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1996-008339 - new septic system
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Old Crystal Bay Road North
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0325 Old Crystal Bay Road North - 33-118-23-31-0003
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1996-008339 - new septic system
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Last modified
8/22/2023 4:49:13 PM
Creation date
3/7/2018 12:36:29 PM
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x Address Old
House Number
325
Street Name
Old Crystal Bay
Street Type
Road
Street Direction
North
Address
325 Old Crystal Bay Road North
Document Type
Septic
PIN
3311823310003
Supplemental fields
ProcessedPID
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MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. FLOW Estimated Sewage Pow in Gallon p7(gppd)Estimated _ � gpd Numberor measured x 1.5 = gpd. aed°oforw ►Ype 1 Type II Type 2 300 225 180 <br /> 218 <br /> B. SEPTIC TANK LIQUID VOLUMES 4 � 375 256 Q. <br /> - ga110nS 6 900 525 35 750 450 32 .�. <br /> 7 1050 600 370 m,m, <br /> 8 1200 675 408 <br /> C. SOILS (refer to site evaluation) Numbs .'"' <br /> 1. Depth to restricting layer= inches of <br /> Bemoan: CM-MY � <br /> 2. Depth of percolation tests = a inches `PLW-4 <br /> 3. Percolation rate m P i 2 3 o<a l,0a leu 0o0 1.lu <br /> 1.500 <br /> 4. Land slope % 7 or 8 2.000 3 i� <br /> ova 9 See 68.C•6 (a 1.5) <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area f rock <br /> layer:A x 0.83 X O0 <br /> gpd x D&sq. ft./gpd = 425_4sq. ft. <br /> 2. Select width of rock layer(10 feet_or_less)_= / ft. <br /> 3. Length of rock layer = area_width= Rock Bed <br /> ft. _ ft. <br /> r•r•r•r r•r��• ti•ti•ti•�•� <br /> ;�•r r•r•r•r•r•r•r• <br /> ti•ti•ti.ti ti.ti ti•ti•ti.ti.ti.ti.ti•ti.ti <br /> r•r.r.r•r.r:r:r•r r•r.r•r•r•r•r• <br /> tdth SIO ft <br /> r•r•r•r•r•r•r•r•r r•r•r•r•r•rY• <br /> ti�ti.ti•yti•ti•ti•ti•ti•�•tiK•Y.ti•ti•tiH <br /> ••r•r•r•t•r•r•r•r r•r•r•r•r•r•r• <br /> E. ROCK VOLUME ' • Length <br /> 1. Multiply rock area by rock depth to get cubic feet f rock; <br /> >'SGaq. ft. x ft. _ Vcu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yard ; <br /> -/, S3)cu. ft. ?-27=_Z�u. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock i tons; <br /> yd. x 1.4 ton/cu. yd. =a3.3 tons. <br /> F. ADSORPTION WIDTH Absorption Width S'min table <br /> 1. Percolation rate in to 12 inches of soil is m 1 Cd • Rt d <br /> P p' ren:olaNen Rate <br /> Minut m per inch Soil'Texture ate a <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 Coarse Sand 120 1.00 <br /> Vie- gpd/ftz 0.1 to Sand iso 1.00 <br /> 0.1 to 5 Fine Sand- 0.60 2.00 <br /> 6 to 15 ndy Loam 0.79 152 <br /> 3. Calculate adsorption width ratio b dividing rock layer 16 to 30 Loam 0.60, 2.00 <br /> y g31 to 45 Silt Loam 050 2.40 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading ra e; 46 to 60 Cla Loam 0.45 2.67 <br /> 1.20 d/ftz . d/ftp = 61 t0120 flay 0.24 5.00 <br /> gP gp _• Slower than 120 Clay <br /> -Sod having 5096 or move of fine or very fin sand. <br /> 4. Multiply adsorption width ratio by rock layer width to et <br /> required adsorption width; <br /> X /O ft=o?0 ft <br /> 71 <br />
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