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1997-009688 - new septic system
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2605 Thoroughbred Lane - 04-117-23-11-0022
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1997-009688 - new septic system
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Last modified
8/22/2023 5:06:39 PM
Creation date
4/24/2019 9:27:42 AM
Metadata
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Template:
x Address Old
House Number
2605
Street Name
Thoroughbred
Street Type
Lane
Address
2605 Thoroughbred La
Document Type
Septic
PIN
0411723110022
Supplemental fields
ProcessedPID
Updated
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' MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. FLOW Estimated Sewage Flow in Gallons per Day(gpd) <br /> Estimated loo gpd Nwnber <br /> or measured x 1.5 = gpd. Bedrofoms ape I Type II Iype III Type iv <br /> 2 300 225 Iso sac <br /> 3 450 300 218 of <br /> B. SEPTIC TANK LIQUID VOLUMES 4 600 375 256 . <br /> 5 750 450 294 <br /> 1 Iowa gallons 6 90 boo 330tha <br /> edamo <br /> 8 1 1200 1 675 408 <br /> C. SOILS (refer to site evaluation) 1 ' <br /> Num ," cm.QW® <br /> 1. Depth to restricting layer= �.ai1 , Sd inches B =�`edrooms Co-" ,� <br /> 2. Depth of percolation tests = 1 a- inches `"t`'' `Pu-) <br /> 3. Percolation rate 1S.0 m i 2 or las 750 1.125 <br /> p 3 or 4 1.000 1.500 <br /> 4. Land slope a % 7 a s 2. 000 00 <br /> over 9 See fig.C-6 (a 1S) <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 /> %2o gpd x 0.83 sq. ft./gpd = 04,) sq. ft: <br /> 2. Select width of rock layer(10 feet or less) _ /D ft. <br /> 3. Length of rock layer = area?-width = Rock Bed <br /> ,Q,Li 9 sq. ft.y 0 ft. 95' ft. ..,.ti.ti.,...ti...,.ti...�.,.�.ti.ti., <br /> r•r•r•�.f•r•r•r•r r•r•r•r•r•r.f• <br /> ti•ti•♦•�•4•ti•�•ti•ti•�•ti•1•ti•ti•ti•ti•'L <br /> r•r•r•r•r•r•r•f'r f'f'f'f•f•r•f' ldth :510 ft. <br /> S•ti•ti•ti•S•t•t•ti.tK•ti•ti•tiM.K•'t•t <br /> ti fti f1 fti fti fti jti f�fti fti fti ft fti ft fti f'�fti <br /> E. ROCK VOLUME 1- Length <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> !14 sq.ft. x .of ft. _ t cu. ft. <br /> 2. Divide cu.ft.by 27 cu. ft./cu. yd. to get cubic yards; <br /> ,-cu. ft. i 27=_ 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. =yQ tons. <br /> F. ADSORPTION WIDTH GL 14 q to 14 N'1 � Width Si=X Table <br /> 1. Percolation rate in top 12 inches of soil is mpi Percolation Rate C.11- a.md <br /> lylny mPPet� Soil Texture �„- 1Odbd <br /> th <br /> � i� � <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 Coarse Sand 1.20 1.00 <br /> d/ft2 0.1 to 5 Sand 1.20 1.00 <br /> -��- 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 30 Loam o.60 2.00 <br /> rp � y g 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 cla Loam 0.45 2.67 <br /> 1.20 gpd/ftzi J 4 S gpd/ft-1= a _ Slower than 120 Clay 024 5.00 <br /> -SW Soil having 509 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 /> .1,2 xyo_ft=aft <br />
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