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2000-P02807 - new septic
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175 Crystal Creek Road - 33-118-23-33-0006
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2000-P02807 - new septic
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Entry Properties
Last modified
8/22/2023 4:50:01 PM
Creation date
6/7/2016 2:01:24 PM
Metadata
Fields
Template:
x Address Old
House Number
175
Street Name
Crystal Creek
Street Type
Road
Address
175 Crystal Creek Road
Document Type
Septic
PIN
3311823330006
Supplemental fields
ProcessedPID
Updated
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� � � MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. �.�w Escimued Sewa6e Flow ia Galla�s pc Day(gpQ) - <br /> Estimated �n�_gpd <br /> or measured - x 1.5 = - gpd. s�� �`I �`�► rya li� ry�rv <br /> 2 30o zas i so � <br /> 3 450 300 218 °r <br /> B. SEP'ITC TANK LIQUID VOLUMES 4 �o0 3�s u� .� <br /> S 750 450 244 � <br /> a -/ L�0 O g�0I1S 6 900 525 332 '^° <br /> 7 IQSO 600 370 �� <br /> 8 12D0 675 408 <br /> C. SOILS (refer to site evaluation) N,�� � � <br /> 1. Depth to restricting layer = ►��'-r� 3�� inches & °� �,�` �,w <br /> 2. Depth of percolation tests = ia''-�� '' inches `'�'"' `''�"' <br /> 3. Percolation rate y - a mpi Z3 a'� ,'o�oo ;•,s o <br /> 4. Land slope � % �a a � s�0000 <br /> we 9 See fis.C-6 (x 1S) <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 /> �oo . gpd x 0.83 sq. ft./gpd = �,�ts sq. ft.+,o�o sy�"� <br /> 2. Select width of rock layer (10 feet or less) _ �o ft. <br /> 3. Length of rock layer = area y width = Rock Bed <br /> �� sq. ft. i �ft. = ss ft. ,.,.�.,.�.�.�...'*'*'^^.^.^*;.,.�.,., <br /> f•t•t•1•1•f•I•t•I•I•t•l•�•t•1•1• <br /> y•'\•ti.1•1•ti•'�•ti•ti•t•ti•ti•ti•ti•ti.�.1 <br /> {fti:�:�:ti:tirtifti:��tif�rtirtirti�r�f{ idth <_10 ft. <br /> �':�:�fti%�:tirti�ti�tirtirti={�ftif�f�f� <br /> .f.f.f.r.r.r.r.f.f.r.r.f.�.r.f.f. <br /> E. ROCK VOLUME ~- �'�'g�' -�' <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> s�� sq.ft. x 1.vSft. = 5�y cu. ft. <br /> 2. Divide cu. ft.by 27 cu. ft:/cu. yd. to get cubic yards; <br /> s�" �.ft. y 2�_��.�u. ya. <br /> 3. Multiply cutiic yards by 1.4 to get weight of rock in tons; <br /> aL cu. yd. x 1.4 ton/cu. yd. _�� tons. <br /> F. ADSORPTTON WIDTH `-L A`� `-'0'�'"� �a,w�aw s�, Lbk <br /> 1. Percolation rate in top 12 inches of soil is �+.-z- mpi PeecoLeonRaee G°�- R�°' <br /> M�m�;,,�, Soil Texture �„�, �,°ed <br /> i.e .�a, <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 /> �'-�S gpd/ft� 0.1 to 5 Fne Sand" 0.60 2.00 <br /> 6 to 15 ndy L,oam 0.79 1�2 <br /> 3. Calculate adso hon width ratio b dividin rock la er 16 to�o t,oam o.bo 2.00 <br /> I'� }' g }' 31 to 45 Silt Loam OSO 2.40 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; �to bo cta Lo�, o.as 2.6� <br /> 1.20�d/ft2i � d/ft�= a•lo Slowerthan 120 Clay 024 5_00 <br /> �� Z-• <br /> "Soil having SQ9L or maee d Fine or vey Cux xnd <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> �.�2 x�ft = ab. ft <br />
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