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1993-005562 - repair septic
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1993-005562 - repair septic
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Last modified
8/22/2023 4:19:53 PM
Creation date
1/17/2019 2:54:49 PM
Metadata
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Template:
x Address Old
House Number
2100
Street Name
6th
Street Type
Avenue
Street Direction
North
Address
2100 6th Avenue North
Document Type
Septic
PIN
2711823310024
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORKSHEET <br /> � � ' � (For Flows up to 1200 gpd) <br /> A. FLOW �v�v'�c' w�� so��=�� <br /> Estimated Sewage Fbws in Gallcxis pa day <br /> Estimated gpd (see pages D-7 or I-3,4,5) «� <br /> -''-t� umber <br /> or measured�� gpd x 1.5 = I 1 ► �1 ��, ��°`�,,,s �YP�I Type�� Type��j iv <br /> B. SEPTTC TANK LIQUID VOLUMES s a� souo ie � <br /> 3 -�u o o �x,x,vb gallons (see pages C-3 or G5) -�-l- �50� s �� as'o i�a "';;' <br /> q,a��e UT�'M'� 3-iGDo^ �1.�A�\G� _ 6 900 525 332 '�t. <br /> '� ' 7 lOSO 600 370 m <br /> C. SOILS (refer to site evaluation) 8 1zoo 6�s ao� �,,�„ <br /> 1. Depth to restricting layer = a�I�'TU 3 0�� inches �p�Tank C�p�citic�in gillwn <br /> 2. De th of ercolation tests = l a" inches "�; Minimum Liquid Liquid capaciry w;ih <br /> P P Gp�ciry tarbaQe d'upowl <br /> 3. Percolation rate �,g m i 2«�y ��o ��u <br /> P 3aa �aoo isoo <br /> 4. Land slope -� �Jp - �48 a 9 ;� ;o� <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 /> �1 i� gpd x 0.83 sq. ft./gpd = ga� sq. ft. <br /> , 2. Select width of rock layer (10 feet or less) _ /� ft. <br /> 3. Length of rock layer = Area 1 Width = <br /> q a� sq. ft. -:- I c� ft. _ _ � 3 ft. Rock Bed <br /> r•r•++•f•f•r.f.f.r��.r•r.r�f <br /> ti•ti•ti•ti•ti•ti•ti•ti•ti ti ti ti ti ti ti• <br /> f•f•l•t•r.t t•t•f.f•f.f•1•1•l <br /> - ti•ti•ti ti•ti•ti•ti•ti•�.ti.ti.ti.ti.ti•ti• ►dth 570ft. <br /> f•t•f.t.t•f•t•t f•f•r•t•t•f•J <br /> '.•ti•'. ti•ti•ti•ti•ti•ti�..U.�,.ti.ti..�. <br /> E. ROCK VOLLJME •f•f�f�����f��'f'•'r�f•j�f�f'f 1 <br /> � Length <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> qa1 sq. ft. x /.v�ft. = '1�3 cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> �cu. ft. i 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. = S� tons. <br /> F. ADSORPTTON WIDTH G�`� � � <br /> 1. Percolation rate in top 12 inches of soil is �f•�i mpi Absorption Width Sizing Tablc <br /> 2. Select allowable soil loadin rate from table on a e E-• ''"`°'s"°"'�` ca��,: Ra��oor <br /> g p g � in Minutesper Soil Texturo per day per Absaption wW�h <br /> .�� p��/r� fnch(MPl) square foa �o Rock I�yer <br /> pt' 1 �Y�dth <br /> 3. Calculate adsorpHon width ratio by dividing rock layer Facter than 0.1• c�s� ____ __.__ <br /> 0.l to S Sand 1.20 1.00 <br /> loading rate oE 1.20 gpd/ft�by allowable soil loading rate; o.��s•• ��sa�a>• o.�0 2.�0 <br /> 1.20 d/ft2 . y d/ft - �.�� 6 w�s s,�ay� 0.�9 ►.sz <br /> � J'_�gP 2- 16 to 30 Loam 0.60 2.0(1 <br /> 31 to 45 Silt L,oam 0.50 2.30 <br /> Check this value on page E-16. �a�o Clay Loam o.4s z_�� <br /> 60 to 1N1 Clay 0.24 S.OU <br /> 4. Multiply adsorption width ratio by rock layer width to get SlowertAan c�ey _____ _____ <br /> i2a•• <br /> required adsorption width; <br /> a,�,� x /o ft = ae.� ft <br />
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