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� � PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> 1. Select number of perforated laterals S arter inch 'erforaHons, I aeed @ 3' ' .2'„ <br /> • �: .:8' of,r�►cic. .. . <br /> 2. Select perforation spacing= . o� ft � �'� ' . <br /> Perf S 3/16"-1/4" <br /> 3. Since perforations should not be placed closer than 1 foot to P�sp�i.s�-s� <br /> the edge of the rocic layer(see diagram),subtract 2 feet from <br /> the rock layer length. �t S'2=3�� '' � 5 E•4: Maxirtwm dbwabie rrur�ber of 1/4inch pe�forations <br /> sa-�. 3+t ��s <br /> __tpsz:_;�...i ��,o per lateral to pu�antee<10%:diachc�e variaHon <br /> -2 ft = -�#1Q-�'?�'�.�1=_�;�_. <br /> � ----_..�.5-�..'i.'-� P9NOfCfl011 <br /> _ -ru-t,ati 1 O 1 ,�� <br /> 4. Determine the number of spaces between perforations. tee 1 inch 1.25 inch . 1.5•inch 2.o inch <br /> Divide the length(3)by perforation spacing(2)and round <br /> ��to nearest whole number. 2;5 8 14 18 28 <br /> Perforation spacing= ft+ ft= spaces �3.0 8 13 17 26 <br /> 5. Number of perforations is equal to or�e plus the number of 3.3 7 12 16 25 <br /> perforation spaces(4)..Check figure E-4 to assure the number of 4.0 7 11 15 23 <br /> perforations per lateral guarantees<10%discharge variation. 5.0 6 10 14 22 <br /> Spaces+1 = perforations/lateral E-6: Porioration Discharpe in gpm <br /> 6. A. Total number of perforations= perforations per lateral(5) perforotion diometer <br /> times number of laterals(1) heod Inches <br /> (feet) 3/16 7/32 1/4 <br /> perfs/lat x lat= 101 perforations 1.Oa 0.42 0.56 0.74 <br /> B. Calculate the square footage per perforation. 2,Ob 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 /> _ �S�,ft X _C, � ft= I`�LSO Sqft a llae l.q toot fa ainple-tamlly homes. <br /> Square foot per perforation=Rock bed area+number of perfs (6) b uae z.o feer ror o►, i� ai�. <br /> ��Sv sqft+ l�perfs=J�_sqft/perf ��p �pq�p AT END OF MIES6l,qE dSTR18UT10N SYSTEM <br /> 7. Determute iequired flow rate by multiplying the total number of <br /> perforations(6A) .by flow per perforation(see figure E-6) „� <br /> /� �'b►� .u�.�a. <br /> )�_perfs x ��m�erfs= S� gpm <br /> �.;,.,._•� <br /> 8. If laterals are connected to header pipe as shown on upper �� ��;��,� <br /> example,to select aLn�*+�+um requu�ed lateraF diameter;enter ��,,,•►�°"'° <br /> figure E-4 with perforation spacing(2)and number of perforations �� <br /> per lateral(5) Select zninimum diameter for <br /> urour a K�raum nr[un�Ks mn <br /> perforated lateral= 7�•� ]11C11eS. •M•*��as,�n�.�a�W wa,No <br /> .w.�.,��...K..� <br /> 9. If perforated lateral system is attached to manifold pipe near �� , ��. x�,,,,�.�° <br /> the center,lower diagram,perforated lateral length{3) and ��` <br /> tr�s <br /> number of perforations per lateral(5)will be approximately one :��,�..�«« <br /> half of thafi in step 8. Using these values,select minimum .,�.�,��; <br /> diameter for perforated lateral= inches. �,�r a <br /> . . ��� <br /> r �. .� <br /> � <br /> I hereby certify that I have completed this work in accordance with applicable ordinances, rules and laws. <br /> ,7� �� � ----=- (signature) �"��_(license#) �o-1S-03 (date) <br />