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1995 Septic System
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2785 White Oak Cir - 04-117-23-42-0014
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1995 Septic System
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Last modified
8/22/2023 5:13:57 PM
Creation date
2/4/2020 9:11:23 AM
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x Address Old
House Number
2785
Street Name
White Oak
Street Type
Circle
Address
2785 White Oak Cir
Document Type
Septic
PIN
0411723420014
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10.11/4_,4 V1 JA-, L a_..)1 LJ1I I I LII•A•Ll1LLL1 <br /> ' (For Flows up to 1200 gpd) <br /> A. FLOW Estimated Sewage Flows in Gallons per day <br /> Estimated (rc:,v gpd (see pages D-7 or I-3;4,5) Number (gpd) <br /> or measured - gpd x 1.5 = -- . Bedro <br /> ofoms Type I Type II Type 111 IV <br /> 2 300 225 180 <br /> B. SEPTIC TANK LIQUID VOLUMES 3 450 300 218 60% <br /> 4 600 375 256 °f she <br /> values <br /> _'-/DUDgallons (see pages C-3 or C-5) 5 750 450 294 i <br /> p g 6 900 525 332 k'';',.1. <br /> 7 1050 600 370 so <br /> 8 1200 675 408 taunts <br /> C. SOILS (refer to site evaluation) I. <br /> 1. Depth to restricting layer = I Ti - I H It inches Septic Tank Capacities,in ga nuns <br /> Number of Minimum Liquid Liquid capacity with <br /> 2. Depth of percolation tests = I inches Bedrooms Capacity garbage disposal <br /> 3. Percolation rate 7 . mpi 2 or less 750 1125 <br /> 3 or 4 IOOD 1500 <br /> 4. Land slope 3. % , 6 <br /> 4"99 ao 22500 <br /> oveD. 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 /> (.,o 0 gpd x 0.83 sq. ft./gpd =LI cl q sq. ft.-r to =5'-)r)' <br /> 2. Select width of rock layer (10 feet or less) = / ) ft. <br /> 3. Length of rock layer = Area ÷ Width = <br /> t-1 '1 sq. ft. 4.- /0 ft. = SS ft. Rock Bed <br /> ' r•r•r•r•r•r•r•r r•r•r•r•r• <br /> ti•ti.i..„y•ti•ti•ti-.,6yi ti•ti•z6 Width 510ft. <br /> �•r•r•r•r•r•r•r•r•r•r•r;f,r;r;r <br /> E. ROCK VOLUME F Length -f <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> S1-11 sq. ft. x 1.os'ft. = 5')41 cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> .5-)Li* cu. ft. 27 = a cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> I cu. yd. x 1.4 ton/cu. yd. = a`3 tons. <br /> F. ADSORPTION WIDTH L1-1I L( Ld4wi <br /> Absorption Width Sizing Table <br /> 1. Percolation rate in top 12 inches of soil is 3..- mpi <br /> Percolation Rale Gallons Ratio of <br /> 2. Select allowable soil loading rate from table on page E-; in Minutes per Soil Texture per day per Absorption width <br /> Li-.S--- <br /> gpd/f l Inch(MPI) square fool to Rock <br /> Layer <br /> 3. Calculate adsorption width ratio by dividing rock layer Fasterthan 0.1 • Coarse Sand --.•. <br /> 0.1 to 5 Sand 1.20 1.00 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; 0.1 to 5•• Fine Sand•• 0.60 2.00 <br /> 1.20 gpd/ft21 H s gpd/ft2= .1., . 6t1 15 Sandy Lons 0.79 12s2 <br /> 16 to 30 Lon0.60 2.1q <br /> 31 to 45 Silt Loam 0.50 2.40 <br /> Check this value on page E-16. 46to60. _Clay Loam 43,45 2.67 <br /> 60 to 120 Clay 0.24 5.00 <br /> 4. Multiply adsorption width ratio by rock layer width to get Slo2erthan Clay <br /> required adsorption width; <br /> a•(,1 ) x IL) ft = qc:,.') ft <br /> I <br />
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