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1180 Willow Dr N - 27-118-23-32-0003
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Septic System Approval
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Last modified
8/22/2023 4:20:03 PM
Creation date
2/11/2020 11:39:56 AM
Metadata
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Template:
x Address Old
House Number
1180
Street Name
Willow
Street Type
Drive
Street Direction
North
Address
1180 Willow Drive North
Document Type
Septic
PIN
2711823320003
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ProcessedPID
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MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. FLOW Euimated Senate Flow in Gallant per Day(gpd) <br /> Estimated L4S0 gpd "obitir <br /> f Type t Type I Type 111 Type ry <br /> or measured - x 1.5 = - gpd. see►ee111a <br /> 2 700 225 Ito ao. <br /> B. SEPTIC TANK LIQUID VOLUMES 3 600 300 2 6 <br /> 00 <br /> 4 600 375 256 <br /> S 750 ISO 294 <br /> -100D gallons 6 9W S25 7 1050 600 370 <br /> t 1200 1 67S 406 <br /> C. SOILS (refer to site evaluation) NltmberAGINkiiiiell <br /> f CA-my <br /> 1. Depth to restricting layer= I ii, inches 2,4" <br /> 2. Depth of percolation tests = ) a. inches `'"' <br /> 3. Percolation rate •S mpi 2or 3 w Wo17,WO 50 1 's o <br /> ori 'ofo0004. Land slope % 7 z 3 <br /> ever 9 Sia 6t.C.6 (a 15) <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 /> 4So . gpd x 0.83 sq. ft./gpd = 373 sq. ft.-+)d-50-y)0 <br /> 2. Select width of rock layer(10 feet or less) = J0 ft. <br /> 3. Length of rock layer = area+width= Rock Bed <br /> 410 sq. ft. + j 0 ft. 14 1 ft. ,•�•. ♦ti M� .. .,,., ,., <br /> Me.rrrr? •r r• dth :510 ft. <br /> •r•r•r•�r•�• r•r •r• r•r• <br /> •r•r•r•r•r•r.r• •r•r•r•r?r.r•r•r• <br /> E. ROCK VOLUME I--- Length -� <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> L4 I o, sq.ft.x ft. =yaD-cu. ft. <br /> 2. Divide cu. ft.by 27 cu. ft:/cu. yd. to get cubic yards; <br /> HzL cu.ft. +27=�JL cu. yd. <br /> 3. Multiply.cubic yards by 1.4 to get weight of rock in tons; <br /> !t,cu. yd.x 1.4 ton/cu. yd. _�. tons. <br /> F. ADSORPTION WIDTH e-L-A,4 LO Wv^ Ab a,Width st:y, T:ble <br /> 1. Percolation rate in top 12 inches of soil is :9.S mpi pfficaatlo,wk ~.4 <br /> )„ti,,,,us, <br /> • <br /> U inch Soil Texture rr a P.&~' <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 oars Sand 1.20 1.00 <br /> 0.1 to 5 Sand 1.20 1.00 <br /> _4<', gpd/ft2 0.1 to 5 Fine Sand" 0.60 2.00 <br /> 6 to 15 Sandy Loam 0.79 152 <br /> 3. Calculate adsorption width ratio b dividing rock layer 16 to 45 Loam 0.60 2.00 <br /> rp y $ y 31 to 45 Silt Loam Oso 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 tO 120 Clay 024 5.00 <br /> 1.20 gpd/ft2+,y gpd/ft2 Slower than 120 Clay - - <br /> "Soll havfnt 50%or man of fir*or very=-d. <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> D.1") x /I) ft aje,,L ft <br />
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