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1994-006375 - new septic system
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2645 Watertown Road - 04-117-23-12-0008
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1994-006375 - new septic system
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Last modified
8/22/2023 5:06:48 PM
Creation date
7/22/2019 2:01:03 PM
Metadata
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x Address Old
House Number
2645
Street Name
Watertown
Street Type
Road
Address
2645 Watertown Rd
Document Type
Septic
PIN
0411723120008
Supplemental fields
ProcessedPID
Updated
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MOUND UL'51UN wulcn�►r�LI <br /> (For Flows up to 1200 gpd) <br /> to b <br /> A. FLOW ' <br /> Eglmated Sewage Fbwa in Cellons pa day • <br /> Estimated _ 2SOgpd (see pages D-7 or I-3,4,5) <br /> Nwnbff <br /> or measured - gpd x 1.5 = _ or Tye I T,v,c n <br /> nea�m. IV <br /> B. SEPTIC TANK LIQUID VOLUMES 3 450 30000 218 <br /> 4 600 375 256 a a■ <br /> 1-la-CO A )-/oo o gallonsIsee pages C-3 or C-5) s 750 450 294 •o <br /> 6 900 525 332 7b.1. <br /> 7 1050 600 370 u°' <br /> C. SOILS (refer to site evaluation) 8 1200 675 408 tdvms <br /> I. Depth to restricting layer = inches Scplk T-k C.p.,iti,in <br /> 2. Depth of percolation tests = i - inches ; , m; ��inm P-Cily id l...b,,:cp 6po-;n <br /> a C,ip.city aub.�c di.pu..l <br /> 3. Percolation rate /. mpi 2orless 750 1125 <br /> 3a 4 1000 1500 <br /> 4. Land slope % 4or6 Isw 2250 <br /> 7,6 or 9 2000 3000 <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: Daily Flow x 0.83 = <br /> -),<O _gpd x 0.83 sq. ft./gpd = 6,aa sq. ft:lioc>v=6,,dL4 <br /> 2. Select width of rock layer (10 feet or less) _ >o ft. <br /> 3. Length of rock layer = Area+Width = <br /> sq. ft. + f o _ft. _ - ft. Rock Bed <br /> t f�r1.•t•1•ti•1,•ti••• • ., ti.:,: 1 <br /> ti'ti'ti•�'ti•ti•ti•ti•;•ti•ti•ti•ti•ti•ti• Width s10ft. <br /> E. ROCK VOLUME '}�f'f+•ti•ti••••ti••• <br /> I ---1 <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; Length <br /> 4,�sq. ft. x Lv t'ft. = )I e cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> 21Vcu. ft. 4.27= cu-'yd. <br /> 3. Multiply cubic yards by:1.4,to get weight of rock in tons,; ; <br /> Q�n cu• yd.x 1.4 ton/cu. yd. =-3V tons. ' <br /> } t L,0, <br /> F. ADSORPTION WIDTH z-L-R4 Lo -w l <br /> 1. Percolation rate in top 12 inches of soil is 9.7 mpi Absorption width Sizing Table <br /> 2• Select allowable soil loading rate from table on page E-- a_Rp"e� so,,Texture CaOons Ratio or <br /> gpd/ftz P g pvdayper AoPockWXftnyon er <br /> � X ,: tneh(MPIj .. ith <br /> tgo,re taa to Rock Lave. I <br /> 3. Calculate adsorption width ratio by dividing rock layer FwterOwn 0.1• com:eSood .__ <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; 0..1 as0• FuicSad•• 0.o 2OD <br /> 1.20 gpd/ft2�+- .L.�gpd/ft2= _ a.t� 6`t'0 t�. 0.6600 i00 <br /> 311o,15 Sib Loam 0.50 2.40 <br /> Check Misvalue on page E-16. 46w6o cIa L.vm oas 2.67 <br /> 4. Multiply adsorption width'ratio by rock layer width to get sD b„a Clay 0`24 s.:U0 <br /> s�tn>nt as <br /> required adsorption width; <br /> a•reg x / o ft =a4 o ft <br /> . i+ F syq• �`�..�.W'++;�'�Y�,� .moi Y;,� :.i r .. :?c r <br />
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