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1994-006537 - new septic system
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1994-006537 - new septic system
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Last modified
8/22/2023 3:16:26 PM
Creation date
2/8/2018 1:08:36 PM
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x Address Old
Address
0820 North Shore Dr W
Document Type
Septic
PIN
0711723220004
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORKSHEET SJc% IV. R <br /> (For Flows up to 1200 gpI) <br /> A. FLOW (} xEvamst:d Scaogc Mo v to Gal:aq <br /> Estimated 46-0 gpd C 38-tAlcct � ,x <br /> �.•mxr fgpdl <br /> or measured x 1.3 = gpd• of I Type I Type U Type m <br /> B ewooma <br /> ISO <br /> B. SEPTIC TANK LIQUID VOLUltiEEC <br /> c `e° J30 213 <br /> /el-v gallons C Sst cL, 1 �� �> 5 no .50 2%4 <br /> 6 900 721 312 <br /> 7 IWO 600 370 <br /> � , <br /> C. SOILS (refer to site evaluation) E ItDO 67J OX <br /> 1. Depth to restricting layer y inches '.P.rOC.P.^--tom-- <br /> 2 Depth of percolation tests = lam. inches ....•_•.� �,..,� � ,.,..,,,,�„ <br /> 3. Percolation rate 3), / mpi a a <br /> 4. Land slope c "0 ,s; I ,o p <br /> D. ROCK LAYER D 11rf E-,N'SI0, S <br /> 1. ?Multiply flow rate by 0.33 to obtain required area of <br /> layer: A x 0.33 = L <br /> gpd x 0.33 sq. ft./gpd = 7.5 sq. f:. .- y <br /> 2. Select wi.jth of rock laver (10 feet or IL-ss) Zc ft. <br /> 3. Length of rock layer - area + width = R.c'.; r?•�� <br /> �S�r• sq. ft. + iu ft. ft. �....,...:..; ;•.:•. •. ...;.... •.:•. ,••; •. <br /> CC :.......•�...: •.:% (( <br /> Y•r•:•r•r•r•:•r•r•r•r•;•r•:•r•r•:•a <br /> F•r•:•r•r•r•:•r•r•r•r•r•r•.•r•r•:.� <br /> E. ROCK VOLUNIE ►- Length --i <br /> 1. !iitultiply rock area by rock depth to get cubic feet of rock; -3s <br /> s q. f t. x / f t. _ �IL cu. f t. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> 3 cu. ft. +27= i 4 cu. yd. <br /> 3. Multiply cubic yards by 1e4 to get weight of rock in tons; <br /> cu. yd. x 1.4 ton/cu. yd. =Qc tons. iay� <br /> F. ADSORPTION WIDTH Absorption Width sizing,772ble <br /> 1. Percolation rata in top 12 inches of soil is mpi Prcolason RA. c.l % ><. <br /> to Minutes per Sal Teuwe`pei day per At.sor. <br /> b.<n cmpl) Isquart foa toRC <br /> 4 <br /> 2. Select allowable soil loading rate from table; Fa,o d,� .1- co Sant �� <br /> O.5 c gpd/ft2 0.ltoJ•• FinoSa..Sandd•• 060 <br /> 6 to 11 Sar.ay ta=m 0.79 <br /> 16 b 30 60 <br /> 1 to a dt lrstm 030 . r <br /> 3: Calculate adsorption width ratio by dividing rock laver 4,to.l, c., o:m 0. 3 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; Sw•.zc:n C,; + ° <br /> 1.20 gpd/f;�+ �, gpd/Et2 = e�r•�,/G . Ito... <br /> Sod too coarse fix msutlanas of a <br /> standard syitem. <br /> See 7,0E0.0170 Sutpp 2.G.J,pale:6. <br /> 4. Multiply adsorption w 6 width ratio by rock layer width to get •• Soil 50. at mom of rum%and <br /> plus very rine und. <br /> required adsorption width; •'•Soil too hea ry for in%uJtation of a <br /> .?. O X /c ft = ft standard system. <br /> Sae:Qil0.U:10 Subp S.A.page 33. <br />
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