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1992-004655 - mound system
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0315 Tonkawa Road - 06-117-23-14-0021
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1992-004655 - mound system
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Last modified
8/22/2023 3:14:47 PM
Creation date
5/13/2019 1:47:28 PM
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x Address Old
Address
0315 Tonkawa Rd
Document Type
Septic
PIN
0611723140021
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/' �' ,�'�F•�t✓JFiv yo/�r F.S l B 6.v.�F rT �6JioF.�-c F� <br /> QFQ�D D/>l,� CE�O�s�GN FG.e 7 BFoeoon.s�E-19 <br /> MOUND DESIGN PROCEDURE <br /> (For Flows up to 1200 gpd) <br /> A. Sewage Flow Rate F. Pressure Distribution System <br /> See D-7 or I-3, 4, or 5, or use <br /> metered value; Flow Rate = 1• Select number of perforated <br /> /0570 gpd laterals _ <br /> 2. Select perforation spacing <br /> B. Septic Tank Liquid Volume = 3 ft <br /> (see C-3 or C-5) gallons As aer09E <br /> 3. Select perforated lateral <br /> C. Soil Characteristics length; Note if manifold is <br /> at end of rock layer, lateral <br /> 1. Depth to restricting layer length is rock layer length <br /> such as seasonally saturated less half a,.perforation <br /> soil, bedrock, coarse soil, spacing. If manifold is in <br /> etc. ; 2 �1 inches center of rock layer, lateral, . <br /> 2. Depth of percolation tests; length is one-half rock layer <br /> Ay inches length less half a perforation <br /> spacing. Perforated lateral ' <br /> 3. Number of percola tion test length = _ 6 f t•Souri;� y8.5'N&*xrW <br /> holes; S holes 4. Divide lateral leng,01 by pul-For- <br /> 4. Ave. percolation rate; ation` spacing to get number of <br /> /2 . 6 mpi perforations per-' lateral <br /> 5. Landslope = 7 % 5 feet 3 feet ,perfs fvHTi, <br /> Note: last perforation must be <br /> D. Rock Layer Dimensions' in <br /> Send ,c p., (see ager-14) <br /> o ^ G ` 7 ' <br /> 1. Multiply gpd by 0.83 to 5. Multiply perforations per <br /> obtain required area of lateral 'by number of laterals <br /> rock layer; to get total number of <br /> /050 gpd. x 0.83 = $75sq ft perforations; <br /> /2 perfs/lat x 3 lats <br /> 2. Select width of rock layer i6 x 31 A",-"( <br /> (10 feet or less) /O feet 6. Determine required flow rata <br /> by multiplying number of 7 � <br /> 3. Length of rock layer = Area perforations by flow per <br /> = Width 875 sq ft IO ft perforation (see page E-17) <br /> = 87. 5 ft 8y perfs xQ,S6gpm/pert = 'Zogpm <br /> E. Rock Volume 7. Select minimum required lateral <br /> diameter from table on Page E-17; <br /> 1. Multiply rock area by rock depth enter table with perforation <br /> to get cubic feet of rock; spacing,, perforation diameter, <br /> $75 sq ft x ./, O f t = 8.75cu ft and..numb,ex..of perforations per <br /> 2. Divide cu ft bylateral. Select minimum .' 27 cu ft/cu yd diameter for perforated lateral <br /> to get cubic yards; 32 , y = ' : 2 inches 4F,razc <br /> 3. Multiply cubic yards by 1.4 to <br /> get weight of rock in tons; G. Basal Width <br /> 3Z_Y cu yds x 1.4 = 3,5. .Y• tons 1. Percolation rate in top 12 <br /> inches of soil is /2. 7 mpi <br /> 2. Select "allowable soil loading <br /> rate from table on page E-16; <br /> GIST7 0, 509 pd/ft2 <br />
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