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1995-007458 - septic system
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1650 North Farm Road - 27-118-23-44-0009
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1995-007458 - septic system
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Last modified
8/22/2023 4:23:16 PM
Creation date
9/29/2017 12:27:54 PM
Metadata
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Template:
x Address Old
House Number
1650
Street Name
North Farm
Street Type
Road
Address
1650 North Farm Road
Document Type
Septic
PIN
2711823440009
Supplemental fields
ProcessedPID
Updated
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, � � � • ' MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. �.•�w Estimucd Sewage Floa in Gallons per Day(gpd) .-1 <br /> Estimated �o � �C� N�� Type I Type fI <br /> or measured -- x 1.5 = -- gpd. $��,,,, �'p�JII �rv�rv <br /> 2 300 2?S 180 sos <br /> 3 450 300 218 °� <br /> B. SEPTTC TANK LIQUID VOLIJMES � 4 �o0 3�s �w '� <br /> s �so eso z� "�°o <br /> 6 soo su 332 � <br /> �-/r�o v g ons � ioso 600 3�o ed1e <br /> �� <br /> �y_r���.� 8 1200 675 408 <br /> C. SOILS(refer to site evaluation) �'°'�'�� �`��� <br /> �, N�mmbc ��t c��.u► <br /> 1. Depth to restricting layer = .-��-�u a�Y inches �� ���,. �,,, <br /> 2. Depth of percolation tests = 1 a inches ``"`"' `"'�'' <br /> � z«� �so �.�u <br /> 3. Percolation rate =�• � mpi s«a �.000 �,soo <br /> 4. Land slope 3 % s�6 �.soo 2aso <br /> 7 a 8 2,000 3.000 <br /> wa 9 See fig.C-6 (z 1S) <br /> D. ROCK LAYER DIlvIIIVSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: A x 0.83 = , <br /> �oo . gpd x 0.83 sq. ft./gpd =��u sq. ft.-t ��`��=Sy�II <br /> 2. Select width of rock layer (10 feet or less) _ �o ft. <br /> 3. Length of rock layer = azea=width= Rock Bed <br /> �y'1 Sq. ft.�- /D ft. = SS ft. ti•ti•�•�...ti.�.�•�•�.�•�.�.�.�.ti.� <br /> J•1.f•f•f•r•f•t•� t•t•1•1•l•t•1• <br /> {•tiH•ti•��ti•ti•ti•ti•ti•ti•ti•ti•ti•ti•\•t <br /> 1•r•f•1•j•1•1•1•t t•�•t•r•t•t•r• �dth 510 ft. <br /> 1.ti•ti•ti•ti.ti•t•ti•ti•b1•tiKKK•ti•ti <br /> . f•t•t•J•1•r•r•/•t t•t•t•t•t•1•f• <br /> 1.4•�•ti•ti.ti•t•ti•t•L•1•�•ti•ti•ti•ti•ti <br /> .r.f.f•J.f•J•r.r.f.�•�.r.f•t.f.j• <br /> E. ROCK VOLUME �- �'g�' -� <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; �- � <br /> �sq. ft. x l,c s ft. _ <�u cu. ft. <br /> 2. Divide cu. ft.by 27 cu. ft./cu. yd. to get cubic yards; <br /> ��u cu. ft. i 27= �. � cu. yd. � <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> a� cu. yd. x 1.4 ton/cu. yd. _ ��> tons. <br /> F. ADSORPTION WIDTH Li-�`� �'o'q�,-� ��►�a�,s�, �bk <br /> � m i r�a�tia,x�� ��. R.m d <br /> 1. Percolation rate in top 12 inches o soi is �� p a x� �,,r ,.��,, <br /> AM;,,u m�n,c}, Soil Texture p�.q,,.� ,,,;,,,,` <br /> (as �b.eroam <br /> Mdd� <br /> 2 Select allowable soil loading rate from table; Faster than 0.1 0�sana 1.20 l.00 <br /> 0.1 to� Sand 120 1.00 <br /> ,y .. � gpd/ffi� 0.1 to� Fine Sand" 0.60 2.00 <br /> 6 to 15 ndy Loam 0.79 iS2 <br /> 3. Calculate adso Hon width ratio b dividin rock la er 16 to�o [.o� o.bo 2.00 <br /> rp Y g Y 31 to 45 s�ic r.�m o� z•� <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; -�.to�o -. ciay t,o�, �0.4$ 2.6� <br /> 61 tU 120 Clay 024 �.00 <br /> 1.20 gPd/ftZ'.- ,�i gpd/ft�_ �.(o'� Slower than 120 Cla - - <br /> "Soil having 509i ar more of fine or vay fine sand <br /> 4. Multiply adsorpdon width ratio by rock layer width to get <br /> required adsorption width; <br /> ,�� x /o ft= ��,,.� ft <br /> ���.J <br />
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