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2009-00270 - septic mound system
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2009-00270 - septic mound system
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Last modified
8/22/2023 5:15:14 PM
Creation date
1/12/2017 1:07:39 PM
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x Address Old
House Number
3285
Street Name
Graham Hill
Street Type
Road
Address
3285 Graham Hill Rd
Document Type
Septic
PIN
0511723110006
Supplemental fields
ProcessedPID
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University of Minnesota Pressure Distribution System Design -10/25/04 <br /> AN boxed 2ctengfes mast be entered,(fie iest will be cakwfated. <br /> ONSIR <br /> .SlWA6! ��� -- <br /> 1. Setect number of perforated taterals: 03 T��•T""�"'T - <br /> PROORAM G'�„�~`" <br /> 2. Select perforation spacing= 03 ft <br /> .-a.t«xtih.f+hrr.- <br /> ��.�. <br /> 3. Since perforations should not be placed closer that 1 foot to ��� ,n�q t <br /> the edge of the rock layer(see diagram),subtract 2 feet from ,.��,�x.k <br /> the rock layer len th <br /> 63 -2ft= 61 ft ��.....�;..,,.,.��,-_��a- <br /> r...r�r.;,,.,.,K i.s-s <br /> 4. Detertnine the number of spaces between perforations. <br /> Divide the length(3)by perforation spacing(2)and round down to nearest whole number. <br /> Perforation spacing= 61 ft! 3 ft= 20 <br /> 5. Select perforation size 1/4 inch <br /> 6. Number of perforations is equaf to one plus the number of perforation spaces(4). <br /> 'Check frgure E-4 to assure the number of pe►fnrations per latera/guaranfees <br /> <?0%discharge varia6on. <br /> 20 spaces+1= 21 perforations/lateral <br /> E-4 Maximum Number of 1/4 inch perforations E-5 Maximum Number of 3116 inch pertorations <br /> r lateral to uarantee<10%dischar e variation r lateral to uarantee<10%dischar e variation <br /> Perforation Perforation <br /> Spacing Pipe Diameter Spacing Pipe Diameter <br /> ft 1 inch 1.25 inch 1.5 inch 2.0 inch feet 1 inch 1.25 inch 1.5 inch 2.0 inch <br /> 2.5 8 14 18 28 2.5 12 19 25 39 <br /> 3.0 8 13 17 26 3 11 18 24 37 <br /> 3.3 7 12 16 25 3.3 10 17 23 36 <br /> 4.0 7 11 15 23 4 10 16 21 33 <br /> 5.0 6 10 14 22 5 9 15 20 31 <br /> 7. A.Total number of perforations=perforations per lateral(5)times number of laterats(1). <br /> 21 perfs/lat x 3 laterals= 63 perforations <br /> B.Calculate the square footage per pertoration. <br /> Recommended value is&10 sqft/perf.Does not apply to at-grades. <br /> 1. Rock bed area=rock width(ft)x rock length(ft) <br /> 10 ft x 63 ft= 630 ft <br /> 2. Square foot per perforation=Rock Bed Area/number of perfs(6) <br /> 630.0 ft� / 63 perfs = 10.0 ft�/perf <br /> 8. Determine required flow rate by muftiplying the total number <br /> of perforations{6A)by flow per perForations see figure E-6) <br /> 63 perfs x 0.74 gpm/perts= 46.6 gpm <br /> E�?erforation Dischar e in GPM <br /> Head Pertorations diameter <br /> feet inches <br /> 3116 7/32 1!4 <br /> 1 0.42 0.56 0.74 <br /> 2° 0.59 0.80 1.04 <br /> 5 0.94 1.26 1.65 <br /> a. Use 1 A foot fw single-famiy homes. --- - -. __. _ <br /> b.Use 2.0 feet for anythi e�se , __ - , '"„` <br /> _"_- r.��• <br /> 9. Determine Minimum Pipe Size �" - <br /> A. Manifold on End. If laterals are connected to header pipe ,, <br /> as shown in Figure E-1,to select minimum required lateral Fwu.E-,:M��a��e�d«s,..� <br /> diameter;enter figure E-4 or E-5 with perForation spacing and <br /> ___ __ __- - _- - -- -' <br /> number of perforations per lateral.Select minimum diameter <br /> for perforated laterals= 2.0 inches <br /> B. Center Manifold. If perforated lateral system is attached to ��»��� ;_-= ���� <br /> manifold pipe near the center,like Figure E-2,perforated lateral length(3) - ;; _--_ `� <br /> and number of perforations per lateral(5)will be approximately F', � = - <br /> one half of that in step A. Using these values,select � " � <br /> minimum diameter for perforated lateral= 1.5 inches • `�_ r`� <br /> I hereby cert� t I have completed this work in accordance with all applicable ordinances,rules and laws. <br /> (signature) 810 (license#) 02/08l09 (date) <br />
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