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1993-005536 - new septic system
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1993-005536 - new septic system
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Last modified
8/22/2023 5:26:11 PM
Creation date
8/28/2017 11:53:52 AM
Metadata
Fields
Template:
x Address Old
House Number
540
Street Name
North Arm
Street Type
Drive
Address
540 North Arm Dr
Document Type
Septic
PIN
0611723310006
Supplemental fields
ProcessedPID
Updated
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'��1'�J'�-� �x��5�n� E 19 <br /> . <br /> • r � � MOUND DESIGN WORKSHEET <br /> � � , � (For Flows up to 1200 gpd) <br /> A. FLOW . � � D-7 <br /> � Esrimated�gpd (see pages D-7 or I-3,4,5) "'�'�"�R°•'����' <br /> u�ra Trrt a�a�ov¢s <br /> or measured — gpd. � � n � � <br /> . � �. �b .�. <br /> . s .eo �oo eu ,�".,, <br /> B. SEPTTC TA�'�TIC LIQUID VOLUMFS : .'o „ � "�.; <br /> , �o.o � .ou s,o � <br /> �= )o n� gallons (see pages C-3 or C-5) � • �� m � «�-» <br /> C-3 <br /> C. SOILS(refer to site evaluation) tERIC TA/MC CA/AdTl�, NI OALLON� <br /> i.� Depth fio r+�stricting layer� 1*6 To ti a�� inches "�••"0' •�•M• •`�' <br /> �nar uew w�an naou► <br /> 2. Depth of percolation tesis = )� inches ,a„» ,.. ,,.. <br /> 3. Percolation rate t_.� mpi •«�� �•« �� <br /> ��. ,... .... <br /> 4. Land slope �v4� 9'0 ,,,,�,. „N .«. <br /> D. ROCK LAYER DIMIIVSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer:A x 0.83 = <br /> �Z�gpd x 0.83 sq. ft./gpd = 45�1 sq. ft.+�u�o= s�� <br /> • 2. Select width of rock layer(10 feet or less)= 10 �ft. <br /> 3.� Length of rock layer=area+width� • ! <br /> Sw� sq.ft.+ �ft. = Ss� ft. Rock Bed <br /> .r.r.�.r.r.r. .►.f.f^^r!w .r. <br /> �.ti•ti•ti•ti.ti•ti.ti ti�•ti•�.ti.ti•ti <br /> r.r.r.f.r.r.r.3.�t•fNH f•r <br /> ti.ti���.ti�ti•�•ti�ti•�•ti�ti•ti.��ti� �1 ���. <br /> . i �'•1'r•l•f•l•t•t•r•t•f.f•�/•/•J <br /> tit'�f:}~I~f~ihi�r}I�ftir~r�fLry! <br /> E. ROCK VOLUME � �— t.«+� <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> �47 sq. ft. x�ft. = s�cu. ft. ' . <br /> 2. Divide cu. ft.by 27 cu.ft./cu.yd.to get cubic yards; <br /> s��cu. f� +27=�_cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> �cu.yd. x 1.4 ton/cu.yd. _�tons. <br /> F. ADSORPT'ION WIDTH <br /> 1. Percolation rate in top 12 inches of soil is l_,u mpi E-16 � <br /> 2. Select allowable soil loading rate from table on page E-16; ��-�a��� <br /> -��.- SPd�� ^...... �...,...... <br /> NN �M YI/� <br /> 3. Calculate adsorption width ratio by dividing rock layer ""` '"'"' """ """ `"'`"'"' <br /> ..,. . ,�. ..« ..» �.» <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; -�. :;: ::N :� ::: :» <br /> ., ... .... ..» ... .... <br /> 1.20 gpd/ft2+ .�g gpd/ft2= 1.S1 : �„ ::� ::� '.:« :» <br /> Check this value on page E-16. <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> �x i.Sa ft=�ft <br />
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