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1 � <br />' � PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> uarter inch erforations aced�3' <br /> 1. Select number of perforated laterals_� 12�� <br /> 2. Select perforation spacing= 3• c7 ft � �"of,rd�` <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 allowabie number of 1/4-inch perforations <br /> (�p perlateral to guarantee<10X discharge varlation <br /> a-t�Gy�T� -2 ft =_�4�ft perforation <br /> spacing <br /> 4. Determine the number of spaces between perforations. <br /> feet 1 inch 1.25 inch 1.S inch 2.0 inch <br /> Divide the length (3)by perforation spacing(2)and roiand <br /> - down to nearest whole number. <br /> 2.5 8 14 18 28 <br /> Perforation spacing= S� ft+�ft=�spaces 3.0 8 13 11 26 <br /> 3.3 7 12 16 25 <br /> 5. Number of perforations is equal to one plus the number of Q� � �� 15 23 <br /> perforation spaces(4). Check figure E-4 to assure the number of 5.0 6 10 14 22 <br /> perforations per lateral guarantees <10%discharge variation. <br /> �spaces+ 1 = Zv perforations/lateral E-b: Penorat�on o�scnarg ne gpm <br /> 6. A. Total number of perforations= perforations per lateral (5) perforation diameter <br /> times number of laterals (1) �r�a�� heod inches <br /> <feet) 3/16 7/32 1/4 <br /> �_perfs/lat x�_lat=�perforations 1.Oo 0.42 0.56 0.74 <br /> B. Calculate the square footage per perforation. 2,pb 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 /> �� ft x�ft= a c�� sqft a u��.0 foot for single-family homes. <br /> Square foot per perforation=Rock bed area+number of perfs (6) b Use 2.0 feet for an nin eise. <br /> 1�v`� sqft+ �'�perfs= ��. sqft/perf MAN,��o �a,,,�, ,T END OF PRESSURE DISTR�BUTION 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 /> �i .�.�„�. <br /> �� perfs x -S b Qv?m�perfs= � � gpm <br /> � <br /> :a <br /> 8. If laterals are connected to header pipe as shown on upper �`� <br /> example,to select minimum required latera�diameter;enter M,,,�^' <br /> �m���` �:;;$�. <br /> figure E-4 with perforation spacing (2) and ntunber of perforations �/``� <br /> per lateral(5) Select minimum diameter for �,�,,,a K�.�.TEo M.F�.TEpp�rop <br /> rRnwne msTnieunor� W MOUNO <br /> perforated lateral= inches. <br /> ,{k�.�.�,K.� <br /> 9. If perforated lateral system is attached to manifold pipe near v`��,�Y����. ��,,,,,,.�� <br /> the center,lower diagram,perforated lateral length,(3) and r,,,,,�, <br /> number of perforations per lateral(5)will be approximately one K.,�;�,�;�.�,�� <br /> half of that in step 8. Using these values,select mirumum '°• �- <br /> '-��s*K•t�:.��.'.� <br /> diameter for perforated lateral=�_inches. <br /> ��e cu b, 0`���� <br /> � � - d �,tpl�Il �n- �.p <br /> r�"" <br /> ���TM <br /> I hereby cerrify�I have c pleted this work in accordance with applicable ordinances, rules and laws. <br /> ���'-'-� ,`� (signature) ����f (license#) �' 4S-O Z (date) <br />