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1993-005185 - septic system
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1993-005185 - septic system
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Last modified
8/22/2023 4:20:00 PM
Creation date
1/18/2019 12:52:18 PM
Metadata
Fields
Template:
x Address Old
House Number
2300
Street Name
6th
Street Type
Avenue
Street Direction
North
Address
2300 6th Avenue North
Document Type
Septic
PIN
2711823320002
Supplemental fields
ProcessedPID
Updated
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. . � ' MOUND DE9IGN WORKSHEET I <br /> � • ` ' � (For Flows up to 1200 gpd) � <br /> A. FLOW e��.�a s�:.��.n��a.a.y I <br />� Estimated " sU gpd (see pages D-7 or I-3,4,5) � -� <br /> or measured — gpd x 1.5 = -- n�� �°� '�°" '!�'P°1° '� �' <br /> � <br /> B. SEPTIC TANK LIQUID VOLUMES � 3 �� 3� � i� �; <br /> :� -10 o a gallons (see pages C-3 or G5) s w �so ` z�e �" ' <br /> 6 900 3?S ' 3�2 ��. <br /> • 7 1050 600 j 770 <br /> C. SOILS (refer to site evaluation) e �ioo bn ; �ae N�. <br /> ,� F�rns�� ; <br /> l. Depth to restricting layer= �a -rv a,ta�� inches �u'' b a�" s''°e7�"�c.'""''�'"'.'w" ; <br /> r/ NwnMr ef Mlnrn�l.Iq�Y Liq�Y a+��elq�M <br /> 2. Depth of percolation tests = �� inches � � ao•�w ..�+�»a� ( �� <br /> 3. PerrnlaHon rate 1� •y mpi � 3;� �� � „� <br /> � .� � <br /> 4. Land slope_ �LA-� 9'0 ;��, ;,,� � ,� , <br /> , <br /> � <br /> D. ROCK LAYER DIMENSIONS � ! <br /> 1. MulNply flow rate by 0.83 to obtain required area of rock ; � <br /> layer: Daily Flow x 0.83 = , ' <br /> ��_gpd x 0.83 sq. ft./gpd = 37 3 sq. f t.-t�o`�-�►�o" � i <br /> 2. Select width of rock layer(10 feet or less) _ /� ft. ; i <br /> 3. Length of rock layer = Area+Width = � ! <br /> u �o sq. ft.+ �_f t. _ �-�1 I f t. 'Rock eed � i <br /> r . . . .�."�n^^r�,^'�f <br /> �;� i��. <br /> ;';';r�ti'•i ?itir r'r'r''"r tb S10 fb� <br /> . �f�.�•�•�_�•�• •�ti•�•�i7�� `} <br /> E. ROCK VOLUME ;f,#;f�r� t f~{i= :::}:#f:� <br /> F-- Length � <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; ' ; <br /> � y�u sq. ft. x�,v s• ft. _�cu. ft. ' ' <br /> 2. Divide cu. ft.by 27 cu. ft./cu. yd. to get cubic yards; � <br /> 430 cu. ft. +27=�_cu. yd. ' i <br /> 3. Mutdply cubic yards by 1.4 to get weight of rock in tons; � <br /> �cu. yd. x 1.4 ton/cu. yd. _�tons. <br /> � � <br /> F. ADSORPTION WIDTH ��° '"' <br /> 1. Percolation rate in top 12 inches of soil is �.� �lp� Abaorpllon Wldlh Sitln��b�e <br /> 2. Select allowable soil loading rate from table on page E-; ��.pa s���e.�. �°a.� ��.�h <br /> •�J� �nd/f� I�eA(MPI) qure fod b R�ak�A yer <br /> t' <br /> 3. Calculate adsorption width ratio by dividing rock layer P�►�a�• c�.s.�e . <br /> loading rate of 1.20 gpd/ft�by allowable soil loading rate; o.�bs s.�e i.io �.00 <br /> o.i b s•• r��s.�a•• o.� �.ao <br /> 6 ro 13 Sadr Lam 0.79 1.32 <br /> 1.20 gpd/ft�+ •4�,gpd/fc�= a. G , ieb:+o i.a.m . 0.6o t.eo <br /> �1 ro�! Slh t.onn O.JO 2.10 <br /> Ch,eck this value on page E-16. . �o b m `"�,.`;'" o:i; suo <br /> 4. Multiply adsorption width ratio by rock layer width to get s�;;�^ �r --- --- <br /> required adsorption width; � , <br /> a.�� x.L ft = a.b,� ft , ; <br /> . . � � <br /> I ; <br /> . . , <br /> ; <br /> � <br /> ' �r <br /> ; I <br /> ' � (e <br />
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