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2008 - 00069 - new septic
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Willow Drive North
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1020 Willow Dr N - 27-118-23-32-0022
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2008 - 00069 - new septic
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Last modified
8/22/2023 4:20:38 PM
Creation date
2/12/2020 9:03:15 AM
Metadata
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Template:
x Address Old
House Number
1020
Street Name
Willow
Street Type
Drive
Street Direction
North
Address
1020 Willow Drive North
Document Type
Septic
PIN
2711823320022
Supplemental fields
ProcessedPID
Updated
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PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> 1. Select number of perforated laterals Quarter inch forations s'aced @ 3' <br /> 12" <br /> .er <br /> 9"of rock <br /> 2. Select perforation spacing = 3ft <br /> Perf Sizing 3/16" 1/9" _ <br /> 3. Since perforations should not be placed closer than 1 foot to Perf Spacing 1.5'- /3 <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 perforations <br /> //It per lateral to guarantee<10%discharge variation <br /> Rock layer length -2 ft = ft perforation <br /> spacing <br /> 4. Determine the number of spaces between perforations. (feet) 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. 2.5 8 1428 <br /> Perforation spacing = 3? ft 4,3 ft=4....9 spaces (3.0 8 13 .17 26 <br /> 3.3 7 12 16 25 <br /> 5. Number of perforations is equal to one plus the number of 4.0 7 11 15 23 <br /> perforation spaces(4). Check figure E-4 to assure the number of 5.0 6 10 14 22 <br /> perforations per lateral guarantees <10% discharge variation. <br /> /3 spaces + 1 = /4Z perforations/lateral E-6: Perforation Discharge in gpm <br /> 6. A. Total number of perforations = perforations per lateral (5) perforation di r <br /> times number of laterals (1) ' / head (inches <br /> //. perfs/lat x lat=`X-,2 3/16 7/3 1/4 J <br /> perforations <br /> 1.0° 0.42 0.56 0.7 <br /> B. Calculate the square footage per perforation. 2.06 0.59 0.80 1,04 <br /> Recommeded value is 6-10 sqft/perf. Does not apply to at-grades. <br /> Rock bed area = rock width (ft) rock length (ft) 5.0 0.94 1.26 1.65 <br /> / ft x C/• ft= '4247c) sqft a 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 anything else. <br /> /c sqft- ."-76: perfs = 7. 731 sqft/perf <br /> 7. Determine required flow rate by multiplying the total number of <br /> perforations (6A) by flow per perforation(see figure E-6) <br /> mantyci pt. <br /> ff pipe PIlan pump <br /> `7 perfs x0.1,‘gpm/perfs =o�O gpmend cop <br /> 8. If laterals are connected to header pipe as shown on upper Vii' atl x,Ia IOCPIke <br /> of Phpe Iron Pump <br /> example, to select minimum required lateral diameter;enter Figure E-1:Manifold Located at End of System <br /> figure E-4 with perforation spacing (2) and number of perforations - - <br /> per lateral (5) Select minimum diameter for <br /> perforated lateral = inches. <br /> Figure E-2 Manifold Located «r+o w <br /> 9. If perforated lateral system is attached to manifold pipe near in the Center of the System gyp <br /> the center, lower diagram, perforated lateral length (3) and =rnanlidd pip.%� �' <br /> - number of perforations per lateral (5) will be approximately one / ; <br /> half of that in step 8. Using these yew,select minimum ���- a ./ e tn�„,,, <br /> diameter for perforated lateral = /2inches. � pie non',imp <br /> /Sa� <br /> 1 h> certify that I have completed this work in accordance with applicable ordinances, rules and laws. <br /> �- 5� � � (signature) � �/l s, <br /> (license#) /�vy (date) <br />
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