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40o'B6.v. .1Frr j (' f1,0rA C6 <br /> 6 BFQIfGGi};.'' (SCE-D���Gv Foe 7 BfO,eGo...s�E-19 <br /> MOUND DESIGN PROCEDURE 149 <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 /> 12.00. .. _gpd laterals 16 <br /> 2. Select perforation spacing <br /> B. Septic Tank Liquid Volume = 3 ft <br /> (see C-3 or C-5) gallons �1T eEroRE <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 y_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 = ,� f t.So-rq ,voX-rW <br /> holes; S holes 4. Divide lateral length by perfor- <br /> 4. Ave. percolation rate; ation spacing to get number of <br /> /2 . 6 mpi perforations per lateral <br /> 5. Landslope = 7 Y -/xx ,4r �u, ENS. -?7• S feet -. 3 feet = /2 perfs Jo'l <br /> Note: last perforation must be <br /> D. Rock Layer Dimensions ' in end cap, (see page E-14) <br /> 62- S r 3 = L! <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 /> /2 O O gpd x 0.83 =/DOOsq ft perforations; <br /> /2 perfs/lat x .3 lats = 36 low-, <br /> 2. Select width of rock layer 71. x -7— <br /> (10 feet or less) Meet 6. Determine required flow rate <br /> by multiplying number of <br /> 3. Length of rock layer = Area perforations by flow per <br /> = Width/QOo sq f t /O f t perforation (see page E-17) <br /> %DO ft perfs x0.5Kgpm/pert <br /> E. Rock Volume 71 Select minimum required latera] <br /> diameter from table on Pale 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 ./, O f t =/QOOcu 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; �7, Q = . 2 inches S,(ofC AJ QFi'aaF <br /> 3. Multiply cubic yards by 1.4 to <br /> get weight of rock in tons; G. Basal Width <br /> ,37. Ocu yds x 1.4 - S/. Ptons 1. Percolation rate in top 12 <br /> inches of soil is /2. 7 inpi <br /> 2. Select allowable soil loading <br /> rate from table on page E-16; <br /> gpd/f t2 <br />