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1992 - Septic System
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2875 Wear Circle - 33-118-23-34-0004
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1992 - Septic System
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Last modified
8/22/2023 4:50:19 PM
Creation date
1/17/2020 11:06:25 AM
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x Address Old
House Number
2875
Street Name
Wear
Street Type
Circle
Address
2875 Wear Circle
Document Type
Septic
PIN
3311823340004
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.7 k- .' IIS\--A, r7 - / -"1 3. <br /> : MOUND DESIGN WORKSHEET • <br /> • <br /> (For Flows up to 1200 gpd) <br /> --1 A. FLOW Estimated Sewage Flows in Galltais per day <br /> Estimated 1 v gpd (see pages D-7 or I-3 4, 5) csf�l) <br /> i r Number t 1 I <br /> or measuredgpd x 1.5 = nr Type I Type II TypeIll Type <br /> Iledrnoma 1V <br /> 2 30) 225180 ' <br /> B. SEPTIC TANK LIQUID VOLUMES 3 450 ,00 218 <br /> 4 600 375 256 de <br /> I - Ia.�cO 1,I-1ocr)gallons (see pages C-3 or C-5) 5 75510 525 2'14 IP <br /> 7 1050 6(10 1 370 ,,, <br /> 8 1200 675 i 408 cda <br /> wn. <br /> C. SOILS (refer to site evaluation) <br /> ,/ Sopttr'(.nk cap....,In a.11rm• <br /> z <br /> 1. Depth to restricting layer = ) -to a inches <br /> umhrr n. himrmum Liquid�(squid a p..•uy with <br /> 2. Depth of percolation tests = t e•� '' inches Ita•Armm• • <br /> I C•�acuy I a�rh:rludl.Iywi <br /> 2 nr Ira 750 1 1 rs <br /> 3. • Percolation rate mpi ,�,, ,lx„ IMO <br /> 4. Land slope 31"' 'S`"' 2250 <br /> 3000 <br /> % 7•a. 9 baa, 7oxa, <br /> over 9 •••••• <br /> • <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain require;; area of rock <br /> layer: Daily Flow x 0.83 = <br /> " <br /> _3_,<_0__ gpd x 0.83 sq. ft./gpd = !a v - <br /> sq. ft.-I- / .)c ( 5.<4 <br /> 2. Select width of rock layer (10 feet or less) = ) o _ ft. • <br /> 3. Length of rock layer = Area Width = . <br /> 2.W-) sq. ft. + J 0 ft. = 6• ' ft. Rock Bed•f• f. T <br /> . f ridth S10(t. <br /> E. ROCK VOLUME t-•- `Lenl;th <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> G,er%Usq. ft. xi•e:sft. = 7l-Ecu. ft. . <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> 0 1 K cu. ft. + 27 = -..). 7 cu. yd. ' <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> D? cu. yd. x 1.4 ton/cu. yd. = 39 tons. <br /> F. ADSORPTION WIDTH (-`0 1_0 411 r <br /> Absorption Width SixinkTuble <br /> 1. Percolation rate in top 12 inches of soil is r7, S rnpi <br /> Percolation Rate Gallons Ratio of <br /> 2. Select allowable soil loading rate from table on page E-; in Minutes Fier Still restart per day per Absorption width <br /> Inch(MPI) Nunn:foot to Roel Layer <br /> •y s <br /> gpd/ft2 wrath <br /> 3. Calculate adsorption width ratio by dividing rock layer 1'aaerthan 0.1• Coarse Sand ••••• <br /> 0.1 to 5 Sand 1.20 1.00 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; 0.1 to 5•• Fine Sand•• 0.60 2.10 <br /> 6 to 15 Sandy Loam 0.79 1.52 <br /> 1.20 gpd/ft2+ ,L) s a gpd/ft2= Com ') . 161°10 Loam 0.60 2. 0 <br /> Check this value on page E-16. c4rn1to 60 Clay S--. ClayLoam <br /> �,na _ 2.67 <br /> 2.67 <br /> page <br /> _� <br /> 60 to Ill -City 0.24 5.00 <br /> 4. Multiply adsorption width ratio by rock layer width to get sly .:;an Clay _... <br /> required adsorption width; -' . <br /> ?. V) x Jo ft = ,a1_,-) ft <br /> 1 . <br /> • <br />
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