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w . PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> ' 1. Select number of perforated laterals � uarter�inch erfotahons s aced�^3 12 �" <br /> _ � � ^ <br /> 2. Select perforation spacing = � ft ::, ... ,? ' ; ��:of,�o�lr. , ;': �; <br /> ,. , <br /> . , , <br /> Perf Sizing 3/16"-1/4" <br /> 3. Since perforations should not be placed closer than 1 foot to Perf Spacing 1.5•-s' <br /> the edge of the rock layer (see diagram),subtract 2 feet from <br /> the rock layer length. E-4: Maximum aUowable number of 1/4-inch perforations <br /> �� per lateral to guarantee<10%discharge variafion <br /> Rock layer length -2 ft = � ft Af�OfOf10fl <br /> P <br /> 4. Determine the number of spaces between perforations. spacing <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 /> down to nearest whole number. <br /> 2.5 8 14 18 28 <br /> Perforation spacing= S�es ft= 3 ft= � spaces 3.0 8 13 17 26 <br /> 5. Number of perforarions is equal to one plus the number of 3.3 7 12 16 25 <br /> perforarion spaces(4). Check figure E-4 to assure the number of 4'0 7 11 15 23 <br /> perforations per lateral guarantees <IO% discharge variation. 5.0 6 10 14 22 <br /> � spaces + 1 = �-I perforations/lateral E-6: Perforation Dischorge 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 3 lat= �.� perforations <br /> 1.0a 0.18 0.42 0.56 0.74 <br /> B. Calculate the square footage per perforation. <br /> Should be 6-10 sqft/perf. Does not apply to at-grades. 2•Ob 0.26 0.59 0.80 1.04 <br /> Rock bed area = rock width (ft) x rock length (ft) 5.0 0.41 0.94 1.26 1.65 <br /> �Q_ft X j� ft=�sqft a 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 onvthin eise. <br /> � sqft=_�perfs = 9.S sqft/perf <br /> i. Determine required flow rate by multiplying the total number of � „�,; � <br /> perforations (6A) by flow per perforation (see figure E-6) � a�r�„o�,� <br /> I <br /> <'o� � <br /> perfs x � 79�,apm/perfs =�gpm e,��� <br /> I � <br /> 8. IE laterals are connected to header pipe as shown on upper � �..• <br /> example,to select minimum required lateral diameter; enter j o G�fe,,�,� <br /> figure E-4 with perforation spacing (2) and number of perforations Figure E•1:Manitold located at End of System <br /> per lateral (5) Select minimum diameter for <br /> perforated lateral = �- inches. �igure E-2:Manifold locofed ,e"0C0p <br /> in the Center of the System <br /> 9. If perforated lateral system is attached to manifold pipe near <br /> the center,lower diagram,perforated lateral length (3) and �robp� <br /> number of perforations per lateral (5)will be approximately one <br /> half of that in step 8. Using these values, select minimum <br /> diameter for perforated lateral= inches. arc�e�,o 0 <br /> ppe irom amw <br /> I hereb certify that I hav co 'pleted this work in accordance with applicable ordinances, rules and laws. <br /> �d��``'� si ature � � license# �o � � ��'� date <br /> � � ) --,-� ) � ) <br />