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1995-006820 - new septic system
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2320 Shadowood Drive - 27-118-23-32-0014
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1995-006820 - new septic system
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Entry Properties
Last modified
8/22/2023 4:20:13 PM
Creation date
8/22/2018 11:10:41 AM
Metadata
Fields
Template:
x Address Old
House Number
2320
Street Name
Shadowood
Street Type
Drive
Address
2320 Shadowood Drive
Document Type
Septic
PIN
2711823320014
Supplemental fields
ProcessedPID
Updated
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J <br /> ENO PERFORATION OF A PERFORATED LATERAL <br /> puFccTTUF nrc�-rzTBU'I'ION SYSTEM <br /> G�ea Covv <br /> 1. Select number of perforated laterals , T°°"° <br /> ��-•.� ... . .. �.. �`.•��l.cyer of Gwteatlle FaWlc (or ta. <br /> ''� LA�'�+b L�Yr .l�ch layr of hay or strar co.erd <br /> 2. Select perforation spacing=�_feet. : .."'�""° ,°`'°°°°'•, <br /> er P�rfora�bn Orllt�d Hwlia�rollr <br /> � • Inlo CaD Neor Top <br /> 3. Since orations should not be laced closer than 1 ft. to ''�P'� '"�'' "' ``°" 12•'° E°°. <br /> � P .�poln Fkld Rod2 . ot Rxk l.arer <br /> � the edpe of the rock laVer (see diap�am),subtract 2 ft. from � �' � �� ' ������«�:'�+�a a� <br /> b J o_ Bonom of Laterol <br /> C{M4$Of14 LOyN <br /> the rock layer length. . <br /> . . SO Orlahol Soll Prea*Ir scorirird <br /> �li,.��,�,-�2 ft. _�feet. a.��..P�«�����a,,. <br /> 4. Detesmine the number of spaces between perforations. �'W��P�'�1O^_�,�.�.`_.,�p,`. <br /> YI��OII/P[1 Q1 YIY1L(�p�/ <br /> _ Divide the length above by perforation spacing and round �.� �,,,�, ,��� <br /> down to nearest whole number. <br /> 1.Oa 0.56 0.74 <br /> Length perf. spacing =�ft.-�- 3 ft. _�_spaces 2.Ob 0.80 1. <br /> �3� �2� a.Use for single family homes <br /> 5. Number of perforations is equal to one plus the number of b.u�ror au«hQ a�i;�ho� <br /> perforation spaces . <br /> Maxirnum number of quarter inch perforations pe <br /> �spaces +1 =�perforations/lateral lateral to guarnantee<10�'o discharge variation <br /> ' ' ' PerfoiaHon ' <br /> 6. Multiply perforations per lateral by number of laterals to 5�'�g 1� l� 2 <br /> get total numbes of perforations. 2,5 14 1 g 2.8 <br /> ,,—� x �/,�„Q,,= -S� perforations. 3.0 13 17 26 <br /> 3.3 12 16 25 <br /> 7. Determine required flow rate by multiplying 4.� 11 15 23 <br /> number of perforations by flow per perforation 5.Q 1 Q 14 22 <br /> � X �/�= 38 gpm. <br /> YYMVLD IOGRD LT C�O O�PRLZ3M d5TR19uiqN Sr57[y <br /> � <br /> 8. If laterals are connected to header pipe as shown on upper �� <br /> example, to select minimum required latesal diameter;enter �'��' <br /> •r'� s�=s^* <br /> table with perforation spacing and number of perforations �,,,.-�' <br /> per lateral. Select minimum d.iameter for �� <br /> perforated lateral =�inches. <br /> �.a„a K,.a,..�..�,.n..��. <br /> w�(1L/�[6flw��li0�w�[ahO <br /> - wrill�AY�K/R <br /> �w�sr�r vr��v' �1/�� <br /> 9. If perforated lateral system is attached to manifold pipe near =L =�� � '�:.�, <br /> the center,lower diagram,perforated lateral length and --r'r,-�:;�-,�-�-- <br /> number of perforations per lateral will be approximately one `j ` -�.�ar�,r <br /> h a l f o f t h a t i n s t e p 8. U si n g these values,select minimu.m l �. ��,,,,,�� � - <br /> diameter for perforated lateral= inches. � ,f�''" <br /> �,`•� <br />
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