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1996-008461 - new septic system
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1920 Sixth Ave N - 27-118-23-42-0002
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1996-008461 - new septic system
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Last modified
8/22/2023 4:21:52 PM
Creation date
1/16/2019 12:09:39 PM
Metadata
Fields
Template:
x Address Old
House Number
1920
Street Name
6th
Street Type
Avenue
Street Direction
North
Address
1920 6th Avenue North
Document Type
Septic
PIN
2711823420002
Supplemental fields
ProcessedPID
Updated
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t-ly <br /> • � � ' MOUND DESIGIV WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. FLOW D-7 . <br /> 1��gP P g �.,�.��R�,,��.�.��.. <br /> Estimated d (see a es D-7 or I-3,4,5) �' <br /> 1W��EA TrP� Of IKS�OFME� <br /> or measured gpd. =°'�'6 t a a s <br /> 7 ]00 �! i�0 iO�t <br /> 3 �!0 ]OO 2H e� <br /> � f00 �1! bt �wuw <br /> B. SEPTIC TA.NK LIQLJID VOLUMES : .'o �n � ''�`� <br /> Z—/000 �allons (see pages C-3 or C-5) ' �� � "� � <br /> a i�oo sie �w cew��w. <br /> C-3 <br /> SEPTIC T�NK CAV�CITES, IN GA�lONS <br /> C. SOILS (refer to site evaluation <br /> ua.��...n.. <br /> 1. Depth to restricting layer = . inches """" � �^^�°'�� <br /> .mwow ��...a.. a..o,., <br /> 2. Depth of percolarion tes s = � Z inches ,a„� ,., ,,,, <br /> ,3. Percolation rate Z mpi •«� ���a �no <br /> �o�� �ue »�a <br /> 4. Land slope % ,...a. <br /> .... �.o. <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: A x 0.83 = <br /> �gpd x 0.83 sq. ft./gpd =�� sq. ft. <br /> 2. Select width of rock layer(10 feet or less) _ /C� ft. <br /> 3. Length of rock layer = area+width = I <br /> � <br /> � sq. ft.+ _�ft. = v�b ft. Rock Bed <br /> 1•f•r•I•r•I•rMM.l.r•r•r•r! <br /> ti�ti t•ti ti ti.ti.�.�.ti.ti.ti.ti.�►. I <br /> •;•r•r.r.f•nr•r �•�•r.r.r•r•r I <br /> �.ti.ti.ti•ti.ti.ti•ti.ti.ti.ti.ti•�•ti• idth 510 <br /> •�•r•r•r•�•�•r•f r•�•r•r•r•�•r <br /> ~•�•~•�•�•�•�•�•�•�•�•�•�•�• 1 <br /> .f.1•j.r.r.f.f.r.f.j.j.f.f.f.f <br /> E. ROCK VOLUME F'- Length --� <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> S�sq. ft. x�ft. �oocu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu.yd. to get cubic yards; <br /> �xu. ft. 127=�cu. yd. <br /> 3. Mulriply cubic yards by 1.4 to get weight of rock in tons; I <br /> _�cu. yd. x 1.4 ton/cu. yd. _�tons. <br /> F. ADSORPTION WIDTH <br /> 1. Percolation rate in top 12 inches of soil is�mpi E-16 <br /> 2. Select allowable soil loading rate from table on page E-16; �����•a���a��•�� I <br /> � (o D SPd�f� .��...� .r.......�,�. � <br /> •...r rr.w� <br /> 3. Calculate adsorption width ratio by dividing rock layer �""" "'"�� `""' �"�" """"'�'" <br /> � <br /> �.�. � �.n i.n �.a �.oe <br /> loading rate of 1.20 �pd/ft2 by allowable soil loading rate; ' " °" '" '�" '" <br /> u ��e. e.w a.�� aa t.00 <br /> 1...�J ���ItG O d/ft2 = .�� , " '•� a.w ew �w �..o <br /> ' KF �- ,�.—KP � :,�a e:.: :;: :::: .�va <br /> Check this value on page E-16. <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; ; <br /> �D x Z.o ft = Zo ft <br />
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