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2100 Webber Hills Road - 03-117-23-34-0012
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1993 Septic System
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Last modified
8/22/2023 4:37:20 PM
Creation date
1/23/2020 9:25:43 AM
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x Address Old
House Number
2100
Street Name
Webber Hills
Street Type
Road
Address
2100 Webber Hills Rd
Document Type
Septic
PIN
0311723340012
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MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. FLOW Estimated Sewage Flows in Gallons pa day <br /> Estimated 6,06 gpd (see pages D-7 or I-3,4,5) cam) <br /> Number <br /> or measured -- gpd x 1.5 = -- of Type I Type II Type Ill Type <br /> Bedrooms 1 V <br /> B. SEPTIC TANK LIQUID VOLUMES 3 300 3000 18025 160% <br /> a. - / 0 0 cc gallons (see pages C-3 or C-5) 5 75063755256 <br /> in <br /> 6 900 525 332 Type I. <br /> 7 1050 600 370 IIm <br /> 8 1200 675 aos <br /> C. SOILS (refer to site evaluation) columns <br /> /I <br /> • <br /> 1. Depth to restricting layer = Jo inches Septic Tank Capacities,in gallons <br /> 2. De th of percolation tests = Number of Minimum Liquid Liquid capacity with <br /> P I inches Bedrooms Capacity garbage disposal <br /> 3. Percolation rate /0.cr mpi 2orleu X50 112 <br /> 3or4 1pOp 1 <br /> 4. Land slope 4``71.7 ¶o 9 % 748 or 9 ,2.250 <br /> 003 <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 /> X00 gpd x 0.83 sq. ft./gpd = yg sq. fta-1o`50-S4O <br /> 2. Select width of rock layer (10 feet or less) = / 0 ft. <br /> 3. Length of rock layer = Area _ Width = <br /> Sy'1 sq. ft. - 10 ft. = ss' ft. Rock Bed <br /> ��J�J�J�J�J:J�J:f:f•f:f:f:f:f:f' <br /> '•J•J•J•J•J•J J•J•J•J•J•J•J•J•J <br /> ti:::p.p ti•ti•ti•ti••••ti•ti•ti.ti•ti••.•:Width _<l0 ft. <br /> l f J J J JJ•J�J•J•J JJ• <br /> E. ROCK VOLUME ,tJ;}_;.• �:: �r=V:J:f:•J:J•�J:J,1 <br /> •• Length 1 <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> Si-Li sq. ft. x/.v,eft. =,c' ' cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> ,'N cu. ft. _ 27= a I cu. yd. • <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> a-1 cu. yd. x 1.4 ton/cu. yd. = ac, tons. <br /> F. ADSORPTION WIDTH -1-a-'-( 1-0A-vi <br /> 1. Percolation rate in top 12 inches of soil is /0., mpi Absorption Width Sizing Table <br /> 2. Select allowable soil loadinrate from table on E-• Percolation Rate Gallons Ratio of <br /> pagein Minutes per Soil Texture per day per Absorption width <br /> •Li gpd/ft2 <br /> Inch(MPI) square foot to Rock Dyer <br /> Width <br /> 3. Calculate adsorption width ratio by dividing rock layer FacterMan 0.l.e coarse Sand ----- <br /> 01 to 5 Sand 1.20 <br /> . 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/ft2 i- ,y gpd/ft2= 0 .(p16 1 . 6 t1 15 Sandy LAW 0.79 1.52 <br /> to 30 Loam 0.60 2.00 <br /> 31 to 45 Silt Loam 0 240 <br /> Check this value on page E-16. a6 cIa s67 <br /> 60 to 12U <br /> 4. Multiply adsorption width ratio by rock layer width to get Sl12erthan Clay ____ <br /> required adsorption width; . <br /> D.tor) x /0 ft = 3Y) ft <br /> • <br /> 1 <br />
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