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PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> ` .:r:,i::+, 'i-;�,=•pr:-.-,--;-•-:=t.:�.:=-p•;+-;i.�_rtri.`sYr_t'. :r- <br /> 1. Select number of perforated laterals 3Quarter inch perforations spaced 0 3' <br /> ._ :. ; " of rocs <br /> 2. Select perforation spacing= 3 .0 ft , ' <br /> Perf S 3/16"-1/4" <br /> 3. Since perforations should not be placed closer than 1 foot to Perf Spadng 1.5'-5' <br /> the edge of the rock layer(see diagram),subtract 2 feet from <br /> the rock layer length. E-4: Madman dawdle number of 1/4•Mch perforations <br /> per laterd to guarantee<10%discharge variation <br /> Rock n -2 ft = (.4 ft perforation <br /> s1uW <br /> 4. Determine the number of spaces between perforations. <br /> (feet) 1 Inch 125 Inch 1.1inch 2.0 inch <br /> Divide the length(3)by perforation spacing(2)and round <br /> down to nearest whole number. 2:5 8 14 18 28 <br /> Perforation spacing= (.00 ft+ 3 ft= 10 spaces 3.0 8 13 17 26 <br /> 3.3 7 12 16 ° 25 <br /> 5. Number of perforations is equal to one plus the number of 4.0 7 11 15 23 <br /> perforation spaces(4). Check figure E-4 to assure the number of 5.0 6 10 14 22 <br /> perforations per lateral guarantees <10%discharge variation. - <br /> a 0 spaces+1 = a) 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 /> (feet) 3/16 7/32 1/4 <br /> _ a t perfs/lat x 3 lat= (03 perforations 1.00 0.42 0.56 s.74 <br /> B. Calculate the square footage per perforation. 2,0b 0.59 0.80 1.04 <br /> Should be 6-10 sqft/perf.Does not apply to at grades. <br /> Rock bed area= rock width(ft)x rock length(ft) 5.0 0.94 1.26 1.65 <br /> /o ft x (.Z ft= 1.• ,0 sqft o 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 onvthInp else. <br /> (oma sqft+ (03 perfs= g•c sqft/perf MANIFOLD LOCATED AT END OF PRESSURE DISTRIBUTION 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 /> JO <br /> perfs x •14. gpm/pperfs= 47 . gpm imm- <br /> 8. If laterals are connected to header pipe as shown on uppervol .,.c <br /> example,to select minimum required lateral diameter;enter ��,M,�'"° <br /> figure E-4 with perforation spacing(2)and number of perforations \,/ <br /> per lateral(5) Select minimum diameter for <br /> LAYOUT a KII�pMTtB nnL FOR <br /> perforated lateral= inches. mi.MISTMMNTION IN PAWKY <br /> P.a Ramie Pot <br /> 9. If perforated lateral system is attached to manifold pipe near44 <br /> . •o,,,,, <br /> the center,lower diagram,perforated lateral length(3) and IND �` �'' • <br /> " „tp <br /> number of perforations per lateral(5)will be approximately one .- ;yr, ••••�•. <br /> half of that in step 8. Using these values,select minimum .-,e.=.e,,µ„ <br /> diameter for perforated lateral= a•0 inches. •.. M <br /> n`P' <br /> I hereby certify that I have completed this work in accordance with applicable ordinances, rules and laws. <br /> ' 5.-&-4•�� (signature) 3''74 (license#) (o-14-03' (date) <br />