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2004-P08025 - new septic
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3330 Watertown Road - 32-118-23-41-0002
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2004-P08025 - new septic
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Last modified
8/22/2023 4:41:17 PM
Creation date
9/25/2019 11:10:48 AM
Metadata
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Template:
x Address Old
House Number
3330
Street Name
Watertown
Street Type
Road
Address
3330 Watertown Road
Document Type
Septic
PIN
3211823410002
Supplemental fields
ProcessedPID
Updated
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PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> 1. Select number of perforated laterals 3 .rt«Bach dons .eea.a 3• <br /> 2. Select perforation spacing L <br /> Perfng 3/16"-1/4" <br /> 3. Since perforations should not be�placed closer than 1 foot to Perf Spacing 1.51-s• <br /> the edge of the rock layer(see diagram),subtract 2 feet from <br /> the rock layer length. E-4: M=k=allowable number d 1/4-inch perforations <br /> SSper-Ideral to t N<10%.dscharpe variation <br /> >er -2 ft s. ft perfora8on <br /> .> spacing <br /> 4. Determine the number of spaces between perforations. ee i'inch .1.25" .1.binch 2.0 inch <br /> Divide the length(3)by perforation spacing(2)and <br /> dj=to nearest whole number. 2.6 8 . 14 18 28 <br /> Perforation spacing= S,'� ft+ 3. ft= spaces 3.0 8 13 11 26 <br /> .3.3 7 12 16 ' 25 <br /> 5. Number of perforations is equal toone 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 1 1422 <br /> perforations per lateral guarantees<10%discharge variation. <br /> spaces+1 s 1 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 /> Iki (feet) 3/16 . 7/32 1/4 <br /> _L(j'__perfs/lat x_j__Jat= 9� -perforations 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 sgft/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 /> 10 ft x r t= '� qft a use t.Q foot tor.�ino�e-ramuv Homes. <br /> Square,foot per perforation=Rock bed area+number of perfs(6) ° se 2•��� n i^ else• <br /> .�.-._sgft+- �pe a /0.Z •sgft/perf .Mo L=ff=AT s=a._.SLAC as" , s,MM <br /> 7. Determine 'required flow rate by multiplying the total number of <br /> perforations-(6A) .by flow per p'erforation'(see figure E=6) <br /> - —perfs x .q gpm/perfs gPm <br /> 8. If laterals-are connected f&header.pipe as shown on upper VEY <br /> example,to select minimum required lateral diameter;enter <br /> figure E-4 with perforation spacing(2)and number of perforations <br /> per lateral(5) Seleo#minimumi diameter for X,,a,,.,.,fT11N 2.,�".I=t a,, <br /> perforated lateral= inches. <br /> 9. If perforated lateral system is attached to manifold pipe near "0,the center,lower diagram,perforated lateral length(3)and " �. <br /> number of perforations per lateral(5)will be approximately one :�• y�• •� <br /> half of that in step S. Using these values,select minimum •a*rr ► <br /> diameter for perforated lateral= f �inches. we w. „► <br /> ✓ w <br /> I hereby certify that I have pleted this work in accordance with applicable ordinances, rules and laws. <br /> tura al (license#) —K:-P ' (date) <br />
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