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1995 - 006973 - new septic system
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2760 White Oak Circle - 04-117-23-42-0018
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1995 - 006973 - new septic system
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Last modified
8/22/2023 5:14:07 PM
Creation date
2/3/2020 12:33:07 PM
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x Address Old
House Number
2760
Street Name
White Oak
Street Type
Circle
Address
2760 White Oak Cir
Document Type
Septic
PIN
0411723420018
Supplemental fields
ProcessedPID
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MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> - i <br /> A. FLOW Estimated Sewage Flow in Gallons per Day(gpd) <br /> Estimated (,o 0 gpd Number <br /> f <br /> or measured - x 1.5 = - gpd. Bedrooms Type I Type II Type III Type IV <br /> 2 300 225 180 60% <br /> 3 450 3(X) 218 of <br /> B. SEPTIC TANK LIQUID VOLUMES • 4 600325 256 •,o <br /> 5 750 450 294 a. <br /> - o 0 gallons 6 900 525 332 the <br /> 7 <br /> 1050 600 370 �� <br /> 8 1200 675 408 <br /> C. SOILS (refer to site evaluation) <br /> , ,i N�ba �mt +r Mw <br /> 1. Depth to restricting layer = a -to ail inches Bedrooms t- �° , <br /> 2. Depth of percolation tests = I a'' inches `p"m'' `"`m`' <br /> 3. Percolation rate /0 . (" mpi 23 or 41750 <br /> 1.125 <br /> ,500 <br /> 4. Land slope To to % ;or sor 1.500 <br /> 2.000 3,000 2,250 <br /> ova 9 See fig.C-6 (x 1.5) <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: A x 0.83 = , <br /> v gpd x 0.83 sq. ft./gpd = L sq. ft..-)-)020, St-t0 L, <br /> 2. Select width of rock layer (10 feet or less) = l D ft. <br /> 3. Length of rock layer = area_width = Rock Bed <br /> St-t'") sq. ft. ± /o ft. = cc ft. eti:ti:ti:; :�:ti:ti:L:tti:c:;:ti.;:;:ti. <br /> rtiflfti~fti76.74i51%tiff Si.S S5.V.: 1 <br /> :tiftiftiftiftiftifti ti tiftiftifti tiftif46:7: ,Width 510 ft. <br /> -%•%•%•%•%•%•%-%-%•%•%•%•%•%•%•%•%.r.e.". ..e." •r I <br /> E. ROCK VOLUME ~ Length i <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> St-il sq. ft. x//Ls/ ft. = S-?1-I, cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> Sr)N cu. ft. --27= a I cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> D ) cu. yd. x 1.4 ton/cu. yd. = ac, tons. <br /> F. ADSORPTION WIDTH �t 4 L°401 <br /> Absorption Width Siring Table <br /> 1. Percolation rate in top 12 inches of soil is /0.ci mpi Percolation RateCatIono Racod <br /> Minutes per inch Soil Texture F.•,n , <br /> (top.) f ab.or ea <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 Coarse Sand 1.20 1.00 <br /> �-1.( gpd/ft'- 0.1 to 5 Sand 1.20 1.00 <br /> 0.1 to 5 Fine Sand" 0.60 2.00 <br /> 6 to 15 Sandy Loam 0.79 1.52 <br /> 13. Calculate adsorption width ratio bydividingrock layer to 30 Loam 0.60 2.00 <br /> � 331 1 to 45 Silt Loam 0 <br /> 50 2.40 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; 46 to 60 Clay Loam 0.45 2.67 <br /> 61 tO 120 Clay 024 x.00 <br /> 1.20 gpd/ft2÷ . L-t,(- gpd/ft2 = D. ') Slower than 120 Clay -- - <br /> -Soil having 50%or more of fine or very fine sand. <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> :).1.,-) x /o ft = ;DL..2 ft <br />
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