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ATGRADE DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) Estimated Sewage Flows in Gallons per day I <br /> Numocr <br /> I.A. FLOWof Type 1 hpe II Type III -... <br /> Bedrooms V <br /> Estimated 3o o gpd . •,, 1 2 300 225 180 <br /> , <br /> or measured •-. .. x 1.5 = gpd. 3 450 300 218 60,< <br /> 4 600 375 256 7 <br /> 5 750 450 294 So <br /> .B. SEPTIC TANK LIQUID VOLUMES 6 900 525 7s I. <br /> loco 370 <br /> Z - 'G% gallons 81 <br /> 8 1200 675 408 whams <br /> C. SOILS (refer to site evaluation) Septic Tank Capacities,in gallons <br /> 1. Depth to restricting layer = ., • inches <br /> Number of Minimum Liquid Liquid capacity with <br /> 2. Depth of percolation tests = -! inches Bedrooms Capacity garbage disposal <br /> 3. Percolation rate . • mpi 2 or less 750 1125 <br /> , 4. Land slope % 3 or 4 1000 1500 <br /> 4 or 6 1500 2250 <br /> D. ROCK WIDTH 7,s or 3 2000 3000 <br /> over <br /> 1. Percolation rate in top 12 inches of soil is mpi <br /> 2 Select allowable soil loading rate from table; ._ .i , Absorpdon Width Sizing Table <br /> 1 ,c, gp(Vft2 Percolation Rare Galton Riad of <br /> in Minutes per Soil Tenutou per day per At orpeierr width <br /> 3. Calculate adsorption width of the rock by dividing web own, Nowa foto to Rot <br /> rock layer loading rate by allowable soil loading rate; Faster than 0.1- Coats*Sand —1 10 — <br /> 4 gpd/ft . Gro gpd/ft2= ') ft. o.i oss. Ftn.ss"'t'and•• 0 0 22.0A0o <br /> 6 to 15 S.nyr turn 0,79 1s2 <br /> E. ROCK VOLUME (.g �.I./,---r-- (,n - /0'4,x,., . . +`s =Y Y 1:_c'Yr. 11:1(51 `ole"ee_m .(0-601 2.200w <br /> e <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; 6 :120 cla tgrarn oi: IQ <br /> )0 '&'-Y A4 l ' 47:I sq. ft. x .t? - ft. = ry " ' cu. ft. Slower than any — <br /> 1 F <br /> e:? 2 Divide cu. ft. by 27 cu. ft./cu. yd. to get cubic yards; <br /> • <br /> `'~ , cu. ft. -t-27= ' .' cu. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 1 cu. yd. x 1.4 ton/cu. yd. = I ') tons. <br /> F. SYSTEM SIS <br /> a. The Height of the System is 1 foot of rock and 1 foot of cover over the System. <br /> 1ft+ lft= a feet ' <br /> b. Enter table with landslope andupslope dike ratio. Select dike multiplier of. <br /> . ..:-z., <br /> c. Multiply upslope dike multiplier by height to find upslope dike width: <br /> •2 ft x C. feet <br /> d. Enter table with landslope and downslope dike ratio. Select dike multiplier of =e e% . <br /> e. Multiply downslope dike multiplier by height to get downslope dike width: <br /> 2ftx t' ;c-' = I feet ..•.a 1 _ <br /> f. The System width is the sum of upslope dike width plus downslope dike width; <br /> 4 ft+ 1's ft= feet <br /> g.The rock layer length is the flow divided by 4: - <br /> .:)o -'-4 = ') feet Li -i - . x -..;•:`;..•:.,..1 , <br /> 1. Total length is the sum of upslope dike width plus rock layer length plus ..-«v,.4.tyyt -r <br /> upslope dike width: <br /> 9 ft + +-1?, ft + '`') ft= r' feet <br />