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1997-008926 - new septic
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2650 Countryside Drive West - 04-117-23-12-0014
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1997-008926 - new septic
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Last modified
8/22/2023 5:07:07 PM
Creation date
5/2/2016 3:50:19 PM
Metadata
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Template:
x Address Old
House Number
2650
Street Name
Countryside
Street Type
Drive
Street Direction
West
Address
2650 Countryside Dr W
Document Type
Septic
PIN
0411723120014
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORKSHEET <br /> , (For Flows up to 1200 gpd) <br /> A.' FLOW Estimatcd Scwagc Flows in Callixis pa dny <br /> (BPd) <br /> Estimated '� c� gpd (see pages D-7 or I-3, 4, 5) �m�� <br /> or measured r-- gpd x 1.5 = - ��o�ms Type I 'fypc 11 Type III �� <br /> 2 300 225 180 <br /> B. SEPTIC TANK LIQUID VOLUMES 3 aso 30o Z�a o� <br /> 4 600 375 256 ���� <br /> j -lasc� dt 1--/Oo� gallons (see pages C-3 or G5) s -so aso Z�a ;o <br /> 6 900 525 332 T��• <br /> 7 1050 600 370 ID <br /> 8 1200 675 408 edumn <br /> C. SOILS (refer to site evaluation) <br /> '1 i I V . ScplicT�nk Cypacilic�,in gylluny <br /> 1. Depth to restricting layer= 1�s,, �,ti a- aa inches N���o� ������um Li �a i.� ���.,��� <br /> 11 q ' y ' I y wIN <br /> 2. Depth of percolation tests = 1 a inches ����m� �.�.�,ry Q+rba�c dupwyl <br /> 3. Percolation rate �.�- mpi z a kf+ ,�o ,�u <br /> 3�a �000 �soo <br /> 4. Land slope S % 'o� 9g � ,� <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: Daily Flow x 0.83 = , <br /> � v gpd x 0.83 sq. ft./gpd = Ga� sq. ft.-�an`� ��4� <br /> 2. Select width of rock layer (10 feet or less) = l D ft. <br /> 3. Length of rock layer = Area+Width = <br /> �sq. ft. 1 �ft. _ �� ft. Rock Bed <br /> r•r•r•r•r.r•r•r•r�r•r•r•f•r�r <br /> �.ti.ti.�.ti.�.ti.ti.ti.�.ti.�.ti.�.�. T <br /> tiftiftir{rtiftifti�tifl�tifti�'tfti�tif{� idth S10ft. <br /> . , r r•r•r•r•r.r.r r•r•r•r•r•r•r <br /> ;.�.�.�.ti.ti.�.�.;.�.�....•..ti.... 1 <br /> .r.f.;.f.f.f.r.f.f.f.f.;.f.l.r <br /> E. ROCK VOLUME � Length <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> l�sq. ft. x�ft. _�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 /> a' cu. yd. x 1.4 ton/cu. yd. = 3� tons. <br /> F. ADSORPTTON WIDTH LL �-�0 �''� <br /> Absorption Width S3zfng 7'able <br /> 1. Percolation rate in top 12 inches of soil is ��'t- mpi <br /> Pucolatlon Ratc Gallons Ratio of <br /> 2. Select allowable soil loadin�rate from table on page E-; in Minute�pu Soil Texture per day per Abswp�ion wiJih <br /> (ocA(MPl) square foot �a Rock Irycr <br /> ��- gpd/ft2 w��n <br /> 3. Calculate adsorption width ratio by dividing rock layer Facter than 0.1• c�sv,a .--- _._._ <br /> 0.1 to S Send 1.20 1.00 <br /> loading rate of 1.20 gpd/ft�by allowable soil loading rate;. o.��s•• Flne Sand�• o.�o z.no <br /> 6 to 15 Sandy Loam OJ9 I.52 <br /> 1.20 gpd/ft�i ,�s� gPd/ftz= Z�1� . 16 to 30 �.,,, o.� z.�, <br /> 11 to 45 Silt l.oam 0.50 2.40 <br /> Check this value on a e E-16. ���o c►�y�,,, o.4s z.�� <br /> p g . 6o w�2o c�sy o.za s.W <br /> 4. MulHply adsorption width ratio by rock layer width to get� s��m;;;�� c�ny _____ ____ <br /> required adsorption--width; � � � - °�� <br /> Z.t,? x�ft = Z�-7 ft <br /> .1:.r ... ... . . . , _� <br />
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