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2001 - Septic System approval
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2001 - Septic System approval
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Last modified
8/22/2023 5:26:56 PM
Creation date
1/24/2020 1:41:04 PM
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Template:
x Address Old
House Number
4760
Street Name
West Branch
Street Type
Road
Address
4760 West Branch Rd
Document Type
Septic
PIN
0611723330002
Supplemental fields
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Updated
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MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> IEstimated Sewage Flows in Gallons pa day <br /> A. FLOW (amd) <br /> Estimated (0 0 6 gpdNumber "'t`' L'`aII �` <br /> " '° Class <br /> Bedrooms <br /> or measured x 1.5 = gpd. <br /> 2 300 225 180 <br /> OK <br /> 3 6(X)450 3CO 2 18 <br /> °l"the <br /> values <br /> B. SEPTIC TANK LIQUID VOLUMES 5 7750 525 294 TY.. <br /> DO O gallons 7 1050 6755 4 e m <br /> columns <br /> C. SOILS (refer to site evaluation) <br /> SeptiaiiideipaaidesA, ) <br /> 1. Depth to Lestricting layer = J " inches 1,1 feet Liquid c,pacity <br /> Number of Minimum Liquid Liquid capacity with with disposal& <br /> 2. Depth of persolation tests = /2 inches Bedrooms Capacity garbage disposal Itfl•nsde <br /> 3. Texture 0.)Ct m Percolation rate /d mpi 2orless 750 1125 1`.(A <br /> ` P 3 or 4 1000 1500 2)M1) <br /> 4. Land slope 6 0 5 or 6 1500 2250 3 lu0/0 <br /> ,,g a 7 2000 <br /> 3000 40W <br /> D. ROCK LAYER DIMENSIONS Mound LLR <br /> 1. Multiply flow rate b 0.83 to obtain required area of rock layer: A x 0.83 = Perc Rate LLR <br /> (IOU gpd x 0.83 sq. ft./gpd =5-00 sq. ft. <120 MPI <_ 12 <br /> 2. Determine width of rock layer = 0.83 sq. ft./gpd x Linear Loading Rate (LLR) <br /> 0.83 sq. ft/gpd x 1 = /0 ft >l20 MP[ 5 6 <br /> 2 <br /> 3. Length of rock layer = area --. width = _ <br /> ;:....;: .....;:-;:t.;72;.'7.:7 DD Ra..6".`''ogp��e0n.ere RQO,dp <br /> S00 sq. ft. - /0 ft. = �S O ft.' ,qoa,oeo- �Qa eate.,:. <br /> 0 o e aD a0 e00. p.� ,Oe A n_�•:[tarAa on�n L`.Y.w^�i <br /> aV.om; d <br /> .9!°.C11.1.50-77°,77°.c. D aD oD n:-�.+U`.yora,Q•,n:0a c.n:n.al.i <br /> Width /O ft ° <br /> ,y.°aaaa..e.dO.QO.POf .t.a+a wo.yea .s <br /> <120mpi <10' Length .5V ft <br /> E. ROCK VOLUME >120mpi <5' <br /> 1. Multiply rock area by rock depth to get cubic feet of rock;,SVOsq. ft. x / ft. = <br /> SdUcu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> 5d 0 cu. ft. _ 27= /',S cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons;/ •S cu. yd. x 1.4 ton/e2(j 1 r7 J <br /> Pdd y tons k'o r ID ss <br /> F. ABSORPTION WIDTH <br /> Absorption Width Sizing Table <br /> 1. Percolation rate in top 12 inches of soil is /0 mpi Pe,,,at,nRre in G.)Inn. Rainnt <br /> Texture A) Q hit Mlnmes per Inch Soil Texture per day per Absorption.teh <br /> IMP)) square tool to Rock Myer <br /> Width <br /> Faster than 0.1 Coarse Sand 120 1.00 <br /> 2. Select allowable soil loading rate from table; °7707 Loaamy$ <br /> and <br /> 120 <br /> gpd0.1 t0 5 Fine Sand 0.60 200 <br /> /f t2 6105 Surly Loam 0.79 1.52 <br /> 615 60 2 OI1 <br /> to Siltam n�+i O - Z00 <br /> r Sill <br /> 16 to 60 Clay Loan)CL) 1).45 2.67 <br /> 3. Calculate absorption width ratio by dividing rock layer Sandy L <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; Sh. CL <br /> 60 ;h;n 21) 6 6) <br /> 1.20 gpd/ft2 _ /& gpd/ft2 = . <br /> 4. Multiply absorption width ratio by"rock layer width to get <br /> required absorption width; . <br /> c;), x ie.) ft =P(' ft <br />
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