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PRESSURE DISTRIBUTION SYSTEM Geotextile fabric p.pT <br /> b ,.� e,,',- J i,•-1 re n.` rJ-Ot 'c `, t YI `yV A. <br /> 1. Select number of perforated laterals 3 Quarter mch,perforahons spaced 0 3 a' <br /> ,`�h CI;,,•' LC,r(lt)Ij.:ay .,p;1',,,,v,.'::'4, .�:I �. <br /> r or 'do ''g° 9"tifa�k i;:1')',.',=-,:;::%:, <br /> 2. Select perforation spacing= 3 . 0 feet. '`�"�`?' ° ` <br /> � r.i� i�o�e,V 4df+9{)�t 15 o,� Vol: <br /> Perf Sizing 7/32"-1/4" <br /> 3. Since perforations should not be placed closer than 1 ft. to Perf Spacing 1.5'-5' <br /> the edge of the rock layer (see diagram), subtract 2 ft. from <br /> the rock layer length. Perforation Discharges in gpm <br /> c/a perforation diameter <br /> kocx layer engtn-2 ft. = 4( O feet. head (inches) <br /> (feet) 1/8* 3/16 7/32 1/4 <br /> 4. Determine the number of spaces between perforations. 1.00 0.18 0.42 0.56 0.74 <br /> Divide the length above by perforation spacing and round 2.0b 0.26 0.59 0.80 1.04 <br /> down to nearest whole number. <br /> 5.0 0.41 0.94 1.26 1.65 <br /> Length perf. spacing = `d Oft. _ 3 ft. = a 9spaces °Use 1.0 foot for single-family homes. <br /> / <br /> (3) l2) bUse 2.0 feet for for plugging else. <br /> • Potential <br /> 5. Number of perforations is equal to one plus the number of <br /> perforation spaces . - -F <br /> Maximum number of quarter inch perforations per <br /> o't') spaces + 1 = 3.5e perforations/lateral lateral to guarnantee<10%discharge variation <br /> Perforation <br /> 6. Multiply perforations per lateral by number of laterals to S(pfeeg 1% M 2 <br /> get total number of perforations. 2.5 14 18 28 <br /> 3 _ '1y�i <br /> p ,a <br /> ,;»A. 3.0 13 17 26 <br /> laterals x per ater i perforations. <br /> 3.3 12 16 25 <br /> Calculate the square footage per perforation (6-10 sqft/perf) 4.0 11 15 23 <br /> System area: )0 x eg a. = a 0 5.0 10 14 22 <br /> azeao •_ per or ons = 9' 7 sqft/perf <br /> MANIFOLD LOCATED AT END OF PRESSURE DISTRIBUTION SYSTEM <br /> 7. Determine required flow rate by multiplying <br /> number of perforations by flow per perforation <br /> fze I\ PIR Lo <br /> �'yths �- <br /> �3 eif= l/Z gpm <br /> pf E,,;j.. <br /> x gp P <br /> `""'N d <br /> E,, tt.t .PR <br /> 8. If laterals are connected to header pipe as shown on upper \` <br /> example, to select minimum required lateral diameter;enter <br /> table with perforation spacing and number of perforations LAYOUT OF PERFORATED PIPE LATERALS FOR <br /> PRESSURE DISTRIBUTION IN MOUND <br /> per lateral. Select minimum diameter for RRPMATED PLASTIC PIR <br /> perforated lateral = inches. eACINd <br /> r°71.1/47414. �,iMA' <br /> ISA NA �nI..A <br /> E <br /> Lo <br /> 9. If perforated lateral system is attached to manifold pipe near --arm-" ; <br /> the center, lower diagram,perforated lateral length and ,,.' - <br /> number of perforations per lateral will be approximately one EI.o DP b. ow" <br /> half of that in step 8. Using these values,select minimum �RFBR°ED \4= ,C.PIPE PP..Of <br /> diameter for perforated lateral= 1/) Z inches. \``"GTN <br />