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septic info including 1995 design
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3225 6th Avenue North - 29-118-23-44-0001
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septic info including 1995 design
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Last modified
8/22/2023 4:27:11 PM
Creation date
1/23/2019 11:40:49 AM
Metadata
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Template:
x Address Old
House Number
3225
Street Name
6th
Street Type
Avenue
Street Direction
North
Address
3225 6th Avenue North
Document Type
Septic
PIN
2911823440001
Supplemental fields
ProcessedPID
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MOUND DESIGN WORKSHEET <br /> , . <br /> (For Flows up to 1200 gpd) <br /> A. �..�W Estimued Sewage Floa in Gailons per DaY(gpd) <br /> Estimated Lo� gpd x�� • <br /> of Type I 'type 1[ '[Ype Ili Type IV <br /> or measured - x 1.5 = �- gpd. ��,,,, <br /> z 300 225 160 � <br /> B. SEPTIC TANK LIQUID VOLUMES a 6 o i°s° zsb '� <br /> s �so aso z9a `�"" <br /> �-���� gc'1llOI1S 6 900 S1S 332 � <br /> � 1050 600 370 ��. <br /> $ 1200 675 408 <br /> C. SOILS (refer to site evaluation) ,, <br /> ,i !��� �� ,�m <br /> of �� G��"° <br /> 1. Depth to restricting layer = 3� - bo inches B�,� �, � <br /> 2. Depth of percolation tests = 1 a " inches `A"°"' `'"�`' <br /> 3. Percolation rate 5.� m i 2«,� no ,.,u <br /> P 3aa t.000 �.soo <br /> 4. Land slope (o % ;a 8 z� 3:� <br /> ova 9 See fig.C-6 (x 1.5) <br /> D. ROCK LAYER DIl�IVSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: A x 0.83 = a, <br /> l00 O gpd x 0.83 sq. ft./gpd = �}q�C sq. ft.-► �o°�c=5y'� <br /> 2. Select width of rock layer (10 feet or less) = /D ft. <br /> 3. Length of rock layer = area-�-width = Rock Bed <br /> Sti7 sq. ft.-�- �ft. = SS ft. ti.,.ti.ti.�.ti.ti.�.�... <br /> �•ti.,...,.,.. <br /> � � !•f•1•�.1�I. r•l r•f�f•f.r.r.f. <br /> . �•�.ti.ti.�.�.�t�.S.ti.ti.�•1.�•t.�K <br /> I•I•!.I•1•t•t•I•t f•t•1•f•t.f.I• <br /> . ti•ti•�.ti.�.ti.�.ti.ti.L.ti.�.ti.ti.ti.ti.ti idth <_10 ft. <br /> 1•t•t•J•I•�•tM�l t•r•/•1•t•t•�• <br /> �•tiN�ti•�•ti•�•ti•�.ti.�.ti.ti.ti•�.�•ti <br /> - . . .�•l.��f•1N.t•1N•I�t.f•f�t.ry <br /> E. ROCK VOLUME �- �n8� -'� <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> 5'-�2 sq. ft. x 1.os f t. = S�� cu. f t. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> �cu. ft. i 27= a �_cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> �cu. yd. x 1.4 ton/cu. yd. = a q tons. <br /> F. ADSORPTION WIDTH �`-'� �o <br /> n «,w,ae�s� tib� <br /> 1. Percolation rate in to 12 inches of soil is m i ��. A.md <br /> P p n��ti�,R�� �.., ..�� <br /> Mirmusperincn SoilTexture yQ,y„� ,,,,�,b <br /> (mpil t.a ,b�v9ea. <br /> .ndd� <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 oarse sana 1.2o i.00 <br /> �.� gpd/ffi 0.1 to� Sand 1.20 1.00 <br /> 0.1 to� Fine Sand" 0.60 2.00 <br /> 6 to 15 ndy Loam 0.79 1 S2 <br /> 3. Calculate adso tion width ratio b dividin rock la er 16 to�o-__ t:.o�► o.-so� 2.00 <br /> rP Y g Y �o as s�ic t.oa�► oso z.ao <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; �-to_bo _ ctay[.our,.__ .4 2.6� <br /> 61 t0120 Clay .24 5.00 <br /> 1.20 gpd/ft�� .y � gpd/ft�_ �,� 7 Slower than 120 Cla - - <br /> ••sdl tv.ti,g�09G ar mar oE ty,e or vay fine sana <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> �,L'� x�ft= a�,'Z ft <br /> � � <br /> A crs+-� � -�o w� . (. ],� o ' ,L 0 = J. 0 �c 1� = p <br />
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