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• PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> 1. Select number of erforated laterals � � � Y� �1� - <br /> p Quarter inch erforahons s aced Co?3 � <br /> 2. Select perforation spacing= 3•o ft £ ' � > ° ' g °f;�b`k `; <br /> t <:: <br /> < � <br /> , . .. ......._..�..;�.,:., <br /> Perf Sizing 3/16"-1/4" <br /> 3. Since perforations should not be placed closer than 1 foot to Perf spa��g 1.5'-5' <br /> the edge of the rock layer (see diagram),subtract 2 feet frorn <br /> the rock layer length. E-4: Maximum ailowable number of 1/4-inch perforafions <br /> I�-y1 ' C �t [ per lateral to guarantee<10%discharge variafion <br /> Rockla� -21t =_ ��/ lt <br /> perforafion <br /> spacing <br /> 4. Determine the number of spaces between perforations. feet) 1 inch 1.25 inch 1.5 inch 2.0 inch <br /> Divide the length (3) by perforation spacing(2) and round <br /> down to nearest whole number. <br /> 2.5 8 14 18 28 <br /> Perforation spacing = ?,`� ft= `� ft=�"�Z spaces 3.0 8 13 17 26 <br /> 5. Number of perforations is equal to one plus the number of 3'3 � �2 �6 25 <br /> perforation spaces(4). Clzecic figure E-4 to assure the number of 4�0 > >> >5 23 <br /> perforatians per lateral guarantees <10% discharge variation. 5.0 6 10 14 22 <br /> ��_spaces + 1 = 1'� perforations/lateral E-6: Perforotion 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) �/8 3/16 7/32 1/4 <br /> 1`' perfs/lat x�_lat=�_perforations <br /> 1.0a 0.18 0.42 0.56 0.7� <br /> B. Calculate the square footage per perforation. <br /> Should be 6-10 sqft/perf. Does not apply to at-grades. 2•0b 0.26 0.59 0.80 1.0� <br /> Rock bed area = rock width(ft) x rock length (ft) 5.0 0.41 0.94 1.26 1.6� <br /> �ft x�! _ft= �t► o sqft a Use 1.0 foot for singie-family homes. <br /> Square foot per perforation= Rock bed area=number of perfs (6) b Use 2.0 feet for an tni� else. <br /> �ll o sqft= �Z..-=perfs = `�'.rI sqft/perf <br /> 7. Determine required flow rate by multiplying the total number of ,,,�,,� � <br /> perforations (6A) by flow per perforation (see figure E-6) � pq���q��P <br /> �`�'� perfs x �7 gpm/pE>rfs =�gpm ��� <br /> � <br /> 8. If laterals are connected to header pipe as shown on upper ��,� <br /> example,to select minimum required lateral diameter;enter olterrafeloca6on <br /> or plpe h«n purt� <br /> figure E-4 with perforation spacing (2) and number of perforations Fi9ure E-1:Manifold Located at End of System <br /> per lateral (5) Select minimum diameter for <br /> perforated lateral = �l �� inches. Figure E-2:Manifold Located e`�`� <br /> in the Cenfer ot the System <br /> 9. If perforated lateral system is attached to manifold pipe near <br /> the center,lower diagram,perforated lateral length(3) and morifoldplpe <br /> number of perforations per lateral(5) will be approximately one <br /> half of that in step 8. Using these values,select minimum <br /> "'�A - anernole location <br /> diameter for perforated lateral = inches. or��n�� <br /> va��«�a�w <br /> I he eby certify that I have completed this work in accordance with applicable ordinances, rules and laws. <br /> / "` "� ��� ��--�.._.._ si nature ��l licen !Z �" '�C7 <br /> , ( S ) ( se�) � )3 '� (date) <br />