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1275 Brown Road South - PID: 10-117-23-31-0024
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Septic design
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Last modified
8/22/2023 3:22:50 PM
Creation date
3/9/2020 3:07:55 PM
Metadata
Fields
Template:
x Address Old
House Number
1275
Street Name
Brown
Street Type
Road
Street Direction
South
Address
1275 Brown Road South
Document Type
Septic
PIN
1011723310024
Supplemental fields
ProcessedPID
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MOUND DESIGN WORKSHEEI' <br /> '• � ' � (For Flows up to 1200 gpd) <br /> � , <br /> A. FLOW S�g_ N o,..�S�- ���,e�a s�.n�e�oM�,c.�i«,s��,r esr� <br /> Tb�- L�v1'�� �r1 D L1 S T� �'6 A�y.a' <br /> Estimated �O 0 gpd � N'o� ,ry�� ��� �.y�!u �ry�rv <br /> or measured — x 1.5 = - gpd• s�e►°°^" <br /> 2 100 2?S 1E0 �o+ <br /> 7 150 300 21E w <br /> B. SII'TIC TANK LTQLTID VOLLJMES � s ��so 4'so � '�` <br /> �- ►ato � �-�n�o gallons N o�5�- 6 90o sas »z ,� <br /> a i� 66is ;oe d.�'. <br /> a.— /000 I-ou^��C' 1-Edvs� . � <br /> C. SOIIS (re�er to site evaluation) � ,� '� �'��a � )}� N�eQ "�-- "� <br /> ''r�. .,.t e�—a-.+ <br /> 1. Depth to restricting layer= o -3� - ��� inches B °� � � <br /> 2. Depth of percolation tests = ..��`_inches <br /> 3. Percolation nte l Z.to mpi 2i«�` i oSOoo i'.'s o <br /> 4. Land slope s�6 �soo uso <br /> �j % 7 a i 2,000 3,000 <br /> - ov�r 9 S�e!Sp C�6 (i 1 S) <br /> D. ROCK LAYER DIIv�.IVSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required azea of rock <br /> layer:A x 0.83 = <br /> �oa . gpd x 0.83 sq. R./gpd = �sq. ft. <br /> 2 Select width of rock layer(10 feet or less) _ �a ft. <br /> 3. Length of rock layer = area+width= Rock Bed <br /> ,�_sq. ft.+ _!Q_ft. =2�_ ft. ,.� ,,. ,.,. ,...�,,., ,., <br /> r•} �•3• r. ~}•3 r.}•r�r•r 3.r• <br /> � � �.�•�•t ti:� �•�ti•ti•ti•ti•t•ti•� <br /> i•'r•r r•r i �•}•�•�•S r•r1•r•r• �dth 510 ft. <br /> �.�.ti•��•ti.�r�•ti•ti•�ti�•> >•�•ti <br /> �♦.�•r•}• r• �•���• ♦•r•�• <br /> .ti•�•�•�'�ti•�fti•�•�•�•�•�•ti•+•�•� <br /> �.�.r.f.�•r•r•�•r�•�•r•r•�•r•r• <br /> E. ROCK VOLUME �— �'�' —� <br /> 1. Multiply rock azea by rock depth to get cubic feet of rock; <br /> ��sq. ft. x -o ft. =2�cu. f t. <br /> 2. Divide cu. ft.by 27 cu. ft:/cu. yd. to get cubic yards; <br /> ��cu.ft. +27=�cu. yd. <br /> 3. Muldply.cutiic yards by 1.4 to get weight of rock in tons; <br /> �,cu.yd.x 1.4 ton/cu. yd. _�0 tons. <br /> F. ADSORP'IION WIDTH G l.�t'� �-b Prvv� Wfdth St�n Lbk <br /> G.Dr► A.i..! <br /> 1. Percolation rate in top 12 inches of soil is 12.c� mpi M�tl�� �ilTexttue �,� �,`: <br /> cmpU '.. .►..��, <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 oarse Sand �.20 l.00 <br /> d/ft� �2o i.00 <br /> 0.1 to 5 Sand <br /> ,� � $p 0.1 to 5 Flne Sand" 0•� 2•� <br /> 6 to 15 ndy Loam ��79 ��2 <br /> 16 to 30 L,oam 0.60 2.00 <br /> 3. Calculate adsorpHon width ratio by dividing rock layer 31 to as stit t,o�►m oso 2.ao <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; 61 io 21 0 �'�l�y�' o�a s:oo <br /> 1.20 gPd/ft�i- .kS gpd/ft� _ ��l�'� Slower than 120 Cla - - <br /> "Shc havins 501i a mae d fvx or ve�fv,e su,b <br /> 4. Multiply adsorption width rario by rock layer width to get <br /> required adsorption width; <br /> .� L'1 x�o ft=��ft <br />
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