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2001-p04520 - new septic
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4245 Chippewa Lane - 31-118-23-42-0001
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2001-p04520 - new septic
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Last modified
8/22/2023 4:32:06 PM
Creation date
4/13/2016 10:49:49 AM
Metadata
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Template:
x Address Old
House Number
4245
Street Name
Chippewa
Street Type
Lane
Address
4245 Chippewa Lane
Document Type
Septic
PIN
3111823420001
Supplemental fields
ProcessedPID
Updated
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� MUUND DESIGN WORK SHEET (For Flows u to 1200 d) <br /> A. Average Design FLOW A-1: Estimated Sewage Flows in Gallons per Day <br /> num er o I <br /> �'stimated � gpd (see figure A-1) bedrooms cioss i cios�n ciass nt cioss�v i <br /> or measured — x 1.5 (safety factor) — gpd 2 300 225 t so � <br /> 3 450 300 218 ofthe <br /> 4 600 375 256 values <br /> B. SEPTIC TANK Capacity 5 750 � 294 in the <br /> 6 900 525 332 Class I, <br /> I-��So � �-��-�no allons (see ure G1) � �050 �'�o s7o u, orm <br /> g �g 8 1200 67� 408 columns. <br /> i <br /> C. SOILS (refer to site evaluation) C-1: Se ticTankCa acitiesfm¢allons i <br /> Liquid capacity <br /> ,� u Number of Minimum Liquid Liquid capacity with W��disposal& <br /> 1. Depth to restricting layer = _1� - a, � feet Bedrooms Capaciry garbage disposal lift inside <br /> 2. Depth of percolation tests = I a `� feet Zo�ieS: �so �izs 15� <br /> Ld��� 3or4 1000 1500 2� <br /> 3. Texture 5 or 6 1500 2.'.50 3� <br /> Percolation rate •�o•o mpi �,a o�9 Z000 '000 <br /> 4. Soil loading rate , �t S 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 layer area. <br /> � SO gpd x 0.83 sqft/gpd = (��,a sqft a ��`°�D=���'�, <br /> 2. Determine rock layer width = 0.83 sqft/gpd x linear Loading Rate (LLR <br /> 0.83 sqft/gpd x 1 � gpd/sqft = �o ft Mound LLR <br /> 3. Length of rock layer = area= width = <br /> c���' sqft (Dl) = i o ft (D2) _ (o� ft < 120 M P I < � 2 <br /> E. ROCK VOLUME > 120 M PI < b <br /> 1. Multiply rock area (D1) by rock depth of 1 ft to get cubic feet of rock <br /> �!�- sqft x 1 ft = (�,�cuft � <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards i, <br /> (.�� cuft = 27 cuyd/cuft = �_cuyd i <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons <br /> a S cuyd x 1.4 ton/cuyd = 3 s� tons. <br /> D-33: Absorption Width Sizing Table <br /> F. SEWAGE ABSORPTION WIDTH PercolationRate LoadingRete <br /> in Mmutes per Soil Tcacse Gallons Absorptwn <br /> Inch per day per Ratio <br /> MPI wro foot <br /> Fester than 5 Coarsc Ser.d 1.20 1.00 <br /> Medium Sand <br /> Absorption width equals absorption ratio (See Figure D-331 '�°�"'ys`° <br /> times rock layer width (D2) 16 io 30 �. - o.�o z.00 <br /> 31 to a5 Silt Loa.-� 0.50 2.40 <br /> �.�� X �� ft = a c�. ft ae�0 6o s�,ar a,� o.as z.6� <br /> Silry Clay Loam <br /> 61 to 120 Silty Qay 0.2a 5.00 <br /> Sandy Cay <br /> I�. <br /> Slower than 120' <br /> •Sy��em duiEned for ihrx wil�rtun be aLer or perfwrncce <br />
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