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2000-P02623 - new septic system
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3745 Jacobs Mill Road - 32-118-23-24-0011
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2000-P02623 - new septic system
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Last modified
8/22/2023 4:40:14 PM
Creation date
3/15/2017 12:32:16 PM
Metadata
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Template:
x Address Old
House Number
3745
Street Name
Jacobs Mill
Street Type
Road
Address
3745 Jacobs Mill Road
Document Type
Septic
PIN
3211823240011
Supplemental fields
ProcessedPID
Updated
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� � MOUND DESIGN WORKSHEET 5 <br /> , (For Flows up to 1200 gpd) <br /> A. �,Q�1 Estimated Sewage Flows in Gallons per day <br /> Estimated ��o d � d, <br /> � Numbu 'fype 1 Type� Type IIl Type <br /> or measured x 1.5 = gpd. °` `v <br /> 2 sao 2zs i so � <br /> B. SEPTIC TANK LIQUID VOLUMES a 6 0 300 zs� °f''� <br /> ��,� <br /> 2�'�O gallons 6 750 su 3�z '° <br /> 7 600 370 Typel. <br /> ❑a <br /> 8 1200 675 408 w <br /> C. SOILS (refer to site evaluation) `°�"'°`� <br /> � lic 7Lnk Ca adUa lin alluns) <br /> 1. Depth to restricting layer = inches�_feet �qwa�����y <br /> 2. Depth of percolation tests = /2 inches "°^��°� MinimumLiyuid I�yuidcapadrywiN �w�,�=:�& <br /> Bednx�ms Capacity garbage dispusal lih iaside <br /> 2�x kss 750 1125 1500 <br /> 3. Texture Percolation rate -Z mpi �,.<< �� ,5� z� <br /> Su�F �500 -�$- �4W�0 <br /> 4. Land slope �/ � % '.R°�y 2°°° <br /> T G : <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock layer: A x 0.83 = <br /> vo gpd x 0.83 sq. ft./gpd = �Z sq. ft. <br /> 2. Select width of rock layer (max 10' if<120 mpi max 5') _ �D ft. <br /> 3. Leng of rock layer = area _width = � a=�n�� a��Q ���a =n� � ¢o � <br /> °.40° o�°bo oon-.e D1� <br /> sq. ft. - 1�ft. -��ft. ��°onDo�aopoC�oa o <br /> 4ApdpQoeQQ oete.e'oo'oo .. <br /> � �o DO'�D�DDD�.pD� tib. <br /> W1�TYl /O I} Q0000> o'Q00ObOD D oDD <br /> 1u1.11 � <br /> <120mpi <10' Length 7�5'- ft� <br /> E. ROCK VOLUME <br /> >120mpi <5' <br /> 1. Multiply rock area by rock depth to get cubic feet of rock;lT�-sq- ft. x / <br /> ft. _��cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get c bic yards; <br /> 7.�cu. ft. =27 =��cu. yd. � <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons,�7�u. yd. x 1.4 <br /> ton/cu. yd. ��tons. <br /> F. ABSORPTION WIDTH � Absorption Width Siziag Table <br /> 1. Percolation rate in top 12 inches of soil is�7 Z mpi �a���� c�„o�s �o�rA�� <br /> Minutes per Inc6 Soil Tezhue per day per width to Rock <br /> Texture �� (MPI) squacc foot Laycr wdth <br /> Fazter Nan 0.1 Coarsc Sand 1.20 1.00 <br /> 0.1 to 5 Sand 1.20 1.00 <br /> 2. Select allowable soil loadin rate from table; o.►�o s Fine Sand o.6o z.00 <br /> � 6 to 15 Sandy L.oam OJ9 1.52 <br /> . lj(� d/� 16 to 30 .jpam._ Oy�_ 2.00 <br /> � 31 to 45 Silt Loam �0 2.40 <br /> 46 to 60 Clay Loam 0.45 2.67 <br /> 60 to 120 Clay 0.24 5.00 <br /> 3. Calculate adsorption width ratio by dividing rock layer s�o,�,a w��zo c►ay o.zo 6.00 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; <br /> 1.20 gpd/ft2=�gpd/ft2= Z •d <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> Z xl�ft= Zo ft <br />
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