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2925 Deer Run Trail - 04-117-23-24-0011
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Last modified
8/22/2023 5:10:50 PM
Creation date
6/20/2016 9:30:53 AM
Metadata
Fields
Template:
x Address Old
House Number
2925
Street Name
Deer Run
Street Type
Trail
Address
2925 Deer Run Tr
Document Type
Septic
PIN
0411723240011
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> � <br /> A. FLOW Estimaltd Scwagc(S�j in Galluns per day � <br /> Estimated ��0 gpd (see pages D-7 or I-3,4, 5) ,,,,,,,,�� <br /> or measured gpd x 1.5 = --- ���ms Type I Typc II Type I!I I� <br /> B. SEPTIC TANK LIQUID VOLUMES Z 30o zzs �xo <br /> 3 450 300 218 � <br /> 4 600 375 256 ofiee <br /> I-1 a �v�1 1-/ooJ gallons (see�ages C-3 ur C-5) s �so aso Z�a "';o` <br /> 6 900 525 332 T��• <br /> 7 1050 600 370 � <br /> 8 1200 675 408 caumn <br /> C. SOILS (refer to site evaluation) <br /> /1 Scplic T.nk Gpxilkr,in gall.�ns <br /> 1. Depth to restr�cting layer = � '�0 3y inches �„r„�,�,� ���n:riomLiqo,� �.,a�,��,;,�:,�Y..,�� <br /> 2. Depth of percolation tests = I a inches ����m= �..a�;:Y �,�b�.���,W,.� <br /> 3. Percolation rate 4S. � mpi zw�u ,� „u <br /> 3aa �an tsoo <br /> 4. Land slope � % °°`6 `x'° Zu° <br /> 7,8 a 9 �pp 7cX10 <br /> over 9 '---- <br /> D. ROCK LAYER DI�IENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: Daily Flow x 0.83 = , <br /> i �� gpd x 0.83 sq. ft./gPd = �,�.� sq. ft.-r �.���.a c�ul-�� <br /> 2. Select width of rock layer (10 feet or less) = t t� ft. <br /> 3. Length of rock layer = Area -:-Width = <br /> � 4�/ sq. ft. � �ft. _ .�u ft. Rock Bed <br /> r•r•r�r•r•f•r.f••:r•r�r•r�r�r <br /> ti••.••..ti•ti•ti•ti••.•::•.••.••.•ti•ti•ti• <br /> l•r•• t•t•t•!•t•f•t•t.t•t.t•: <br /> ti•ti•ti�ti•ti�ti•ti•ti••.•ti-ti•ti•ti•ti•ti• idth S10ft. <br /> .f.f.f.f.f.f.f.r.f.r.f.f.f.r.f <br /> . �....�.�.�.ti.�.............�..�..�.� 1 <br /> •.f.:.:.:.f.:.:.:.f.:.:.:.:.: <br /> E. ROCK VOLUME Length --� <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; � <br /> !.�c� sq. ft. x .�ft. __^,��cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> �cu. ft. - 27 = a� 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 <br /> 1. Percolation rate in top 12 inches of soil is `✓�� iriPl Absorption WiJth Sizing Table <br /> Percola[ion Ra�c Gallons Ratio of <br /> 2. Select allowable soil loadin�; rate from table on page E-; in Minutes p:r Soil Texmre per day per Absorplion wid�h <br /> Inch(MP!) square foot to Rock Laycr <br /> �� gpd/ft� W��h <br /> 3. Calculate adsorption width ratio by dividing rock layer Fa��erthan0.1• c�sa„� ----- ._.__ <br /> o.��s sa�a �.zo i.00 <br /> loading rate of 1.20 gpd/ft�by allowable soil loading rate; o.�as•• FineSand•• o.�o z.�o <br /> 6 to 15 Sandy Lt�n 0.79 1.52 <br /> 1.20 gpd/ftz?- ,�,i gpd/ftz= � , l�� �b�o�+o �em o.6o z.cx� <br /> 3�ro as si�t Loam o.so 2.ao <br /> Check this value on page E-16. 46��o ClayLavn o.4s z.�� <br /> 6o a�zo ct�y o.za s.� <br /> 4. Multiply adsorption width ratio by rock layer width to get Slo���an c�ay ----_ _____ <br /> required adsorption width; <br /> .�:�" x ft = `�� " ft <br /> I <br />
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