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1998-010218 - new septic system
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0725 Sixth Ave N - 26-118-23-44-0006 Spring Hill Golf
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1998-010218 - new septic system
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Entry Properties
Last modified
8/22/2023 4:19:08 PM
Creation date
1/9/2019 11:07:03 AM
Metadata
Fields
Template:
x Address Old
House Number
725
Street Name
6th
Street Type
Avenue
Street Direction
North
Address
725 6th Avenue North
Document Type
Septic
PIN
2611823440006
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORKSHEET . , <br /> (For Flows up to 1200 gpd) ��,,� �� �� �� �� <br /> A. FLOW Esttmated SeMte Flo.v In G•tlon�pa D,Y(gp� <br /> Estimated 3 o O �d N'o°e .n,y�� .q,�i� ry�iti .n,y�rv <br /> or measured — x 1.5 = —' gpd. a�a►� <br /> s �oo us i ao .o� <br /> 3 �30 300 31E •r <br /> B. SEPTIC TANK I.IQLTID VOLUMES � `s � ;'so � ." <br /> a—/v O c� gallons � ia��o � iio � <br /> .� <br /> w.a. <br /> ,�����.� d f200 675 �Os <br /> C. SOILS (refes to site evaluation) ►� r,,mea ,�,,,, ��, <br /> 1. Depth to restricting layer= a.�s'�3 a.�' inches s °�„m �..., �, <br /> 2. Depth of percolation tests =_�, inches `'""° `'""' <br /> 3. Percolation rate �•b mpi �i��` i oSOoo i'.s'o <br /> 4. I.and slope S % s«s tsoo �so <br /> 7 R i Y.000 ].00O <br /> ovs 9 Sa�!Sp C-6 (a ls) <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 /> . <br /> 3ov gpd x 0.83 sq. ft./gpd = �.�� q sq. ft.tw�'o = a�y° <br /> 2 Select width of rock layer(10 feet or less) = i o ft. <br /> 3. Length of rock layer= azea+width= Rock Bed <br /> a�y SC�. �.+ �_ft. _�_ f t. �rti}ti��::3"n�'r 3�}?r2r�3'�`f� <br /> �•�.�,N.•L•ti.�•{•ti•ti•�•ti.�.�•, <br /> .r.i•i i•}•r}���•�• •r r•r•r•i• �dth 510 ft. <br /> �.ti.'►.{1+•ti.y.L•ti•�•��.;.�•ti•ti•ti <br /> {fti:ti:'s�s��t�►�ti:tir�:tiNri�{r��� <br /> .r.r.r��•r•r.r•r•r�•�•r•i•�•r•�• <br /> E. ROCK VOLUME ~— 1'�'�' —� <br /> 1. Multiply rock azea by rock depth to get cubic feet of rock; <br /> a�y sq.ft. x �•us'ft. _�cu. f t. <br /> 2. Divide cu.ft.by 27 cu. ft:/cu. yd. to get cubic yards; <br /> �cu.ft. +?7=J I _cu. yd. <br /> 3. Multiply.cutiic yards by 1.4 to get weight of rock in tons; <br /> �cu.yd.x 1.4 ton/cu. yd. _ ��' tons. <br /> F. /�►uJ�.J��N YY iL� Ll.�' � � NM1dth SlsSn Lbk <br /> CrO� Iw�r <br /> 1. Percolation rate in top 12 inches of soil is .� mpi M�tl�� �il'Texture �,� �,': <br /> i„� �_ .:..�., <br /> � <br /> 2. Select allowable soil loading rate from table; Faster than 0.� oarse Sand 1.2o i.00 <br /> �- ' d/ft� oa cos s�na � �.20 �.00 <br /> , l J $P 0.1 to 5 Flne Sand �•� Z•� <br /> 6 to 15 ndy Loam 0.79 1�2 <br /> 3. Calculate adso tion width ratio b dividin rock la er tb co ao Loam o.bo z.00 <br /> � }r $ y 31 to 45 Silt Loam OSD 2.40 <br /> loaciing rate of 1.20 gpd/ft2 by allowable soil loading rate; 61 eo 21 o cta�l�°�a' 0�4 s:o� <br /> 1.20 gj�d/ft2i •���gpd/ft� = a- Lo�_. Slower than 120 Cla - - <br /> "Soll l+�rin6 50'R a man d fu+e or�er fine sand <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> �,L7 x1Q_ft= a�•� ft <br />
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