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2003-P06605 - new septic system
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2003-P06605 - new septic system
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Entry Properties
Last modified
8/22/2023 5:13:24 PM
Creation date
10/21/2016 1:44:37 PM
Metadata
Fields
Template:
x Address Old
House Number
2530
Street Name
Fox
Street Type
Street
Address
2530 Fox St
Document Type
Septic
PIN
0411723410010
Supplemental fields
ProcessedPID
Updated
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' ' , PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> 1. Select number of perforated laterals�_ �..--.--:--�---�---� 12" <br /> � ��a..«„ �Nu�v.auviu spacea�r j' � �- <br /> 9''of,.rock <br /> 2. Select perforation spacing = 3. o ft <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 ollowable number of 1/4-inch perforotion: <br /> �� r � per lateral to guarantee<10%discharge variotion <br /> Roc a�ngth '2 ft = V�o ft <br /> perforation <br /> 4. Determine the number of spaces between perforations. spacing <br /> Divide the length (3) by perforation spacing(2) and round teet t inch 1.25 inch 1.5 inch 2.0 inch <br /> wn to nearest whole number. <br /> 2,5 8 14 18 28 <br /> Perforation spacing= (�c, ft T 3 ft=a��spaces 3,0 8 13 17 26 <br /> 5. Number of perforations is equal to one plus the number of 3'3 > 12 �6 25 <br /> perforation spaces(4). Check figure E-4 to assure the number of 4'0 � >> 15 23 <br /> perforations per lateral guarantees <10% discharge variation. 5.0 6 10 14 22 <br /> o�.� spaces + 1 =,�3 perforations/lateral E-6: Perforation 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 /> a�; perfs/lat x�_lat=��perforations (feet) 3/16 7/32 1/4 <br /> 1.Oa 0.42 0.56 OJ4 <br /> B. Calculate the square footage per perforation. <br /> Should be 6-10 sqft/perf. Does not apply to at-grades. 2•0b 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��_ft = lo�O SClft ° 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 /> Cd�v sqft=�perfs =��sqft/perf <br /> MANIFOID LOCATED AT EN� pF PpE55URE DISTRIBUTION SYSTEI <br /> 7. Determine required flow rate by multiplying the total number of <br /> perforations (6A) by flow per perforation (see figure E-6) W��„ <br /> �� <br /> .��perfs x , Sb gpm/perfs=�� m �"�°� <br /> � <br /> , <br /> 8. If laterals are connected to header pipe as shown on upper � <br /> �,,� .��;��� <br /> example, to select miniinum required lateral diameter;enter d,�.��" � <br /> figure E-4 with perforation spacing (2) and number of perforations ��`"`M <br /> per lateral (5) Select minimum diameter for <br /> PA Y }pp111� c II�YWT 0!►ERIqUTEO►III L�TERA�S!pX <br /> I.I Ol�ll� ��YI.T�l— yy��QJ, PME35URE D�STRiBUT10N W wOUHD <br /> c W[0 R/Jl�c�M( <br /> 9. If perforated lateral system is attached to manifold pipe near �,�„a,,,,,M,,,�,x• ,,,�,� <br /> the center, lower diagram,perforated lateral length (3) and °"� ��'"'"��'`R^�� • �""�°��� <br /> Yu�w�o <br /> number of perforations per lateral (5)will be approximately one Kw�;,,o,��,.,,,�,� <br /> .�.�,�.,.� <br /> half of that in step 8. Using these values, select minimum '°• d- <br /> diameter for perforated lateral = t ��z inches. ��T"""'"'� <br /> -nw cu b. Y�^�� <br /> . r/"",I�p� <br /> d <br /> K w+..�..- <br /> ` ��*M <br /> \Ci <br /> I hereby certify that I have completed this work in accordance with applicable ordinances, rules and laws. <br /> �� �h �, , <br /> �/�r /! ��`�—_ (signature) �"�� (license#) � —1�—U2 (date) <br />
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