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1994-006171 - new septic
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4545 Watertown Road - 31-118-23-24-0004
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1994-006171 - new septic
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Last modified
8/22/2023 4:30:23 PM
Creation date
12/16/2019 10:27:31 AM
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x Address Old
House Number
4545
Street Name
Watertown
Street Type
Road
Address
4545 Watertown Road
Document Type
Permits/Inspections
PIN
3111823240004
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. E-19 <br /> MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) . <br /> rki S'rp..1 e„ c}e <br /> r = e FLOW • D-7 <br /> 1 , istimated (200 gpd (see pages D-7 or I-3, 4, 5) ESTIMATED SEWAGE rLOWS IN GALLONS PEA CAT <br /> 14.44310104- TrK OF RESIDENCE° <br /> or measured gpd. fIEDROGAIS I 12 <br /> a 122 <br /> 2 300 223 16010; <br /> 3 450 300 214 •, <br /> 4 600 375 236 "O1"" <br /> B. SEPTIC TANK LIQUID VOLUMES 33 ' z6 450 <br /> 0'. 1 c-`4(? gallons (see pages C-3 or C-5) 1050 <br /> � 675 Gehonn. <br /> 370 <br /> C-3 I.• <br /> C. SOILS (refer to site evaluation) SEPTIC TANK CAPACITIES, IN GALLONS , <br /> 1. Depth to restricting layer = / NL606R o. 1.wa* .T,1 D.R..0*inches 1I1N` 911404GAPACIT <br /> @10010 UOOm OAPACRT D12P08•L <br /> 2. Depth of percolation tests = ID- inches <br /> 2 011 22311 781 1121 <br /> 3. Percolation rate Li, 7 mpi 6•o t l/ / i3,), "3 10111 ,... 1140 <br /> 4. Land slope % <br /> .0 $ <br /> T.\ 148 11 <br /> OR 2 !. 2217.• 7077 <br /> • <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 = 1 <br /> 1,-,(-0 gpd x-0:8.1 sq. ft./gpd =/r(-,C sq. ft. <br /> 2. Select width of rock layer (10 feet or less) = IC, ft. <br /> 3. Length of rock layer = area+ width = <br /> (-)C, sq. ft. + i 0 ft. = (()(1 ft. Rock Bed <br /> 157r•r•r•0 r•r•r•r r•r•r•r•r•r•r <br /> ti.ti.ti.ti.ti.ti.ti.ti.�.ti.ti.ti.ti.t.ti.. <br /> r.r•r•r•r•r•r•r•r•r•r.r•r•r•r <br /> 1.ti•ti.%...%.%.% %.•. .. .. .. '. :Width 51°ft- <br /> 4•r•r•r•r•r•r•r•r•r•r•r•r•r•r1 � <br /> .,r•r•r•r•r_r.r.r•r•r_r•r•r•r•t• <br /> E. ROCK VOLUME Length <br /> 1. Multiplyrock area by rock depth to get cubic feet of rock; <br /> fi'(, sq. ft. x I ft. = i‘,6()cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> tt() cu. ft. j-27= cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> d9-.Zcu. yd. x 1.4 ton/cu. yd. = 3/1 tons. <br /> 1. <br /> F. ADSORPTION WIDTH <br /> 1. Percolation rate in top 12 inches of soil is 141 7 mpi E-16 <br /> 2. Select allowable soil loading rate from table on page E-16; ALLONIMMUI, NAM 0,WWI uroc2MOuMDS <br /> tl P••••w.w A�.d.1••01 :: <br /> 1 gpd/ft2 -'IWO «�� <br /> 3. Calculate adsorption width ratidividingrock layer " " °""• `""' '•"•' `"`• z <br /> 1 by 2_, ° 110 280 2.01 1.00 <br /> loading rate of 1.20 gpd/ft2 by allowable Toil loading rate; 4 21 °" "' 3.421.12 <br /> 11 20 ..10 011 2.w 2,00 <br /> 1.20 gpd/ft2 1 gpd/ft2 = 21 41 0.20 •N 2w 2..7 <br /> .1 II 0.•1 0.72 ,.02 <br /> .� 12. 011 /A .... 2.47 <br /> Check this value on page E-16. <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> I x /0 ft = IC) ft <br />
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