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14 1 C4-0 co <br /> ONSIT! R Ct( o1Z � U 12�t C' M �) <br /> seve,—-! / [- 'j[ ' Geotextile fabric <br /> YRlATMI IT �� •O �� �� `�' `�) <br /> PROGRAM <br /> PRESSURE DISTRIBUTION SYSTEM quarter inch perforations.s aced A 3' 12 <br /> 1. Select number of perforated laterals 9"of rock. <br /> j - A/Tdr <br /> 2. Select perforation spacing= 3 ft Perf Sizing /1 ' - 1/4" 16 <br /> Perf Spacing <br /> 3. Since perforations should not be placed closer than 1 foot to <br /> the edge of the rock layer (see diagram), subtract 2 feet from E4: Maximum allowable number of 1/4-Inch perforations <br /> per lateral to guarantee<10%discharge variation <br /> the rock layer4engt1-r-- perforation <br /> /�) spacing <br /> C = L <br /> ft feet 2f1 Inch 1.25 inch 1.5 inch 0 Inch1Rock lav -2 ft er length l <br /> '{-0' Zr Zr 3�; <br /> 4. Determine the number of spaces between rforations. 2.5 a 1a 18 2s <br /> P p 3.0 8 13 17 26 <br /> Divide the length (3)by perforation spacing (2) and round 3.3 7 12 16 25 <br /> down to nearest whole number. 4.0 7 11 15 23 <br /> ,f � � �a 5.0 s 10 14 zz <br /> �)Perforation spacing= (r' t_ ft= spaces <br /> 5. Number of perforations is equj?to one plus tFie number of E-6: Perforation Discharge in gpm <br /> perforation spaces(4). Check figure E-4 to assure the number of <br /> perforations per lateral guarantees <10% discharge variation. perforation diameter <br /> head inches <br /> spaces + 1 = 23 perforations/lateral (feet) F 3/16 7/32 1/4 <br /> �s1 <br /> If IL42 0.56 0.74 <br /> 6. A. Total number of perorations= perforations per lateral (5) <br /> times number of laterals (1) 2.0b 0.59 0.80 1.04 <br /> Z _perfs/lat x 3 lat= perforations T r 5.0 0.94 1.26 1.65 <br /> � <br /> ., .r �- ,(p I S}�S. <br /> B. Calcu ate the square�Otage per �f�ration. _ D use 1.0 foot for single-family Homes. <br /> Recommeded value is 6-10 sqft/perf. Does not apply to at-grades. usez.ofeetforan anything else. <br /> Rock bed area = rock width(ft)it rock length (ft) <br /> ft x ft= sgft <br /> Square foot per perforation =Rock bed area _number of perfs(6) <br /> sgft_ perfs = sgft/perf <br /> -)'l Dipe <br /> 7. Determine required flow rate by multiplying the total number of pipe born pump <br /> X L))�,� perforations (6A) by flow per perforation (see figure E-6) <br /> end np <br /> 3`� <br /> 11 �� 95 perfs x 0 s`t`c)pm/perfs = 46 gpm <br /> // allelna{e w aaon <br /> of DiDe Iron pump <br /> SAY8. If laterals are connected to header pipe as shown on upper Figure E-1:Manifold Located at End of System <br /> V example,to select minimum required lateral diameter;enter <br /> figure E-4 with perforation spacing(2) and number of perforations <br /> per lateral (5) Select minimum diameter for Figure E-2:Manifold Located 'p <br /> In the Center of the System <br /> perforated lateral = inches. <br /> 9. If perforated lateral system is attached to manifold pipe near <br /> the center, lower diagram, perforated lateral length (3) and <br /> number of perforations per lateral (5) will be approximately one Pw eom vw <br /> half of that in step 8. Using these values,select minimum <br /> diameter for perforated lateral = inches. <br /> I hereb certify that I have completed this work in accordance with applicable ordinances, rules and laws. <br /> (signature) '3 r' `}L (license#) / (date) <br />