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1995-007181 - new septic system
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2855 Somerset Lane - 04-117-23-24-0020
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1995-007181 - new septic system
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Last modified
8/22/2023 5:11:04 PM
Creation date
2/22/2019 1:53:00 PM
Metadata
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Template:
x Address Old
House Number
2855
Street Name
Somerset
Street Type
Lane
Address
2855 Somerset La
Document Type
Septic
PIN
0411723240020
Supplemental fields
ProcessedPID
Updated
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� ' MOUND DESIGN WORKSHEET " <br /> (For Flows up to 1200 gpd) <br /> A: �.�w Estimued Sewage Flow in Gailons pa Day(gpd) <br /> Estimated 6�O �C� N o{� 'ryp�I 'ryp�II lype III 7ype IV �� <br /> or measured x 1.5 = gpd. a��� <br /> 2 300 225 180 6oac <br /> 3 450 300 218 � <br /> B. SEPTTC TANK LIQUID VOLUMES s �so 4'so Z�va `r°" <br /> /L S� g3llOriS t� 1000 G.�GCO�✓ ZN� T,/N,r 6 90� S1S 332 � <br /> f /250 G./CtG,� �lqM//Nt''�r r,/NK. � 1Q50 600 370 �,m„ <br /> 8 1200 675 408 <br /> C. SOIIS (refer to site evaluation) N,�� �� .�� <br /> 1. 'Depth to restricting layer= 26 inches �,,,� �,, � <br /> 2. Depth of percolation tests =_ /`� inches ``�'"' <br /> � 3. Percolation rate //. 2 mpi Za� �so �ass <br /> 3 or a 1.000 1.500 <br /> 4. Land slope 3 % s«6 i.soo z.aso <br /> 7 a 8 2,000 3.000 <br /> wa 9 Sea fip C-6 (x 1.� <br /> D. ROCK LAYER DIlvIENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer:A x 0.83 = <br /> 60o gpd x 0.83 sq.ft./gpd = Soo Sq. ft. <br /> 2. Select width of rock layer(10 feet or less) _ /� ft. <br /> 3. Length of rock layer= area+width= Rock Bed <br /> 0 o sq.ft.+ /D ft. = S o ft. �.�.�., �.....�.�.ti..�......,.� <br /> ��f.��� ����r�tN t I��.��r r•r• <br /> • ti�ti 1 �ti��y�y.�.ti����•����Z��.♦ <br /> I•J'•�•r rrl•t.r•f J��•r.�•tr/�•�• <br /> i.ti.ti.�•t ti.�.ti.�.ti.ti.�.�•�•�•ti•� idth S10 ft. <br /> �.�.r.�.i.r•r•r•f r•r•�•r•r•�•r• . <br /> ti.ti.ti.ti.�.ti.ti•ti.ti•ti•ti•�•�•�•1•�•ti <br /> ,r.�.r.�.r.f.r.f.r.r.r.�.f.f.�.�. <br /> �- jRAgt}l <br /> E. ROCK VOLLTME �.� <br /> 1. Multiply rock azea by rock depth to get cubic feet of rock; <br /> soo sq.ft. x / ft. =SUo cu, ft. ' <br /> 2. Divide cu.ft.by 27 cu. ft./cu. yd. to get cubic yards; <br /> Svo cu,ft. +27= / ,S cu.yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> /�S cu, yd. x 1.4 ton/cu. yd. = Z 6 tons. <br /> F. ADSORPTION WIDTH n •a,w�a�,s�, vbk <br /> 1. Percolation rate in top 12 inches of soil is �� L mpi �a s�«,w h � �� <br /> Soil Texture <br /> • (m� �.r .�oam <br /> w�dih <br /> 2. Select allowable soil loading rate�from table; Fascer chan o.l oarse saaa 120 l.00 <br /> - �.�F' o. 60 d/f� 0.1 to5 Sand 1.20 1.00 <br /> $� 0.1 to 5 Fine Sand" 0•� 2•� <br /> 6 to 15 ndy Loam 0.79 152 <br /> 3. Calculate adso hon width ratio b dividin rock la er 16 to�o t,oam o.bo 2.00 <br /> rP ' y g Y si co as siic c.�� oso z.4o <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; 61�1� ��t-y� o.4s 2b� <br /> y 024 SAO <br /> 1.20 gPd/ft2i- O. 6o gpd/ft�= Z . O Slowerthan 120 Cla -- - <br /> , »Soil having 8o9i or mQe d&r ar vey 6ae sand. <br /> 4. Multiply adsorpHon width ratio by rock layer width to get <br /> required adsorption width; <br /> Z .o X �o ft= 2O ft � <br />
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