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Geotexttle fabric
<br /> PRESSURE DISTRIBUTION SYSTEM
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<br /> 3 uarter inch erforatlons e aced�3' ,
<br /> 1. Select number of perforated laterals , ,. ;:;;;;,,..::-..;;,.;, , �.:..:. ..::.. .
<br /> • • _;;.`::i ;:;�• ,,;J•: .:�,g:•.':,pfs',o'r�c' . ,
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<br /> 2. Select perfora{ion spacing= •?�,y�-�--ft • ,.'Perf.Sizing 3�/16"-�1/4"
<br /> Perf Spacing 1.5'-5'
<br /> 3. Since perforations should not be�placed closer than 1 footto ,
<br /> the edge of the rock layer (see diagram),subtract 2 feet from
<br /> the rock layer length• E•4: Maxlmum a�owable number of 1/4•inch perforotions
<br /> per laferd fo Quarantee<IOX.discharge varfatlon
<br /> �– -2 ft = �� ft . . perioiation ,
<br /> y�r �,8 –— . . ,
<br /> _:.1. ,�aP�lnp
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<br /> 4. Determine the number of spaces between perforations. feet 1 Inch 1.251nch .1:5•inch 2.0 inch
<br /> Divide the length (3)by perforation spacing(2) and r�un� '
<br /> �q�to nearest whole number. 2.;5 � g . �14 18 28
<br /> Perforation spacing= �0 ft+�_ft= Z o spaces 3.0 8 13 1 Z 26
<br /> �3.3 7 12 16 � 25
<br /> 5, Number of perforations is equal to one plus the number.of 4� � >> 15 2s
<br /> perforation spaces(4)..Check figure E-4 to assure the number of �.0 6 l0 14 22�
<br /> perforations per lateral guarantees <10% discharge variation.
<br /> . �SpaCeS + 1 = a 1 . PerfoTatiOns/lateTal E-6: Pertorotfon Discharge fn gpm
<br /> 6. A. Total number of perforations = perforations per 1�iteral (5) --i Perforatlon dlameter
<br /> heod inches
<br /> � times number of Iaterals (1} � ��3� �r (feet) 3/16 7/32 1/4
<br /> ��p e r f s/l a t x�._lat=��perforations �,pa 0,42 0.56 0.74
<br /> B. Calculate the square footage per perforation. 2,0b 0.59 0.80 1.04
<br /> Should be 6-10 sqft/perf. Does not apply to at-grades. 5� p,q4 1 .26 1.65
<br /> Rock bed area = rock width��ockf{ngth(ft)
<br /> �ft x��t = q � a Use 1.Q toot fpr slr�le-family homes.
<br /> Square�foot per perforation=Rock bed area+number of perfs (6)
<br /> b Usa 2,0 teet for an hin else,
<br /> �sqft+�.._.perfs= `�. �_sqft/perf ,..,,,,«., �«,�,�, ,.T �� p ►NESSUiE dSTRIBU7lON rnE�
<br /> 7. Determine required flow rate by multiplying tlze �otal number of W�
<br /> perforations (6A) .by flow per perforation'(see figure E=6)
<br /> 11
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<br /> �e�perfs x -S lo�Qv.�m/.perfs= � $Pm � "•,
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<br /> 8. If laterals are connected to header.pipe as shown on upper � �,,��� ,�,a,,�
<br /> example, to selec��iiniintun ret�uired latera��diameter;enter ��„�J'`
<br /> figure E-4 with perforation spacing (2) and number of perforations \�/
<br /> per lateral (5) S.elect m;nlmurii diameter for ,.�,��;,�o„;,w�;;�„;;T�;,Na'°�
<br /> perforated lateral= inches.
<br /> ►�►•a..m w�K�+c
<br /> 9. If perforated lateral system is attached to manifold pipe near ��.�..���g,� � ��'"`"'
<br /> � �:�w K�s:�, �,�'°"
<br /> the center,lower diagram,perforated lateral length(3) an
<br /> r�ko.o
<br /> number of perforations per lateral (5)will be approximately one ��,c��,�;��w� �
<br /> half of that in step 8. Using�these v�lues;select mulimum , ''� � "'d'�'%�
<br /> diameter for perforated lateral=� �zT��es. � � ��
<br /> . . , . �a� ��+'�r. M r,r,,.."'`��. �'"'
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<br /> I hereby certify that I have completed this work in accordance with applicable ordinances, rules and laws.
<br /> ��� -�i;� ( �Sn ��=�-----�license#) �_L}--`(��`}'� (date)
<br /> si ature)
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