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2002-P05639 - new septic
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3051 Farview Lane - 04-117-23-33-0009
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2002-P05639 - new septic
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Last modified
8/22/2023 5:12:37 PM
Creation date
8/4/2016 3:21:27 PM
Metadata
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Template:
x Address Old
House Number
3051
Street Name
Farview
Street Type
Lane
Address
3051 Farview La
Document Type
Septic
PIN
0411723330009
Supplemental fields
ProcessedPID
Updated
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f <br /> � ' <br /> ' ' PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> 1. Select number of perforated laterals�_ uarter inch erforations s aced�3� 12�� <br /> 9"of rock <br /> 2. Select perforation spacing = 3.c� 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-a: Maximum aliowable number of 1/4-inch perforations <br /> �.,� J per lateral to guarantee<10%discharge variation <br /> Roc ayer eng �2 ft —�Ft perioration <br /> spacing <br /> 4. Detennine the number of spaces between perforations. <br /> Divide the length (3)by perforation spacing (2) and r un }eet 1 inch 1.25 inch 1.S inch 2.0 inch <br /> down to nearest whole number. <br /> 2,5 8 14 18 28 <br /> Perforation spacing= ' 3 ft= � ft= a� 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 �5 23 <br /> perforations per lateral guarantees <10% discharge variation. 5,0 6 10 14 22 <br /> �`-_spaces + 1 =��perforations/lateral E-6: Perforotion Dischorge in gpm <br /> 6. A. Total number of perforarions = perforations per lateral (5) perforation diameter <br /> times number of laterals (1) head inches <br /> (feet) 3/16 7/32 1/4 <br /> � � perfs/lat x�_lat= /0 0 perforations 1.0° 0,42 0.56 0.74 <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 /> a�� ft X � `� ft = �J���' sqft ° 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 on n�� else. <br /> 1 J� � sqft+ !✓ perfs = 1`. (.� sqft/perf M4NIfOlO LOCATED AT ENO OF PRESSURE DISTRIBUTION SYSTEM <br /> 7. Determule required flow rate by multiplying the total number of <br /> perforations (6A) by flow per perforation (see figure E-6) wn;� <br /> � )�,, <br /> .�.�„a., <br /> ��v perfs x ..S l� gpm/perfs =�gpm <br /> 8. If laterals are connected to header pipe as shown on upper �'1 <br /> o�^" �:;.�'�n. <br /> example, to select minimum required lateral diameter;enter ,.�M°"'`° <br /> figure E-4 with perforation spacing (2) and number of perforations ��/``�'� <br /> per lateral (5) Select minimum diameter for ,,.o�,pF►ERlOMTEO VIPE L<TEPPLS�A <br /> PR[SSWE DISTPiBUT10N w MOUHO <br /> perforated lateral = in es. <br /> /[IIIOYT[0 RbTIC I�K <br /> 9. If perforated lateral system is attached to manifold pipe near �,�„�,.,,��[,,Q.�,,,• y-�� <br /> V� OPS E•WY�K Lnr hi�� �.A/l�K��T10'� <br /> the center,lower diagram,perforated lateral length(3) and f„wpao <br /> number of perforations per lateral (5)will be approximately one K�.�;=��,«,�bT.�.a <br /> half of that in step 8. Using these values, select minimum '°� ''- <br /> •-����•:s����o: <br /> diameter for perforated lateral = "a.�J inches. _�„,�., b� <br /> ��,���� <br /> ,�,,,�•,�� ;;�� <br /> \ ��,�m <br /> a ^ <br /> �� <br /> I hereby certify that I have com�leted this work in accordance with applicable ordinances, rules and laws. <br /> � � �• --'r.,--�/ � (signature) ���r��' �"� (license#) d - �� - n�-- (date) <br />
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