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09/21/18 Septic Compliance Inspection
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2245 French Lake Rd - 10-117-23-22-0004
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09/21/18 Septic Compliance Inspection
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Last modified
8/22/2023 3:20:53 PM
Creation date
11/29/2018 3:25:29 PM
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Address
House Number
2245
Street Name
French Lake
Street Type
Road
Address
2245 French Lake Road
Document Type
Septic
PIN
1011723220004
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' IVIUUIVU U,C,J1lalN VVUKt1/451-ILL1 <br /> • (For Flows up to 1200 gpd) <br /> ' <br /> A. FLOW <br /> Estimated '>t 0_ Sewage Nowa in Gallons l <br /> gpd(see pages D-7 or 13,4,5) (end) jadiy <br /> or measured gpd x 1.5= <br /> __. Bedroom <br /> _-._.___.._. r Irl I Type n Type III � <br /> v <br /> �� <br /> 13. SEPTIC TANK LIQUID VOLUMES <br /> 2 300 225 180 <br /> -l�'S-03 430 300 218 boa <br /> •�1,tt»» ,7 gallons+. (see pages C-3 or C-5) 4 600 375 256 aM <br /> /-/ITO Ir4fiV)tinK 5 7511 450 <br /> rawer <br /> v <br /> 6 1°90 525 32 k <br /> Iff <br /> C. SOILS(refer to site evaluation) ' S0 `41 370 �" <br /> 8 i 1200 675 408 aslant* <br /> 1. Depth to restricting layer=____AS__ �1�r���. <br /> 2. Depth of percolation tests= inches j�2�j�p_�.�yyI■�,�y�` <br /> __a__inches oorne^66 a M_IoCapa Liquid .iquW1 `,purity with' <br /> 3. Percolation rate 13 mpi Capacity �ieY•4„�a,, <br /> 4. Land slope -$ % 23 q.' 750 Ilu <br /> eas tan Ilri <br /> DS <br /> 7,II or 9 2001 30W <br /> •S «uG. Meta., /elf-Sft.514 it,„,. ove,9 ...... <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 /> ______12___gpd x 0.83 sq.ft./gpd= 623 sq.ft.-ea-.d we I. ,Iis,..76S , <br /> 2. Select width of rock layer(10 feet or less)= <br /> 3. Length of rock layer=Area+Width = ft. <br /> 4n sq. ft.+ _,_J12_ft. --43-It. <br /> ock Bed <br /> E. ROCK VOLUME <br /> 5pso.+'a...a•v:•:•`.`t!"a'•' <br /> ••,..........:ti.y:..;;{!{{{i,*.f.t ,so.width 5IUtf. <br /> l.f J_2.lali.'e,a4 1 <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; �C061h <br /> Gia sq.ft.x_.j_ft. <br /> 2. Divide Cu. ft.by 27 cu. ft./cu,y get cubic yards; <br /> nen cu.f t. +27= d.3? cu.yd. <br /> 3. Multiply cubic yards by 1,4 to get weight of rock in tons; <br /> . cu.yd.x 1.4 ton/cu.yd._ tons. <br /> F. ADSORPTION WIDTH-1 <br /> 1. Percolation rate in top 12 inches of soil is_f mpi Select allowable soil loadingm 1 AAsori Ion WIdIh Shing Toble <br /> rate from table on page E-; in„j,�iaflk k■ie Gelbnn <br /> pee it awn per do kelio of <br /> --_lam_ gpd/f tZ Inch IMP!) I y Pea Muni-Minn width <br /> 3. Calculate adsorption width ratio by dividing rock layer P,ae than 0.1• ase Ml Nu' row lo Rock Liver <br /> Width <br /> loading rate of 1.20 gpd/ft2 by allowable soil loadingrate; 0.1 to s•• • <br /> 2.00 <br /> 0.10 5 Send 1.20 <br /> L20 gpd/ft2+ • In gpd/ft2= Z.a 610 3s dy Lo o.w 2.00 <br /> -----_• 16 to 30 0.79 1.52 <br /> Check this value on page E-16. 46 io " 04; 22,00 <br /> 6; <br /> 4. Multiply adsorption width ratio by rock layer width to get Slowerth 0 Clay m 0.:4 <br /> [A l0 60 CI y Lau" 0.45 2.67 <br /> required adsorption width; s:'"' <br /> 120••• <br /> _Q. x 2. ft=_:2O ft <br />
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