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PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> -ar-..--=.t-. -at-► :a..0'v-eq. . :;t-• -_.-•s „. <br /> 1. Select number of perforated laterals 3 - _ 0' 12'" <br /> 2. Select perforation spacing = ";,.O ft of rock <br /> Perf Sizing 3/16"- 1/4" <br /> 3. Since perforations should not be placed closer than 1 foot to Perf Spacing 1.5'-5' <br /> the edge of the rock layer (see diagram),subtract 2 feet from <br /> the rock layer length. E-4: Maximum allowable number of 1/4-inch perforations <br /> gS <br /> per lateral to guarantee<10%discharge variation <br /> Rock layer length -2 ft - c'I ft <br /> perforation <br /> 4. Determine the number of spaces between perforations. spacing <br /> Divide the length (3)by perforation spacing (2) and round (feet) 1 inch 1.25 inch 1.5 inch 2.0 inch <br /> down to nearest whole number. <br /> 2.5 8 14 18 28 <br /> Perforation spacing = 6'a ft+ IS ft= 1`) spaces 3.0 8 13 17 26 <br /> 5. Number of perforations is equal to one plus the number of 3.3 7 12 16 25 <br /> perforation spaces(4). Check figure E-4 to assure the number of 4'0 7 11 15 23 <br /> perforations per lateral guarantees <10% discharge variation. 5.0 6 10 14 22 <br /> I'1 spaces + 1 = 1' 1 <br /> perforations/lateral E-6: Perforation Discharge in gpm <br /> 6. A. Total number of perforations = perforations per lateral (5) perforation diameter <br /> times number of laterals (1) head (inches) <br /> )E6 perfs/tat x 3 tat= S V perforations (feet) 3/16 7/32 <br /> 1.00 0.42 0.56 0.74__, <br /> B. Calculate the square footage per perforation. <br /> Should be 6-10 sqft/perf. Does not apply to at-grades. 2 0b 0.59 0.80 1.04, <br /> Rock bed area = rock width (ft) x rock length (ft) 5.0 0.94 1.26 1.65'' <br /> 1 0 ft x Sd‘ ft = SiS/0 sqft <br /> a Use 1.0 foot for single-family homes. <br /> Square foot per perforation = Rock bed area +number of perfs (6) b Use 2.0 feet for onvthing else. <br /> Sv sqft+ 61-1. perfs = /9.I sqft/perf <br /> MANIFOLD LOCATED AT ENO OF PRESSURE DISTRIBUTION i SYSTEM <br /> 7. Determine required flow rate by multiplying the total number of <br /> perforations (6A) by flow per perforation (see figure E-6) <br /> S y perfs x . 7%.1 gpm/perfs = R gpm \; <br /> 8. If laterals are connected to header pipe as shown on upper / <br /> /i .. <br /> example, to select minimum required lateral diameter;enter ,DM„D` ^`:°w��" <br /> figure E-4 with perforation spacing (2) and number of perforations �``"°M d <br /> per lateral (5) Select minimum diameter for <br /> LAPRDSPRRARBUPpN EWtlROperforated lateral = inches. <br /> ry�/'P[PlDneEED RASI.E PIPE <br /> 9. If perforated lateral system is attached to manifold pipe near `PE.,Dn.T,DN3 S..E ,.,< <br /> the center,lower diagram,perforated lateral length (3) and "D '",E"'^E `4^`'� "" "°" <br /> number of perforations per lateral (5)will be approximately one r "--. <br /> PP y PERE. o�,P;.E.D„DN DE "- �-�' <br /> half of that in step 8. Using these values, select minimum ti=�_ <br /> _ '-"=4"42“..g.% <br /> diameter for perforated lateral = <br /> inches. <br /> END EAP .0 <br /> 1 iP RAO'l''''" <br /> / - d I.PIo _n <br /> I hereby certify that I havempleted this work in accordance with applicable ordinances, rules and laws. <br /> (i... <br /> (signature) 3 l L (license#) 9 -P -a,c)}0 (date) <br />