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2009-00588 - new septic
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3420 High Lane - 05-117-23-12-0026
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2009-00588 - new septic
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Last modified
8/22/2023 5:16:20 PM
Creation date
1/26/2017 2:19:53 PM
Metadata
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Template:
x Address Old
House Number
3420
Street Name
High
Street Type
Lane
Address
3420 High La
Document Type
Septic
PIN
0511723120026
Supplemental fields
ProcessedPID
Updated
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• � MOUND DESIGN WORK SHEET(For Flows u to 1200 d) <br /> A. Average Design FLOW A-1: Estimated Sewage Flows in Galions per Day <br /> number of <br /> Estimated ��v gpd (see figure A-1) bedrooms Class I Class 11 Class ill Class iV <br /> or measured -- x 1.5 (safety factor) _ -- gpd 2 300 225 180 60% <br /> 3 450 300 218 ofthe <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 /> 2,—J�6c� allons (see i ure C-1} � 1050 600 37o II,or III <br /> �� g Gl����f S . 8 1200 675 408 columns. <br /> ��eo ��1 � wl� �� - <br /> C. SOILS (refer to site evaluation) C-1: Se ticTankCa acities(in allons <br /> Liquid capacity <br /> � � Number of Minimum Liquid Liquid capacity with W��disposal& <br /> 1. Depth to restricting layer = I•� 1.Co �i.� feet Bedrooms Capacity garbagedisposal ��ftinside <br /> 2. Depth of percolation tests = /- o feet zo�iess �so �tzs 15� <br /> 3. Texture L��`� �---����1 3or4 t000 isoo 20� <br /> 5 or 6 1500 2250 3� <br /> Percolation rate I S�(� ri1p1 7,8 or 9 2000 3000 <br /> 4. Soil loading rate -�-t-� gpd/sqft (see figure D-33) <br /> 5. Percent land slope 3 % <br /> D. ROCK LAYER D�IMENSI01�1S <br /> 1. Multiply average design flow (A) by 0.83 to obtain required rocl< layer area. <br /> �-isc� gpd x 0.83 sqft/gpd = 3�'3 sqft��o°�p-����� - <br /> 2. Determine rock layer width = 0.83 sqft/gpd x linear Loading Rate (LLR <br /> 0.83 sqft/gpd x 1� g-pd/sqft = 1 c� ft �ound LLR <br /> 3. Length of rock layer = area = width = <br /> !O sqft (Dl) = 10 ft (D2) _ �I�ft < 120 M PI < � 2 <br /> E. ROCK VOLUME > 120 (VI PI < C� <br /> 1. Multiply rock area (D1) by rock depth of 1 ft to get cubic feet of rock <br /> y)� sqft x 1 ft = u 10 cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> �t)0 cuft = 27 cuyd/cuft = I� cuyd <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons <br /> J Sr cuyd x 1.4 ton/cuyd = a I tons <br /> D-33: Absorption Width Siaing Teble <br /> ]F. SEWAGE ABSORPTION WII�TH PercolationRate LoadingRate <br /> in Minu�es per Soil Tezture Gallons Absorption <br /> Inch per day per Ralio <br /> MPI s uarc foot <br /> Faster than 5 Coarse Sand 1.20 1.00 <br /> Medium Sand <br /> Absorption width equals absorption ratio (5ee Figure D-33) F�s�dd <br /> times rock layer width (D2) 16 io 30 L�m o.�o z.o� <br /> 3i�o as s�u t.� oso 2.ao <br /> a,� X 1 o ft = ac�,� ft 46 to 60 s�Y QaY� 0.45 z.6� <br /> Siiry Clay Loam <br /> 61 to 120 SiltyClay 0.24 5.00 <br /> Sandy Clay <br /> Cle <br /> Slowcrthan 120• <br /> •Sysiem dcsi�ned tor Ncse aoils m�s�Dc aLer or perlomunce <br />
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