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2001-P04408 - new septic system
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385 Turnham Road - 31-118-23-31-0006
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2001-P04408 - new septic system
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Last modified
8/22/2023 4:30:34 PM
Creation date
7/3/2019 11:55:44 AM
Metadata
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Template:
x Address Old
House Number
385
Street Name
Turnham
Street Type
Road
Address
385 Turnham Road
Document Type
Septic
PIN
3111823310006
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORK SHEET(For Flows up to 1200 d) <br /> A.. Average Design FLOW A-1: Estimated Sewage Flows in Gallons per Day <br /> number o <br /> Estimated (�aQn gpd (see figure A-1) bedrooms Class I Class II Class III Class IV <br /> or measured -- x 1.5 (safety factor) = gpd 2 300 225 180 60% <br /> 3 450 300 218 of the <br /> B. SEPTIC TANK Capacity 4 600 375 256 values <br /> 5 750 450 294 in the <br /> 6 900 525 332 Class I, <br /> 0 l✓ gallons (see figure C-1) 7 1050 600 370 II, or III <br /> 8 1 1200 675 1 408 1 columns. <br /> C. SOILS (refer to site evaluation) C-1: Septic Tank Capacities(in alions <br /> caaci <br /> Number of Minimum Liquid Liquid capacity with withldisposal& <br /> 1. Depth to restricting layer = _ eco - a.3 feet Bedrooms Capacity garbage disposal lift inside <br /> 2. Depth of percolation tests = feet 2 or less 750 1125 1500 <br /> 3. Texture 3or4 1000 1500 2000 <br /> 5 or 6 1500 2250 3000 <br /> Percolation rate / —,,.1,,Z mpi 7,8 or 9 2000 3000 <br /> 40W 1 <br /> 4. Soil loading rate 9 gpd/sqft(see figure D-33) <br /> 5. Percent land slope Cl % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow (A) by 0.83 to obtain required rock layer area. <br /> `r; gpd x 0.83 sqft/gpd = y'� V sqft -110`'-x-) : `u� ` '' ` . <br /> 2. Determine rock layer width = 0.83 sqft/gpd x linear Loading Rate (LLR <br /> 0.83 sqft/gpd x )D gpd/sqft= /0 ft Mound LLR <br /> 3. Length of rock layer = area _width = <br /> L) q sqft (D1) = / ft (D2) = 5,5- ft < 120 MPI < 12 <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 /> ,'L ' sqft x 1 ft = Sal'? cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> S'-i ') cuft + 27 cuyd/cuft = -�0 cuyd <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons <br /> "> cuyd x 1.4 ton/cuyd = t tons <br /> D-33: Absorption Width Sizing Table <br /> F. SEWAGE ABSORPTION WIDTH in MiuRate Loading Rate <br /> n Minnuttee s per Soil Texnue Gallons Absorption <br /> loch per day per Ratio <br /> MPI square foot <br /> Faster than 5 Coarse Send 1.20 1.00 <br /> Medium Sand <br /> Absorption width equals absorption ratio (See Figure D-33) L6 to oamy Sand <br /> dy Loam 0.79 1.50 <br /> times rock layer width (D2) 1 0 0 Loam 0.60 2.00 <br /> 31 to 45 Silt Loam 0.50 2.40 <br /> Silt <br /> X ft = ft 46 to 60 Sandy Clay LoajT 0.45 2.67 <br /> 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 d«igned for tMae soils nest be otter ar perf-wc <br />
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