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. . . �� ��c��.�..� O <br /> i �u nX�, i�'t, �... <br /> DISTRIBUTION SYSTEM � �°��� <br /> " PRESSURfi , . .., . :.� <br /> : - ' � ��-`o c� ',p'�g4�;?D'°A,4°�:. :��;'?c?ci��c_i'i;.. '%y� _, . _ <br /> L�.Oe Gi dv o c�.. a� 0 5-::. : . �2:�... <br /> 1 '�j� ��7��7�+�''Of�OM�I1t@Y�S Inch tions a oed�3' . ,. .� <br /> t'� ar�' � ' <br /> i. n".���� �:�O;:�A,; paPOe'l?Q. b9Q¢�x�'4 N417G'��L.~f O.;.u4.�. ,;' . <br /> � VA'sv?4��Y� 'O• � o � �I. <br /> _� . � `r. �Y(s�f►ovev:•b,�,'. A v�cQ� 30�� . _ <br /> � o�A e s • ..e J 4 a. n <br /> �n r'ac�w�'.cy. a;_,��o"��'P,�,n�ly�lj:�� �`J??p•�4 'n._o,.' <br /> ,2. $P�@Ct�OT81'IOII S�A $ Perf Sizing 3/I6"-1/4" <br /> f Perf Spadng 1.5'-5' <br /> 3. Since perforations should not be placed closer than 1 foot#o <br /> the edge of the rock layer(see diagram),subtract 2 feet from E�e: Mmdr�m apowable mar�be►ot U4-�ch perrom�oc� <br /> the rock layer length. ��b�<tox dischar�e v�iadon <br /> � -2 ft =J�ft c,erfo►��on <br /> � <br /> 4. Deterpnine fihe number of spaces between perforations. �� y,�q� 1.5(nch 2.0 h�ct! <br /> Divide the iength(3)bY Perforation spaci,ng(2)and�und <br /> d�wn to nearest whole number. 8 � 14 28 <br /> Perforation spacing= �3 ft+ `�-3 ft=��� � 8 13 17 26 <br /> �,3 7 )2 16 25 <br /> 5. Number of perforations is egual to one plus the number of 4A 7 11 15 23 <br /> per£orataon space.s(4). Check figure E-4 to assure the nscmber of �0 6 tU �4 � <br /> :- perfnrations per Iaterai guurantee.s<2096 discharge variation. . — <br /> . `. .,*�. - �•�. spaces-r 1 =___�_P�orations/lateral f-6: Partorallon DUcharye h►�m . <br /> 6. A. Total number of perforations= perfarations per la.teral(5� perforutlon diameter <br /> times umber of laterals(1) heod . <br /> (feet 3/16 7/ 2 1/4 - <br /> � �perforations �T�� <br /> / perfs/lat x�_lat= 1.0� � 0.42 0.56 ��•'15.�4 : <br /> . . B. .Caleulate the square foatage PeT Perforation. 2,Ob p,59 0.80 1.G4 <br /> ---- -- R�commeded.value is 6-ZO sqft/pei'f.�t oes not apply to at-grar�s. <br /> �-- Rock bed.area= rock width(ft)x rock length(ft) 5.0 0.94 1.26 1.65 <br /> _.._ �'o ` ft X�_�_ �.�;� sqff a use t.o root t«�amw�• <br /> .. Square#oot per perforation=Rock bed area+number of perfs(b) ''u�s.o reer r�or �. <br /> � sh.5"�- sqft+�P�-1l�..--Sqft�P� <br /> 7. 'Determine required flow rate by multiplying the total number of <br /> perforations(5A) by flow per perforation(see figure E-6) �� P,�• � <br /> p'°.°°""""'° � <br /> � � � � <br /> ��f -;���:,� m/perfs= SPm «b� . <br /> . <br /> 8. If lat aLs�e c������'��� ,���Oi�""n upper ��, i <br /> ._ _ . example,to select minimum required Iateral diameter;enter �.�.►:�aa w�a.a a dw a sr� ' <br /> figure fi-4 with perforation spacing t2)and number of perforations <br /> per..la,teral.(5) Select minimum diameter for �- <br /> - per#orated latera�i=�._..�inches. . <br /> 9, If perforated Iateral system is attached to man�fold pipe near �g�«"a'��°� «�� <br /> �'°•�*'"' the�center,lower diagram,perforated lateral length(3}and ,^°^'�°°"• <br /> _?`:;�'� . numb�,of perforations per Iateral(5}will be approximately one . <br /> half of tliat in step 8. Using these va�ues,SP.�eCt minim�7(t ,�{'p n �� <br /> '`'�. dis,m�eter for perfor�ted latera1= '��- inches. ,S� .�, °".�","..� <br /> .. .. . C�c� ` ` <br /> .�-,' �-^'�'j <br /> � ' rules and laws. <br /> ,��� Z e� ' that I haye pleted this work in accordance with applicable ordinanoes, . <br /> �--/'���'�i� (signature) / � � Z (license#> � 2 G {u (date? <br />