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04/21/1999 - letter re: Septic System Site Plan
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04/21/1999 - letter re: Septic System Site Plan
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Last modified
8/22/2023 3:22:16 PM
Creation date
3/27/2019 2:46:41 PM
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x Address Old
House Number
2125
Street Name
Carriage
Street Type
Lane
Address
2125 Carriage Lane
Document Type
Septic
PIN
1011723240037
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Zof-- 6 <br /> MOUNT) I)l.SI(,N lvt)J is W l I /4/47>i/9 s' <br /> (For Flows up to 12(10 gpd) <br /> I A. H.()W --� I Estimated Sewage Flows in Gallons pet day ' <br /> 1' <br /> Estimated gpd ;Numter ITypc 1 Tme 0 Typc III T_ <br /> of lV <br /> or measured x 1.5 = __ gpd <br /> 2 300 225 180 <br /> 450 <br /> B. SEPTIC TANK LIQUID VOLUMES 3 600 375 218 . <br /> a 600 375 256 of Ar <br /> 131 <br /> /..—CV __ gallons r 900 5� 5 750 450 �'� <br /> 25 132 I ypx 1 <br /> 2 ' 1050 1 600 370 u <br /> ut <br /> H 12(X) 675 I 408 111 <br /> C. SOILS (refer to site evaluation) - ��� <br /> Septic Tank Capaut.e I u/.11Am <br /> L Depth to restricting layer = 3,41 inches p7- g feet Laycalla. <br /> Numbs,..I W n won L.gwd L go.d caI'..1y r h Wo h m,raw& <br /> Z <br /> 2. Depth of percolation tests = / inches B4r...rt Cap,oty gutmgc mspcnat 1,h muck <br /> test 750 1125 ;500 <br /> 3. Texture 0-e '- �3.3 <br /> ) Percolation rate ) mpi �. woo 1500 _020 <br /> 5 A;n 1500 2250 <br /> X <br /> 4 an slope 6 % '7a..tv 20111000 ,�,000� <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock layer: A x 0.83 = <br /> 75c gpd x 0.83 sq. ft./gpd = 630 sq. ft. <br /> 2. Select width of rock layer (max 10' if <120 mpi max 5') = /U ft. <br /> 3. Length of rock layer = area ± width = <br /> 463 o sq. ft. - IC) ft. = 6 3 ft. e.i"S7ao` o`�° Qoa „> , a _ <br /> do7.1'2 QE a c o v n a <br /> e�ooboDooaDeaDn aD--&;...-W10° o a�..d°:r�:d°4°:n.�ia.q'1a <br /> Width /D ft <br /> p-o-n o.e�se�Gp00ea0D ea o�ononea of <br /> <120mpi <10' Length 4.3 ft <br /> E. ROCK VOLUME >120mpi <5' <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; sq. ft. x <br /> ft. = cu. ft. <br /> 2. Divide cu. ft.by 27 cu. ft./cu. yd. to get cubic yards; <br /> cu. ft. -27 =.-3 cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of.rock in tons; cu. yd.x 1.4 <br /> ton/cu. yd. = ,'3 tons. <br /> F. ABSORPTION WIDTH Absorption Width Sizing Table <br /> 1. Percolatio/n rate in top 12 inches of soil iso �3Dmpi lcrcolauoo Gallows Rano to GalloRano of Absorpoon <br /> Texture ///--yi <br /> Minutes pet loch Soil 7cpa width per day pwidth to Rock <br /> (MPI) square toot Layer Width <br /> Faster than 0 1 Coarse Sand 1.20 1 00 <br /> 2. Select allowable soil loading rate from table; o; Fato 5 x d 060 200 <br /> gp d/f t' 6 t 15 Sandy Loans 0 60 52 <br /> I t�to 30 Loam 0.60 2 z 00 <br /> 31 10 45 Silt Loam 0.50 240 <br /> 46 to 60 Clay Loam 045 267 <br /> 3. Calculate adsorption width ratio by dividing rock layer SiowQ than60it' Clay 020 600 <br /> 120 Clay 0.20 6 OU <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; <br /> 1.20 gpd/ft2_ gpd/ft2 = . 2. O <br /> a Multiply adsorption width ratio by lock layer width to get <br /> required adsorption width, <br /> x /6 ft =0RO ft <br />
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