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1995-006980 - new septic system
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North Arm Dr W
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4590 North Arm Drive West - 06-117-23-24-0016
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1995-006980 - new septic system
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Last modified
8/22/2023 5:25:59 PM
Creation date
9/21/2017 1:50:31 PM
Metadata
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Template:
x Address Old
House Number
4590
Street Name
North Arm
Street Type
Drive
Street Direction
West
Address
4590 North Arm Dr W
Document Type
Septic
PIN
0611723240016
Supplemental fields
ProcessedPID
Updated
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, , ` MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. �..�w Estimated Seavagt Flow in Gallons per Day(gpd) <br /> Estimated o o gpd <br /> Nianber <br /> of Type I Type II Type iII Type IV <br /> or measured - x 1.5 = - gpd.� s��� <br /> 2 soo Zu �so � <br /> 3 450 300 21g "r <br /> B. SEPTTC TANK LIQIJID VOLUMES `� <br /> 4 600 375 256 <br /> 5 750 450 294 �;�� <br /> a- �OO C� g�0I1S 6 900 525 332 � <br /> 7 1050 600 370 otlir <br /> �a�,�. <br /> 8 221'10 6'S 4138 <br /> t�S� :>,;a. <br /> C. SOIIS (refer to site evaluation) N�� �� �� <br /> 1. Depth to restricting layer= �o�' -�o a t�'inches B�� �,,. �, <br /> 2. Depth of percolation tests = 1 a " inches `""m'' `p'�`' <br /> 3. Percolation rate U.a mpi z�� �so �.�u <br /> 3 or 4 1,000 1,500 <br /> 4. Land slope (o `70 ;a s zo�oo 3:0�0°0 <br /> ova 9 See fig.C-6 (z 1.5) <br /> D. ROCK LAYER DIlviIIVSIONS <br /> 1. Multiply flow zate by 0.83 to obtain required area of rock <br /> layer:A x 0.83 = � � <br /> �o p gpd x 0.83 sq. ft./gpd =��sq. ft.�-�o�=5`��� <br /> 2. Select width of rock layer (10 feet or less) = I v ' ft. <br /> 3. Length of rock layer = area=width= Rock Bed <br /> �sq. ft.i �ft. = Ss� ft. ti.�.�.�.ti.w.ti.�.�.�.ti.ti.,.ti.�.ti.� <br /> , tif;?{:ti{{:{:'��:tiftifti tfti ti tiftifL <br /> ti:ti�ti:ti:Y~:��1 ti ti��t ti 1�tifti idth �l��t. <br /> tirt�ti tiKft�tirti•ti:tirti�ti�tifti�tifti ti <br /> .f.f.f.f.f.f.f.r.r.f.f�r.r•f.r.f. <br /> E. ROCK VOLUME �- �ng�` -� � <br /> 1. Multiply rock azea by rock depth to get cubic feet of rock; , <br /> �2 sq. ft. x ,vS ft. = S�cu. ft. <br /> 2. Divide cu. ft.by 27 cu. ft./cu. yd. to get cubic yards; <br /> �cu. ft. -�- 27= a L cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 2,1 cu. yd. x 1.4 ton/cu. yd. =3�t tons. <br /> F. ADSORPTTON WIDTH A� ti«,w�au s�, Table <br /> Gilw Raeo d <br /> 1. Percolation rate in top 12 inches of soil is N- mpi M�L�ti�R� �il Texture �.�:e '°;,,m <br /> (mpil f� ,e.o.�aa <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 oarse Sand 1.2o i.00 <br /> 0.1 to 5 Sand 1.20 1.00 <br /> ,�-! gpd/ft� 0.1 to5 FineSand" 0.60 2_00 <br /> 6 to 15 ndy Loam 0.79 1.�2 <br /> 3. Calculate adso hon width ratio b dividin rock la er 16 to 30 Lo�, o.bo 2.00 <br /> I'� }' $ }' 31 to 45 Silt Loam OSO 2.40 <br /> loadin rate of 1.20 d/ft2 b allowable soil loadin rate; �to bo cta Loam o.45 2_6� <br /> g gP Y g bi �o i2o �iay o.24 �_oo <br /> 1.20 �C�/ft2�- �`is� gpd/ft�_ �.(i� Slowerthan 120 Cla -- - <br /> �soil tv.�ix,g�ox or more of Eic�e a�erv f�e su,a. <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> a•��_x�_ft=a�•7 ft <br />
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