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1996-008441 - new septic system
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2335 Shadowood Drive - 27-118-23-32-0019
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1996-008441 - new septic system
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Entry Properties
Last modified
8/22/2023 4:20:36 PM
Creation date
8/22/2018 12:09:50 PM
Metadata
Fields
Template:
x Address Old
House Number
2335
Street Name
Shadowood
Street Type
Drive
Address
2335 Shadowood Drive
Document Type
Septic
PIN
2711823320019
Supplemental fields
ProcessedPID
Updated
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., � P�,�,UR,B-D IBUTION SY3TEM'. � °�0 �T10N °���°D uTeau. <br /> � l. Select number of perforated laterals � TOpwN �' <br /> a a�w�aaa t«iwr- <br /> "' • waw toM w.r � « «;r�r aww <br /> 2. Select perforation spa�ing=�_feet �ow�,��, <br /> 3. Since perfora�ions shoutd not be plac�ed c��er than 1 f�to .., . . ..:... . � <br /> .��i'rr�.."t�o. <br /> the edge of the rock Tayer(see diagram),subtract 2 f�from ': ««.�.�. �" �" <br /> the rock layer 1e�ngth. , ... .,.::: .:. . .,.., <br /> �'.�.�.�. �«""' <br /> ���-2 ft._��feet <br /> TABLB OP P�iRPOItATION DI9C�IARCBS]N GPM <br /> 4. Determin�the number of spaces between perforaHons. H«d p�ra„ao�,a�„+�a,►�.� <br /> Divide the length above by perforaHon spacing and round ��� ��• <br /> down to nearest whole number. <br /> �.� 0.56 094 <br /> Length perf.spacing a 3y,� ft.+ 3 ft.��_spaces z�y a� 1� <br /> (#3) (#2) zs o.a9 �a� <br /> 5. Number of perforaHons is equal to one plus the number of . �A �� �� <br /> perforation spaces. s.o �.ab i.6s <br /> I� s aces+1 a �2 .v��.o raoe�r�a r��a�„w�,y.�. <br /> p perforat�ons per Iateral bu�e�.ot�eo��a r��r�.e��ts <br /> 6. MulNply perforations per lateral by number of laterals to <br /> get total nwnbe�r of perforations. ' <br /> � Table 2 <br /> � x.��_� perforations M • ,,�,'';w'�'°''",°;;%�b,e�.w�rt.l�`P°' <br /> �'�"� l.�ItlC11 �.s 1DC�1 Z.�11IC�1 <br /> 7. Deterniine required flow rate by muIHplying Z,S 14 ,'( -18 . 28 <br /> number of perforations by flow per perforaHon 3.0 . 13 �� 17 26 <br /> • _ . 3.3 12 16 23 <br /> 4.0 11 , lS 23 <br /> 7� x � ��gpm. S.0 10 14 22 . <br /> 8. If laterals are conn to h der pipe as shown on upper " <br /> example,select minimu r uired Iateral diameter from .,�. <br /> . table 2;enter table with foration spaang and number <br /> .of perforations per Iate l. lect minimum diameter for <br /> . .perforated Iateral� in es �,,,✓ ��. <br /> , , �/,,.• <br /> 9. If perforated lateral sys is attac to manifold pipe near <br /> the center,as in lower ple, rforated lateral length and ""���s^� <br /> nu:nber of perforations per ral will be approximately one �'"""""""` <br /> half of that in#6. Using values,select minimum ��� �` �'''" <br /> d�ameter for perforated teral m table 2 � �m�^-� <br /> perforated�ateral�+„_, inches e� ._. '" '� <br /> ' »M �� Y�M <br /> usc. 2 �� /.D. Pv� % � �i�.',�,� � � <br /> �� <br /> � <br />
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