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�%• 7FA41F.v yds"'r E..S <br /> l BFA.-• .'FTT <br /> BFQ/lGGiF;.`� (,�E-p��iG✓ Foe 7BFORGOMS,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 /> 1200 ^gpd laterals <br /> 2. Select perforation spacing <br /> B. Septic Tank Liquid Volume - 3 ft <br /> (see C-3 or C-5) gallons AS efFGRE <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 _inches center of rock layer, lateral <br /> 2. Depth of percolation tests; length is one-half rock layer <br /> _inches length less half a perforation <br /> spacing. Perforated lateral <br /> 3. Number of percolation test length = •? 6 ft.Sourir� 61 'Noxrtl <br /> holes; S holes 4. Divide lateral length by pc:rfor- <br /> 4. Ave. percolation rate; ation spacing to get number of <br /> /2 . 6 mpi perforations per Lateral <br /> 5. Landslope = 7 % A1,4X 4T P. -?/' S feet -. 3 feet = /2 perfs fP# <br /> Note: last perforation must be <br /> D. Rock Layer Dimensions ' in end cap, <br /> p, (see page E — 4) <br /> 4',Z. S ZAl <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 /> 1200 gpd x 0.83 -/OOOsq ft perforations; <br /> /2 perfs/lat x 3 lats = 36 loch; <br /> 2. Select width of rock layer 7 /. x -7- _ -,vdRrc: <br /> 6. Itetermine required flow rate <br /> (10 feet or less) Meet by multiplying number of <br /> 3. Length of rock layer = Area perforations by flow per �'72 <br /> Width/pOo sq ft - /O ft perforation (see mage E-17) <br /> /DO ft perfs x0,569pm/perf = 5g1>in <br /> E. Rock Volume 7.. Select minimum required 1aCcral <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 /> /000 sq f t x ./, 0'f t =/,040cu f t and number of perforations per <br /> 2. Divide cu ft by 27 cu ft/cu yd lateral. Select minimum <br /> diameter for perforated lateral <br /> to get cubic yards; ,37, Q = 2 inches 5w,4fFrORF <br /> 3. Multiply cubic yards by 1.4 to <br /> get weight of rock in tons; G. Basal Width <br /> .?7. Ocu yds x 1.4 = S/, Ptons 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 /> L/;F O, 50 gpd/ft2 <br />