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1996-008262 - new septic system
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640 Orchard Park Road - 32-118-23-22-0005
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1996-008262 - new septic system
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Last modified
8/22/2023 4:39:30 PM
Creation date
4/30/2018 1:33:09 PM
Metadata
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Template:
x Address Old
House Number
640
Street Name
Orchard Park
Street Type
Road
Address
640 Orchard Park Road
Document Type
Septic
PIN
3211823220005
Supplemental fields
ProcessedPID
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• <br /> MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) - <br /> , <br /> /-, . A. FLOW Estimated Sewage flows in Gallons per day <br /> • <br /> Estimated (coo gpd (see pages D-7 or I-3,4,5) Number � ') <br /> or measured - gpd x 1.5 = BedroomsTypc 1 Type II Type HI TIV <br /> t 2 300 225 180 1i <br /> B. SEPTIC TANK LIQUID VOLUMES 3 45° 300 218 60% <br /> 4 600 375 256 h. <br /> a, -/0 0 0 gallons (see pages C-3 or C-5) 6 7550 45025 3994 VI <br /> I- <br /> 7 1050 600 370 IT <br /> 8 1200 675 4011 rause <br /> C. SOILS (refer to site evaluation) <br /> 1. Depth to restricting layer = 1 .'-i-o a v/,inches Septic Teak Capacities.hi galloon <br /> ids <br /> 2. Depth of percolation tests = /:Z ' inches ;ed, s' Mm Capacityber oimum Lial l.qus disponi' <br /> 3. Percolation rate . /O. 1-1 mpi 2;eI ss ,p 15tH <br /> 4. Land slope --- % 74. a 9 Wit° 3000 <br /> ' <br /> over 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 /> bA <br /> C.oD gpd x 0.83 sq. ft./gpd =y9cel sq. ft.4 I Dc.-if, = 6111 <br /> 2. Select width of rock layer (10 feet or less) = /D ft. <br /> 3. Length of rock layer = Area 1 Width = <br /> 54 '? sq. ft. 1 /0 _ft. = .__Sft. Rock Bed <br /> �:r:;:iigi r• •r• ::::P r.�**4' <br /> : .I.•..ti.•..ti.,,•ti..,.? t•!4 <br /> ••.•ti•ti•ti•ti••.• 510 ft. <br /> f•1•r•t•f•t•t• f•t•t•r•t• I <br /> E. ROCK VOLUME '•f►- f '-ff fn f•• f f f <br /> - • •_r f, <br /> Length <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> 414? ,sq. ft. x/.o-S ft. = S7L cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> S')Y cu. ft. 127= a 1 cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> _al_ cu. yd. x 1.4 ton/cu. yd. = tons. <br /> F. ADSORPTION WIDTH <br /> 1. Percolation rate in top 12 inches of soil is /D. 4 mpi Absorption Width Sizing Table <br /> Percolation Rate Gallons Ratio of <br /> 2. Select allowable soil loading rate from table on page E-; in Minutes per Soil Texture per day per Absorption width <br /> Inch(MPI) square foot to Rock Layer <br /> . LI gpd/f t2 Width <br /> 3. Calculate adsorption width ratio by dividing rock layer Faster than 0.1• Coarse Sand _... <br /> 0.1 to 5 Sand 1.20 1.00 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; 0.1 to 5•• Fine Sand•• 0.60 2.no <br /> 6 to 15 Sandy Loam 0.79 1.52 <br /> 1.20d/f t2 1 .LI J d/f t2= �, . C,o-) . 16 to 30 Loam 0.60 2.00 <br /> gP gP 31 to 45 Silt Loam 0.50 2.40 <br /> Check this value ona(�eE-16. .46to60 _Clay_t.a,n, 0.45", 167 <br /> p b 60 to 120 Clay 0.24 5.00 <br /> 4. Multiply adsorption width ratio by rock layer width to get Stie;;an Clay --- -- <br /> required adsorption width; <br /> .C, ix /(-) ft -c)! ,- ft <br />
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