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1993-005570 - new septic
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1290 Orono Oaks Drive - 35-118-23-34-0018
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1993-005570 - new septic
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Last modified
8/22/2023 4:59:19 PM
Creation date
5/4/2018 11:15:04 AM
Metadata
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Template:
x Address Old
House Number
1290
Street Name
Orono Oaks
Street Type
Drive
Address
1290 Orono Oaks Drive
Document Type
Septic
PIN
3511823340018
Supplemental fields
ProcessedPID
Updated
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• <br /> MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. FLOW <br /> Estimated m ted Sewa`e Flows in Gallons per day <br /> �(11P40I <br /> 6,00 <br /> Estimated gpd (see pages D-7 or I-3,4, 5) <br /> Number <br /> or measured gpd x 1.5 = Bedrof ooms 1 type 11 Type III Tipe <br /> v <br /> B. SEPTIC TANK LIQUID VOLUMES 3 45000 300 Al 14 <br /> 4 600 373 256 'h <br /> -10 o 0 gallons (see pages C-3 or C-5) 5 750 450 294 1. <br /> 6 900 525 332 lts•1, <br /> 7 1050 600 370 1f°r <br /> C. SOILS (refer to site evaluation) <br /> 1/ <br /> 8 1200 675 408 °,,r,- <br /> 1. Depth to restricting layer= I cQ11-+0 1 'e inches Septic Task Cap/palm In,Huss <br /> 2. Depth of peicolation tests = I a, " inches <br /> Number, „ Minimum,scity id I.�gYleeadipaWhk <br /> C�+eNy Iliad d.P...t <br /> 3. Percolation rate I y .L-1 mpi ale.. 730 1125 <br /> 4. Land slope / anini , '210 'w0 <br /> 1� % .co 6 1500 2230 <br /> 7,!29 apOp MOO <br /> over9 .._.. <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 /> („U0 gpd x 0.83 sq. ft./gpd = 4,/ c sq. ft.-}- iO 'u= 541 <br /> 2. Select width f rock layer(10 feet or less) = /0 ft. <br /> 3. Length of r k layer = Area+ Width = <br /> s <br /> Li sq. ft.+ / 0 ft. = ss' ft. Rock Bed <br /> ;•r.r•r le r r•r•r•r•r•r r r r <br /> L L•'L•ti•{•ti•ti•ti•ti•ti.S.ti.{•{•ti.J <br /> {{{?•r•r•r•r•r•r•r•r•r.r.r•r•r <br /> .r. ti•ti•tijtif;•+•+ti•ti•ti•ti•ti•ti•1Nidth s10ft. <br /> r•r•r• tir`r•r.r•r•r.r.r•r� <br /> E. ROCK VOLUME f f f f r_r_ngth r rrr_r; <br /> Length <br /> 1. Multiply rook area by rock depth to get cubic feet of rock; ' <br /> ,Sc-i 9 sq. ft. x (., ft. = ') ' cu. ft. 1 <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> • <br /> <-)L4 cu. ft. +27= a I cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> a I cu. yd. x 1.4 ton/cu. yd. = ate, tons. • <br /> F. ADSORPTION WIDTH i✓l.A t._o lea 1i-1 <br /> 1. Percolation rate in top 12 inches of soil is ►u.y mpi <br /> Absorption Width Sizing'ruble <br /> 2. Select allowable soil loading rate from table on page E-; in Minu e.Rate pr soil Texture Gallonser aper Ratio°t <br /> „ Inch IMP!) y Absorption width <br /> Li. gpd/f t2 square foot to RWi l oyer <br /> Ct' WNNh <br /> 3. Calculate adsorption width ratio by dividing rock layer Past.,than 0.1• Coarse Sand -- <br /> 0.1 <br /> ._ <br /> loadingrate of 1.20gpd/ft2 byallowable soil loadingrate; 0'tos•• Sand 1.20 2.00 <br /> Pi=Sand•• 0.60 2.00 <br /> 1.20 gpd/ft2',+ •Li,�' gpd/ft2= 0 .L.'7 . 6to30 Sandy Loam 0.79 2.(11 <br /> Check this value on page E-16. • 46106; Silt Laarmtt ((IV 2.677 <br /> 4. Multiply adsorption width ratio by rock layer width to get war io 12`1 Clay °'2' *-- <br /> required <br /> s1l20 c1. <br /> required adsorption width; <br /> ?.to') x /u ft = M(,.7 ft <br /> . <br /> L_ <br />
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