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2009-00641 - new mound septic
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2250 French Lake Road - 10-117-23-22-0012
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2009-00641 - new mound septic
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Last modified
8/22/2023 3:20:59 PM
Creation date
12/6/2016 2:05:30 PM
Metadata
Fields
Template:
x Address Old
House Number
2250
Street Name
French Lake
Street Type
Road
Address
2250 French Lake Road
Document Type
Septic
PIN
1011723220012
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORK SHEET(For Flows up to 1200 gpd) <br /> A. Average Design FLOW A-l: Estimoted Sewoge Flows in Gallons per Day <br /> num er o <br /> Estimated �o�a gpd (see figure A-1) bedrooms Class I Class II Closs III Closs IV <br /> or measured -- x 1.5 (safety fnctor) _ ---� gpd 2 300 225 �8o bo% <br /> 3 450 300 218 ofthe <br /> 4 600 375 256 values <br /> B. SEPTIC TANK Capacity 5 750 450 294 in the <br /> 6 900 525 332 Class I, <br /> �, — J c�c� v gallons (see figure C-1) �"���:���� , � �o� boo s�o u, or�u <br /> �,,i '��� 1 �i'4� l,�a ��, "��°" �, <br /> 8 1200 675 408 columns. <br /> OOC� �•. * "'R%.{, !�1'iS-,l) <br /> C. SOILS (refeY t0 Slte CUCiIl�Cif1011� C-]: Se ticTankCa acities(in allons <br /> Number o( Minimum Liquid Liquid capacity with Liquid capacit� <br /> 1. Depth to restricting layer = l.�) �, � �,� {eet Bedrooms Capacity garbage disposal �'ith disposal& <br /> -� �"'-T— lifl inside <br /> 2. Depth of percolation tests = /.0 . feet zo�iess �so �i25 15� <br /> 3. Texture G�/��� �.c>�� 3°�4 10°° 1500 z000 <br /> 5 or 6 I500 2250 3� <br /> Percolation rate �,t�, f) mpi �,s o�9 z000 3000 <br /> 4. Soil loading rate � �},� gpd/sqft (see figure D-33) <br /> 5. Percent land slope � % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow (A) by 0.83 to obtain required rock layer area. <br /> G.,c�c� gpd x 0.83 sqft/gpd = � � r sqft <br /> 2. Determine rock layer width = 0.83 sqft/gpd x linear Loading Rate (LLIZ <br /> 0.83 sqft/gpd x J a, gpd/sqft = �0 ft Mound LLR <br /> 3. Ler�gth of rock layer = area = width = <br /> �{� `C sqft (D1) = /u ft (D2) _ �v ft < 120 M PI < � 2 <br /> E. ROCK VOLUME > 120 M PI < C� <br /> l. Multiply rock area (Dl) by rock depth of 1 ft to get cubic feet of rock <br /> �`�sqft x 1 ft = � ` �' cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> �!`;�''t' cuft = 27 cuyd/cuft = 1 � cuyd <br /> 3. Multiply cubic yards Uy 1.4 to get weight of rock in tons <br /> ��; cuyd x 1.4 ton/cuyd = �, c� tons <br /> D-33: Absorptinn Widlh Sizing Table <br /> F. SEWAGE ABSORPTION WIDTH PercoleiionRa�c LoadingRele <br /> in Minutea per $oil lezturc Gallons AbsorpGon <br /> ]nch per da}'ptr Ralio <br /> MP� s uare foot <br /> Faster�han 5 Coerse Send 1.20 1.00 <br /> Mcdium Sand <br /> Absorption width equals absorption ratio (See Figure D-33) LoemySand <br /> Finc end <br /> times rock layer width (D2) ` <br /> ]6 io 30 Loem 0.60 2.00 <br /> 31�l0 45 Silt L.oam 0.50 2.a0 <br /> ,,lr�) X i c� Et = ��,{,...`) f t 46 to 60 s�,ay aeY� o.as z.6� <br /> Silty Clay Loem <br /> 6]to 120 Sil�y Clay O.Z4 5.00 <br /> Sandy Cley <br /> Cle <br /> Slowerthan 120• <br /> •Sys�em dui�Md!or m�,c.oil.�,.�.�tx oa,er or�,ro�,,,a,K< <br />
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