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2005-P09239 - septic
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Entry Properties
Last modified
8/22/2023 5:11:43 PM
Creation date
11/16/2016 11:27:07 AM
Metadata
Fields
Template:
x Address Old
House Number
3060
Street Name
Fox
Street Type
Street
Address
3060 Fox St
Document Type
Septic
PIN
0411723320002
Supplemental fields
ProcessedPID
Updated
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t-ly <br /> MOUND DESiGN WORKSHEET <br /> (For Flows up to 1200 gpd) ' � <br /> A. FLOW D-� _ � <br /> . <br /> Estimated - '-J�:_ gpd (see pages D-7 or I-3, 4, 5) `3"""�"�R°�'���"°'y. <br /> w..eci+ �.n or��oexc• <br /> or measured gpd. �°� � ° II s <br /> : wo a� �w tox <br /> 3 a�p 7pp =�� ai <br /> � � {00 )T� ISi <br /> B. SEPTIC TANK LIQtTID VOLUMES : .'o �� � ''�` <br /> , ,� .� �� <br /> > � . �� .,� .o. ��„ <br /> _ �:�� ;- r, �� gallons (see pages C-3 or C-S) <br /> C-3 <br /> C. SOILS (refer to site evaluation) SE77iC r.HK c�v.c�nEs, �H GALLONS <br /> tpuo C✓�cm <br /> 1. Depth to restricting layer = %'•;•�' inches ""`"°' ��� ""'°"'�°' <br /> �[Di1CY�Id lJQl�O CU�QTt Otf10111 <br /> 2. Depth of percolarion tests = � i inches ,a,�.. ,,, ,,,, <br /> 3. Percolarion rate �;s mpi ��� �••• ,�.� <br /> � --� <br /> �o,. ,.e. �,�o <br /> 4. Land slope � - i .�_. °'o ,.�a, <br /> ,... �,ae <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock � <br /> layer: A x 0.83 = <br /> =' �-� g-pd x 0.83 sq. ft./gpd = � � f sq. ft. <br /> 2. Select width of rock layer (10 feet or less) _ ?:"% ft. <br /> 3. Length of rock layer = area i width = I <br /> �' -; _3 sq. ft. y _ ft. _ �� �' �ft. Rock Bed <br /> . �f.f•f•f•r•f•f.f r•f•f•r•r•f•f , <br /> .�.�.,.�.�.�.�.�.�.�.�.�.�.�. T <br /> f•r•r.r•f.f•r•r �•r.r•f•f•�•r � <br /> L•ti•ti.ti.ti.ti.ti.�.ti•ti•ti•ti•�•ti.ti• idth 510 <br /> /•f•/•f•f.l•t.l f•r•J•l•t.r•r <br /> �•ti•�•�•ti•ti•ti•`•ti�ti.�ti•ti•ti•ti•ti• <br /> .r.r.r.f.f.�.f.f.�r.r.f.�.�./.r <br /> E. ROCK VOLUME �-- Le�g�h --; <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> � �` � sq. ft. x ft. =� '-�' cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> � � cu. ft. y 27 = � -j cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; I <br /> � _� cu. yd. x 1.4 ton/cu. yd. _;�:�_� tons. � <br /> F. ADSORPTION WIDTH I <br /> 1. Percolation rate in top 12 inches of soil is ' ��� mpi E-16 <br /> 2. Select allowable soil loading rate from table on page E-16; ��..•��n,d.���ou��•ot I <br /> ��� —I � <br /> gpd/ft2 � �......._. I <br /> 3. Calculate adsorption width ratio by dividing rock layer � "�� �� ��� � ������' ' <br /> ..,. . ,., ..a �.. ,oa <br /> loadin� rate of 1.20 gpd/ft2 by allowable soil loading rate; .� ,a ::e e:i '.:: e>: <br /> -„ .., a„ ... ,., ,.o <br /> 1.2� gpd/ft2-� _�_RPd/ft2 = "' - �� :� :;_, ::: a:�: .� ato <br /> Check fhis t�aii�e on page E-16. <br /> 4. Multiply adsorption width ratio by rock layer width to get � <br /> required adsorption w-idth; � <br /> x /,,: f t — -' '-� f t <br />
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