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1993-12-11 Septic System Design Report
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1993-12-11 Septic System Design Report
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Last modified
8/20/2025 2:59:42 PM
Creation date
8/20/2025 2:53:51 PM
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x Address Old
House Number
2905
Street Name
Deer Run
Street Type
Trail
Address
2905 Deer Run Trail
PIN
0411723240010
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MOUND DESIGN V'3RKSHEET <br />(For Flows up to 1200 gpd) <br />A. FLOW <br />Estimated sQ gpd (see pages D-7 or 1-3,4, 5) <br />or measured _ gpd x 1.5 = - <br />B. SEPTIC TANK LIQUID VOLUMES <br />t 1-j oo u gallons (see pages C-3 or C-5) <br />C. SOILS (refer to site evaluation) <br />1. Depth to restricting laver = at; I -Io 'S a inches <br />2. Depth of percolation tests = 1 inches <br />3. Percolation rate `6 .3 mpi <br />4. Land slope -:Z <br />EWmlawd Sewage Flows in Gallons pa day <br />(gpd) <br />mbcr <br />ot <br />Type I <br />Type It <br />Type III <br />Type <br />Bedroom, <br />( V <br />2 <br />300 <br />223 <br />IRO <br />3 <br />e50 <br />3W <br />219 <br />as <br />e <br />600 <br />375 <br />256 <br />ie i <br />•to <br />5 <br />730 <br />e�0 <br />29e <br />6 <br />900 <br />525 <br />332 <br />�L <br />7 <br />1050 <br />600 <br />370 <br />v <br />1200 <br />675 <br />408 <br />.�,.. <br />Sg1fc I« a•16... <br />Nu ha -1 ?lin.m.m Lya.J 1 q.a..i -.iy . u <br />Ildruom• C.pway a«e..c J..y..•.I <br />2 ar Ira <br />730 <br />I l ls <br />3are <br />lOD <br />l4D <br />ear 6 <br />I hD <br />2230 <br />7, t or 9 <br />EIW <br />11D0 <br />owr1 <br />..-•• <br />D. ROCK LAYER DIMENSIONS <br />1. Multiply flow rate by 0.83 to obtain required area of rock <br />layer: Daily now x 0.83 = <br />�0 gpd x 0.83 sq. ft./gpd = as sq. ft.-,r io% _ -�-4 <br />2. Select width of rock layer (10 feet o► less) _ �_ ft. <br />3. Length of rock layer = Area + Width = <br />GcZ4 _ sq. ft. + / G ft. _ ft. Rock Bed <br />t•I•I•t•I•I t•t•�•J•�•t•t <br />dthS1011. <br />� � ti •�i��i�3tiitiS���.�tii <br />E. ROCK VOLUME 1-- Length <br />1. Multiply rock area by rock depth to get cubic feet of rock; <br />4,y4 sq. ft. x, ft. _ J Lj� cu. ft. <br />2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br />9jig cu. ft. + 27 = 3 1 cu. yd. <br />3. Multiply r ubic yards by 1.4 to get weight of rock in tons; <br />a 1) cu. yd. x 1.4 ton/cu. yd. = 2,- <br />g� tons. <br />F. 4 DSORPTION WIDTH C.vAq L o A wl <br />1. ercolation ,.�-te in top 12 inches of soil is I.3 mpi <br />2. Select allowable soil loading rate from table on page E-; <br />L1�;' gpd /ft2 <br />3. Calculate adsorption width ratio by dividing rock layer <br />loading rate of 1.10 gpd/ft2 by allowable soil loading rate; <br />1.20gpd/ft2+ as gpd/ft2= <br />Check this value on page E-16. <br />4. Mv!tiply adsorption width ratio by rock layer width to get <br />required adsorption width; <br />�2.to -) x L ft = s i ft <br />Absorption Width Sizing Table <br />rwrohill mRole <br />is M <br />IAMN <br />(Me� <br />Soil Tcuula <br />Calk)" <br />=paer <br />Rri001 <br />AhsopioridlA <br />w Rat l.yer <br />Wxhh <br />Fagar don 0.1 • <br />Corse Said <br />0.1105 <br />Sand <br />1.20 <br />1.00 <br />oS•• <br />0.1 IDS*. <br />Fine Sand.. <br />0.60 <br />2.00 <br />610 13 <br />Sr10y Lamm <br />0.79 <br />1.52 <br />16 ao.10 <br />Loam <br />0.60 <br />2110 <br />311043 <br />Sik Loam <br />0.30 <br />2.40 <br />e6 a 60 <br />C4r Lavn <br />0.e3 <br />2.67 <br />60 to 120 <br />Clay <br />0.24 <br />S.00 <br />Slower t . <br />Clay <br />..... <br />..... <br />120••• <br />
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