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1994-006632 - new septic system
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2860 Deer Run Trail - 04-117-23-24-0014
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1994-006632 - new septic system
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Last modified
8/22/2023 5:10:54 PM
Creation date
6/16/2016 3:07:15 PM
Metadata
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Template:
x Address Old
House Number
2860
Street Name
Deer Run
Street Type
Trail
Address
2860 Deer Run Tr
Document Type
Permits/Inspections
PIN
0411723240014
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORKSHEET � � <br /> (For Flows up to 1200 gpd) <br /> , <br /> A. FLOW [stimatcJ Scwagc Fbws in Gallcxis per day <br /> Estimated �J C� gpd (see pages D-7 or I-3, 4, 5) ,,,�m�< <s�� <br /> or measured — gpd x 1.5 = — ���mS Typc I Type 11 Type III �� <br /> B. SEPTIC TANK LIQUID VOLUl�1ES 3 4°0 3 o zgs p«� <br /> 4 600 375 256 ���� <br /> ►-la�o� � -�v�� gall�ns (see pages C-3 or C-5) s �so oso Z�o ,o <br /> 6 900 52S 332 Tb�t. <br /> 7 1050 600 370 m <br /> 8 1200 675 408 cdumn <br /> C. SOILS (refer to site evaluation) <br /> 1. Depth to restrictin� layer = o���0 3 ��r inch�s ti�V�MT.nkCayacitw��ying.ti,,,,. <br /> 2. Depth of percolation tests = )a '' inches �����m= "'°�a;a�`y�"° `�`�ai���'<`'�;:��,e�'' <br /> 3. Percolation rate �{ . I mpi z�ku �� ��u <br /> 3aa tan tsc,� <br /> 4. Land slope O �v � °Jo ,°8 a y �� � <br /> over 9 ..__. <br /> D. ROCK LAYER DIMENSIONS <br /> l. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: Daily Flo�v x 0.83 = <br /> �,�gpd x 0.83 sq. ft./gpd = ���, sq. ft.rt�o'=�- ��'� <br /> 2. Select width of rock layer (10 feet or less) _ /� ft. <br /> 3. Length of rock layer = Area = Width = <br /> �_ sq. ft. = _ ft. = 1��' ft. Rock Bed <br /> . t•t•f•.'•f%t�f.f::�l.J:f:f:l,• <br /> ti••.••.•ti•ti•ti•ti••.•'.•'.••.••.••.•'.•`.• <br /> •f•f•;•l.l.f.f.f,f.f•1•,I,f,f,� <br /> . ti•ti••.•ti•ti•ti••.•• ti••.•ti.ti••.••.• Width <_l0 ft. <br /> .f.f.f.f.r.f.f.f;f.f,f,f,;,f,f <br /> . .•t••.•ti••.••..•,.•,.•,..�.•,. <br /> E. ROCK VOLiJME' •'f'''''''''''''''f'''''''''' <br /> � Length -� <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> ��a sq. ft. x ' _ ft. _ ��K cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> '�1 u"cu. ft. - 27 =�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. _ ?��% tons. <br /> F. ADSORPTION WIDTH _'�� ���� Y%i <br /> Absorption Width Sizing 7'abic <br /> 1. Percolation rate in top 12 inches of soil is�� mpi <br /> Percolation Ratc Gallons Ratio of <br /> 2. Select allowable soil loading rate from table on page E-; in Atinutes per so��T�x��� ���,�y a� Absorpiion widiA <br /> � 4 \ d/ft� Inch(MP!) square foot io Rock Lnycr <br /> gp Width <br /> 3. Calculate adsorption width ratio by dividing rock layer Facte�than 0.1• c�s�„� ..___ ._.__ <br /> o.�ws s��a i.zo i.00 <br /> loading rate of 1.20 gpd/ft�by allowable soil loading rate; o.��s•� Fine Sand•► o.�o z.�o <br /> 6 to]5 Sand L.oam 0.79 1.52 <br /> 1.20 gpd/ftz= ,�+ ' gPd/ft2= ?, �'�1 �6�o�o �A�, o.�o z.��, <br /> 31 ro SS Silt Loam 0.50 2.s0 <br /> Check this value on page E-16. ���o ClayLoam o.4s 2.�� <br /> 60 to 120 Clay 0.24 S.W <br /> 4. Multiply adsorption width ratio by rock layer width to get Sl;���an c�ay _____ _____ <br /> required adsorption width; <br /> � �� x �% ft = � '� ft <br /> ; <br />
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