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1993 Septic System Approval
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2700 White Oak Cir - 04-117-23-42-0021
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1993 Septic System Approval
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Last modified
8/22/2023 5:14:18 PM
Creation date
2/3/2020 9:14:39 AM
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x Address Old
House Number
2700
Street Name
White Oak
Street Type
Circle
Address
2700 White Oak Cir
Document Type
Septic
PIN
0411723420021
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' MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. FLOW Estimated Sewage Flows in Gallons per da I <br /> (gPd) y <br /> Estimated ((o 0 gpd (see pages D-7 or I-3,4,5) Number <br /> or measured - gpd x 1.5 = . BedroomsType I Type 11 Type III 1 pe <br /> B. SEPTIC TANK LIQUID VOLUMES 3 45000 3000 tie <br /> 4 600 375 256 t <br /> lues <br /> -/o o O gallons (see pages C-3 or C-5) 5 750 450 294 ryi�sa <br /> 6 900 525 332 T t. <br /> 7 1050 600 370 m <br /> C. SOILS (refer to site evaluation) 8 1200 675 408 columns <br /> II <br /> 11 <br /> 1. Depth to restricting layer = D\L) -I'D a c inches Sepik Tank Capacities,in gallons <br /> t Number of Minimum Liquid Liquid capacity with <br /> 2. Depth of percolation tests = inches Bedrooms Capacity garbage disposal <br /> 3. Percolation rate P-1. ' mpi 2 or less 750 1125 <br /> 3«4 1000 1500 <br /> 4. Land slope % 4«6 15W 2250 <br /> 7.8«9 2000 1000 <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 /> (220 o gpd x 0.83 sq. ft./gpd = 4*''i sq. fta 1 coo=540 <br /> 2. Select width of rock layer (10 feet or less) = /o ft. <br /> 3. Length of rock layer = Area+ Width = • <br /> 54 1 sq. ft. _ / 0 ft. = ,s--g- ft. Rock Bed <br /> tr.f••.%ti.f.f.j.ff : ff-f.f <br /> :titif?ti.•;fti. .titfti•tirtirr+rti•.frti.:rrti•f •rr.S•rrr•tif <br /> 1 <br /> N <br /> kith 570 ft. <br /> '• f•• f- -f-� gt�-r-ffelE. ROCK VOLUMEi = - Length -i <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> S'-1') sq. ft. x /.oSft. = S1L Cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> Sly cu. ft. -27= a ► cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> D, l cu. yd. x 1.4 ton/cu. yd. = 2. i tons. <br /> F. ADSORPTION WIDTH - w fri-VV 1 <br /> 1. Percolation rate in top 12 inches of soil is i y-c, mpi Absorption Width Sizing Table <br /> 2. Select allowable soil loading rate from table on p in a g e E-; iPn inon Minutes per Soil Texture per dRate Gad <br /> y ps Ratio of <br /> ay per Absorption width <br /> J/ d f t2 Inch(MPI) square foot to Rock Layer <br /> • <br /> bi / Width <br /> 3. Calculate adsorption width ratio by dividing rock layer Faster than 0.1• Coarse Sand <br /> loadingrate of 1.20gpd/ft2 byallowable soil loadingrate; 0.1 to Sand •• 1.20 2.00 <br /> Fine Sand 0.60 2.00 <br /> 1.20 gpd/ft2 i .1-1S/ gpd/ft2= a-La16 t <br /> ') . 6tt 1s Sandyo 30 Loam nsLo0.79 1.52 <br /> 0.60 2.00 <br /> 31 to 45 Silt Loam 0.50., 2.40 <br /> Check this value on page E-16. 46 vLoam 2.67 <br /> i <br /> ie <br /> Slower <br /> 4. Multiply adsorption width ratio by rock layer width to get 640 <br /> 0we,th,oan Clay 6° "'Clay <br /> required adsorption width; <br /> a.(,01 x JO ft = ai..r) ft <br />
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