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1995-006820 - new septic system
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2320 Shadowood Drive - 27-118-23-32-0014
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1995-006820 - new septic system
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Entry Properties
Last modified
8/22/2023 4:20:13 PM
Creation date
8/22/2018 11:10:41 AM
Metadata
Fields
Template:
x Address Old
House Number
2320
Street Name
Shadowood
Street Type
Drive
Address
2320 Shadowood Drive
Document Type
Septic
PIN
2711823320014
Supplemental fields
ProcessedPID
Updated
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� �PK6 .�y�rn 6etii�a F,�r <br /> + t�rl, Bco�k i I//Aoa�✓000 F.I�cnt <br /> MOUND DESIGN WORKSHEE'� ' � ' <br /> � (For Flows up to 1200 gpd) `��R H��'�E <br /> A. �.OW Estimued Sevage Aov in Gallanz per Day(gpd) <br /> Estimated 6 Oo �d �'or `/ BFo�aor^ <br /> N�anber <br /> of 'Iypc I 'Iype II 'Iype 11I 7ype IV <br /> or measured x 1.5= gpd• B��� <br /> 2 300 Z7S I80 sos ' <br /> 3 450 300 218 •� <br /> B. SFPTIC TANK LIQLJID VOLUMES ° 6°° "s 's� •�, <br /> D�vf /2 s o gallons r o,v E iaoo G�[,c o,✓ � � �o i o ^�• <br /> � ONE /DOO G.�tta� Pu�r�"�G T�1"'K s i2ao 6�s aos °�'m" <br /> C: SOIIS (refer to site evaluation) rr�� �� �_„� <br /> 1. �Depth to restricting layer= inches . s�� � w� <br /> 2. Depth of percolation tests = inches <br /> z«t� �sa i.�zs <br /> 3. Percolation rate mpi saa �,000 �.soo <br /> 4. Land slope % s a 6 i.soo z,sso <br /> �a s z,000 s.000 <br /> wer 9 See fig.C-6 (:1S) <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 /> 00 gpd x 0.83 sq. ft./gpd =So o sq. ft. <br /> 2 Select width of rock layer(IO feet or less) _ �� ft. <br /> 3. Length of rock layer= area y width= Rock Bed <br /> S�� sq. ft.i �,o ft. = SO ft. ,.,.�...�.,.......�.,.........�.� <br /> l�1.l.I•I•�•I.l•� I•I•�•!•I•f•f• <br /> ti.�.�.�.�•�•�.�•�•�•\•ti•ti•�•ti•ti•1 <br /> !•�•I•�•l•I•r•l•t•t•I•!•I'•!•r•f• L <br /> ~/�t�l~!~l�t�I�I��I�/�I t I�f! Idt�1 �1� t. <br /> ~!~I~!~I,1�J:t�I�/�I,/�J%f�r~�� �Q <br /> E. ROCK VOLUME �-- �ength -� <br /> 1. Multiply rock azea by rock depth to get cubic feet of rock; s� <br /> So D sq.ft.x / ft. = Soo cu.ft. ' <br /> 2. Divide cu.ft.by 27 cu. ft./cu.yd. to get cubic yards; <br /> Soo cu.ft. +27=/�S cu.yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> /�5 cu,yd.x 1.4 ton/cu.yd. = 2 6 tons. <br /> F. ADSORP'ITON WIDTH A� o�w�a�s�, Table ' <br /> 1. Percolation rate in top 12 inches.of soil is 7-Smpi r�a,e«,x�� G°"' `'°"` <br /> 1,�;,,uy,�;,�, Soil Texture p�'.,N:� p a,,": � <br /> � �m� I.a •b.Qvaa. <br /> dJ� <br /> 2. Select allowable sbil loading rate from table; ' Fascer'than oa oa�sana 1.2o i.00 <br /> i.2o i.00 <br /> ulf O. S gpd/ft� OicoS FineSand�- 0.6o Z.00 <br /> 6 to 15 ndy Loam 0.79 la"2 <br /> .3. Calculate adso tion width ratio b dividin rock la er 16 co�o t,o�n o.60 2.00 <br /> rp y g Y 31 to 45 s�ic t.�m oso z.ao <br /> � loadin rate of 1.20 d/ft2 b allowable soil loadin rate; 46 to 60 ci� t,o�► o.as 2.6� <br /> g 8P Y g 61 co lzo �iay o.�a s.00 <br /> 1.20 gpd/ft2�-�SD gpd/ftZ= 2.YD Slower than 12o C�a -- — <br /> «4d1 hwin�SOi ar man d flrr ar vory Ri»rnd <br /> �4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> �O X 2.y ft=�ft <br />
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