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1997-009024 - new septic system
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2430 Thoroughbred Lane - 04-117-23-11-0013
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1997-009024 - new septic system
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Last modified
8/22/2023 5:06:00 PM
Creation date
4/22/2019 11:11:53 AM
Metadata
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Template:
x Address Old
House Number
2430
Street Name
Thoroughbred
Street Type
Lane
Address
2430 Thoroughbred La
Document Type
Septic
PIN
0411723110013
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. FLOW Estimated Sewage flow in Gallons per Day(gpd) <br /> Estimated O gpd <br /> Number <br /> Tylie <br /> or measured x 1.5 = - gpd. Bedrooms I Type II Type III Type IV <br /> 2 300 225 l80 boa, <br /> 3 450 300 218 of <br /> the <br /> B. SEPTIC TANK LIQUID VOLUMES 4 600 375 2% . <br /> 5 750 450 294 <br /> I-) ,Sy 4 1-1000 gallons 6 900 525 332 dw <br /> 7 1050 600 370 °dw <br /> col� <br /> 8 1 1200 1 675 1 408 <br /> C. SOILS(refer to site evaluation) NumberST; s <br /> 1. Depth to restricting layer = ](I inches sea° <br /> Of C.P.,,Y <br /> 2. Depth of percolation tests inches `Pu-4 `"am'' <br /> 3. Percolation rate 4s.S- mpi 23 al4 17050 1.'12 <br /> 4: Land slope - % 7 or s 2000 30000 <br /> over 9 See fig.C-6 (11-5) <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer:A x 0.83 = <br /> 7 Sb gpd x 0.83 sq. ft./gpd sq. ft-v 1012o=lo" <br /> 2. Select width of rock layer(10 feet or less) _ >0 ft. <br /> 3. Length of rock layer=area_width = Rock Bed <br /> (o_l6�sq. ft._ )D _ft. = (,<I ft. <br /> ti.ti•ti.ti•ti.ti•1•ti•��•�•1.ti•ti•ti•ti•ti <br /> !•r•!•!•!•!... . .!•!•!•r•r.!. . idth 510 ft. <br /> ti•ti•ti•ti•ti•ti•ti•ti• ti•ti•L•1•ti•ti•ti•ti <br /> ~!~!~!1!t 1~!�l~!~!�!•t f~J•!1~ <br /> E. ROCK VOLUME l- Length i <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> (may sq.ft. x i.dS ft. =,7)q cu.ft. <br /> 2. Divide cu.ft.by 27 cu. ft./cu. yd. to get cubic yards; <br /> 0)41 cu. ft. _27=_1)cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> a r) cu.yd. x 1.4 ton/cu. yd. _ tons. <br /> F. ADSORPTION WIDTH wA`v-) <br /> Absorptionw,achsian Tab,e <br /> Gua,• It-41 <br /> 1. Percolation rate in top 12 inches of soil is mpi e o>��Raw Soil Texts �f,� d <br /> W <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 Coarse Sand 120 1.00 <br /> 0.1 to S Sand 120 1.00 <br /> 'Li!� gpd/f t2 0.1 to S Fine Sand- 0.60 2.00 <br /> 6 to 15 Sandv Loam 0.79 1.52 <br /> 3. Calculate adsorption width ratio b dividing rock laver 16 to 3o Loam 0.60 2.00 <br /> rp }' g 31 to 45 Silt Loam 0.50 2.40 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; 46 to 60 Clav Loam 0.45 2.67 <br /> 61 t0120 flay 024 5.00 <br /> 1.20 gpd/ft2�- .Ligpd/ft2= a.u Slower than 120 Clav - - <br /> ••Soil having 3D%or more of fine or very fine sand <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> ,Z_ .6-7 x_ 0 eft <br />
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