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septic info including older design
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1940 Sixth Ave N - 27-118-23-42-0003
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septic info including older design
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Last modified
8/22/2023 4:21:57 PM
Creation date
1/16/2019 12:36:59 PM
Metadata
Fields
Template:
x Address Old
House Number
1940
Street Name
6th
Street Type
Avenue
Street Direction
North
Address
1940 6th Avenue North
Document Type
Septic
PIN
2711823420003
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ProcessedPID
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" . . PRESSURE DISTRIBLJTION SYSTEM Geotextile fabric <br /> 1. Select number of perforated laterals�� uarter inch erforations s aced�?3' 12 <br /> 2. Select perforation spacing=��ft 9"of rock <br /> Perf Sizing 3/16"-1/4" <br /> 3. Since perforations should not be placed closer than 1 foot to Perf Spacing 1.5'-5' <br /> the edge of the rock layer(see diagram),subtract 2 feet from <br /> the rock layer length. E-4: Maximum allowabie number of 1/4-inch perforations <br /> per laterd to guarantee<10�discharge variation <br /> R«��el,g�, -2 ft = ��ft ertoration <br /> P <br /> 4. Determine the number of spaces between perforations. spacing <br /> Divide the length (3)by perforation spacing (2) and ro u1 feet 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= •�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 �2 16 25 <br /> perforation spaces(4). Check figure E-4 to nssure the nu��tber of 4.0 7 11 15 23 <br /> perforations per laterc�l gzcnrantees <10% disclic�rge variation. 5.0 6 10 14 22 <br /> ��spaces + 1 =�Z perforations/lateral E-6: Perfo�ation 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 <br /> �7 perfs/lat x��lat= -�/ perforations a <br /> 1.0 0.18 0.42 0.56 0.74 <br /> B. Calculate the square footage per perforation. b <br /> Should Ue 6-10 sqft/perf. Does not apply to at-grades. 2•0 0.26 0.59 0.80 1.04 <br /> Rock bed area = rock width(ft) x rock length(ft) 5.0 0.41 0.94 1.26 1.65 <br /> �_ft x So • ft= �s'oo SCift ° 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 hin ei5e. <br /> �"�(Zsqft=��perfs= Q.�.3 sqft/perf M4NIFOl.D IOCATED AT END OF PRESSURE DISTRIBUTION SYSTEM <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 ,7 m/perfs =�gpm <br /> 8. If laterals are connected to header pipe as shown on upper � � <br /> `'�[�µ urc o�°a a.c <br /> W i�d:i�•" <br /> example, to select minimum required lateral diameter;enter d�°"'` <br /> figure E-4 with perforation spacing(2) and number of perforations ��``�M <br /> per lateral(5) Select m;n;mum diameter for <br /> 1Y� L4YOUT Of PEFiORnTED vW[LGTEP�lS NN <br /> perforated lateral yy����. PAESSURE O�SiAIBUf10N W MOulq <br /> vc�so�•rEo nes*ic nv[ <br /> 9. If perforated lateral system is attached to manifold pipe near �,E..,�.,,,��,x• ,,,,,.�^" <br /> the center, lower diagram,perforated lateral length(3) and �w �^`^'"'°`� � "r�„�, <br /> P�K <br /> number of perforations per lateral(5)will be approximately one ��;w�;;�;��.�..�a <br /> half of that in step 8. Using these values, select minimum '°• °" ..,a,�,�.�, <br /> diameter for perforated lateral= inches. �\b ' � <br /> �„�� <br /> rt�'if0 Yart sror <br /> a�..���,.�a <br /> � ��M 0� <br /> /l[ <br /> I hereby certify that I have completed this work in accordance with applicable ordinances, rules and laws. <br /> � (signature) �9�3' (license#) -Z-O (date) <br />
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