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1994-006377 - septic system
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1994-006377 - septic system
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Last modified
8/22/2023 4:29:12 PM
Creation date
12/3/2019 3:04:45 PM
Metadata
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Template:
x Address Old
House Number
4345
Street Name
Watertown
Street Type
Road
Address
4345 Watertown Road
Document Type
Septic
PIN
3111823130003
Supplemental fields
ProcessedPID
Updated
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t <br /> ... .7 <br /> MOUND DESIGN WORKSHEET t;, 4•-13 lig Gv at,, a''>Rd <br /> • • (For Flows up to 1200 gpd) <br /> A. FLOWt Et..matai Sewage Flo..'in Gallons per dal <br /> Estimated to O4 gpd (. �B 'n�J)ocrn, (mit <br /> ".mxr <br /> or measured x 1.3 t gpd. sevof I Type 1 Type ll Typo III T'. <br /> - <br /> 2 300 • 3 110 <br /> B. SEPTIC TANK LIQUID VOLUMES 3 n 300 211 <br /> 6(70 . 375 J6 <br /> _c !toad gallons 6 LLoo » 3332 <br /> 7 I0!0 600 370 ". <br /> • • 1 I 1:1X) 673 arid .a..•. <br /> C. SOILS (refer to site evaluation) _ __ - <br /> 1. Depth to restricting layer = c.. 8- 3c) inches kms.rem c+.-•nm.apia. <br /> 2. Depth of percolation tests = . �•t.tttr '"_`'""`""""'-" <br /> /c?• inches "--____',.._2.' u� ,......•..,_. <br /> 3. Percolation rate '7'i r 7 mpi =•�- "U <br /> 4. Land slope 5,0 "o '" 'X° L"' <br /> 7 r er 1 attrs r..l] <br /> trey. <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 ft./gpd =„oo sq. h.i- 1090 ' St d \i‘ Z <br /> 6o o gpd x 0.33 sq. <br /> \ 2. Select width of rock laver (10 feet or less) = It) _ ft. <br /> 3. Length of rock layer = area y width = Rcc'.< P,'2'1 <br /> sso sq. ft. + Io ft. = 55 ft. \.•..\.............. ...\.•.. .......\. ..•. Ie <br /> ,...,...,.,.....,.,......•.•..: width Si ft. <br /> • <br /> • <br /> .:, .•...::....::.. ...::•tett• <br /> E. ROCK VOLUME 1 Length - -3 <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; 5 -Ait - - <br /> 550' sq. ft. x i ft. = 550 cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; • <br /> 550 cu. ft. .+ 27 = ao• 9 cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> .?o. N cu. yd. x 1.4 ton/cu. yd. = a9 tons. 4 to 1)0 <br /> F. ADSORPTION WIDTH Absorption Width Sizing,Table <br /> Prreotaaon Rate l Galion. LI <br /> 1. Percolation rate in top 12 inches of soil is a'1,7 mpi ,n%sinuses per Sat Tcu.uaday per Ab+orpoue <br /> loot(`TPU (square foot to Rock t. <br /> wad <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 • coarse sane -- -- <br /> 0.Or0 gpd/ft2 0..Iits stand 110 1.00 <br /> to 3•• FM Sand•• OW LCC <br /> ,soli Stud tarn I_31 <br /> 16 to)-0 � ` 0.60 faX <br /> 3. Calculate adsorption width ratio by dividing rock layer .4,0 c:43 !Loam Dees 26 <br /> KJ a 110 clay 0.24 1 7 CC <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; st„� car 1 <br /> 1.20 gpd/ft2+ O,Go gpd/ft2 = o?.Co _ Sar.,too coarse for anataltaa.onora <br /> standard sy.tem. <br /> Sc.7olo.ot7o S.tbp 2.G.3. age:6. <br /> 4. Multiply adsorption width ratio by rock layer width to get " soil hau.n4 sa•r.or more of floe rand <br /> plus very fine land. <br /> required adsorption width; •••sc..'too heavy for Installation or a <br /> standard system. <br /> ' 0/ 0 0 x /0 ft = c>?d ft sm 7080.0_10 Sabi,3.A.peso 33. <br />
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