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septic design - no permit on file
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507 Ferndale Road North - 36-118-23-13-0007
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septic design - no permit on file
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Last modified
8/22/2023 5:01:19 PM
Creation date
8/16/2016 11:43:20 AM
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Template:
x Address Old
House Number
507
Street Name
Ferndale
Street Type
Road
Street Direction
North
Address
507 Ferndale Road North
Document Type
Septic
PIN
3611823130007
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NIOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. �OW ��� Estimatcd Scwagc Flows in Galluns per day <br /> Estimated �vv gpd (see pages D-7 or I-3, 4, 5) ��� <br /> � um ber <br /> or measured gpd x 1.5 = �� Ty�� Type II Typc III �ry� <br /> I�ed.00m S �v <br /> 2 300 225 I80 <br /> B. SEPTIC TA��1K LIQUID VOLUMES 3 4so 30o z�a � <br /> 4 600 ]75 256 ���� <br /> allons (see a es C-3 or C-5) <br /> g P g 6 900 °sz° 33i �Q'. <br /> 7 1050 600 370 m <br /> 8 1200 675 408 ca„m„ <br /> C. SOILS (refer to site evaluation) <br /> 1. Depth to restricting layer = Z� inches �9�xTankCapaci�ic�;�g.���,. <br /> Vum6eY of Minimum Liquid L.iquid capYciry wiih <br /> 2. Depth of percolation tests = �Z inches ��m= �.�o�;ry tubayedispocal <br /> 3. Percolation rate� mpi 2or1C3 750 "� <br /> 3 a a ia�o �soo <br /> Q p1 a a b �soo zuo <br /> 4. Land slope C� /� 7,8 or 9 za,o 3000 <br /> o.�.9 __.. <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> lay2r: Daily Flow x �c� = r ':: <br /> �s-;; � V gpd x 0.�3 sq. ft./gpd = � sq. ft. <br /> �3-- <br /> 2. Se��c: �vidth of rock layer (10 feet or less) _ /o ft. <br /> 3. Length of rock layer = Area = Width = � <br /> � -= �'r sc�. ft. - /r.` ft. _ ;;�� ft. Rock Bed I <br /> .r.J:,•.f:::::::::::::::::r:f:f I <br /> ..�.�..�..�........�.�.... T <br /> ��ftiJ:ftil'.%'.%:j:�.f:f:?tif•.•:� Nidth 510 ft. <br /> •:f.r.f.t•:•l•:•:•f•t•!•J�J.J <br /> ,ti,.�.ti..�,.�,.�,.�,.�,.�,.,.�.. <br /> E. ROCK VOLUME '�f.. . f f . . . f . . . f.r l <br /> F--- Length —� <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> � � �,a sq. f t. x / f t. _ -S"r,�% cu. f t: " <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 /> = ; fo=S cu. yd. x 1.4 Eon/cu. yd. _ � tons. <br /> F. ADSORPTTON WIDTH <br /> Absorption Width Sizfng Table <br /> 1. Percolation rate in top 12 inches of soil is�mpi <br /> Pacolation Rete Galions Retio of <br /> 2. Select allowable soil loading rate from table on page E; ��M���u� Soil Texturc per day per .�n��«�W.d�n <br /> (nc�(MPI) squux foot io Rock laycr <br /> r,�� g�C��ft� w�a�n <br /> 3. Calculate adsorption width ratio by dividing rock layer Fiqer dan 0.1• c�sy,a ----- --- <br /> o.�ws se�a i.zo i.00 <br /> loading rate of 1.20 gpd/ft�by allowable soil loading rate; o.�,�s•� P;M s�•• o.�o z.00 <br /> 1.20 gpd/ft2-�- 0- gpd/ft - /,s2 6�o�s s�,ay�, o,�s.. �,sz <br /> Z 16 io 30 L.oam 0.60 2.W <br /> 31 to 43 Sitt Loam 0.50 2.40 <br /> - Check this value on page E-16. �e� c,�y c.�, o.4s �.�� <br /> 60 tn 12p Clay 0.24 S.W <br /> 4. Multiply adsorption width ratio by rock layer width to get s�o�.,w� c�sy — -- <br /> required adsorption width; <br /> SZx �ft = J5,2ft <br />
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