My WebLink
|
Help
|
About
|
Sign Out
Home
Browse
Search
1999-011851 - new septic system
Orono
>
Property Files
>
Street Address
>
W
>
Watertown Road
>
3745 Watertown Road - 32-118-23-34-0012
>
Septic
>
1999-011851 - new septic system
Metadata
Thumbnails
Annotations
Entry Properties
Last modified
8/22/2023 4:40:58 PM
Creation date
10/21/2019 2:14:52 PM
Metadata
Fields
Template:
x Address Old
House Number
3745
Street Name
Watertown
Street Type
Road
Address
3745 Watertown Road
Document Type
Septic
PIN
3211823340012
Supplemental fields
ProcessedPID
Updated
There are no annotations on this page.
Document management portal powered by Laserfiche WebLink 9 © 1998-2015
Laserfiche.
All rights reserved.
/
25
PDF
Print
Pages to print
Enter page numbers and/or page ranges separated by commas. For example, 1,3,5-12.
After downloading, print the document using a PDF reader (e.g. Adobe Reader).
View images
View plain text
MOUND DESIGN WORKSHEET " :� 3711 �JZZVp-� RCP <br /> (For Flows up to 1200 gpd) OA /W 61 <br /> A. Ow Fes,T..c�Scr.Y:Fi...'s of rj"_w 1,ct S.y <br /> Estimated y-52 gpd / N,== -T1Ycj 7ywti T)yc W Ti�, <br /> or measured x 1.5 = gpd. i <br /> B. SEPTIC TANK LIQUID VOLUMES �g y <br /> gallons �� ' 025 ,2 <br /> 6 bW 32S �3= i Ts1F c. <br /> u � <br /> i t�J I 075 4c" i w <br /> C. SOILS (refer to site evaluation) <br /> 1. Depth to restricting layer= a inches feet <br /> 2. Depth of percolation tests = /Z- inches <br /> 3. Texture C Percolation rate -3&- `f mpi 2""" , <br /> 4. Land slope 1 q•0 % U A.%- q(o - (Q T.:"r <br /> J <br /> D. ROCK L.AYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock layer. A x 0.83- Z <br /> 14 lo gpd x 0.53 sq. ft./gpd = 394 sq. ft.-., to °=q1 C <br /> 2. Select width of rock layer (max 10' if<120 mpi max 5') _ /0 ft. <br /> ;. Length of rock layer = area width= gyp• �- --.-• _.___.. <br /> 44—sq. ft. - _/6 ft. ft. <br /> Q�°oOp C�000a�. sD r0 .D lavCv.^r"."R <br /> Width Al ft z"=" <br /> <120mpi <10' Length 'Y/ ft <br /> E. ROCK VOLUME >120mpi <5' <br /> 1. Multiply rock area by rock depth to get cubic feet of rock;4110 sq. ft. x <br /> ):o ft. = 410 cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> tLI.0—cu. ft. 27 =o cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; cu. yd. x i.-I <br /> ton/cu. yd. =Z 3 tons.- 10 al <br /> u <br /> F. ABSORPTION WIDTH A wurpuuu WW bwu j T-+ <br /> 1. Percolation rate in top 12 inches <br /> //of soil is 38•x/ mpi ��1�R."W c..,. <br /> '7t0—�DC+ Maw%aa P"loch SW T"", Pc+a.,r s� •.ui w h :a <br /> Texture C M p R^^^�� 1MY1, 6%—.:W" <br /> Fawcr Yua 0.1 coame S.,4 l._] 1 W <br /> 0.1 w s Sid 1_o W I <br /> 2. Select allowable soil loading rate from table; a1 is s.Fac l J : � i s' <br /> ggpd/ftp 16 W 30 L & 00 2 w <br /> a1 IQ 41 s. - <br /> +e w w <br /> 3. Calculate adsorption width ratio by dividing rock layer slara0..a 1ao clay <br /> - loading rate of 1.20 gpd/ft2 by allowable soil loading rate; i <br /> 1.20 gpd/ft==0,y3gpd/ft== <br /> 4. Aluldply adsorption width ratio by rock layer width to get i <br /> required adsorption width; <br /> 7 x /0 ft =..-a7 ft <br />
The URL can be used to link to this page
Your browser does not support the video tag.