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1996-008510 - new septic system
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3065 Watertown Road - 04-117-23-22-0033
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1996-008510 - new septic system
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Last modified
8/22/2023 5:09:48 PM
Creation date
7/29/2019 9:38:19 AM
Metadata
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x Address Old
House Number
3065
Street Name
Watertown
Street Type
Road
Address
3065 Watertown Rd
Document Type
Septic
PIN
0411723220033
Supplemental fields
ProcessedPID
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MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A% FLOW Estimated Sewage Flows in Gallons Per day <br /> Estimated o o gpd (see pages D-7 or I-3, 4,5) umber (bpd) <br /> or measured gpd x 1.5 = neaof Type I Type II Type"' tv <br /> 2 300 225 180 <br /> B. SEPTIC TANK LIQUID VOLUMES 3 450 300 218 <br /> 4 600 375 256 .ria <br /> - <br /> /00 0 gallons (see pages C-3 or C-5) 59 0 45550 294 ,n <br /> 7 1050 600 370 m <br /> 8 1200 675 408 cdamw <br /> C. SOILS (refer to site evaluation) <br /> 1. Depth to restricting layer= I! -10 inches septic Tank Capacities,Inaaltuos <br /> Number of ,Minimum Liquid Liquid capacity with <br /> 2. Depth of percolation tests = =rl inches Bodrmxns Capacity garbagedispoad <br /> 3. Percolation rate .�;4 mpi �L �r' % "; 2;«ss ;50 ;� <br /> 4. Land slope (� % 8a9 ---� _._3_. <br /> over 9 -.. <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 /> I <br /> ✓oo O gpd x 0.83 sq. ft./gpd = yg56 sq. ft.+sy?Q <br /> 2. Select width of rock layer (10 feet or less) _ /0 ft. <br /> 3. Length of rock layer = Area-i-Width = <br /> Sy sq. ft. % ft. = 5S ft. Rock Bed <br /> r•r•r•r•r r•r•r•r•r•r•r.r r•r <br /> ti•ti•ti•ti•ti•ti•ti•+•ti.. ...•+•ti•ti• T <br /> r•r•r•r•r•r•r•r•r•r•r•r.r•r•r <br /> ti•ti•ti•ti•ti•ti•ti•ti•�•ti•ti•ti•ti•ti•ti• idth 510 ft. <br /> •r•r•r•r•r•r•r•r r•r•r•r•r•r•r <br /> ti•ti•ti•ti•ti•ti•ti.1.ti•'t•ti.ti•1.•'.•b• - <br /> E. ROCK VOLUME •r•r•r•r•r•r•r•r•r•r•r•r•r•r•r <br /> I-- Length <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> Sx-n sq. ft. x,/.c,.-"ft. =L2-�-cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> . -2�-cu. ft. --27= -.� ► cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> cu. yd. x 1.4 ton/cu. yd. = D,cy tons. <br /> F. ADSORPTION WIDTH L L A L� �x 4n'1 <br /> 1. Percolation rate in top 12 inches of soil is mpi Absorption Width Sizing Table <br /> Per oladon Rate 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 /> Lil gpd/f t2 tnch(MPl) square foot to Rock Layer <br /> 3. Calculate adsorption width ratio by dividing rock layer Faaterthen 0.1• Coarse sand .-- _. <br /> 0.1 to 5 Sand 1.20 1.00 <br /> loading rate of 1.20 gpd/f t2 by allowable soil loading rate; 0.1 to 5•• I=Sand•• <br /> 0.60 2.00 <br /> 1.20 gpd/f t2 .Ll gpd/f t2= �� (� 6 to 15 Sandy Loam 0.79 1.52 <br /> 16 to 30 Loam 0.60 2.00 <br /> 31 to 45 Silt Loam 0.50 2.40 <br /> Check this value on page E-16. 4 t 1m Clay Loam °oii 2,65.i17i <br /> 4. Multiply adsorption width ratio by rock layer width to get s than Clay -_.- -_._ <br /> required adsorption width; <br /> a •�q x /0 ft = ! .� ft <br />
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