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1994-06-14 Septic System Design Report
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1994-06-14 Septic System Design Report
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Last modified
4/23/2025 12:24:00 PM
Creation date
4/23/2025 12:18:55 PM
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x Address Old
House Number
1003
Street Name
Cox farm
Street Type
Road
Address
1003 Cox Farm Road
PIN
2711823330014
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MOUND DESIGN WORKSHEET <br />(For Flows up to 1200 gpd) <br />A. FLOW <br />Estimated 6'00 gpd <br />or measured x 1.5 = gpd. <br />B. SEPTIC TANK LIQUID VOLUMES <br />12-5-0 gallons f- /000 6W 4 0A" Z N0 riX.O& <br />t /2?o G,/i.tor✓ NMI.,04 re",r. <br />C. SOILS (refer to site evaluation) <br />1. 'Depth to restricting layer = Z 8 inches <br />2 Depth of percolation tests = y inches <br />3. Percolation rate 9_ D mpi <br />4. Land slope -5 % <br />Estimated SeonCc <br />Nm <br />&b <br />af� <br />Type i <br />Bedroom <br />2 <br />300 <br />3 <br />430 <br />4 <br />5W6 <br />1-$ <br />7900 <br />6 <br />ova 9 <br />7 <br />1050 <br />I <br />1200 <br />NAW <br />h4- <br />&b <br />C.A- <br />rte <br />ea <br />4� <br />2 or im <br />75 <br />3 ar 4 <br />1.0 <br />5W6 <br />1-$ <br />Tor a <br />zo <br />ova 9 <br />Sem 64 <br />D. ROCK LAYER DMIENSIONS <br />1. Multiply flow rate by 0.83 to obtain required area of rock <br />layer: A x 0.83 - <br />6oU gpd x 0.83 sq. ft./gpd = Soo sq. ft. <br />2. Select width of rock layer (10 feet or less) _ /0 ft. <br />3. Length of rock layer = area + width = Rock Bed <br />50,0 sq. ft. + /0 ft. = So ft.4. <br />J.;,.,.,...,...,.,.,...,.,.,., . <br />.....,.,.J.J.J J.J.J.J.J.J.;. <br />J.J.J•J•/•J•J.J•J J.J./•J.J.J.J• idth 510 ft. <br />J•J•J•J.J•J•J• <br />,J.I.J.l•J./•I.I•J. J. J.J.I.J./.J. <br />E. ROCK VOLUME Lwgth --' <br />1. Multiply rock area by rock depth to get cubic feet of rock; <br />Soo sq. ft. x / ft. =SUo cu. ft. <br />2. Divide cu. ft. by 27 cu. ft. /cu. yd. to get cubic yards; <br />Soo cu. ft. + 27 - /.F. S cu. yd. <br />3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br />/& S cu. yd. x 1.4 ton/cu. yd. = Z 6 tons. <br />F. ADSORPTION WIDTH <br />1. Percolation rate in top 12 inches of soil is 9. o mpi <br />2. Select allowable soil loading rate'from table; <br />4IE o. 60 gpd/ft2 <br />3. Calculate adsorption width ratio by dividing rock layer <br />loading rate of 1.20 gpd/ft2 by allowable soil loading rate; <br />1.20 gpd/ft2+ o. 0.0 gpd/ft2 = Z . o <br />4. Multiply adsorption width ratio by rock layer width to het <br />required adsorption width; <br />z. o x /o ft..,20 ft <br />A W%dthSisn Lbit <br />PVMV1AtMRar <br />Q6,; „,h <br />Soil Texture <br />C.A- <br />rte <br />a...♦ <br />s <br />Faster thin 0.1 <br />Coarse Sand <br />120 <br />1.00 <br />0.I to 5 <br />Sand <br />120 <br />1.00 <br />0.1 to 5 <br />Fine Sand- <br />0.60 <br />2-00 <br />6 to 15 <br />Sandy Loam <br />079 <br />152 <br />16 to 30 <br />Loam <br />0.60 <br />2.00.. <br />31 to 45 <br />Silt Loam <br />OSO <br />2.40 <br />46 to 60 <br />Clay Loam <br />0.45 <br />2.67 <br />61 t0120 <br />Clay <br />024 <br />5.00 <br />Slower than 120 <br />C1a <br />"sal ha" sm W was of am w vwy Am aar.& <br />
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