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' ' ' ' PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> � 1. Select number of perforated laterals � uarter inch erforadons s aced�`3' 1?`' <br /> 2. Select erforahon s acin � '� 9"of.ro�tk <br /> , P P g =�-ft . . <br /> I' 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: Maxirrwm allowable number o(1/4-inch pedorations <br /> ,� 5,5 �►� per laterd fo guarantee<10'�discharge vadation <br /> I Roc ayer ng -2 ft --`�_ft <br /> , per(oration <br /> , 4. Determine the number of spaces between perforations. spxing <br /> I Divide the length (3)by perforation spacing(2)and r un feet 1 inch 1.25 inch 1.5 inch 2.0 inch <br /> down to nearest whole number. <br /> 2.5 8 14 18 28 <br /> Perforation spacing=�ft=�ft=,�_spaces 3.0 8 �13 17 � <br /> 5. Number of perforations is equal to one plus the number of 3'3 � � 12 w �6 25 <br /> ' perforaHon spaces(4). Check figure E-4 to assure the number of 4'0 > >> 15 23 <br /> j perforations per lateral guarantees <10%discharge variation. 5.0 6 10 i4 22 <br /> , �_spaces + 1 =��perforaHons/lateral E-6: Per(oratton Dischorge in gpm <br /> I6. A. Total number of perforations = perforations per lateral (5) <br /> i�imes number of laterals (1) perfo iat�i he diameter <br /> head <br /> r� perfs/lat x v lat=�perforations �feet� 3��b ��32 1/4 <br /> 1.Oa 0.42 0.56 0.7 <br /> B. Calculate the square footage per perforation. <br /> I, Should be 6-10 sqft/perf. Does not apply to at-grades. 2•0b 0.59 0.80 1.04 <br /> � Rock bed area = rock width (ft) x rock length (ft) 5.0 0.94 1.26 1.65 <br /> _L�ft x�_ft= 55� sqft a Use 1.0 foot for single-family homes. <br /> Square foot per perforaHon = Rock bed area +number of perfs (6) b Use 2.0 feet for an nir, eise. <br /> �S� sqft�__$�_�perfs = .2-sqft/perf <br /> MANIFO�D LOCnTED 4T END OF pqE55UqE DISTRIBUTION SYSTEM <br /> , 7. Determine required flow rate by multiplying the total number of <br /> ' perforations (6A) by flow per perforaHon(see figure E-6) ,,,,,,,,, <br /> ��perfs x�_gprn/perfs=.�_gpm � <br /> 8. If laterals are connected to header pipe as shown on upper %ti:-� <br /> , example,to select minimum required lateral diameter;enter M d��o"���"`� �`='w="� <br /> ', figure E-4 with perforation spacing (2) and number of perforations ��`"" <br /> per lateral (5) Select minimum diameter for <br /> perforated lateral= �l�� 1riC�'leS. �..o�.w►enraurEo/IPF lsT[I1�L5.oa <br /> PRESSURE D177N1lVTtON w MpuNO <br /> �[ OY�0 Rtl1K��K <br /> 9. If per,forated lateral system is attached to manifold pipe near � <br /> •[MOIYTiOM t/K[0 y' WC� <br /> the c�nter,lower diagram,perforated lateral length (3) and �= �'�""`"��`�''°" ^"'"°"�� <br /> Yr�xK o <br /> number of perforations per lateral (5)will be approximately one K,„�;,��,�.,,,�,� <br /> 'i, half of that in step 8. Using these vatlues,select minirnum '^• �- _ <br /> ' diameter for perforated lateral = 1 �'� inches. � '�'��"�" <br /> -���.. b. ���,,� <br /> ' d��,,,a,T`° �.�.• -� <br /> ��[�M <br /> '' I hereby certify that I have ompleted this work in accordance with applicable ordinances, rules and laws. <br /> �.�-���, � ���.�.. <br /> ��, (signature) zv`�1 (license#) /D - � -(,?� (date) <br />