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i 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 u e <br /> metered value; Flow Rate =� 1 . Select number of perforated <br /> 9c gpd laterals 1�5 ✓ <br /> 2. Select perforatioty spac'iII ; <br /> B. Septic Tank Liquid Volume / _ ft .� <br /> (see C-3 or C-5) ,2)n gallons ✓ <br /> 3. Select perforated lateral <br /> C. Soil Characteristics length; Note if manifold is <br /> at end of rock layer, lategral <br /> 1. Depth to restricting layer length is rock layer lent <br /> 41511, <br /> such as seasonally saturated less half a perforation <br /> soil, bedrock, coarse soil, / spacing. If manifold is in <br /> etc. ; -.:) LJ inches ✓ center of rock layer, laterral <br /> 2. Depth of percolation tests length is one-half rock layer <br /> _inches length less half a perfori.ltion <br /> spacing. Perforated latera/ <br /> 3. Number of percolation test / length = 3 6 ft. ✓ <br /> holes; S holes ✓ <br /> 4 . Divide lateral length by pc�rfor- <br /> 4. Ave. percolation rate; ation spacing to get number ofd / <br /> /2 . - mpi / perforations per lateral J <br /> 5. Landslope = 7 % ?2, 5 feet . 3-f eet <br /> Note: last perforation must be <br /> D. Rock Layer Dimensions ' in end cap, (see page E-14) <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 /> 900 gpd x 0.83 = 750 sq ft ✓ perforations; / <br /> /2 perfs/lat x fats = -7 ' J <br /> 2. Select width of rock layer <br /> (10 feet or less) = /0 feet 6. Determine required flow r. e <br /> by multiplying number of <br /> 3. Length of rock layer = Area perforations by flowper <br /> - Width 750 sq f t - /O f t / perforation (see page E-1 / <br /> 75. 0 ft perfs xQS,>gpm/pert = /'.-i'g[nn ✓ <br /> I or— <br /> E. Rock Volume 7. �ect minimum' requireZ Ia t�al <br /> diameter from table on Page E-1.7; <br /> 1. Multiply rock area by rock depth / enter table with perforation <br /> to get cubic fe of rock- / spacing, perforation diameter, <br /> 750 sq ft x 0, Sft =✓F. cu ft and number of perforations per <br /> 2. Divide cu ft/ <br /> Sy 27 cum cu yd lateral. Select minimum <br /> to get cubic yards; - `.`f, diameter for perfora later< 1. <br /> �i� = y '•':! inches �e! <br /> 3. Multiply cubic yards by 1.4 to <br /> get weight of rock in tons; G. Basal Width <br /> 20.E-2cu yds x 1.4 = 79, /%tons <br /> 1. Percolation rate in tap 1 ? <br /> inches of soil is 1. 7 mpi <br /> 2. Select allowable soil loading <br /> rate from table on page E-16; <br /> 2 <br /> gpd/ft <br />