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1996-008290 - new septic system
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Old Crystal Bay Road South
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0850 Old Crystal Bay Road South - 09-117-23-12-0005
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1996-008290 - new septic system
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Last modified
8/22/2023 3:18:01 PM
Creation date
4/2/2018 12:43:17 PM
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x Address Old
Address
0850 Old Crystal Bay Rd S
Document Type
Septic
PIN
0911723120005
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ProcessedPID
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MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> t <br /> A. FLOW Estimated Sewage Row in Gallons per Day(gpd) <br /> Estimated leo gpd Nianba <br /> of or measured -- x 1.5 = - gpd. Bedrooms ape I IN/pe II ..lY>x 111 Type IV <br /> 2 300 225 180 604 <br /> 3 450 300 218 of <br /> B. SEPTIC TANK LIQUID VOLUMES 4 600 375 256 ,`eoco <br /> 5 750 450 294 in <br /> a.-10 0 0 gallons 6 900 525 332 ur <br /> 7 1050 600 370 ,, <br /> i, 8 1200 675 408 <br /> C. SOILS (refer to site evaluation) aw-ss�� <br /> 5 E `� 1 •,-so 3a'' NttmbcMIMMONIS <br /> � . <br /> 1. Depth to restricting layer= 14"-To a vS inches , �,, Dogma <br /> 2. Depth of percolation tests = la" inches `..11-4 `""°°'' <br /> • <br /> 3. Percolation rate /0, to mpi 2 or less <br /> 750 1.100 <br /> 3 a 4 <br /> 1.000 1.500 <br /> 4. Land slope 4 % s or 6 1.500 <br /> 2.0032.250 <br /> 7 ors 3,000 <br /> ova 9 See fig.C-6 (x 1.5) <br /> i <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 /> loo gpd x 0.83 sq.ft./gpd = 49 u sq. ft.4-10%=54i." <br /> 2. Select width of rock layer(10 feet or less) = /a ft. <br /> 3. Length of rock layer = area=width = Rock Bed <br /> St-i r) sq. ft.i /o ft. = 5S ft. ti...�.ti.ti.ti...�.•...�.ti.,.�.�.�.ti.ti. <br /> ti,1•ti,1•ti,�•1 1•S•ti•ti•1•ti•ti•ti•ti•'ti• <br /> :•r•r•r•r•r•r•r•r•t f•f•J•t•t•t•1•fyyidth 5510 ft. <br /> ti.ti.1.ti.1•ti•�.t.t•'�.�.ti•t•ti•b•ti•'t• I <br /> •ti•%•%.%•. •ti•1•ti•ti•s•ti•ti•ti•%.s.reeee.f�f�lfaf•rlf�t_r�r'_fJ <br /> E. ROCK VOLUME 1......____ <br /> Length <br /> -___I ( <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> >4-w1 sq. ft. x ha.: ft. = 51y cu. ft. <br /> 2. Divide cu. ft.by 27 cu. ft./cu. yd. to get cubic yards; <br /> • <br /> 51y cu.ft. s 27= a ) cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; , <br /> a ) cu. yd. x 1.4 ton/cu. yd. = aei tons. <br /> F. ADSORPTION WIDTH G.L.AL( LO/IYV1Absorption WiidthSizing Table <br /> 1. Percolation rate in top 12 inches of soil is i o,L. mpi Percolation Rate Callen.Minutes per inch Soil Texture e°,„ d <br /> (mpi) h., b...wem <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 Coarse Sand 120 1.00 <br /> 0.1 to 5 Sand 1.20 1.00 <br /> Li gpd/ft2 0.1 to 5 Fine Sand" 0.60 2.00 <br /> 6 to 15 Sandy Loam 0.79 1.52 <br /> 163. Calculate adsorption width ratio bydividingrock layer 31 to 40 Loam 0.60 2.00 <br /> rp y 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 Clay Loam 0.45 2.67 <br /> 61 t0120 Clay 0.24 5.00 <br /> 1.20 gpd/ft2y ,t.1 5 gpd/ft2= z.t. ? • Slower than 120 Clay - - <br /> Soil having 50%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 /> a.t,? x Jo ft=a1..0 ft <br />
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