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1993-005483 - repair septic
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1993-005483 - repair septic
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Last modified
8/22/2023 4:16:46 PM
Creation date
1/14/2019 2:39:57 PM
Metadata
Fields
Template:
x Address Old
House Number
1540
Street Name
6th
Street Type
Avenue
Street Direction
North
Address
1540 6th Avenue North
Document Type
Septic
PIN
2611823320005
Supplemental fields
ProcessedPID
Updated
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� ' MOUND DESIGN WORKSHEET <br /> • ' (For Flows up to 1200 gpd) <br /> A. FLOW Estima�ed Sewage Flows in Gallons per dey <br /> Estimated �-1�v gpd (see pages D-7 or I-3,4,5) ���� <br /> � �� <br /> or measured -- gpd x 1.5 = - ���m: Tra�I ry�u �y�ii� �� <br /> B. SEPTIC TANK LIQUID VOLUMES i aso 3a'�o ii°s � <br /> 4 600 375 256 �r�. <br /> � - �D o v gallons (see pages C-3 or C-5) s �so aso Z�a ;. <br /> 6 900 525 332 '�t. <br /> 7 1050 600 370 Q <br /> C. SOILS (refer to site evaluation) $ 1200 6's °os `°'�" <br /> l. Depth to restricting layer = ao��fo y y �� inches �p�KLnkGpseilic;inplMms <br /> 2. Depth of percolation tests = " Numberof MinimumLiquid l.iquideaprcitywitA <br /> I a inches ��. c.�x�ry �..�du�w► <br /> 3. Percolation rate_ �-I , la m i 2«�» ��o ��u <br /> p �a. ,� ,� <br /> 4. Land slo e S % '�6 �xp � <br /> P �,a«9 � � <br /> over 9 .._.. • <br /> D. ROCK LAYER DIMENSIONS <br /> 1. MulHply flow rate by 0.83 to obtain required area of rock <br /> layer: Daily Flow x 0.83 = <br /> y v gpd x 0.83 sq. ft./gpd = ��3 sq. ft.-j-�u:b y�o sa.��. <br /> 2. Select width of rock layer (10 feet or less) _ _ /c� ft. <br /> 3. Length of rock layer = Area i Width = <br /> y 1 n sq. ft. -:- �U ft. __y�ft. Rock Bed <br /> r�t t r•l f•J•t•f:f.f:f.f.f:r T <br /> ti••ti•ti•ti•ti•ti•ti••.•'. • '. ti ti ti• I <br /> ;f•r•f•f•f•f•t•l�t.f:j.f•f.!•l <br /> •�•ti•ti•ti•ti•ti•ti•. ti•ti.ti•ti•ti•ti• Nidth SlOft. <br /> f�r•t�f.f.r�f.f�f.r�f�r.f.f�f� <br /> �ttif�•�tftiltif::tif:•�r�1•:lf:J:.' <br /> E. ROCK VOLUME � � Length <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> �sq. ft. x .1�_s ft. _�cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> 3o cu. ft. s 27= �cu. yd. � <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> �cu. yd. x 1.4 ton/cu. yd. _� tons. <br /> F. ADSORPTTON WIDTH L� Ld�-v►� <br /> 1. Percolation rate in top 12 inches of soil is mpi Absorption Widtb Siz(ng7ablc <br /> Peaeolation Raic Gallons Ratio of <br /> . lect allowable soil loading rate from table on page E-; inrM�A`��pu so�,r�x�U� per day per A�,�;�,W;a,n <br /> . k� gpd/f t� � squarc foot �o Rock l�yer <br /> WiJ�h <br /> 3. Calculate adsorption width ratio by dividing rock layer Faettrthan0.l• c�xsa„a ...__ __.._ <br /> 0.!to S SanJ 1.20 1.00 <br /> loading rate of 1.20 gpd/ft�by allowable soil loading rate; o.�as•• FincSand•• <br /> o.eo z.no <br /> 1.20 d/ft21 u ' d/ftZ= a �L 7 . 6�0�s Snndy Loam o.�9. �.sz <br /> gP �_gp �b�o,o �e�, o.�o z.�,� <br /> 31 l0 45 Sil�Loam 0.50 2.� <br /> Check this value on page E-16. 46�0�o ClsyLavn o.4s 2.�, <br /> 60 to 120 Clay 0.24 S.W <br /> 4. Multiply adsorption width ratio by rock layer width to get s��W�;;r� c�aY ...__ .____ <br /> izo <br /> required adsorption width; <br /> .Z,(n1x��ft =��ft <br /> I <br />
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