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1997-009539 - new septic
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1530 Tanglewood Road - 26-118-23-32-0011
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1997-009539 - new septic
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Last modified
8/22/2023 4:17:10 PM
Creation date
4/17/2019 10:55:21 AM
Metadata
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Template:
x Address Old
House Number
1530
Street Name
Tanglewood
Street Type
Road
Address
1530 Tanglewood Road
Document Type
Septic
PIN
2611823320011
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> l T <br /> A. FLOW Eattmated Sewate Flow in Gallons per Day(Vd) <br /> Estimated (-Oo gpd "`o°ef 7ype t 1pe a Type in Type ry <br /> or measured - x 1.5 = - gpd. see►ooma <br /> 2 700 ns 160 boa <br /> 7 Oso 700 216 •� <br /> B. SEPTIC TANK LIQUID VOLUMES `s 7600s0 iso 294 256 <br /> -i Do a gallons 6 , iso <br /> m� <br /> a 1 1200 1 675 1 406 <br /> C. SOILS (refer to site evaluation) N mbar ►�-- <br /> 1. Depth to restricting layer= 1-a 1 1 -6�� inches of <br /> sed�aaa <br /> 2. Depth of percolation tests = 1 a" inches <br /> 3. Percolation rate mpi 23«� 1.0 1,'s 0 <br /> 4. Land slope L� . % 7 or i 12.000 7 0 0 <br /> ev.9 Soa tit.C.6 (a►s) <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 /> i-oo . gpd x 0.83 sq. ft./gpd = 494S sq. ft. <br /> 2. Select width of rock layer(10 feet or less) = / o ft. <br /> 3. Length of rock layer = area+width= Rock Bed <br /> x-19 4l sq. ft.+ .Lp_ft. _ -<0 ft. <br /> �•,M•; .;•;•;•,•,4 �•,.,.,., Idth 510 ft. <br /> /.Hs•}? M.S./Hs/�f././././. <br /> E. ROCK VOLUME ~- Leto -� <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> 49 y sq.ft. x /.o.0t. _ rak cu. ft. <br /> 2. Divide cu.ft.by 27 cu.ft:/cu. yd. to get cubic yards; <br /> .Qa,cu.ft. +27=,a D cu. yd. <br /> 3. Multiply.cubic yards by 1.4 to get weight of rock in tons; <br /> _as2 cu.yd. x 1.4 ton/cu.yd. _w. tons. <br /> F. ADSOFYnON WIDTH �`-'` �- Width Slain Dbit <br /> 1. Percolation rate in top 12 inches of soil is 30l mpi M� Rate C4-soil Texture �,� Ma: <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 oam Sand 1.20 1.00 <br /> d/ffi� 0.1 to Sand 1.20 1.00 <br /> $P 0.1 to 5 Fine Sand" 0.60 2.00 <br /> 6 to 15 Sandy Loam 0.79 1-52 <br /> 3. Calculate adsorption width ratio b dividing rock layer 16 to 45 Loam o.60 2.00 <br /> rp y $ y 31 to 45 Silt Loam 050 2.40 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; 46 to 60 Cla Loam o.as 2.67 <br /> 1.20 gpd/ft2+ , 61 t an May 0sa - <br /> _�gpd/ftz = a" Le� Slower than 120 Clay <br /> "SW ha"50%of men d fine orvcy fw au�d <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> �L7 x 10 ft = ft <br />
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