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1994-006523 - new septic system
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3515 Sixth Ave N - 29-118-23-43-0002
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1994-006523 - new septic system
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Last modified
8/22/2023 4:26:56 PM
Creation date
1/23/2019 1:58:56 PM
Metadata
Fields
Template:
x Address Old
House Number
3515
Street Name
6th
Street Type
Avenue
Street Direction
North
Address
3515 6th Avenue North
Document Type
Septic
PIN
2911823430002
Supplemental fields
ProcessedPID
Updated
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� MOUND DESIGN WORKSHEET <br /> , (For Flows up to 1200 gpd) <br /> � � : <br /> A. �.�w Estimuod Sewage Flow in Gallons pet Day(gpd) <br /> Estimated �t SO gpd � <br /> Number <br /> or measured x 1.5= - gpd. ' B�°�,,,, �'a i �'��i �YP�I11 �y�rv � <br /> 2 300 21S 180 soa� <br /> �J'>,`-i 3 450 300 218 � <br /> B. SEPTIC TANK LIQUID VOLUMES -�'• � � 4 �o0 3�s zs� .�,�. <br /> s �so aso 29a a <br /> �--��(� U g�OIIS 6 900 57S 332 <br /> � <br /> � 7 1050 600 370 odir <br /> �.�. <br /> 8 1200 675 408 <br /> G SOIIS (refer to site evaluation) ,, ,� N,�� �� �� <br /> 1. Depth to restricting layer = � I y To aa inches B�� �,,, �, <br /> 2. Depth of permlation tests = 1 a'' inches `""�'' `"�°'' <br /> 3. Percolation rate 3.� mpi Z3 a`� ,'� ;•� <br /> 4. Land slope % ;a 8 �,'� 3�° <br /> ova 9 5ea fig.G6 (x 1.5) <br /> D. ROCK�LAYER DIIv�NSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer:A x 0.83 = � � <br /> y�0 . gpd x 0.83 sq. ft./gpd = 3�� sq. ft-+��`�o W lu" <br /> 2. Select width of rock layer (10 feet or less) = I D ft. <br /> 3. Length of rock layer= azea i width= Rock Bed <br /> y 1 D sq. ft.i /0 ft. _ ��- ,�ft. �.,.,.�.�.ti.,.�.,�.�.ti.,.�....., <br /> ti�tifti�ti•ti:�:1�1�ti�ti•ti ti ti t���{:ti <br /> ti{:ti:ti:tirtifti ti=ti:tifti{�ti:tirtirtir� idth <_10 ft. <br /> 1fti�l�ti�ti�lfl�ti ti�tf��ti tirti�1?ti ti <br /> •f.�.t•f.f.f.f.l.t.�.t.f.f.l.�.l. <br /> E. ROCK VOLUME ~- �ng�' -� <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> y�sq. ft. x .�ft. _ �o cu. ft. <br /> 2. Divide cu. ft.by 27 cu. ft./cu. yd. to get cubic yards; <br /> �cu. ft. -:-27=�!�cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> �_cu. yd. x 1.4 ton/cu. yd. _� tons. <br /> F. ADSORPTION WIDTH Gt��( LDY��v� n� �a,w,a�,s�, rabk <br /> 1. Percolation rate in top 12 inches of soil is 3'� mpi r�,,,e�,R,� �°� R�°' <br /> �trtinuee,prr inci, Soi1 Texture �,� �„ <br /> (mpU �� .�� <br /> 2. Select allowable soil loading rate from table; Fascer chan o.i oarse Sand i.2o i.00 <br /> �-I S d/ft� 0.1 to� Sand ].20 1.00 <br /> ' � 0.1 to� Fne Sand"' �•� 2•� <br /> 6 to 15 ndy Loam 0.79 1S2 <br /> 3. Calculate adso hon width ratio b dividin rock la er 16 to 30 t,oam o•� 2•� <br /> Ip }' g }' 31 to 45 Silt Loam OSO 2.40 <br /> load.ing rate of 1.20 gpd/ft2 by allowable soil loading rate; �co bo ��ay Loan, o.4s 2.6� <br /> 67 t0120 Clay 0.24 �.00 <br /> 1.20 �d/ft2�-�gpd/ft2= a� (0 7 Slower than 12o Clav -- - <br /> ••Soil having�09�or more oE Eine or very fine und <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> �,t�r x�ft =�L•� ft <br />
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