> 6. Multiply perforations per lateral by number of laterals to <br /> get total number of perforations. <br /> Table 2 <br /> 3 i y Ma><imum alio+abk number of quater inch paTfuralivn.per <br /> lateral a x penis/lateral= y? perforations latero)to Euaranle.<la'k 1F.ehortc variation <br /> "e(hon lk .p.tl"sl1 1.25 inch I 1.5 inch 2.0 inc <br /> 7. Determine required flow rate by multiplying 2.5 14 18 28 <br /> number of perforations by flow per perforation 3.0 13 17 26 <br /> 3.3 12 16 25 <br /> • 4.0 11 15 23 <br /> 11a. . 9u 5.0 10 14 22 <br /> penis x gpm/perf - 7-7, . gpm. <br /> MA..a 0,6 LDCAtl4 a, t+.V nl Mr.(tK,w...o.tin• <br /> 8. If laterals are connected to header pipe as shown on upper <br /> example,select minimum required lateral diameter from -.rr- <br /> table 2; enter table with perforation spacing and number <br /> of perforations per lateral. Select minimum diameter for <br /> perforated lateral = inches ,--•••%•J• <br /> N.------'..-./, t 14ru- <br /> 9. If perforated lateral system is attached to manifold pipe near <br /> the center, as in lower example, perforated lateral length and 1--an 1=••===•• <br /> number of perforations per lateral will be approximately one cc�""'" <br /> half of that in # 6. Using these values, select minimum m.r 'Lz' <br /> diameter for perforated lateral from table 2 f y -111211.11-124 perforated lateral = I '/ inches ,r is "' � <br /> 2'" ` <br />