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2012-01197 - septic mound repair
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2012-01197 - septic mound repair
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Last modified
8/22/2023 5:05:58 PM
Creation date
4/29/2016 3:58:44 PM
Metadata
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Template:
x Address Old
House Number
2590
Street Name
Countryside
Street Type
Drive
Address
2590 Countryside Dr
Document Type
Septic
PIN
0411723110010
Supplemental fields
ProcessedPID
Updated
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' • , PRESSUIZE DISTRIBUTION SYSTEM Geotextile fabric <br /> 1. Select number of perforated laterals 3 _ 12�� <br /> 2. Select perforation spacing = 3�t� ft of ro�k <br /> Perf Sizing 3/16"-1/4" <br /> 3. Since perforations should not be placed closer than 1 foot to Perf Spacing 1.5'-s� <br /> the edge of the rock layer (see diagram),subtract 2 feet from <br /> the rock layer length. E-4: Maximum allowable number of 1/4-inch pertorcrfions <br /> � �`, per lateral to guarantee<10�o discharge variation <br /> r� <br /> Rock a- Yer� -Z ft - t6' ft perforotion <br /> spacing <br /> 4. Determine the number of spaces between perforations. teet 1 inch 1.25 inch 1.5 inch 2.0 inch <br /> Divide the length (3) by perforation spacing (2) and round <br /> down to nearest whole number. <br /> 2.5 8 14 18 28 <br /> Perforation spacing=�ft=�ft= '�� spaces 3.0 8 13 17 26 <br /> 5. Number of perforations is equal to one plus the number of 3.3 7 12 16 25 <br /> perforation spaces(4). Check figure E-4 to assure the number of 4.0 7 11 15 23 <br /> perforations per lateral guarantees <10% discharge variation. 5.0 6 10 14 22 <br /> �,l-? spaces + 1 = � "�� perforahOns/lateral E-6: Perforotion Discharge in gpm <br /> 6. A. Total number of perforations = perforations per lateral (5) perforation diameter <br /> times number of laterals (1) head inches <br /> (feet) 3/16 7/32 1/4 <br /> ���`� perfs/latx 3 lat= ���'/ perforations <br /> 1.0a 0.42 0.56 0.7a <br /> B. Calculate the square footage per perforation. b <br /> Should be 6-10 sqft/perf. Does not apply to at-grades. 2•0 0.59 0.80 1.04 <br /> Rock bed area = rock width (ft) x rock length (ft) 5.0 0.94 1.26 1.65 <br /> �ft X (O?� ft = �.�41 U SC�ft ° Use 1.0 foot for single-family homes. <br /> Square foot per perforation = Rock bed area =number of perfs (6) b Use 2.0 feeT for an n�� else. <br /> �/''(!�_7 S(]1L T `''��� perfs = �� sqft/perf MnNIFOLD �OCATED AT END Oi PHESSURE DISTRIBUTION SYSTEM <br /> i <br /> 7. Deternune required flow rate by multiplying the total number of <br /> perforations (6A) by flow per perforation(see figure E-6) w�;� <br /> '���✓ <br /> C�`� perfs x ��g-pm/perfs = .�� gpm �� <br /> 8. If laterals are connected to header i e as shown on u er �':�1� <br /> P P PP �,,,F"' �,,.,��„d�.� <br /> example, to select minimum required lateral diameter;enter M,K�°" <br /> figure E-4 with perforation spacing (2) and number of perforations ��``� <br /> per lateral (5) Select minimum diameter for <br /> LGTOUT OF PERfORATEO P�iE LnTEH<LS�ON <br /> perforated lateral � yy��QS. YRESSURE Oi5TA1luT10N W MOUNO <br /> c w.rco n.snc nn <br /> 9. If perforated lateral system is attached to manifold pipe near �,�.,,�..,,,�M y.�„�• y�� <br /> ENO D^�/�wi�x 4��ory;ior� /` ^K�pMIN� <br /> the center,lower diagram,perforated lateral length (3) and °'Ew <br /> r.�w�ao <br /> c <br /> number of perforations per lateral (5)will be approximately one K.�;�;:;�,�;�b«�.o. <br /> .. <br /> half of that in step 8. Using these values, select minimum '°' °' - <br /> �p�I�n�PC�iE1+o���PMv1 <br /> diameter for perforated lateral = inches. �„�„ b. <br /> � - � ��rto��rt�,� �rr..� _p <br /> d� �� <br /> ��`*M <br /> I hereby certify that I have c mpleted this work in accordance with applicable ordinances, rules and laws. <br /> r` / <br /> /� f�[ <br /> f <br /> -�� �J`-' -' >��-'F '~��----- (signahzre) �G'/ �� (license#) /f' �1 c� �lZE (date) <br />
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