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........ .. . <br /> E-19 <br /> MOUND DESI N PROCEDURE <br /> (For Flows u to 1200 gpd) <br /> A. Sewage Flow Rate F. Pressure Distrib " <br /> See D-7 or I-3, 4 or 5 ution System <br /> or use <br /> metered value; Flow Rate = 1. Sele'ct'number of perforated <br /> �lS O gpd - - erals 6 <br /> B. Septic Tank Liquid Volume 2• Select perforation spacing <br /> - 3 f <br /> t <br /> (see C-3 or C-5) /000 gallons - <br /> 3. Select perforated lateral <br /> 'C. Soil Characteristics 'length; Note if manifold is <br /> at end of'"tnck layer, lateral, <br /> 1. Depth to restricting layer <br /> such as seasonally saturated length is rock layer l.engtt,r <br /> soil, bedrock, coarse soil, less half a perforation <br /> etc. ; spacing. If manifold is in <br /> 12. inches center "of.'rock layer, lateral <br /> 2. Depth 'of percolation tests; length ;is,one-half rock layer <br /> _ /8 inches length less half a perforation <br /> 3. Number of percolation test spacing. Perforated lateral <br /> length = / 2 <br /> holes; - Z holes -_,7 5 f t. <br /> 4. Divide lateral lcng'th by perfor- <br /> 4. Ave. percolation rate; ation spacing to. get number .of <br /> /91-7 1Api perforations per lateral <br /> 5. Landslope _ S / /7.25 feet . 3 feet = 6 perfs <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 /> ySO gpd x 0.83 = ?75 sq ft perforations; <br /> 2. Select width of rock layer .Per.f.g/lat x 6 . lats = 36 <br /> (10 feet or less) .- /D feet 6. Determine required -flow rate <br /> 3. Length of rock layer = Area by multiplying number of <br /> perforations by flow per A ��® <br /> = Width 37S sq ft 10—f t perf oration <br /> �7.S ft (see page E-17): <br /> _perfs x ,_?Ygpm/perf =26,6gpm <br /> E. Rock Volume 7. Select minimum required lateral <br /> 1. Multiply rock area b rock depth diameter-ter -em table on Page E-17; <br /> y P enter table with perforation , <br /> to get .cubic feet• of rock; spacing, perforation diameter, <br /> �7 S sq f t x / ft = 3 7,5 cu ft and, number of perforations per <br /> 2. Divide cu ft by 27 cu f t/cu yd lateral. Select minimum <br /> to get cubic yards; 13. 9 . diameter for perforated lateral. <br /> _ <br /> 3. Multiply cubic yards by 1.4 to in'Aes <br /> get weight of rock in tons; G. Basal Width - <br /> /7.9 cu yds x 1.4 � /9.Y tons <br /> 1. PeredLation rate in top 12 <br /> inches of 's'oil is AKS mp i <br /> 2. Select allowable soil loading, <br /> rate from table on page F.-16; <br /> G/IF O. CO gpd/f t2 <br />