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1991-003690 - septic system
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1991-003690 - septic system
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Last modified
8/22/2023 4:34:23 PM
Creation date
7/16/2019 1:43:57 PM
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x Address Old
House Number
2165
Street Name
Watertown
Street Type
Road
Address
2165 Watertown Rd
Document Type
Septic
PIN
0311723210026
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ProcessedPID
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.., <br /> is; E-19 <br /> MOUND DESIGN PROCEDURE <br /> (For Flows up to 1200 gpd) <br /> A. Sewage Flow Rate F. Pressure Distribution System <br /> See D-7 or I-3, 4, or 5, or use 1. Select number of perforated <br /> metered .value; Flow Rate = laterals <br /> �O d gpd <br /> 2. Select perforation spacing <br /> B. Septic Tank Liquid Volume = .3 ft <br /> (see C-3 or C-5) 1000 gallons <br /> 3. Select perforated lateral <br /> C. Soil Characteristics length; Note if manifold is <br /> at end of rock layer, lateral <br /> 1. Depth to restricting layer length is rock layer length <br /> such as seasonally saturated less half a perforation <br /> soil, bedrock, coarse soil, spacing. If manifold is in <br /> etc. ; Z 8 inches center of rock layer, lateral <br /> 2. Depth of percolation tests; length is one-half rock layer <br /> /8-20 inches <br /> length less half a perforation <br /> spacing. Perforated lateral <br /> 3. Number of percolation test length = 2?, S ft. <br /> holes; _� holes 4. Divide lateral length by perfor- <br /> 4. Ave. percolation rate; ation spacing to get number of <br /> mpi �r perforations per lateral <br /> 5. Landslope = y% J-/—,: ZZfeet = 3 feet = s7 perfs <br /> "Note: last perforation must be* <br /> D. Rock Layer Dimensions <br /> in end cap, (see page E-14) <br /> 1. Multiply gpd by 0.83 to 5. Multiply,perforations per <br /> obtain required area of lateral by number of laterals <br /> rock layer; to get total number of <br /> d'00 gpd x 0.83 =SOOsq ft perforations; <br /> perfs/lat x o' lats <br /> 2. Select width of rock layer <br /> (10 feet or less) _ /O ,feet 6. Determine required flow rate <br /> by multiplying number of <br /> is <br /> 3. Length of rock layer = Area perforations by flow per <br /> Width_f sq ft - jaft perforation (see page E-17) <br /> _ ..SD ft /Z perfs x .71/gpm/perf =�S:sgpm <br /> E. Rock Volume 7. Select minimum required lateral <br /> diameter from table on Page E-17;' <br /> 1. Multiply rack area by rock depth- enter table with perforation <br /> to get cubic feet of rock; spacing, perforation diameter, <br /> SSOO sq ft x / f t- =S'00cu ft and number of perforations per <br /> 2. Divide cu. ft by 27 cu ft/cu yd lateral. Select minimum <br /> diameter for perforated lateral <br /> to get cuk�ic.yards.; /8 S = / " inches % " <br /> 3. Multiply cubic yards-by 1.4 to <br /> get weight of rock in tons; G. Basal Width <br /> Scu yds x 1.4 7X. 9 tons 1. Percolation rate in top 12 <br /> inches of. soil is ,, Smpi <br /> 2. Select allowable soil loading <br /> rate from table on page E-16; <br /> G/IF 0.3'0—gpd/f t2 <br />
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