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2006-P09892 - new septic
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2006-P09892 - new septic
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Entry Properties
Last modified
8/22/2023 4:29:29 PM
Creation date
12/3/2019 11:31:03 AM
Metadata
Fields
Template:
x Address Old
House Number
4100
Street Name
Watertown
Street Type
Road
Address
4100 Watertown Road
Document Type
Septic
PIN
3111823140011
Supplemental fields
ProcessedPID
Updated
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' PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> .,•`r:. ..►ariiiO-+'rs. .VC.0.4.•Orf. .• .e-?T. C. "..•. <br /> 1. Select number of perforated laterals Quarter inch perforation<_paced 0 3' f 12 <br /> 9 of rock <br /> 2. Select perforation spacing= •.'F -' 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'-5' <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 /> G per lateral to guarantee<10%discharge variation <br /> Rock layer length -2 ft - (,'L l 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 14 18 28 <br /> Perforation spacing= `- .. ) ft+ " ' ft= , `` 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 <br /> 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 /> '. spaces + 1 = .(.4 / 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 /> (feet) 3/16 7/32 1/4 <br /> Ts 1 perfs/lat x lat= ,:. perforations <br /> 1.00 0.42 0.56 0.74 <br /> B. Calculate the square footage per perforation. 2.Ob 0.59 0.80 1.04 <br /> Should be 6-10 sqft/perf. Does not apply to at-grades. <br /> Rock bed area = rock width (ft) x rock length (ft) 5.0 0.94 1.26 1.65 <br /> /o ft x V% ft= f,.' ' 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 /> f,: sqft+ perfs = /,',// sqft/perf MANIFOLD LOCATED 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) <br /> >'�_ perfs x .121-1 gpm!/perfs= Lis gpm <br /> 8. If laterals are connected to header pipe as shown on upper ,,,,� W:; {, <br /> example,to select minimum required lateral diameter;enter ,- M,KF'DP' <br /> figure E-4 with perforation spacing (2) and number of perforations \, '- <br /> 'so <br /> per lateral (5) Select minimum diameter for <br /> LAYOUT OF PERFORATED PIPE LATERALS FOR <br /> perforated lateral= inches. PRESSURE DISTRIBUTION w MOUND <br /> r{PEMOMTEO PLASTIC PIPE <br /> 9. If perforated lateral system is attached to manifold pipe near `�PE�(Pa,,IBN1Ey�E.p.���Bx „a,y-°"` <br /> VI w ol1 K �K+'w�Ni K� <br /> the center,lower diagram,perforated lateral length(3) and ,...a,.... <br /> number of perforations per lateral (5)will be approximately one PIntSTrPiP�.RTTAM <br /> half of that in step 8. Using these values, select minimum1:.,. 0 ---- ..,o�, , <br /> diameter for perforated lateral= I /)^ -- inches. A TAP <br /> 10, <br /> yATEO/0°' <br /> rie <br /> NAVY , <br /> I hereby certify that I have completed this work in accordance with applicable ordinances, rules and laws. <br /> :, <br /> &- - ,' (signature) '' ``I (license#) (o -).S�-0 to (date) <br />
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