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2000-P02333 - new septic
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195 Crystal Creek Road - 33-118-23-32-0004
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2000-P02333 - new septic
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Entry Properties
Last modified
8/22/2023 4:49:43 PM
Creation date
6/7/2016 2:42:02 PM
Metadata
Fields
Template:
x Address Old
House Number
195
Street Name
Crystal Creek
Street Type
Road
Address
195 Crystal Creek Road
Document Type
Septic
PIN
3311823320004
Supplemental fields
ProcessedPID
Updated
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� � MOUND DESIGN WORKSHEET <br /> � (For Flows up to 12U0 bpc�) <br /> A. �'�W E�[unucd ScaaRc Flow in C:allc�s per Day(gp�) <br /> Estimated �SO gpd N,a„�, — <br /> or �Yr�i ryv�ii iy�tii <br /> or measured __ x 1.5 = --- �;j'd. �i�«,m, �y��v <br /> 2 lcxl 2�5 I BO � <br /> B_ SI�:I'TIC TANK LI(2UID VOLUMES e bai iis zs6 � <br /> 5 7Sq �50 29d <br /> a-►300 A��0�1 5e j g�]p� 6 900 525 332 � <br /> 1 - r Soo 9allo•� P�v,,�p t.,�1� > >aso � 3�0 o�,e <br /> 8 1200 675 408 �`�' <br /> C SOILS (refer to site evaluation) <br /> 1. Depth to restricting layer = o� � �� inches h��� �� �,"" <br /> 2. Depth of pereolation tests = / Z " inc-hes `""�`' `""°"`' <br /> 3. Percolation rate �,q- ��, mpi �«� �so �.,u <br /> 4_ Land slope � % s a v i:so zz�so <br /> �«a z.000 3.000 <br /> wQ 9 See fig.C-6 (z 1.5) <br /> D. ROCK LAYER DIl�IEI�TSIONS <br /> 1. Multiply fl�w rate by 0.83 to obtain required area of rock <br /> layer: A x 0.83 = <br /> �So gpd x 0.83 sq. ft./gpd = �sq. ft. (6 30� <br /> 2 Select width of rock layer (10 feet or less) _ _ 1 U ft. <br /> 3_ Len�th of rock layer = area =u7idth = Rock Beei <br /> �� sq. ft. y �0 ft. � <br /> - ��__ f� ti-.�ti•�•�•�•.•.-ti•ti•.•�•.-�•.•�-�� <br /> 1'.{:t:tir�:�r'f'r''''"'r'r'''''r'� <br /> ti•ti•ti•ti•ti•�•1•1•ti•ti•ti� <br /> ti:L:ti:{:ti:ti:L:ti:ti:ti:ti:�,fti:tifti:ti:ti idth <_10 ft. <br /> ti ti�ti�ti�ti�ti�t�tifti�tfti�ti��fti 1fti?t <br /> .f.f.t.l.f.t.f.f.f.f.r.f.f.f.f.f. <br /> E. ROCK VOLUME � �S� � � <br /> 1. Multiply rocic area by rock depth to get cubic feet of rock; <br /> 6a3 �. ft. X�ft. = 6�3 cu. ft. (630� <br /> 2. Divide cu. ft. by 27 cu_ ft./cu. yd. to get cubic yards; <br /> �a3 �. ft. �- 2� = a3 �u. ya_ <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> �_cu. yd. x 1.4 ton/cu. yd. = 3 a tons. <br /> F. ADSORPTTON WIDTH � �,w,a�,s�, Tible <br /> 1. Percolation rate in top 12 inches of soil is �9 16 mpi Pelro,,ti�,R,� �a. A.md <br /> t�t;�,�,u,�II,a, Soi17'exture �,� '�`,a,�`e <br /> (mpJ ra.a .b�.wa. <br /> ad, <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 oarse Sand i2o i.00 <br /> 0.6o gpd/fr� o.� co� sana izo �.00 <br /> 0.1 to� fiine Sand" �.� z-� <br /> 6 to 15 ndy Loam 0.79 1.�2 <br /> 3. Calculate adsorption width rario by dividing rock layer 37 to 4 ssi�m o50 2-ao <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; �to�� Clav Loam o.as 2.6� <br /> 1_20 gpd/ftZ1 0•6Q d/ft� = a .0 p 6� co�Zo C�ay o.za �_oo <br /> gp Slower than 120 Clav <br /> ^5«!}uvv�t 5QR ar more d fu+e or very fine sand. <br /> 4. Multiply adsorption width ratio b}' rock layer width to get <br /> required adsorption width; <br /> .00 x t� ft = � ft <br />
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