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2004-P08193 - new septic system
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2945 Jamestown Road - 28-118-23-31-0005
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2004-P08193 - new septic system
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Last modified
8/22/2023 4:23:45 PM
Creation date
3/21/2017 11:05:57 AM
Metadata
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Template:
x Address Old
House Number
2945
Street Name
Jamestown
Street Type
Road
Address
2945 Jamestown Road
Document Type
Septic
PIN
2811823310005
Supplemental fields
ProcessedPID
Updated
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, � • MOUND DESIGN WORKSHEET <br /> . , • • (For Flows up to 1200 gpd) <br /> A. FL.OW Estimued Sew�aae Flow in Gallons pa DaY(gpd) <br /> ,. <br /> Estimated�,�gpd <br /> or measured x 1.5 = gpd. �°f,�,,,, �'�`I ��I TY�rn �ry�rv <br /> 2 300 225 160 sos <br /> 3 430 300 218 er <br /> B. SEPTIC TANK LIQUID VOLUMES � 4 �oo ;�s zs� � <br /> s �so �so ss� "m� <br /> a-,�oo o gauo� 6 � S� 332 � <br /> � inw soo 3�0 �,�,. <br /> g 1200 675 408 <br /> C. SOIIS(refer to site evaluation) � � <br /> 1. Depth to restricting layer= ►l.�' d 1 �d �� inches NB °�� �,"�` �,,� <br /> 2. Depth of percolation tests = 1 a. '� inches `'"`'' `'""' <br /> 3. Pereolation nte l��3 mpi Z3'4` 1'0�0o i',s o <br /> s or 6 1.500 2.250 <br /> 4. Land slope �-I °lo �a a �.000 s.000 <br /> �Q 9 su e�.c.s �x i s� <br /> D. ROCK LA,YER DIlVIENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer:A x 0.83 = , <br /> l�oo . gpd x 0.83 sq. ft./gpd =�t�sq. ft-���o=sy�° <br /> 2 Select width of rock layer(10 feet or less) = i o ft. <br /> 3. L,e�gth of rock layer=area i width= R�� <br /> S�t� sq. ft.+ �_ft. _�ft. 1•tih.ti•tiKK•�1��•ti•ti•tiK.ti ti.ti <br /> .r•r.r.f•f•r•rr•r•r•r•r•r•r•f.r• <br /> yrtftifNyl�t:f1�r~Itir�f�Mr�f~fT <br /> ti.�.ti•�.ti.ti.ti.ti.ti.ti.ti.ti•�•ti•> >dth 510 ft. <br /> f•Mht•f•f•Mt•f fY•r�l�t•r•1• <br /> �•ti•tiK•ti•1�Y�ti•�•MtiKw�1•ti•ti <br /> . .f./.f.r.f.t.f.f.f.j�t.t.r.f.r.f. <br /> E. ROCK VOLUME ~- �'�' -� <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; � <br /> Sy� sq.ft. x .�u s ft. =S�4 cu. ft. <br /> 2. Divide cu. ft.by 27 cu. ft:/cu. yd. to get cubic yards; <br /> �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. = a��- tons. <br /> F. ADSORPTTON WIDTH L►-A� �.o v✓1 ,� oa,w�ae,s;�, r,bk <br /> 1. Percolation rate in top 12 inches of soil is !�•3 mpi ��a,ti«,�� � �;,; <br /> r�,,,,m�tt„�, Soil Texture �„q„R ,,,;,,,,` <br /> i.. .e.e. a <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 oarse Sand 1�o l.00 <br /> d/f� 0.1 to5 Sand 1.20 1.00 <br /> •y S � 0.1 to 5 Fine Sand" 0.60 2.00 <br /> 6 to 15 ndy Loam 0.79 1 S2 <br /> 3. Calculate adso hon width ratio b dividin rock la er 16 co 30 [.oam o.bo 2•00 <br /> � � }� $ }' 31 to 45 Silt Loam �� 2•� <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; �to bo c� L.oa�, o.4s 2.6� <br /> 1.20 gpd/ft2i ,►�S gpd/ft�_ �.�� Slowerthan 120 ciay oz4 �.00 <br /> "Sal havin6 509G or maee d fine or vcy fine sv+d <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> a•�� x �o ft=a�ft <br />
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