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2009-00641 - new mound septic
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2250 French Lake Road - 10-117-23-22-0012
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2009-00641 - new mound septic
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Last modified
8/22/2023 3:20:59 PM
Creation date
12/6/2016 2:05:30 PM
Metadata
Fields
Template:
x Address Old
House Number
2250
Street Name
French Lake
Street Type
Road
Address
2250 French Lake Road
Document Type
Septic
PIN
1011723220012
Supplemental fields
ProcessedPID
Updated
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� � <br /> PRESSURE DISTRIBUTION SYST�M <br /> Geotextile fabric <br /> 1. Select number of perforated laterals 3_ , <br /> Quarler inch perforaHons s aced�3' 12 <br /> 2. Select perforation spacing -�_ft . 9"of rock <br /> 3. Su�ce perforations should not be placed closer than 1 foot to Perf Sizing 3/16��- 1/y�� <br /> the edge of the rock layer (see diagram),subtract 2 feet from Perf Spacu-tg 1.5'-5' <br /> the rock layer lengtli. <br /> E-4: Maximum allowable number of 1/d-inch perforations <br /> — �� -2 ft �..�� per laferal to guarantee<10%discharye variation <br /> Rock layer IengUi -- �� (; f(� <br /> perloration <br /> 4. Determine the number of spaces between perforations. SPacing <br /> Divide the length (3)by perforation spacing (2) and round �Qe} 1 inch L25 inch 1.5 inch 2,0 inch <br /> do�vn to nearest whole number. <br /> Perforation spacing = �-)�r it- -� 2�5 8 14 18 28 <br /> v ft= ) �� spaces 3.0 g 13 �� <br /> 5. Number of perforations is equal to one plus ille number of 26 <br /> 3.3 7 12 ib 25 <br /> perforation spaces(4). Gieck figure E-4 to assure the nu�liber of 4'0 > >> 15 23 <br /> j�e�foratio�is per-lateral gitnra�ltees <10% discharge variatio�i. 5�0 6 10 �q 22 <br /> !! s aces + 1 = <br /> P �___'Z___perlorations/latera] <br /> E-6: Perforation Discharg� in gpm <br /> 6. A. Total numUer of perforations = perforations per lateral (5) '—' <br /> times number of laterals (1) perforation diameter <br /> head inches <br /> I � perfs/lat x U lat= S/ perforations (feet) 3/16 7/32 1/4 <br /> B. Calculate the square footage per perforation. �'�� �•42 0.56 �0.74�, <br /> Should be 6-10 sqft/perf. Does not npply to at-grndes. 2�0b 0.59 0.80 1.04 <br /> Rock bed area = rock width (ft) x rock length (ft) � <br /> �ft x�c�ft = !;<��,� sqft <br /> ��0 0.94 1.26 1.65 <br /> Square foot per perforation = Rock bed area +number of perfs (6) b Use 2A feet for n9lhina e�i5ehomes. • <br /> r�,-,,. , <br /> sqEt=_ �,-/ perfs = �.4� sqEt/perf <br /> 7. Determule required flow rate by multi 1 in M�"�F�`° `°`"TE° °' `"° � p�E55ME DISTRIBUTION nSrEM <br /> perforations (6A) by flow per perfora on (ee figu�e E-6�lmUer of <br /> �_perfs x ,�_g�m�perls = 3�; m . <br /> gP �. <br /> 8. If laterals are connected to header pipe as shown on upper ` <br /> example, to select i�unirrium required lateral diameter; enter d��w„���`"`�"'`� <br /> i�:.�;�..;.,. <br /> figure E-4 with perforation spacing (2) and number of perlorations <br /> per laieral (5) Select miniinum diameter for �� �M <br /> perforated lateral= 2,,c� ir��les. �.o�. o, YEn�DW.T[p„�E w.,E„A�,�a <br /> PACSSUnE OiSiHia�l�ON uou D <br /> 9. If perforated lateral system is attached to manifold i e near � �rc <br /> � p �,�.,,�,:�, ,,,�.� <br /> the center, lower diagram,perforaied lateral length (3) aiid <br /> ICw oqt�.:'f e['L,",',4,°'!'w \ �Nwon.rw� <br /> number of perforations per lateral (5) wil] be approximately one .,. �- <br /> half oF that in step 8. Using these values, select minimum K �`�_"`"" � [ -�" <br /> /`Y �. �_� �� <br /> diameter for perforated lateral = ,-�--�-- inches. � �`""�a�.,K•:;,�,;;�;, <br /> b ��rrN���= .. <br /> rro"u[� <br /> d� <br /> �``w�M ...,.. „ <br /> I her by certify that I have comp]eted this work in accordance with applicab]e ordinances, rules and laws. <br /> r � <br /> �__-- (�f' � � : -- '�c� ( <br /> — (signaeure) .� i � (license#) _`I —)j—r�� <br /> (date) <br />
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