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2005-P08908 - septic
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280 Hollander Road - 25-118-23-43-0012
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2005-P08908 - septic
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Entry Properties
Last modified
8/22/2023 4:15:34 PM
Creation date
2/24/2017 1:32:32 PM
Metadata
Fields
Template:
x Address Old
House Number
280
Street Name
Hollander
Street Type
Road
Address
280 Hollander Road
Document Type
Septic
PIN
2511823430012
Supplemental fields
ProcessedPID
Updated
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'� MOyfJND DESIGN WO�IC SHEET(For Flows u to 12�0 d. . . <br /> A. Average Desigl��F'LOW�� .� :A-1.�fstimatad Sawo�o FIQw;1n Gc11o�u,per D�ry <br /> nu r o <br /> �.�c• bedrooms Clau I Clau 11 Clas;lil Class IV <br /> Estimated �=gpd(see figure�A-1)� : 2 � �22s � t eo � <br /> or measured -- x 1:5 (safety factor)=-- gpd 3 � � 2�$ ot tne <br /> q b00 � 375 � 256 values <br /> B. SEP'TIC TANK Capadty 5 �50 450 294 in the <br /> 6 9q0 625 332 Class I, <br /> 7 1p5p 600 370 II,or ill <br /> ►�SO �+- �000 gallons(see figure C-1) g :� : 12pp:. . 675 408 colur�ns. <br /> 4r laSOg'�.Y�w.T c�Ar+�6IF � ' <br /> C. SOILS (refer to site evaluation) � . ai:S �IC�OkCY CI� IIOOS . <br /> .. . .. : liquid'capacir <br /> Na.mba d� Miuimwn Liquid �.Liqufd capulry with .with disposal8 <br /> 1. Depth to restricting layer= �/.� � a.� feet �°'°� . �'�' 8��R� lift inside <br /> 2. Depth of percolation tests= �•0 feet z«� �so llu �soo <br /> 3 a 4 1000 1300 200p <br /> 3. Texture ��� 1��v� sa6 �sao � � 3000 <br /> Percolation rate �t�.•� mpi � • � . ��$°r9 . �0° 3°°° <br /> 4. Soil loading rate -4S gpd/sqft(see figure D-33) <br /> 5. Percent land slope I � % � <br /> D. ROCK LAYER DIMENSIONS � � <br /> 1. Multiply average design flow (A)by 0.83 to obtain required rock layer area. <br /> �S o gpd x 0.83 sqft/gpd = �a�, sqft <br /> 2. Determine rock la.yer width= 0.83 sqft/gpd x liriear Laading Rate:(I:LI� <br /> 0.83 sqft/gpd x 1=Qud/sqft= �;_ft I�lound LLR <br /> 3. Length of rock layer=area+.width= <br /> 'c�_sqft(Di)+ �ft(D2) _�_ft� ��. � � 20 M P I <1.2 <br /> E. ROCK VOLUME . . � . . � � 2O M�PI < 6 <br /> . � - - . <br /> 1. Multiply rock area.(Dl)by rock depth of 1 ft to get cubic feet of rock , <br /> �_��sqft x 1 ft= �t�� .cuft <br /> 2. Divide cuft by 27 cuft/c�xyd to get cubic yards � <br /> t��_cuft +27 cuyd/cuft.= �� _:cuyd <br /> 3. Multiply cubic y�rds by 1.4 to get weight of rock in tons - <br /> �_cuyd x Y.4 ton/cuyd = a a tons <br /> ' n�33:.Ab.orPuoe vViath si�Ir�T1.6�s <br /> Ptirod�Uan Rat� -1�d�s�. <br /> F. SEWAGE ABSORT'TTON vVIDTH �� � ±���� �Tµ� �;,�;�^' "R�``�' <br /> _ , . � F..a.w�s cu.�.s.�a �ao �.00 <br /> : M.diums.na. , <br /> Absorption width equals absorption ratio (See Figure D-33) � � 'p'"''�' , <br /> times rock layer width(D2) " � . , a � . <br /> �. ��x I� ft= �10•� ft f ,b a bo s�iidya.°"'r i.o.m o.4s s.6r..:>.,. <br /> . 61 a 1� S�•nd Q.y 0.24 3.00 <br /> . � •f�erfP���e�otbmn�Mab�ror� emrow � <br />
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