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PRESSURE DISTRIBUTION SYSTEM Geote�tile fabric <br /> l. Select number of perforated laterals � ; �U,r�er��r,rerr-or,hon5 SL,a�ea c�� 1' . <br /> 2. Select perforation spacing= � ft 9"�f rock <br /> Perf Sizing 3/16"- 1/-�" <br /> 3. Since perforations should not be placed closer than 1 foo to Perf Soacing 1.5'-�' <br /> the edge of the rock layer (see diagram), subtract 2 feet m <br /> the rock la�rer length. E-4: Maximum allowable number of 1/4-inch perforotions <br /> �G per lateral to guarantee<10%discharge vanation <br /> '�- - <br /> 2ock lacer icngu� � ft -_�ft perforation <br /> spacing <br /> 4. Determine the nLunber of spaces between perforations. <br /> Divide the length (3) by perforation spacing (2) and roun (feet) i inch 1.25 inch 1.5 inch 2.0 inch <br /> down to nearest whole numUer. <br /> 2.5 8 14 18 28 <br /> Perforation spacing =�.�ft= � ft =�(�spa es 3.0 8 �3 �7 26 <br /> 3.3 ? 12 16 25 <br /> 5. Number of perforations is equal to one plus the number f 4 0 7 11 15 23 <br /> perforation spaces(4). Check figecre E-4 to assrcre the nianbe•of <br /> pe�forations per Interal guarantees <10% dischar�e variation. 50 6 10 la 22 <br /> l C�. spaces + 1 =�perforations/lateral E-6: Perforation Discharge in gpm <br /> 6. A. Total munber of perforations = perforations per later 1 (5) perforation diameter <br /> times number of laterals (1) head inches <br /> > (feet) ��8 3/16 7/32 1/4 <br /> �perfs/lat x_� lat= � / perforations <br /> 1 0� 0.18 0.4 0.56 i 0.74 <br /> B. Calculate the square footage per perforation. 2.Ob 0.26 0.59 0.80 1 .04 <br /> Should be 6-10 sqft/perf. Does rtot apply to at-gracles. <br /> Rock bed area = rock width (ft) x rock length (ft) 5.0 0.41 0.94 1.26 1 .65 <br /> ���ft X c5 [,% ft = `S G�� SC�ft ° Use 1.0 foot for single-family homes. <br /> SquJ'�are foot per perforation = Rock Ued area =mtmber f perfs (6) b Use 2.0 feet for onv*hinq else. <br /> �-+v � sqft=�_perfs = C�.�: sqft/perf M4NIFC.O IOCATEO 4T ENO OF PqESSURE �715TRIBUTION SYSTEM <br /> 7. Determine required flow rate by multiplying the total nt ber of <br /> perforations (6A} Uy flow per perforation (see figtcre E-6) �W;,� <br /> � <br /> ��perfs x �S gpm/perfs =2� Lgpm � �� <br /> 8. If lateraLs are connected to header pipe as shown on uppe �� �,�,���E 1 K <br /> example, to select minunum requued lateral diameter;en er d.� <br /> figure E-4 with perforation spacing (2) and number of per orations ,�/tf�M <br /> per lateral (5) Select miniinum diameter Eor <br /> perforated lateral - ,� �<rvur or Penronnrco Prve L<tEAa�S roa <br /> �_LeS. rr+ess�ne o�s+n�aur�oN �N MouHo <br /> 11L <br /> ocnro+uro.insnc>�ac <br /> 9. If perforated lateral system is attached to manifold pipe n ar `S.�„o,.,,o�,,..�,,,. �,S,Kw� <br /> the center, lower diagram,perforated lateral length (3) an �E �'a`��"::°`�y' s "Y.����, <br /> number of perforations per lateral(5) will be approximate y one ��a.,.;,;�=,a,�aa�,a�� � <br /> half of that in ste 8. Usin these values, select minimum � <br /> 'a. 'V.. <br /> P g �, •-�..,���•;:����sx <br /> diameter for perforated lateral = inches. ��`'� �- <br /> _ . �,a. � <br /> _ ��rta�IfCR/L V�.�r�E r�ou <br /> �y �.�� w.. �,,...�w <br /> � � ���„ <br /> � <br /> I hereby certify that I have completed this work in accordance with a plicabie ordinances, rules and Iaws. <br /> � `; �;�u��� (signature) ��/�-% (license#) �`� '�� "`� Y (date) <br /> i <br />