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, • MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> 1 A. FLOW I: tim.ted Sewage(lows in Gallons pa day <br /> Estimated 4,SJ gpd (see pages D-7 or I-3, 4, 5) Number Cod) <br /> or measured --- gpd X 1.5 = -- • le4°m, Type 1 Type 11 Type 111 Tge <br /> . 1 <br /> B. SEPTIC TANK LIQUID VOLUMES 2 450 2250218 aea <br /> 4 600 375 256 ti„se <br /> se <br /> .2 ---.o e d gallons (see pages C-3 or C-5) 56 9 0 45025 3 32 ty��.L <br /> 7 1050 600 370 0In <br /> e 1200 675 40t1 .r.,.. <br /> C. SOILS (refer to site evaluation) <br /> ,. <br /> 1. Depth to restricting layer = :;1,`4N-10 a,e inches SgUeTook GptMn".legallon <br /> �� NumI of Minimum liquid I•evil e.pecey well <br /> 2. Depth of percolation tests =_ 12. inches i t Copeck), Drew diakonal <br /> 3. Percolation rate 7; 0' mpi \_line_ �Cl��'�'T 2orleu 750 a 1 n <br /> 4. Land slope I-}• % ->�\ Lia v Sc 1 L 74or <br /> a 9 15(1) <br /> 2150 <br /> MILT 9 ...... <br /> i <br /> D. ROCK LAYER DIMENSIONS -- <br /> 1. <br /> •-•1. Multiply flow rate by d.83 to obtain required area of rock <br /> layer: Daily Flow x 0.83 = , <br /> Li,_,;() gpd x 0.83 sq. ft./gpd = 37 sq. ft.+1('� ' = 4 I O <br /> 2. Select width of rock layer (10 feet or less) = /0 ft. <br /> 3. Length of rock layer = Area +Width = <br /> 4 1 b sq. ft. + /c' ft. = 4 I ft. Rock Bed • <br /> •.•ti.;:.ti•ti•ti••.•ti•=.,.v.•ti.4Nldth S10 ft. <br /> :::::5:;45:555554.4:5::::455:a1::::•4 I <br /> E. ROCK VOLUME 1-- Length <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> ylo ,sq. ft. x 1cs' ft. = 4:;_ cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> 1i3__ cu. ft. +,27= 1 (, cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> l c. cu. yd. x 1.4 ton/cu. yd. = ) tons. <br /> F. ADSORPTION WIDTH _,;.r-. 11.4'3o >~-t,1 „1 <br /> 1. Percolation rate in top 12 inches of soil is 7-{ mpi Abspuon Width st:tnx'lbbIe <br /> � Percolation RateGallons Iloilo Select allowable soil loading rate from table on page E-; n Minutes pa Soil per day per AlwntMinn <br /> width <br /> Inch(MPI) equate fans � l <br /> g d/f <br /> ukb <br /> 3. Calculate adsorption width'ratio by dividing rock layer Faster than 0.1• Corse Send . -_ <br /> oat sa (_.601.00loadin rate of 1.20 gPd/ft2b allowable soil loadingrate; o.l b 5• Pine Sand•• 060 1 <br /> o <br /> nd <br /> 6 to 15 Sandy Own 0.79 132 <br /> • 1.20 gpd/ft2+ .t.- gpd/ft2= ..) . . 161n 30 Loam '0.601 2110o <br /> 31 to 43 Sill Lon 0.30 2.40 <br /> Check this value on page E-16. 6o o . c41-1L". m 0.45 2.M <br /> 4. Multiply adsorption width ratio by rock layer width to get s;;;an clay --- -- I <br /> required adsorption width; <br /> •' - x /0 'ft = AU ft <br />