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1810 Lakeview Terrace - 27-118-23-43-0005/8
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septic info including design
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Last modified
8/22/2023 4:22:28 PM
Creation date
4/25/2017 11:56:37 AM
Metadata
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Template:
x Address Old
House Number
1810
Street Name
Lakeview
Street Type
Terrace
Address
1810 Lakeview Terrace
Document Type
Septic
PIN
2711823430005
Supplemental fields
ProcessedPID
Updated
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. • , MOUND DESIGN WORKSHEET <br /> - � (For Flows up to 1200 gpd) <br /> A• ��W Esdmated Scwage Flowa in Gallcx�a pa day <br /> Estimated �v o gpd (see pages D-7 or I-3,4,5) «� <br /> � a <br /> or measured gpd x 1.5 = - ���5 7YP�� TYP��I Type llt �� <br /> B. SEPTIC TANK LIQUID VOLUMES i as�o 3ouo ia � <br /> a �ao s�s zsb �,,,�� <br /> "1, -/oo U gallons (see pages C-3 or C-5) s �so aso Z�a ;o <br /> 6 900 525 332 '�i. <br /> 7 1050 600 370 m <br /> 8 1200 675 408 ed�s <br /> C. SOILS (refer to site evaluation) <br /> 1. Depth to restricting layer = 34�'-rv ti o�� inches <br /> ScplkT�ak Gpscilic;in grlluns <br /> Numbr o[ Minimum Liquid Liquid ceprciq wiih <br /> 2. Depth of percolation tests = i � '� inches ��• a��ry e��d��► <br /> 3. Percolation rate � . ; mpi z3 4u �� �� <br /> 4. Land slope � % `�6 1� u� <br /> �.s a 9 �000 woo <br /> over9 .---- <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: Daily Flow x 0.83 = , <br /> C�co gpd x 0.83 sq. ft./gpd = �� sq. ft.��u��=�y�� <br /> 2. Select width of rock layer(10 feet or less) _ �� ft. <br /> 3. Length of rock layer =Area 1 Width = <br /> �� sq. ft.1 __�v ft. = SS ft. Rock Bed <br /> f•r•t.r•f•t•t•f t•t•f.f•1�r•r <br /> � ti•ti•t•ti•ti.ti•ti•ti•S•ti•'. ti ti ti•ti• <br /> r•r•rN•r•1 �t t•f•�•t•r•t•f <br /> . . ti•ti ti ti•ti•ti•ti b�•ti•ti•4•ti•ti•ti• �dth <_]Oft. <br /> f•f•r•l•t•f•t•t t•t•f•r•t•f�t <br /> ti•ti '.ti•ti.ti.ti.ti.ti.�..�..'..ti•ti•1• I <br /> E. ROCK VOLUME •f•f•r•f•f•f•r•f r•f•r•f.r•f•f 1 <br /> 7 Length --� <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> S47 sq. ft. x .1 v 5 ft. = s�u cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> �� cu. ft. s 27= a I 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. = a� tons. <br /> F. ADSORPTTON WIDTH ���r w�� <br /> 1. Percolation rate in top 12 inches of soil is ti./ mpi ADsorption Width SizingTablc <br /> 2. Select allowable soil loadin rate from table on a e E-• '�°'en�Re� c'"°"' R°"°°f <br /> g p g � in Minula per Soil Texturo per day per Abswp�ion wid�h <br /> .��, ��/r� IncA(MPl) square foa to Rw�clth y� <br /> 1 <br /> 3. Calculate adsorption width ratio by dividing rock layer Faaterthsn0.1• c�s�,e ----_ __.__ <br /> o.�ws sa�a i.zo l.00 <br /> loading rate of 1.20 gpd/ft�by allowable soil loading rate; o.�a s�• F'u�e Sand•• o.�o z.�o <br /> 6 to 15 Sandy Loam 0.�9 1.52 <br /> �.20 gpd/ft�= ►-�,�gPd�ftZ= � -t�� t6to30 �am o.�0 2.��, <br /> 31 to as sih Loam o.50 z.40 <br /> Check this value on page E-16. �a;u, �'gciy�° o:io s��io <br /> 4. Multiply adsorption width ratio by rock layer width to get s��;:8� c�ey ___._ _____ <br /> required adsorption width; <br /> a.t�� x iu ft = a�.� ft <br />
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