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1994-006290 - replace septic
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0540 Old Crystal Bay Road South - 04-117-23-42-0030 - New PID
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1994-006290 - replace septic
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Last modified
8/22/2023 3:12:25 PM
Creation date
3/26/2018 3:15:29 PM
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x Address Old
Address
0540 Old Crystal Bay Rd S
Document Type
Septic
PIN
0411723420030
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ProcessedPID
Updated
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, MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> 1 A. FLOWr Estimated Sewage Flows in Gallons pix day <br /> Estimated 600 gpd (seq pages D-7 or I-3, 4, 5) Lig-,41.4.1..1. <br /> " ) ' Number (GO) <br /> or measured gpd x11.5 = of Type I Type II Tyle III Type <br /> Bedrooms I V <br /> 2 300 225 180 <br /> B. SEPTIC TANK LIQUID VOLUMES 3 450 acro 218 61:1% <br /> 4 600 375 256 er'he <br /> 'rims <br /> 2-/cC° 6 I tr 4 gallons (see1pages C-3 or C-5) 5 750 450 294 n <br /> 9(X) 525 332 Ty at. <br /> I-wooi}:t,.1) :.:4.,:i, 7 1050 600 370 <br /> 8 1200 675 408 columns <br /> C. SOILS (refer to site evaluati n) <br /> 1. Depth to restricting layr = / (' inches Septic Tank Capacities;iu gallon <br /> Number of Minimum Liquid Liquid capacity with <br /> 2. Depth of percolation to is = /i) inches Ikdrnoms Capacity garbage disposal <br /> 1125ss <br /> k <br /> 3. Percolation rate , 750;/A mpi a 2 or k 100 1125 <br /> 4. Land slope 3 % 4`x 6 <br /> 7,R or 9 1501 2250 2100 3410 <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 = <br /> (v0 gpd x 0.83 Sq. ft./gpd = litig sq. ft. <br /> 2. Select width of rock layr (10 feet or less) = /0 ft. • <br /> 3. Length of rock layer = rea _ Width = <br /> 'IC1 sq. ft. - /0 ft. = L19/ ft. Rock Bed <br /> -T:T <br /> rJ . f idthJtl <br /> E. ROCK VOLUME 1--•• iYiLe�.'ti) • 1 <br /> 1. Multiply rock area by z1ock depth to get cubic feet of rock; _S‘ ' <br /> 500 sq. ft. x l ft. =500 Cu. ft. <br /> 2. Divide cu. ft. by 27 cu. t./cu. yd. to get cubic yards; <br /> 51)0 Cu. ft. -s 7 = /5,6'i cu. yd. <br /> 3. Multiply cubic yards b 1.4 to get weight of rock in tons; <br /> f '. t cu. yd. x 1.4 tOn/cu. yd. = d to tons. <br /> i <br /> F. ADSORPTION WIDTH <br /> 1. Percolation rate in top 12 i ches of soil is 4g mpi Absorption Width Sizing rabic <br /> Percolation Rare Gallons Ratio of <br /> 2. Select allowable soil loadi g rate from table on page E-; in Minutes per Soil Texture per clay per Absorption width <br /> Inch IMM) square fait to Rock Layer <br /> a < gpd/f t2 Width <br /> 3. Calculate adsorption width ratio by dividing rock layer Faster than 0.1• Coarse Sand <br /> 0.1 to 5 Sand 1.20 1.00 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; 0.Ito5•• rine Sand•• 0.60 2.110 <br /> 1.20 d/ft2 r to d/ft2= a,OO 6to36 10 SandyLoamm 0.79 1.52 <br /> .i � 16 m.lU Loam 0.60 2.1x) <br /> gP 31 to 45 Silt Loans 0.50 2.40 <br /> Check this value on page E-16. ; o 12Clay <br /> l y Loam 0.45 <br /> 2.67 <br /> 60 ; 1 0.24 <br /> 4. Multiply adsorption width atio by rock layer width to get Sl .!h.an Clay <br /> required adsorption width; <br /> ,,00 x /0 1ft = cl0 ft <br /> Li <br />
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