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1993-12-18 Septic System Design Report
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1993-12-18 Septic System Design Report
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Last modified
4/16/2025 1:51:32 PM
Creation date
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x Address Old
House Number
2735
Street Name
Countryside
Street Type
Drive
Street Direction
West
Address
2735 Countryside Drive West
PIN
0411723120018
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MOUND DESIGN WORKSHEET <br />(For Flows up to 1200 gpd) <br />A. FLOW <br />Estimated d gpd (see pages D-7 or I-3, 4, 5) <br />or measured - gpd x 1.5 = - <br />B. SEPTIC TANK LIQUID VOLUMES <br />�A t -/aoo gallons (seepages C-3 or C-5) <br />C. SOILS (refer to site evaluation) <br />I. Depth to restricting layer <br />2. Depth of percolation tests = inches <br />3. Percolation rate S .3 mpi <br />4. Land slope 2 % <br />D. ROCK LAYER DIMENSIONS <br />1. Multiply flow rate by 0.83 to obtain required area of rock <br />layer: Daily Flow x 0.83 = <br />")_N gpd x 0.83 sq. ft./gpd = ' sq. ft. -- <br />2. <br />t. 2. Select width of rock layer (10 feet or less) = i ft <br />3. Length of rock layer = Area + Width = <br />�_ sq. ft. + 10 ft. =� ft. <br />Gslim&" Sewage (3pws in Galluns pa dry <br />6p d) <br />\umber <br />Rua of <br />Abaorprion widen <br />so Rock Layer <br />WkItk <br />Faster that 0.1 • <br />Coni Sand <br />-- <br />of <br />T�pc I <br />Type 11 <br />TSpc III <br />Type <br />Iledrn•ms <br />Fine SwA •• <br />0.60 <br />2.n0 <br />IV <br />2 <br />300 <br />225 <br />150 <br />[ wm <br />3 <br />450 <br />300 <br />218 <br />9'+ <br />4 <br />600 <br />375 <br />256 <br />2.67 <br />167 <br />5 <br />750 <br />450 <br />294 <br />`• <br />6 <br />9W <br />525 <br />332 <br />TU. 1. <br />7 <br />1050 <br />600 <br />370 <br />m <br />1200 <br />675 <br />409 <br />.d .. <br />2a1.. <br />750_- <br />1123 <br />3aalUu <br />190 <br />4.6 <br />ISTD <br />223(1 <br />7.6 a 9 <br />20W <br />3LLU <br />.,vy <br />E. ROCK VOLUME <br />1. Multiply rock area by rock depth to get cubic feet of rock; <br />G 5L4 sq. ft. x ). o r ft. = ')) -?cu. ft. <br />2. Divide cu. ft. by 27 cu. ft. /cu. yd. to get cubic yards; <br />cu. ft. + 27 = D �)_ cu. yd. <br />3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br />2a cu. yd. x 1.4 ton/ cu. yd. = 35 tons. <br />F. ADSORPTION WIDTH L,_ _At -LON M <br />1. Percolation rate in top 12 inches of soil is _� mpi <br />2. Select allowable soil loading rate from table on page E-: <br />'i4-<, gpd/f t2 <br />3. Calculate adsorption width ratio by dividing rock layer <br />loading rate of 1.20 gpd/fe by allowable soil loading rate; <br />1.20gpd/ft2+gpd /ft2= a.LI) <br />Check this valae on page E-16. <br />4. Multiply adsorption width ratio by rock layer width to get <br />required adsorption width; <br />P.tJ-' x ) 0 ft = D,;r,.? ft <br />Rock Bed <br />J•I•!•I•J•J•J•J•J•J•J•J•J•J•J <br />1rf1,flr� ��♦�ti�l�hftif,,f�ftfti of dth S10ft. <br />!•!•!•I •!•!•!•!•!•!•!•J•!•I •! <br />J•J•. •J•J•!•J•/•JrJ•J•J•yJ•J <br />�- Length <br />Absorpilon Wldth SI:In=7bble <br />14rcolatlon Ras <br />in Minules {w <br />Inch IMF <br />Soil Teuum <br />Gallons <br />pa day per <br />square foa <br />Rua of <br />Abaorprion widen <br />so Rock Layer <br />WkItk <br />Faster that 0.1 • <br />Coni Sand <br />-- <br />---- <br />MI u SSand <br />1.20 <br />1.00 <br />0.1 to S •• <br />Fine SwA •• <br />0.60 <br />2.n0 <br />6 to 15 <br />Smody Liam <br />0.79 <br />1.52 <br />1610 30 <br />[ wm <br />0.60 <br />2.110 <br />311045 <br />SiMLoam <br />M. <br />2.40 <br />On LO&UL <br />-q!-q!D <br />2.67 <br />167 <br />�i-10 12(1 <br />Clay <br />!D <br />Slowcr than <br />Clay <br />-•- <br />1211• <br />
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