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1995-007167 - new septic
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1870 Sixth Ave N - 27-118-23-42-0022
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1995-007167 - new septic
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Last modified
8/22/2023 4:22:22 PM
Creation date
1/16/2019 11:41:48 AM
Metadata
Fields
Template:
x Address Old
House Number
1870
Street Name
6th
Street Type
Avenue
Street Direction
North
Address
1870 6th Avenue North
Document Type
Septic
PIN
2711823420022
Supplemental fields
ProcessedPID
Updated
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� � MOUND DESIGN WORKSHEET <br /> , , (For Flows up to 1200 gpd), � <br /> A. �OW Estimued Seaage Flow in Gallons per Day(gpd) <br /> Estimated Ce O o gpd N�� 'Iype I 'Iype II 'Iype III 'I)+pe IV <br /> or measured - x 1.5 = - gpd. a��� <br /> 2 300 27S 180 � <br /> 3 450 300 218 °� <br /> B. SEPTIC TANK LIQUID VOLUMES 4 �ao 3�s z.� .� <br /> s �so aso 29a � <br /> a-ioo o gallons 6 90o Su 332 � <br /> „ ., � taso boo 3�0 �,�,. <br /> '�1="��4=�-:i,�--�, � g 1.'!X? 575 408 <br /> C. SOILS (refer to site evaluation} �, Y� x�� �� �,,�„ <br /> 1. Depth to restricting layer =1� �f o a� inches B�� �, �, <br /> 2. Depth of permlation tests = i a inches ``'�'' `A`�"`' <br /> 3. Percolation rate 4.l� m i 2�� �so �.�zs <br /> p s a a �.000 i.soo <br /> 4. Land slope � % s«6 i.soo z.zso <br /> �a a z.000 s.000 <br /> mc 9 See 6g.G-6 (:1.5) <br /> D. ROCK LAYER DIl�IINSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer:A x 0.83 = Q � <br /> �DO gpd x 0.83 sq. ft./gpd =��sq. ft-�►�70= s�►� <br /> 2. Select width of rock layer (10 feet or less) _ /v ft. <br /> 3. Length of rock layer= area=width = Rock Bed <br /> 5y'� sq. ft.y /D ft. = SS lt. L•ti.ti.ti.ti•w.,.�.ti.ti.ti.��.ti.ti.,.ti <br /> :�::�f�l.r. l•t•�.l�f J•r•t•f• <br /> . , ti �ti ti•ti-�• {S•ti.ti•ti•ti•ti-ti.ti.ti•ti <br /> 1•l•J•l•1•t�r•l•l•1•1•t•l�J•l•�• <br /> ti•L•ti�ti•ti•t•ti•1•ti•ti•ti•t•L.ti•ti•ti•ti idth �IQ ft_ � � <br /> 1j1KK���Yti ti•ti�tifti�tiK lftfti�t <br /> .f.f.l•1•r•l.t.f.f.f.t.l.f.f.f.f• <br /> E. ROCK VOLUME �- �ng�' -� � <br /> 1. Multiply rock azea by rock depth to get cubic feet of rock; <br /> �sq. ft. x ,v 'f t. =5� cu. f t. <br /> 2. Divide cu. ft.by 27 cu. ft./cu. yd. to get cubic yards; <br /> 5�cu. ft. i 27=�cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> a�cu. yd. x 1.4 ton/cu. yd. = a�i tons. <br /> F. ADSORI'TTON WIDTH G�a' ��r� A� a,w�ac�s�, iable <br /> CJIw R�eo d <br /> 1. Percolation rate in top 12 inches o soil is mpi M��ti�,R� �iiTexture �,�� '�� <br /> � <br /> (mpil t` `�a' <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 oarse Sand 1�o l.00 <br /> d/ft� 0.1 to� Sand 1.20 1.00 <br /> .y� � 0.1 to� Fine Sand" 0.60 2.00 <br /> 6 to 15 ndy Loam 0.79 la2 <br /> 3. Calculate adso tion width ratio b dividin rock la er 16 to�o Lo� o.bo 2.00 <br /> � y g y 31 to 45 Silt Loam o.� 2-� <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; �to bo ciay[.oan, 0.4� z.6� <br /> 61 tU 120 Clay 0.24 5.00 <br /> 1.20 gPd/ft23- .,ti gpd/ft�_ �.l0 7 Slower than 120 Cla -- - <br /> "Soil having 509G or more of fine or very fine sand <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> �x�_ft=�2ft <br />
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