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PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> 1. Select numUer of perforated laterals '�? i2•• 'i <br /> 2. Select perforation spacirig = �,5:> ft of.ro�k , <br /> Perf Sizing 3/16"-1/4" <br /> 3. Since perforations should not be placed closer than 1 foot to Perf SpacinU 1.5'-s� <br /> the edge of the rock layer (see diagram),subtract 2 feet from <br /> the rock layer length. F-4: M�ximum allowable number of 1/4-inch periorcy6ons <br /> --�-�— -2 ft = � per lateral to guarant��a<10%dischorge variaiion ' <br /> �_ft <br /> Rock layer length perforafion <br /> 4. Detennine the number of spaces beiween perforations. spacing <br /> Divide the length (3) by perforation spacing (2) and round feet 1 inch 1.25 inr:h 1.5 inch 2.0 inch <br /> down to nearest whole munUer. <br /> 2.5 S 14 18 2E� <br /> Perforation spacing =_ ;�',`l ft= ;'> ft=_p'-; spaces 3.0 8 13 17 �6.,,....._ <br /> 5. NumUer of perforations is equal to one plus the numUer of 3'3 � �2 �b ?5 <br /> perforation spaces(4). Check figiire E-4 to assure the number of 4'0 > >> 15 23 <br /> pe�forntions per Interal guarantees <10%discharge vnric�tion. 5.0 6 10 14 22 <br /> � <br /> !,� spaces + 1 = f� perforations/lateral E-6: ParforaUion Dischorge in gpm <br /> 6. A. Total numUer of perforations = perforations per lateral (5) � perforation diameter <br /> times number of laterals (1) head inches ; <br /> I�I perfs/lat x__'a lat- (feet) 3/16 7/32 �/4 �� <br /> - "I..- perforations ,.. <br /> 1.Oa 0.42 0.56 ( 0.74�J <br /> B. Cal�culate the square footage per perfaraiion. '"'"""� <br /> Should be 6-10 sqft/perf. Does not nppl� to at-g►�ades. 2�0b �.59 0.80 1.04;, <br /> Rock Ued area = rock width (fi) x rock length (ft) 5.0 0.94 1.26 1.65! <br /> �'�:° ft x �_,�_ft = r f r 4� s ft ' <br /> 9 ° Use 1.0 foot for single-family homes. <br /> Sc�uare foot per perforation = Rock bed area =number of perfs (6) b Use 2.0 teet for on hin eise. i <br /> �1 l�:) sqft= �--6"4'.. perfs = `�.�1 sqft/perf ' <br /> MAMFOL� LOCATED AT ENO OF PpE55URE 0157RIBUTIONiSI'STEM <br /> 7. Deternune required flow rate by multiplying the total munber of <br /> perforations (6A) by flow per perforaHon (see figure E-6) <br /> �-)`Z.. perls x �1 :gpm/perfs =�gpm � �_:o. <br /> 8_ If]aterals are coiulected to header pipe as shown on upper � � �(� <br /> �:,,= � <br /> example, to select minimiun required lateral diameter;enter �„�,�"`��"` I°q�';�Ea."` <br /> figure E-4 with perforation spacing (2) and number of perforaHons `"`���d <br /> per lateral (5) Sel�ct mini.intun diameter for �/ <br /> ,.,7 LnTUUT OF pFNfOHAIEO PIPE IaTEPPLS fOR <br /> per orated lateral = �� ]nLyZeS. PnE55uNE OiSTP�Bui�ON W uouu0 <br /> .� <br /> 9. If perforated lateral system is attac]Zed to manifold pipe near ��p„�,.fo,�,,.�,�,. �y,�,�� <br /> the center,lower diagram,perforated lateral length (3) and �Ew �'a�a ;s '^�`�'°��� , <br /> r..�po,o _ <br /> number of perforations per lateral (5) will be approximately one ���a,�,.,,o�,o„.o„o�„ <br /> half of that in step 8. Using these values, select minimum - \'°� ��s����� a- ���'-'"'� <br /> diameter for perforated lateral = �""" inches. ,� �� ""i"'"°"";":; <br /> lo�., b, <br /> � '^��R�l �z i. -li <br /> I�P�p'ILO <br /> `t' �`�6rH d ��. <br /> I <br /> I hereUy certify that I have completed this work in accordance with applicable ordinances, rules and laws. ' <br /> � � <br /> ��r'�. . � ,�� "'� ------- <br /> . - (signah�re) ��'1� (license#) /� -'� - %�,<'�f�J (date) <br />