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1999-012089 - new septic system
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1999-012089 - new septic system
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Entry Properties
Last modified
8/22/2023 4:09:34 PM
Creation date
10/6/2016 1:58:56 PM
Metadata
Fields
Template:
x Address Old
House Number
1465
Street Name
Fox
Street Type
Street
Address
1465 Fox St
Document Type
Septic
PIN
0211723330002
Supplemental fields
ProcessedPID
Updated
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� � MOUND DESIGN WORKSHEET <br /> (For Flows up to I200 gpd) <br /> a. �Qw Estimued Scwtigc Flow in Gailo�s pa D�y(gpC) .:� <br /> Estimated '' :��' gpd ` N�,� <br /> of 'type I Type]I 'lypc I1I 7Ypc N <br /> , or measured - x 1.� _ - gpd. s��� <br /> 2 300 Z.25 180 �s <br /> 3 450 300 218 °t <br /> B. SEPTTC TANK LIQUID VOLLZviES � 4 �o0 3�s zs� .�,� <br /> s �so �so z�s ; <br /> ; - I '�� gdllOTLS 6 900 SZS 332 � <br /> 7 1Q50 600 370 ��, <br /> 8 12D0 675 a08 <br /> C. SOILS (refer to site evaluation) , <br /> ' h''�� s�� c�,=W <br /> 1. Depth to restricting layer = �y � a�f,� inches �,� �,,.�,. �, <br /> 2 Depth of percolation tests = I�." inches `'"°"' `A"�`' <br /> 3. Percolation rate mpi Z3 a'i` ,'000 i','so�o <br /> 4. Land slope % ;Q 8 z� ;�° <br /> wa 9 Sce fig.C-6 (z 1.� <br /> D. ROCK LAYER DIIviENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer:A x 0.83 = <br /> o' <br /> ���o gpd x 0.83 sq. ft./gpd = ��_sq. ft.-�ro`�C- (��`� <br /> 2 Select width of rock layer (10 feet or less) = I o ft. <br /> 3. Length of rock layer = area=width = R�g� <br /> �4 sq. ft. = > � ft. _ �� �l ft. �.�.�.,...ti.�.ti.�.ti.ti.�•,......., <br /> !•r•J•�•��f•r•/•f•r•t•r•t•1•.r•t• <br /> ti.1•ti•\•ti•ti•ti•ti•ti•1•5r•t•�•�•ti•\•1 <br /> f•f'f•'•f•''f'r•�'''f'f'r'r'�'f' idth S10 ft. <br /> ti.�•�.ti•�•ti•ti,.ti•�•�.ti•�•ti•ti•ti•ti•ti <br /> �f�f��ti��rti�,���tir`�ti�`f`f,��f,�; <br /> •f•r•r•r.f•r•r.�•r•r•r•�•f•r.r•�• _, <br /> E. ROCK VOLL�ME ~- �'g�' --' � ' <br /> � . <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; �- <br /> r,uw sq, ft. x /, ��ft. _� �� cu. ft. <br /> 2 ` Divide cu. ft.by 27 cu. ft./cu. yd. to get cubic yards; <br /> -�! � cu. ft. -27= �� cu. yd. ' <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> =�, cu.yd. x I.4 ton/cu. yd. _�tons. <br /> F. ADSORI'lTON WIDTH �-��`� L.o'-1-m .a,w��,s�, vb�� <br /> 1. Percolation rate in top 12 inches of soil is mpi r�„�,,,b«,�� G'-� '�°�J <br /> �;n„���,�;, Soil Texture �„",;R a�� <br /> (r <br /> ..no�h <br /> 2 Select allowable soil Ioading rate from table; Faster than 0.1 oane sana l�o l.00 <br /> �'-?� d/ft� 0.1 to� Sand 120 1.00 <br /> � 0.1 to� Fne Sand" 0•� 2-� <br /> 6 to 15 ndy Loam 0.79 1 S2 <br /> 3. Calculate adso tion width ratio b dividin rock la er 16 to 3o t.oam o�� 2�� <br /> rP Y g Y 31 to 45 s�ic�m o-� 2•� <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; �to bo c�a c.oam o.45 2.67 <br /> 1.20 gpd/ft2'- ,y.�gpd/ft� = a, l��L. Slower�haa 120 Clay 0.24 �.00 <br /> `�Soil}uving 509G or ma�e d fine a vey fine sand <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> a-�� x �o ft =�:,.� ft <br /> � � <br />
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