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1992-004740 - septic system
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Old Crystal Bay Road North
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0245 Old Crystal Bay Road North - 33-118-23-31-0012
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1992-004740 - septic system
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Last modified
8/22/2023 4:49:21 PM
Creation date
3/7/2018 10:41:34 AM
Metadata
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Template:
x Address Old
House Number
245
Street Name
Old Crystal Bay
Street Type
Road
Street Direction
North
Address
245 Old Crystal Bay Road North
Document Type
Septic
PIN
3311823310012
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORKSHEET <br /> o (For Flows up to 1200 gpd) <br /> M •A. FLOW tistimated Sewage`&lorw�s In Gallons per day <br /> Estimated `o o gpd (seepages D-7 ori I-3,4,5) <br /> or ., I �1 In <br /> v <br /> or measured gpd x 1.5 � <br /> = i, ,t,o,t,s <br /> 2 :300 .225 ISO <br /> B. SEPTIC TANK LIQUID VOLUMES 3 Oso 300 218 <br /> _-zv 6b allons'(see pages C-3 or -5) 4 � 373 236 <br /> •.lw <br /> 3 730 430 25w n <br /> g P g 6 .900 szs 332 <br /> 7 A050 6a, :370 to <br /> C. SOILS (refer to site evaluation) e. '• 12M 075 40e , <br /> 1. Depth to restricting layer= 14+46 ay inches <br /> umt,n of Minimrm U�id tjquid�y...y.,u <br /> 2. Depth of percolation tests <br /> inches t am,. c,>reky z e:a <br /> 3. Percolation rate mpi„ 3o SM W;u_ 75001Y <br /> 4 nt 6 1S1tU 12250 <br /> 4. Land slope % ;E dz tc1 " 7,It«9 xlY, xw <br /> aet 9 »» <br /> D. ROCK LAYER DIMENSIONS. <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock - <br /> layer, Deily Flow Xi 0.91 :- <br /> 4,oy gpd x 0.83 sq. ft./gpd = • <br /> 2. Select width of rock layer (10 feet or less) _ lo ft. <br /> 3. Length of rock layer = Area+ Width = <br /> s� 7 sq. ft.+ )u ft. = S5 ft. Rock Bed <br /> r;f,'.•fir;.••'� r r;.•:r r�r r r•:T <br /> f;: :j:f:.+�: ;f•f.�tifti`}`�;rl%Idth SION <br /> E. ROCK VOLUMELengt::r:,:;;:r:r• gt `:: :•:;; l <br /> F-- h• <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> sem-sq. ft. x l US ft. =s � cu. ft. <br /> 2. Divide cu. ft.by 27 cu. ft./cu. yd. to get cubic yards; <br /> s,),cu.-ft. +27= Ate_cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight-of rock in tons; <br /> a i cu, yd. x 1.4 ton/cu. yd-. _ tons. <br /> F. ADSORPTION WIDTH <br /> 1. Percolation rate in top 12 inches of soilis G,.S mpi Absorption YAdth Slziny.7bble <br /> i•ercolatlon Rate Gallons Ratio of <br /> 2. Select allowable soil loading rate from table on page E; inMinutetpa SoilYimwe per day per Absorptanwhhh <br /> -��-- gPd/f Inch(MPI) squats font to R0N Lays <br /> 3. Calculate adsorption width ratio by div ding rock layer F&4wdaut0.1• Coarse Sand <br /> 0.1 tr 5 swul 1.20 1.00 <br /> loading rate of 1.20 gpd/f 6 by allowabl soil loading rate; 0.1 r.�•• Fine sans•• 0.60 2.M <br /> 6--.13 Stualy two 0.79 1.32 <br /> 1.20 gpd/ft2+ ,y r gpd/ft2= fata3o . L=m 0.60 2.00 <br /> 31 1043 Slit Loam_ 0 SO 2.40 <br /> Check this value on page E-1 5. C46'1060-1 -Clay latvn OAS 2.67 . <br /> 60 to 1'tl _ Of - 0.24 S.W <br /> 4. Multiply adsorption width ratio by rock layer width to get Slowc;; <br />
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