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septic info including 1993 septic design
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511 Ferndale Road North - 36-118-23-13-0012
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septic info including 1993 septic design
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Last modified
8/22/2023 5:01:38 PM
Creation date
8/16/2016 12:53:32 PM
Metadata
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Template:
x Address Old
House Number
511
Street Name
Ferndale
Street Type
Road
Street Direction
North
Address
511 Ferndale Road North
Document Type
Septic
PIN
3611823130012
Supplemental fields
ProcessedPID
Updated
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. . �. MOUND DESIGN WORKSHEET <br /> . . (For Flows up to 1200 gpd) <br /> A. FLOW Estimatcd Sewage Flows in Gallons per day <br /> Estimated (�o o gpd (see pages D-7 or I-3,4,5) �g�� <br /> � umbcr <br /> � or measured — gpd x 1.5 = ��o�TS Type I Typc 11 Type II( �� <br /> B. SEPTIC TANK LIQUID VOLLTMES 3 Qso 30�o ia � <br /> 4 600 375 256 ���� <br /> a — / 0� � gallons (see pages C-3 or C-5) s �so aso z� , <br /> 6 900 525 332 T�x I. <br /> 7 1050 600 370 Q <br /> S 1200 675 4p8 �d,m� <br /> C. SOILS (refer to site evaluation) <br /> 1. Depth to restricting layer = 3���0 3�ti � inches <br /> Seplic Tank C�pxilics,in grlk.na <br /> �Iumba of Minimum Liquid l.iquid cepYcity with <br /> 2. Depth of percolation tests = I a inches �.�m• ��.��ry Q.rb:ycdispusyl <br /> 3. Percolation rate ! 1 . 3 mpi zak3f ,� „u <br /> 3 a a �an isoo <br /> 4. Land slope 3 % °or6 ,x'° zu° <br /> 7.8 or 9 �p }ppp <br /> over 9 ----- <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 /> �C�O gpd x 0.83 sq. ft./gpd = 4°��S sq. ft.-3-�0�„5'-�'� <br /> 2. Select width of rock layer (10 feet or less) = 1 D ft. <br /> 3. Length of rock layer = Area = Width = <br /> S�sq. ft. _ �U ft. = Ss' ft. Rock Bed <br /> t•J•t•f•f•f�f:r:;:f.f:f,f:f;f T <br /> ti��'��•�ti�1'�ti'ti�ti'�• �• �• � ti'ti'�•' � <br /> 'f•f•f•f•t•t•f•f f•f•J:!•f•l•: <br /> tititititititi � ti titi.•, •, Vidth SlOft. <br /> �f•f•t l.l f l l:f•r�r.f :•r�f� <br /> •;f��,;f;f~f~��:��;f;l����:���� <br /> E. ROCK VOLUME ���� Length ---+ <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> �2 sq. ft. x >.��' ft. _��y cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> S�y_cu. ft. = 27 = �, I 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. ADSORPTION WIDTH �L�}-`�' t.-n�h'1 <br /> 1. Percolation rate in top 12 inches of soil is � I ,3 mpi Absorption Width Sizing7'able <br /> Percolacion Ratc Gallons Ratio of <br /> 2. Select allowable soil loadi�rate from table on page E-; in Minules per so��T�x��� per day per Absorp�ion width <br /> . � gpd/ft� fnch(MYI) squ�rc foot �o Rock Layer <br /> Width <br /> 3. Calculate adsorption width ratio by dividing rock layer F���«tha�o.,� c�us�� ----- _:.._ <br /> o.��as s��a i.zo i.00 <br /> loading rate of 1.20 gpd/ft�by allowable soil loading rate; o.��os•• FineSand•� o.�o z.�o <br /> 1.20 gpd/ftz 1- • `'��gpd/ftz= a.L�� 6 to]5 STndy Loa�n o.�9 i.sz <br /> 16 to 30 Loum 0.60 2.00 <br /> 31 to SS Silt Loam �,50_ 2.10 <br /> Check this value on pab>e E-16. �6��� ____�aX�Q �o�s z.�, <br /> 60 �o izo ci3y �:za � s.��u <br /> 4. Multiply adsorption width ratio by rock layer width to get Slowtrthan c�ay ___._ __.__ <br /> ,�o... <br /> required adsorption width; <br /> a, t,�x �ft = al�.7ft <br /> I I <br />
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