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PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> . 1. Select number of perforated laterals 3 Quarter inch perforations spaced e 3 12 <br /> I <br /> 2. Select perforation Spacing= 0 ft of rdcl` <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 /> per lateral to guarantee<10%discharge variation <br /> Rock ll yerlength -2 ft - ft perforation <br /> spacing <br /> 4. Determine the number of spaces between perforations. <br /> (feet) 1 inch 1.25 inch, 1.5 inch 2.0 inch <br /> Divide the length(3)by perforation spacing(2)and round <br /> clown to nearest whole number. <br /> 2.5 8 14 28 <br /> Perforation spacing= y K ft+ L ft= 1 6 spaces 30 8 13 1 <br /> .„2§.._ <br /> 3.3 7 12 16 25 <br /> 5. Number of perforations is equal to one plus the number of <br /> 4.0 1 11 15 23 <br /> perforation spaces(4). Check figure E-4 to assure the number of <br /> 5.0 6 10 14 22 <br /> perforations per lateral guarantees <10%discharge variation. - <br /> I L ' <br /> E I ,.,,)i�j . <br /> /(' spaces +1 = 1'? 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 /> If? perfs/lat x _lat=__Si_perforations (feet) 3/16 7/32 1/4 <br /> 1.00 0.42 0.56 0.74 <br /> B. Calculate the square footage per perforation. 2,Ob 0.59 0.80 1.04 <br /> Should be 6-10 scift/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 /> /0 ft x .SO ft= O C sqft a Use 1.0 foot fpr single-family homes. <br /> Square foot per perforation =Rock bed area +number of perfs (6) b Use 2.0 feet for onythinp else. <br /> ,S'(9 t3 sqft+ S'/ perfs= 'J p sqft/perf <br /> MANIFOLD Lapp AT END OF PRESSURE DISTRIBUTION SfSTEM <br /> 7. Determine required flow rate by multiplying the total number of <br /> perforations(6A) .by flow per perforation(see figure E-6) W <br /> S I perfs x a r)4 gpmYperfs= 355 gpm <br /> 8. If laterals are connected to header pipe as shown on upper ,00 .�,rrw,,.... <br /> example,to select minimum required lateral diameter;enter e,o°""°Lo <br /> '�"" <br /> figure E-4 with perforation spacing (2) and number of perforations N./ <br /> per lateral (5) Select um diameter for <br /> perforated lateral 1�i u,an OFn PERFORATED PIPE eN uTu•u Fon <br /> '�• inches. d"�e��t Minium ouTRIwTION w NO1VN0 <br /> RIM01YTtO PLASTIC•K <br /> 9. If perforated lateral system is attached to manifold pipe near �n.,a,,,,, yope . <br /> the center,lower diagram,perforated lateral length (3) and <br /> number of perforations per lateral (5)will be approximately one --KrorA..,'a • <br /> half of that in step 8. Using these values,select minimum s <br /> diameter for perforated lateral= inches. 0. e• <br /> ,' _ 0` <br /> el <br /> I hereby certify that I have completed this work in accordance with applicable ordinances, rules and laws. <br /> Z,WG/ (signature) -7 (license#) • S~'1-5- 0 S (date) <br />