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2004 Septic system
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2550 Woodhaven Drive - 33-118-23-41-0007
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2004 Septic system
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Last modified
8/22/2023 4:51:13 PM
Creation date
2/24/2020 12:45:20 PM
Metadata
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Template:
x Address Old
House Number
2550
Street Name
Woodhaven
Street Type
Drive
Address
2550 Woodhaven Drive
Document Type
Septic
PIN
3311823410007
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' MOUND DESIGN WORK SHEET (For Flows up to 1200gpd) <br /> A. Average Design FLOW A-1: Estimated Sewage Flows in Gallons per Day <br /> t-re714- Prt7P1)4\, 0& VF- number of <br /> Estimated (' 0 0 gpd (see figure A-1) bedrooms Class I Class II Class III Class IV <br /> or measured - x 1.5 (safety factor) =- gpd 2 300 300 28 225 1 o 0% <br /> 4 600 375 256 values <br /> B. SEPTIC TANK Capacity 5 750 450 294 in the <br /> / 6 900 525 332 Class I, <br /> D. - ) -J 6allons (see ure C-I) 7 1050 600 370 II, or Ill <br /> g $ 8 1200 675 408 columns. <br /> C. SOILS (refer to site evaluation) C-1: Septic Tank Capacities(in;allons). <br /> Liquid capacity <br /> Number of Minimum Liquid Liquid capacity with with disposal& <br /> 1. Depth to restricting layer = 1• 46 + a.0 feet Bedrooms Capacity garbage disposal lift inside <br /> 2. Depth of percolation tests = )• 0 feet 2orless 750 1125 1500 <br /> 3. Texture 6L-Pic 1.4A-.).-/‘ 3 or 4 1000 1500 20001 5 or 6 1500 2250 8000 <br /> Percolation rate L6, mpi 7,8 or 9 2000 3000 4040 <br /> 4. Soil loading rate ,y .1' gpd/sqft(see figure D-33) <br /> 5. Percent land slope <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow (A) by 0.83 to obtain required rock;ayer area. <br /> 9 0 D gpd x 0.83 sqft/gpd = Iy') sqft-i-te"?a -- baa <br /> 2. Determine rock layer width = 0.83 sqft/gpd x linear Loading Rate (LLR) <br /> 0.83 sqft/gpd x )-z gpd/sqft = 10 ft Mound LLR <br /> 3. Length of rock layer = area- width = <br /> 4 :."Z., sqft (D1) _ 10 ft (D2) = 4dZ ft , < 120 M PI < 12 <br /> o,C So 1 v.- )5'T1,.b Abp 32 (-3c9 sol'F-C. - <br /> E. ROCK VOLUME > 120 MPI < 6 <br /> 1. Multiply rock area (D1) by rock depth of 1 ft to get cubic feet of rock <br /> 3 aC sqft x 1 ft = 3zJ cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> 3 cuft _ 27 cuyd/cuft = 1-7-- cuyd <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons <br /> )-7...- cuyd x 1.4 ton/cuyd = 1 7 tons <br /> D-33: Absorption Width Sizing Table <br /> F. SEWAGE ABSORPTION WIDTH Percolation Rate Loading Rate <br /> in Minutes per Soil Texture Gallons Absorption <br /> Inch per day per Ratio <br /> (MPI) square foot <br /> Faster than 5 Coarse Sand 1.20 1.00 <br /> Medium Sand <br /> Sand <br /> Absorption width equals absorption ratio (See Figure D-33) Loamyoine Sand <br /> 6 to 15 Sandy Loam 0.79 1.50 <br /> times rock layer width (D2) 16 to 30 Loam 0.60 2.00 <br /> 31 to 45 Silt Loam 0.50 2.40 <br /> Silt <br /> 0 Z 1l?- 46 to 60 Sandy Clay Loam 0.45 2.67 <br /> a••(� )( 1 ft = ft Silty Clay Loam <br /> Clay Loam <br /> 61 to 120 Silty Clay 0.24 5.00 <br /> Sandy Clay <br /> Clay <br /> Slower than 120' <br /> •System designed for these soils oust be other or performance <br />
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