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2008-00230 - new septic
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3765 Watertown Road - 32-118-23-34-0013
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2008-00230 - new septic
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Last modified
8/22/2023 4:41:01 PM
Creation date
10/23/2019 1:47:10 PM
Metadata
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Template:
x Address Old
House Number
3765
Street Name
Watertown
Street Type
Road
Address
3765 Watertown Road
Document Type
Septic
PIN
3211823340013
Supplemental fields
ProcessedPID
Updated
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1V10LIND DESIGN WORK SHEET (For Flows up to 1200 d) <br /> A. Average Design FLOW ( u �1 <br /> / A-1: Estimated Sewage Flows in Gallons per Day <br /> Estin-lated 100 gpd (see figure A-1) number of a., <br /> measured _ x 1.5 (safe actor) = d bedrooms Class I Class 11 Class III Claw <br /> or <br /> (safety f gpd 2 300 225 180 60`w <br /> 3 450 300 218 of the <br /> i. SEPTIC TANK Capacity 4 600 375 256 values <br /> 09 t>)a `o"qe 5 750 450 294 in the <br /> r 61 � 525 332 Class I, <br /> gallons (see figure C-1) � � ``� 1050 . 600 370 11, or III <br /> 8 1200 675 408 columns. <br /> SOILS (refer to site evaluation) C-1: Septic Tank Ca cities in allons <br /> Liquid capacity <br /> Number of Minimum Liquid Liquid capacity with with disposal&1. Depth to restrictinglayer feet — �� Bedrooms Capacity garbage <br /> disposal lift inside <br /> 2. Depth of percolation tests =_> feet _,�.L�f 2 or less 750 1125 1500 <br /> 3. 1 extureAr,4 =a :_14y � ���f/,�0�� 3or4 1000 1500 <br /> ^�Q � , p A(n or X1,500 2250 � <br /> Percolation rate—:3 9. mpi "T 6. 60 p� %� Sor9 2000 3000 3000 <br /> 4. Soil loading rate 6 v 2 _gpd/sgft (see figure D-33) <br /> 5. Percent land slope 'T-.0 % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow (A) by 0.83 to obtain required rock layer area. <br /> _- Go gpd x 0.83 sgft/gpd = ���sqft <br /> 2. Determine rock layer width = 0.83 sgft/gpd x linear Loading Rate (LLR) <br /> 0.83 sqft/gpd x /2 )14, gpd/sgft = /0 ft <br /> 3. Length of rock layer = area=width= Mound LLR <br /> 75c sft (D1) _ /b ft (132) _ 23 ft �- , < 1 <br /> E. <1 2 <br /> �ROCK VOLUME ���' 10 x 3 �� S <br /> > 120 MP < b <br /> -1. Multiply rock area (D1) by rock depth.of 1 ft to get cubic feet of rock <br /> 7y sgftx1ft = `', & cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> __ `9 t cd cuft =27 cuyd/cuft = Z9 cuyd <br /> 3. .Multi ly cubic yards by 1.4 to get weight of rock in tons <br /> _ cuyd x 1.4 ton/cuyd Yc) tons v-la v/ <br /> D-33: Absorption Width Sizing Uble <br /> F. SEWAGE ABSORPTION WIDTH Percolation Rate Loading Rate <br /> in Minutes per Soil Texture Gallons Absorption <br /> Inch per duy er Ratio <br /> NPI) s.oars topot <br /> Faster than 5 Course Sand 1.20 I.00 <br /> Absorption width equals absorption ratio (See Figure D-33) Mediu-SSpnandd <br /> times rock layer width (D2) 6 to <br /> Fine Sand <br /> 6 to 30 Loam _ 0.60._ 2.00 <br /> 31 to 4 Silt Ltwm 0.511 <br /> Z: x /a ft = 2��ft ------- _. _.---- <br /> ___ 46 to 60 S >ry Clay Lor 0.45 2.67 <br /> ty Clay Loa <br /> - �._. <br /> - - --- ---_ - <br /> dt,L 61 to 120 - °1—�''��-•Cl 0.24 S.W <br /> /! Sandy Clay <br /> Fc-�cj,-� <br /> 4 \v —(n� m'1 ( _ Cls <br /> •System deaignW for axle soils moat 6a udu:r or perfonnaiu:e <br />
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