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1990-04-27 Septic System Design Report
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1990-04-27 Septic System Design Report
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Last modified
4/24/2025 2:24:49 PM
Creation date
4/24/2025 2:13:05 PM
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x Address Old
House Number
1090
Street Name
Cox Farm
Street Type
Road
Address
1090 Cox Farm Road
PIN
2711823310022
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�'.I. C a7' d &e r /, <br />t / S <br />•lr <br />, E-19 <br />POUND DESIGN <br />PROCEDURE <br />(For Flows up to 1200 gpd) <br />/GTfr1l-A iF T/TE , <br />A. Sewage Flow Rate <br />F. Pressure Distribution System <br />See D-7 or I-3, 4, or 5, or use <br />1. Selec; number or perforated <br />metered value; Flow Rate <br />laterals 6 <br />7S0 gp d <br />2. Select perforation spacing <br />B. Septic Tank Liquid Volume <br />,3 ft <br />(see C-3 or C-5) /LSO gallons <br />3. Select perforated lateral <br />C. Soil Characteristics <br />length; Note if manifold is <br />at end of rock layer, lateral <br />1. Depth to restricting layer <br />length is rock layer length <br />such as seasonally saturated <br />less half a perforation <br />soil, bedrock, coarse soil, <br />spacing. If manifold is in <br />etc.; / 8 inches <br />center of rock layer, lateral <br />2. Depth of percolation tests; <br />length is one-half rock layer <br />1 r' inches <br />length less half a perforation <br />spacing. Perforated lateral. <br />3. Number of percolation test <br />length - 29.7 ft. <br />holes; Z holes <br />4. Divide lateral length by perfor- <br />4. Ave. percolation rate; <br />ation spacing to get number of <br />.7, 3 mpi <br />perforations per lateral <br />5. Landslope �_% <br />3�feet _ 3 feet - �_perfs <br />Note: last perforation must be <br />D. i,-ck.Layer Dimensions' <br />in end cap, %see page E-14) <br />Y <br />5. Multiply perforations per <br />1. Multiply gpd by 0.83 to <br />lateral by number of laterals <br />obtain required area of <br />to get total number of <br />rock layer; <br />perforations; <br />7S0 Rpd x 0.83-!?I.S_ sq ft <br />/O perf s/la t x 6' la is - 4CO <br />2. Select width of rock layer <br />6. Determine required flow rate <br />(10 feet or less) - /0 feet <br />by multiplying number of <br />3. Length of rock layer - Area <br />perforations by flow per <br />= WidthQjsq ft Lift <br />perforation (see page E-17) <br />• C2, -7 ft <br />IL perfs xo,7ygpm/perf - JT.Ygpm <br />7. Select minimum required lateral <br />E. Rock Volume <br />diameter from table on Par -7-17; <br />1. Multiply rock are_ by rock depth <br />enter table with perforation <br />to get cubic feet of rock; <br />spacing, perforation diameter, <br />621. S so f t x Q, 7S f t - W7 cu f t <br />and number of perforations per <br />lateral. Select minimum <br />2. Divid u ft by 27 cu ft/cu yd <br />diameter for perforate,' 'ateral <br />to get cubic yards; 17-3 <br />inches utt 110 <br />3. Multiply cubic yards by ?.4 to <br />get weight of rock in tons; <br />G. Basal Width <br />/Z L_cu yds x 1.4 - jy. 2 tons <br />1. Percolation rates in top 12 <br />inches of soil is -zt..Lmpi <br />2. Select allowable soil•loading <br />'- <br />rate from table on page E-16; <br />t2 <br />O. ngpd/f <br />/*6/l NI6 O. rO <br />
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