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1996-007706 - new septic system
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Deer Run Trail
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2965 Deer Run Trail- 04-117-23-24-0012
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1996-007706 - new septic system
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Last modified
8/22/2023 5:10:51 PM
Creation date
6/20/2016 11:03:53 AM
Metadata
Fields
Template:
x Address Old
House Number
2965
Street Name
Deer Run
Street Type
Trail
Address
2965 Deer Run Tr
Document Type
Septic
PIN
0411723240012
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORKSHEET , <br /> (For Flows up to 1200 gpd) - i <br /> A. FLOW Gstimated Scwage Flows in Gallcxis pa day � <br /> Estimated �S o gpd (see pages D-7 or I-3,4, 5) h�m�< <g�� <br /> or measured —' gpd x 1.5 = -- ��°�ms Typc I Typc II Typc Il1 �� <br /> z 300 225 I80 <br /> B. SEPTIC TANK LIQUID VOLUNIES 3 450 30o z�a o� <br /> 4 600 375 25G ���� <br /> 1-)a.s o � 1 - 1 o a � gallons (seE pages C-3 or C-5) 6 90o szs° 3�i =��. <br /> 7 1050 600 370 m <br /> 8 1200 675 408 cd�n <br /> C. SOILS (refer to site evaluation) << <br /> 1. Depth to restrictin� layer = �-o inches SCP�KT�nKCaU�f�lt�lllK,��,.,,, <br /> x,r,��.�r ���n_T�umLiau:,1 i.�:������a�,.�:�Y,.��, <br /> 2. Depth of percolation tests = I a ii:ches ���Wms �ti��::y �.���b��;..;�.a� <br /> 3. Percolation rate �- � mpi Z;a;' '�°, ;x <br /> 4. Land slope S p� '�6 `x'° Zu° <br /> /0 7,8 or 9 2(XA 7000 <br /> mer 9 '•—" <br /> D. ROCK LAYER DILIE\1SIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: Daily Flo�v x 0.83 = ,� ' <br /> ��o gpd x 0.83 sq. ft./gpd = a� sq. ft-'r�o'��- b�y <br /> 2. Select width of rock layer (10 feet or less) _ / o ft. <br /> 3. Length of rock layer = Area = Width = <br /> �'-�sq. ft. = /0 ft. _ �� ft. Rock Bed <br /> r.r•r•r•r�f�f•r�:�r•r�r•r�r�r <br /> tif....tifti.tifti.tif�f.f�f.ftiftifti• <br /> ti.ti ti ti ti ti ti ti �. ti•�. ti•ti : ti� idth SlO ft. <br /> .f.f.r•r�r•r•r.r.r.;.r.r.r•r•f <br /> . ti.ti.ti.•,.•,.•,.•,.•,.•..•.. <br /> �:'r�:�:••:•f.:•:�:•:•:•:•:•:•: <br /> E. ROCK VOLUME 1-- Len gth —� <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> !�sq. ft. x /,o�ft. _ �cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> �Ccu. ft. = 27 = _a2__ cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> a� cu. yd. x 1.4 ton/cu. yd. _ � tons. <br /> F. ADSORPTION WIDTH �-��-`� L-°4m <br /> / Absorption Width Sizing 7'ablc <br /> 1. Percolation rate in top 12 inches of soil is �._5 mpi <br /> PercolationRatc Galbns Ratioof <br /> 2. Select allowable soil loading rate from table on page E-; in Minutes per Soil Tczwre ����Y�� Abswpiion widih <br /> fnch(MPI) squ�re foot to Rock Laycr <br /> -`� gPd�f� w�d�n <br /> 3. Calculate adsorption width ratio by dividing rock layer Fa�tcr than 0.1• c�u s��,� ----- ----- <br /> 0.1 W 5 Sand 1.20 1.00 <br /> loading rate of 1.20 gpd/ft�by allowable soil loading rate; o.��s•� Fine Sand'• o.�o z.no <br /> 6 to 15 Sandy Loam 0.79 1.52 <br /> 1.20 gpd/ft�� .-�gpd/ftZ= � .1.7 3i;oas° s�i�°ioam o:o z`a�'o <br /> Check this value on a e E-16. ���o ClayLoam o.ss z.�� <br /> P 8 6o a�w ci3Y o.za s.uo <br /> 4. Multiply adsorption width ratio by rock layer �vidth to get Sli;���an c�ay ----- ----- <br /> required adsorption width; <br /> �.>�x i �- ft = ��.7ft <br /> ; <br />
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