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2001-P04453 - septic
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2601 Rainey Road - 04-117-23-44-0004
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2001-P04453 - septic
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Last modified
8/22/2023 5:15:06 PM
Creation date
7/12/2018 2:26:20 PM
Metadata
Fields
Template:
x Address Old
House Number
2601
Street Name
Rainey
Street Type
Road
Address
2601 Rainey Rd
Document Type
Septic
PIN
0411723440004
Supplemental fields
ProcessedPID
Updated
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, : <br /> � MO ND DE IGN WORK SHEET(For Flows u to 1200 d) <br /> R.. A erage De ign FLOW A-1: Eatimated Sewage Flowa in Gallons per Day <br /> num r o <br /> Es ated�CZ.gpd (see figure A-1) bedrooms ciass� c�au u Gasa u� c�ass�v <br /> or easured x 1.5 (safety factor) _ — gpd 2 �o0 225 �eo � <br /> . 1-l0•,� 3 . 450 300 218 of the <br /> 5 �"�-- la��sT 1-}O�S� 4 600 375 256 values <br /> B. S PTIC TAA1K Capacity 5 750 �.50 294 in the <br /> 6 900 525 332 Class I, <br /> M��� 7 600 370 II,or III <br /> a- � � o �. gallons (see figure C-1) N oUs� <br /> �_ �a��,� b�V,'�s-�1-a o ti..1 S�E:. 8 1200 675 408 columns. <br /> C. S ILS (refer to site evaluation) Gl: Se tkTauk Ca acitia(in Ilons <br /> l A 7,l4 5�• Liquid capacity <br /> Number of Minimum Liquid liquid capacity with W��dis & ' <br /> 1. Depth to restricting layer= �.�`�� 7l�`� ���� feet Bedrooms Capaciry garbagedisposal liftins�d <br /> 2. Depth of percolation tests = ► a" feet Za� �so i�zs 15� <br /> 3. Texture �v�M �5�. c.L��(Lv,av� 3�a i000 �soo � <br /> Percolation rate �,� mpi �Sg�6g � � 300° <br /> 4. Soil loading rate .=1 gpd/sqft(see figure D-33) <br /> 5. Percent land slope 7 % <br /> D. OCK LA�cIER DIMENSIONS <br /> 1. Multiply average design flow (A) by 0.83 to obtain required rock layer area. <br /> 10 5�� gpd x 0.83 sqft/gpd = �6 � a sqft I <br /> 2. Determine rock layer width = 0.83 sqft/gpd x linear Loading Rate (LLR <br /> 0.83 sqft/gpd x I a gpd/sqft= io ft Mound LLR <br /> 3. Length of rock layer= area=width = <br /> 4{� �. sqft(D1) � J�ft (D2) _�_ft < 120 M PI <12 . <br /> E. CK VOLUME > 120 MPI < 6 <br /> 1. Multiply rock area (D1)by rock depth of 1 ft to get cubic feet of rock ! <br /> ��. sqft x 1 ft=�4 cuft � <br /> � <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> . 7�_ �uft +27 cuyd/cuft=�,_cuyd <br /> 3. Multiply�ubic yards by 1.4 to get weight of rock in tons <br /> . . �_ yd x 1.4 ton/cuyd = U� tons <br /> D-33: Abeorptlon Width SizinQ Table <br /> F. S WAGE A;BSORPTION WIDTH �•���� L.oadinaRate <br /> in Minutes per Sal Teainue Gallont Absorption <br /> Ineh per day per Ratio <br /> MPI wre foot <br /> F�sur than S Coersc Sand 1.20 1.00 <br /> Medium Sand <br /> Abs rption width equals absorption ratio (See Figure D-33) `A'°''S'"° <br /> time rock laye�width (D2) � � <br /> ��o< ;� . :< <br /> .t,� x 1Q—ft- � �•�� f t ��o�o s.�Y a.y c�<.s_� - 2.6� <br /> — sci� c�,y i.o.m <br /> 61 to l20 Silty pay 0.24 5.00 <br /> , S�ndy p�y <br /> S ower ihan 120• <br /> . �`: ����'' � •Spaem daiEnd for�ue wils m�a be a6rr w ptfomreca <br /> �� � <br /> � <br />
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