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1996-007851 - replacement system
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4045 Watertown Road - 31-118-23-41-0003
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1996-007851 - replacement system
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Last modified
8/22/2023 4:31:52 PM
Creation date
12/2/2019 1:10:45 PM
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x Address Old
House Number
4045
Street Name
Watertown
Street Type
Road
Address
4045 Watertown Road
Document Type
Septic
PIN
3111823410003
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Updated
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MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. FLOW Estimated Sewage Flow in Gallons pa Day(gpd) <br /> Estimated O 0 gpd Number j• - <br /> or measured=x 1.5 = -tid. °� �Ya 1 Glx 1t Type n 1)' <br /> 1 •`` ;•, <br /> gP Bedroom, pe IV • <br /> 2 300 225 180 so% <br /> B. SEPTIC TANK LIQUID VOLUMES 3 450 300 218 •r <br /> 600 375 256 <br /> I- )000 t3 I-7000 gallons r%1i i' • 54 750 Iso 294 ie <br /> 6 900 sat 332 G. <br /> 7 1050 600 370 edur <br /> C. SOILS(refer to site evaluation) -F-N. --0-11t-' -- 8 1200 675 408 <br /> 1y, IL; au <br /> 1. Depth to restricting layer= ,�f •_)'', 41 x v inches Nuof <br /> mber :'` '1 <br /> 2_ Depth of percolation tests = a" inches •Bedrooms <br /> `m'' °`P°"' <br /> 3. Percolation rate 5, L-I mpi 2 or less 750 1,125 <br /> 4. Land slope "4 % 5 or 6 1,000 1.000 <br /> 7or8 .500 <br /> 2000 00 <br /> over 9 See fig.C-6 (x 1$) <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 /> X00 • gpd x 0.83 sq. ft./gpd = L1 t' sq. ft.'-toho sq 7°. <br /> 2. Select width of rock layer (10 feet or less) = lc.) ft. <br /> 3. Length of rock layer = area_width = Rock Bed <br /> Sy') sq. ft. i l0 ft. = Sc ft. E.;:;:-•-•-.i •-�............... <br /> ti�ti•ti'q'ti•ti•S•1:tiiti�t�fti��l�l�l�f I <br /> ti•ti.ti r.ti.ti.ti.; ti ti•ti•ti.{.{.• ZO ft. <br /> 'rtilti�.l.l•!•!•l•f.f.l•r•!•!•lti�ti}}r idth S <br /> !=!.!:!:!�l:t:t:lel:l:t!�J <br /> E. ROCK VOLUME t Length ____ .....1 <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> - 2 sq.ft. x 1.vc ft. = S7 4 cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> 5)4 cu.ft. ÷27= I cu. yd. <br /> • <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. _ 1 tons. <br /> i <br /> F. ADSORPTION WIDTH -LA-LI LOA-Y 1 <br /> ^ °°^Width s <br /> 1. Percolation rate in top 12 inches of soil is .S.y mpi i t Table <br /> peccohtien Rate �� Retie d e <br /> Minutes per inch Soil Texturedr '°d'e•e <br /> 2. Select allowable soil loading rate from table; (n pi) p"t' a1�, <br /> Faster than 0.1 Coarse Sand 1.20 1.00 <br /> 1-1S gpd/ft2 0.1 toy Sand 1.20 1.00 <br /> 0.1 to Fine Sand" 0.60 2.00 <br /> 3. Calculate adsorption width ratio by dividing rock layer 6 to 15 Sandy Loam 0.79 2.00 <br /> 161 to 30 Loam 050 2.40 <br /> •y <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading 46 rate; 46 to 45 Silt Loam 0o 2.40 <br /> to 60 Clay Loam 0.45 2.67 • <br /> 1.20 gpd/ft2y ,u gpd/f.2 = Q.�,I 61 to 120 Clay 0.24 s.00 <br /> Slower than 120 Clay -- <br /> • <br /> 4. Multiply adsorption width ratio by rock layer width to get -Soil having 509Gormore e(fineorvery SA,sand- <br /> required adsorption width; <br /> a.i, x Jo ft = Dt,., ft 3 <br />
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