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2003-P06629 - new septic system
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4700 North Arm Drive West - 06-117-23-23-0007
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2003-P06629 - new septic system
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Last modified
8/22/2023 5:25:34 PM
Creation date
9/25/2017 1:06:48 PM
Metadata
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Template:
x Address Old
House Number
4700
Street Name
North Arm
Street Type
Drive
Street Direction
West
Address
4700 North Arm Dr W
Document Type
Septic
PIN
0611723230007
Supplemental fields
ProcessedPID
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• I ' � � <br /> � Mf�U�TD DESIGN WORK SHEET(For Flows u to 1200 d) <br /> A. , Average Design FLUW A-1: Eatimated Sewage Flowa in Gallons per Day <br /> num er o <br /> � Estimated o d gpd(see figure A-1) bedrooms c�as:t c�au u c�ass��� c�ass�v <br /> or measured x 1.5 (safety factor) = gpd 2 �oo �225 ��o � <br /> 3 . 450 300 218 of the <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 /> a-/o o O allons (see re C-1) � » � 37o u,or m <br /> g .�� 8 1200 675 408 columns. <br /> C. SOILS (refer to site evaluation) ai:Se ticTaakCa acides(in uo� <br /> liquid capaciry <br /> Numba of Minimum liqnid liquid capaciry with w����� <br /> 1. Depth to restricting layer= /.�.i..a.o��.s feet B� ���' ��� uft��a� <br /> 2. Depth of percolation tests= /o feet . 2«� �so �izs �� <br /> 3. Texture G�-�-`� ��-w� 3o�a �000 �soo � <br /> S a 6 1500 2250 300p <br /> Percolation rate � a•0 mpi , �,8 a 9 ?,000 3000 <br /> 4. Soil loading rate •4� gpd/sqft(see figure D-33) <br /> 5. Percent land slope 3 % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow (A)by 0.83 to obtain required rock layer area. <br /> (oe c� gpd x 0.83 sqft/gpd = � � sqft <br /> 2. Determine rock layer width= 0.83 sqft/gpd x linear Loading Rate (LLR <br /> 0.83 sqft/gpd x l� gpd/sqft= J o ft Mound LLR <br /> 3. Length of rock layer= area+width= <br /> � � sqft(Dl)+ �� ft(D2) _�ft < 120 M PI <� 2 <br /> E. ROCK VOLUME > 120 (VI PI < 6 <br /> 1. Multiply rock area (D1)by rock depth of 1 ft to get cubic feet of rock . <br /> �4s sqft x 1 ft= '�1� cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> � `7 cuft +27 cuyd/cuft=�_cuyd <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons <br /> �_cuyd x 1.4 ton/cuyd = a.3 tons <br /> D-33: Abeorption Width Slzing�Lble <br /> percol�tion Rate l,o�dins Rate <br /> F. SEWAGE ABSORPTION WIDTH inMinmesper Sa�T.�� c��«� A�o�;� <br /> �h pa�y ptt Ratio <br /> 1 u�re foot <br /> Futer Uwn S Co�rx Sud 1.20 1.00 <br /> . Madium Sud' <br /> Absorption width equals absorption ratio (See Figure D-33) `A""'S"� <br /> times rock layer width (D2) ,o ;, 2� <br /> d..�J X 1 � lt- o��o. lt 46 to 60 S,ady�Q��m 0.45 2.67 <br /> 61 to 120 S ilty p�a�� 0.24 5.00 <br /> ud <br /> . ower <br /> . •Srn„n daiprA tor tlws rofb m�p De otL�r a yafomrses <br />
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