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PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> 1. Select number of perforated laterals 3 ,u���,�h �rano�� a«a�s� ?�" <br /> 2. Select perforation spacing= 3. o ft .: � ._ . . -� �.s��o�s� <br /> Perf Sizing 3/16"-1/4" <br /> 3. Since perforations should not be placed closer than 1 foot to pe�spacins 1.5'-s�. <br /> the edge of the rock layer(see diagram),subtract 2 feet from <br /> the rock layer length. E-4: Mmcimum anowable number of 1/4•inch perforations <br /> �Z . per laterd b guarantee<107G discharge variation <br /> c ayez eng '2 ft = �o� ft pA�01QflOfl <br /> aP��9 <br /> 4. Determine the number of spaces between perforations. fee 1 inch 1.25 inch 1.5 inch 2.0 inch <br /> Divide the length(3)by perforation spacing(2)and round <br /> � �lown to nearest whole number. <br /> 2.5 8 14 18 28 <br /> Perforation spacing= �� ft+ 3 ft=,�o 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 a.o 7 i 1 15 23 <br /> perforation spaces(4). Check figure E-4 to assure the number of <br /> perforations per lateral guarantees<10%discharge variation. 5.0 6 10 14 22 <br /> � o spaces+1 = a�_perforations/lateral E-6: Perioratton Discharge in gpm <br /> 6. A. Total number of perforations= perforations per lateral(5) perforation diameter <br /> times number of laterals(1) heod inches <br /> (feet) 3/16 7/32 1/4 <br /> a, � perfs/lat x 3 lat=�perforations �,po 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 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 /> !c� ft X �o Z ft= a� sq ft a u��.0 foot tor sfngle-family homes. <br /> Square foot per perforation=Rock bed area+number of perfs (6) b Use 2.0 feet for on ni� else. <br /> (o� sqft+��perfs= `�• � sqft/perf �M�p �«pTW �T END OF ppESS�E OISTRIBUTION 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 /> �0 3 perfs x •7 ��mN/perfs= �-7 gpm ".� <br /> 8. If laterals are connected to header pipe as shown on upper ��'� <br /> ,m`"""" s`.�:,.`�'• <br /> example,to select mi.nimum required latera�diameter;enter ,,�^' <br /> figure E-4 with perforation spacing(2) and number of perforations ��`"°M <br /> per lateral(5) Select�n�*�um diameter for <br /> uvart a raroiuno rn[urc�us son <br /> perforated lateral= inches. .�«�asT��,^„„w�„� <br /> .a.�..��..,�..� <br /> 9. If perforated lateral system is attached to manifold pipe near �„�,�„ „�„�,,.�.� <br /> the center,lower diagram,perforated lateral length(3) and �= �'�"��� <br /> rr�ae <br /> number of perforations per lateral (5)will be approximately one .n.�,.,�,�.�,�„ <br /> half of that in step 8. Using these values,select minimum '° �- _ <br /> diameter for perforated lateral= I ��-� inches. �� a <br /> " •-���K�� <br /> . „��•� � <br /> . ,t,,,,""°` <br /> ..- <br /> d "".' " <br /> �� <br /> I hereby certi that I ha completed this work in accordance with applicable ordinances, rules and laws: <br /> � - % `�`- ' (signature) ���'l (license#) � -��-�Z (date) <br />