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/' • • S r�,•.s?es . <br /> � .� � BI�D�l oOr�. 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. 5el�ct' number of perforated <br /> l�jrp �pd � later.als 6 " . <br /> 2. Select perforation spacin�; <br /> B. Septic Tank Liquid Volume = ,3 ft <br /> (see C-3 or C-5) /Ov0 gallons " . <br /> 3. Seiect 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 len�th <br /> - such as seasonally saturated •• �t'ess half a perforation � <br /> soil, bedrock, coarse soil, spacing. If manifold is ii� <br /> etc. ; 2 y inches center of rock layer, lateral, <br /> 2. Depth of percolation tests; Iength "is� �one-half rock layer <br /> /.$� inches , length less half a perfarrici��i <br /> ' spacing. Perforated lateral <br /> 3. Number of percolation test length = /7, 2S ft. <br /> holes; ��j.r holes �,` • <br /> 4, D.ivtde later�l lcngth by perfor- <br /> 4. Ave. percolation rate; ation spacing to �et nt►mber of <br /> �7. ] mpi perforations per lateral ' ' <br /> S. Landslope = � / /7.25 feet = �_feet = 6 per.fs <br /> Nofe: last perforation must be <br /> _ D. Rock Layer Dimensions ' in e�d cap, (see page 1:-14) <br /> 1. Multiply gpd by 0.83 to ..5. Multiply perforations per <br /> obtain required area of late.r�.l. b.y'•number oE laterals <br /> rock layer; to get total number of <br /> yS0 gpd x 0.83 = ,37Ssq ft perforations; - <br /> 6 perfs/lat x 6 lats = ;3�6 <br /> 2. Select width of rock layer 6. Determine required flow rate <br /> (10 feet or less) _ /D feet by multiplying numbsr of <br /> 3. Length of rock layer = Area perforations by flow per �y �PFf'� <br /> � Width37S sq ft = /p ft ', •• perforation, (see page E-17� <br /> _ .�7.5 ft �perfs x ,7'�Qpm/perf =L6,6�Prn <br /> E. Rock Volume 7. Select minimum required la�eral <br /> diameter from table on Pa�;e E-J.7; <br /> 1. Multiply rock area by rock•depth enter table with perforntion � <br /> to get cubic feet. of rock; spacing, perforation diameter, <br /> �7S sq ft x / ft = j7Scu ft and ,nunb'er 'of perforations per <br /> lateral.� Select minimum . <br /> 2. Divide cu ft by 27 cu f t/cu yd diameter f or perforated lateral <br /> to get cubi� yards; /,3• 9 � � . . <br /> _ �inches <br /> 3. Multiply cubic yards by 1.4 to . <br /> get weight of rock in tons; - G. Basal Width � <br /> /�,9 cu yds x 1.4 � /9,y tons 1. Pezcoiatiotr rate in top 22 <br /> inches o.f soil is �7 7 mpi � <br /> 2. Selec`t'allowable soil 'lo:adin�; <br /> - rate from table on pa�c� F.-lF,; <br /> O. �O gpd/f t2 . <br />