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2002-P05242 - new septic
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165 Crystal Creek Road - 33-118-23-33-0005
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2002-P05242 - new septic
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Entry Properties
Last modified
8/22/2023 4:49:59 PM
Creation date
6/7/2016 12:38:04 PM
Metadata
Fields
Template:
x Address Old
House Number
165
Street Name
Crystal Creek
Street Type
Road
Address
165 Crystal Creek Road
Document Type
Septic
PIN
3311823330005
Supplemental fields
ProcessedPID
Updated
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� + � / <br /> MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. FLOW Estimited Sew�aae Flow�in Gallo�s pa D�y(gp� <br /> Estimated l_o v gpd <br /> Nianba <br /> or measured � x 1.5 =- gpd. �°�n„ �'p`I �Yx It 'iype III �ry�rv <br /> 2 300 225 IBO bos <br /> 3 450 300 218 •r <br /> B. SEPTIC T�NK LIQtIID VOLLTMES 4 �o0 3�s zs� ,� <br /> � -/l��� g'�OI15 6 900 525 332 u,° <br /> 7 IUSO 600 370 �m, <br /> 8 1200 675 108 <br /> C. SOILS (refer to site evaluation) N,�� � � <br /> or �_. �.d,.r, <br /> 1. Depth to restricting layer = ��" - 3�.'' inches �e,� �„�, �, <br /> 2. Depth of percolation tests = �i" �� � inches `""`' `'""' <br /> 3. Percolation rate � •� mpi 2s a'� ,'o�oo i'.so <br /> 4. Land slope Co % �Q a � �o <br /> wa 9 See fi`.C-6 (z IS) <br /> D. ROCK LAYER DIlv�TSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer:A x 0.83 = , <br /> C.av . gpd x 0.83 sq. ft./gpd = .��sq. ftt�n°��=��° <br /> 2. Select width of rock layer (10 feet or less) _ /� ft. <br /> 3. Length of rock layer= azea y width= Rock Bed <br /> S�7 sq. ft.+ �_ft. = ss� ft. ,.,.,.,.�.,.,.�.�.,.ti.,.,.,.,.,., <br /> l�t.t.l.l•J.J•1.t r•r•1•!•r•f•f. <br /> \•ti•ti�1•�•ti•ti•t•ti•ti�ti•\•tiK•�•ti•1 <br /> ti`4r1:ti�ti:'ti:�ti���{fti ti�'yr�r���r� idth <_10 ft. <br /> y1.ti•ti <br /> t•t•1•f•t•f•t•t•r•�•�•t•I•1•t•!• <br /> '�•ti.ti•ti•1.1.1•tiH•ti•tiK.�•ti.�•1•� <br /> .I.�.t.f.f.f.f��•f.r.r.�.f.r.r.f. <br /> E. ROCK VOLUME ~- �`�' -� <br /> 1. Multiply rock azea by rock depth to get cubic feet of rock; <br /> �sq:ft. x l.os ft. = s� cu. ft. <br /> 2. Divide cu. ft.by 27 cu. ft:/cu. yd. to get cubic yards; <br /> S��1 cu. ft. j 27=�,L cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> �.L cu. yd. x 1.4 ton/cu. yd. _� tons. <br /> F. ADSORPITON WIDTH "�`� �0�"''� �,w,a�,s�, Lble <br /> 1. Percolation rate in top 12 inches of soil is �•S mpi ��,,,ti�,�,� ��. �.�d <br /> M;,,t,4,pe�;xx1, Soil Texture p�,y„�� � '"`"ed <br /> ea�r <br /> (�� Ir �b.syeo� <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 oarse Sand 1.20 l.00 <br /> � gpd/ft� 0.1 to 5 Sand 1.20 1.00 <br /> 0.1 to 5 F'ine Sand" �•� Z•� <br /> 6 to 15 ndy Loam 0.79 1�2 <br /> 3. Calculate adso tion width ratio b dividin rock la er 16 co�o �,o�„ o.bo 2.00 <br /> rP Y g Y 31 to 45 s�ie�m oso 2.ao <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; �to bo ctay[.o�, o.as 2.6� <br /> 61 t0120 Clay 024 �.00 <br /> 1.20 gPd/ft2; .�gpd/ft� _ �.!.'� Slower than 120 Cla - - <br /> �soil tv.�in�sox a ma�e d fi:x a�Qy fux au,d <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> �x�ft =aL•� ft <br />
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