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1994-006102 - septic system
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1994-006102 - septic system
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Last modified
8/22/2023 4:27:22 PM
Creation date
6/27/2019 1:15:31 PM
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Template:
x Address Old
House Number
1040
Street Name
Town Line
Street Type
Road
Address
1040 Town Line Road
Document Type
Septic
PIN
3011823320004
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. FLOW F-Wrnated Sewage IFlows in Gallons per day <br /> Estimated f-t-�gpd (seepages D-7 or I-3,4,5) (gpd,n„ <br /> tter <br /> or measured gpd x 1.5 = Of Type I Type Il Type III Type <br /> Bedroom: IV <br /> 2 300 225 180 <br /> B. SEPTIC TANK LIQUID VOLUMES 3 450 300 218 <br /> 4 600 375 256 Of <br /> •dna <br /> 0 o 0 gallons (see pages C-3 or C-5) 50 Su 3z mal. <br /> 7 1050 600 370 to <br /> C. SOILS (refer to site evaluation) 8 1200 675 408 C01115t , <br /> 1. Depth to restricting layer = as -(o inches Septic Tank Capacities,ho gallons <br /> ,� -Number of Minimum Liquid Liquid capacity with <br /> 2. Depth of percolation tests = l a inches Bedrooms Capacity aarba a dib <br /> C pusal <br /> 3. Percolation rate L-) , mpi 2,«" o o Im <br /> 4. Land slope (� % 7.8orr9 1RM <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 = t <br /> HZo gpd x 0.83 sq. ft./gpd = 302 sq. ft.-0 y 10 <br /> 2. Select width of rock layer (10 feet or less) = I a ft. <br /> 3. Length of rock layer = Area+ Width = <br /> 0 sq. ft. + / 0 ft. = L) _ ft. Rock Bed <br /> r;r•r•r•r•r•r•r•r•r•r•r•:•r�: <br /> r�r•r•r•r•r tir•r•r•r•r•r•r•r•r.ti.ti ti.ti.ti.{ ti•ti.ti..,.ti.ti.ti.ti. Width S]0 ft. <br /> r•r•r•r•r•r•r•r r;r;r;r;r;r;r <br /> E. ROCK VOLUME ti•ti•••ti.,•ti•ti•ti.... . <br /> r•r•r•r•r•r•r.r•::r:r::•::::r <br /> Length --� <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> 4Ly sq. ft. xLf ft. _ LLT <br /> z2 cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> ti?�) cu. ft. +27 =- I (.-. cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> cu. yd. x 1.4 ton/cu. yd. tons. <br /> F. ADSORPTION WIDTH L�'� �{ <br /> 1. Percolation rate in top 12 inches of soil is '%I mpi Absorption Width Sizing Tabic <br /> Percolation Rate Gallons Ratio of <br /> 2. Select allowable soil loadin�rate from table on page E-; in Minutes per Soil Texture per day per Absorption width <br /> ��- gpd/f Inch(MPI) square footto RVAdth ser <br /> 3. Calculate adsorption width ratio by dividing rock layer Faster than 0.1• Coarse Sutd <br /> 0.1 to 5 Sand 1.20 1.00 <br /> loading rate of 1.20 gpd/fF by allowable soil loading rate; 0.1 to 5•• Fine Sand•• 0.60 2.00 <br /> 6 to 15 Sandy Loam 0.79 1.52 <br /> 1.20 gpd/ft2+ gpd/ft2= I6 to 30 Loam 0.60 2.(10 <br /> 31 to 45 Silt Loam 0.50 2.40 <br /> Check this value on page E-16. 60 , C my L a'n oia 2.67 <br /> 4. Multiply adsorption width ratio by rock layer width to get Slower clay _. <br /> required adsorption width; <br /> D•1,0 x /0 ft =ate•0 ft <br />
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