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1996-008221 - new septic system
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2643 Thoroughbred Lane - 04-117-23-12-0021
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1996-008221 - new septic system
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Last modified
8/22/2023 5:07:30 PM
Creation date
4/24/2019 10:18:37 AM
Metadata
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Template:
x Address Old
House Number
2643
Street Name
Thoroughbred
Street Type
Lane
Address
2643 Thoroughbred La
Document Type
Septic
PIN
0411723120021
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> 7 <br /> A. FLOW Estimated Sewage Flow in Gallons per Day(gpd) <br /> Estimated gpd Number <br /> or measured x 1.5 = gpd. of .IYpe I Type 11 Type III Type IV <br /> ooms <br /> 2 300 225 180 60% <br /> of <br /> B. SEPTIC TANK LIQUID VOLUMES a 7500 37755 256 6. <br /> - ) <br /> a,§0 4 )-)OOL) gallons 5 750 450 294 <br /> 6 900 525 332 � <br /> 7 1050 600 370 <br /> 8 1 1200 1 675 1 408 <br /> C. SOILS (refer to site evaluation) it z, " 3 Nwnber „®® <br /> 1. Depth to restricting layer= ail-yo �ie' inches of � I ~Uh <br /> Bedrooms C---y (s-i <br /> 2. Depth of percolation tests = a inches <br /> 3. Percolation rate 15S, L m i 2 or� 750 1.125 <br /> P 3 or 4 1.000 1.500 <br /> 4. Land slope (o % 7 or 6 2.� 20 <br /> over 9 See fig.C-6 (z 1.5) <br /> D. ROCK LAYER DIMENSIONS <br /> I. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer:A x 0.83 = <br /> r)!Sy gpd x 0.83 sq. ft./gpd = to z sq. ftt►o'>v 1,q4° <br /> 2. Select width of rock layer(10 feet or less) _ /o ft. <br /> 3. Length of rock layer= area_width = Rock Bed <br /> (.V y sq. ft. /1L ft. _ /.,V ft. ti.,...�.,.ti.ti.�.,.ti...%.,.�.�.�., <br /> J•J•JM•J•J•hJ•J•I•J•J•J•J•J•J• <br /> ti•\•ti•�.ti.ti•ti•ti.ti•ti.S• •t•ti•�•ti.♦ <br /> tiftifti ti ti ti..:ti ti..t11 tiltiKKrt idth 510 ft. <br /> J•J•J•J•J•J•t•J•�J•J•!•J•J•J•J• <br /> ti•1•tiN•1•�•�•t•1 t•ti.ti•�•�•1.ti•ti <br /> •J•J•J•J•J•J•J•J•J•J•J•J•J•J•J•J• <br /> E. ROCK VOLUME 1 Length -� <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> 4,-Vq-sq.ft.x LQ,<ft. =21S/' cu. ft. <br /> 2. Divide cu.ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> r?1-? cu.ft. i 27= �0 _cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> �2 cu. yd. x 1.4 ton/cu. yd. = 3 V tons. <br /> F. ADSORPTION WIDTH GLA LO-AW1 Abso tion widthSizing Table <br /> 1. Percolation rate in top 12 inches of soil is 1�.(- mpi Pemolation Raw tea• a.�d <br /> .Minutes perinch Soil Texture pf d y '.&b d <br /> P •� <br /> 2. Select allowable soil loadingrate from table; <br /> Faster than 0.1 Coarse Sand 1.20 1.00 <br /> gpd/ft2 0.1 to S Sand 120 1.00 <br /> 0.1 to 5 Fine Sand- 0.60 2.00 <br /> 6 to 15 Sandy Loam 0.79 1..52 <br /> 3. Calculate adsorption width ratio b dividing rock la 16 to 30 Loam 0. 2.00 <br /> y g layer 31 to 45 Silt Loam 0550 0 2.40 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; 46 to 60 Clav Loam2.67 <br /> 61 to 1 4 5.00 <br /> 1.20 gpd/ft21 .�.I.� gpd/ftp_ p,L o Slower than 120 Clav - I - <br /> --soil having 50%or more of fine or very fine Sana. <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> D,(.9 x _ft=aft <br />
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