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Mound Design Worksheet (For flows up to 1200 gpd) <br /> All boxed rectangles must be entered,the rest will be calculated. , <br /> A. FLOW i Al:Estimated Sewage Has InGal=perDay <br /> Estimated 750 gpd(see figure A-1) ' <br /> or measured x 1.5(safety factor)= 0 gpd I bedroom Oda t Ma II Clay IN I Clan IV 1 <br /> 2 300 225 160 1 6016 <br /> B. SEPTIC TANK LIQUID VOLUMES 3 450 300 216 ollhe <br /> Septic tank capacity 2x1000 gallons(see figure C-I) d 600 375 254 votes I <br /> 5 150 +SO 244 In the <br /> C. SOILS(Site evaluation data) 6 900 525 332 CM 1, I <br /> 1. Depth to restricting layer= 1.5 feet 7 1050 6110 370 ii.orll <br /> $ 1200 675 406 corurms. <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture I loam <br /> 4. Soil loading rate(see Figure D-33) 0.5 gpd/ft2 <br /> Percolation rate 15-30 MPI <br /> 5. %Land Slope 1 5 % 1).33: AbsarpNen Width Shing Tabk <br /> pereolauad Rue i. ns Rate <br /> - 14 Mtttatet per Ste f T UM,: Getkietpter Alttoe�oon <br /> C-1: Se k'Taiik Capacities tin jteUOnala i th Izo efa�t . 4(afr <br /> Number of Minimum Liquid Liquid capacity with Liquid capacity 1 vetoer um s coshe hand 1.20 l o0 <br /> wish disposal& , Tethily sand <br /> Bedrooms Capacity garbage disposal lift inside _ - _ 1att Noe Rood <br /> --- � -I <br /> - =�: .3 - _:Zeit=._ <br /> 2 or leas 750 1125 --1e to ao ._Y-- Lou _ _-- 0.6152 vo -- <br /> 1500 31 t R111 loom 0-",10 240 <br /> 3 or 4 1000 1500lila <br /> •Q� 46 to 60 t 0.45 262 <br /> 5 or 6 1500 2250 5i>;1i) 7`' t�w,n <br /> 7.8 or 9 2000 3000 .4000 r<i to 120 _ the Delay 0.24 a on <br /> Gadd-(3m' <br /> ---tigiaer-khoit 120' <br /> ci <br /> 'sreem d.tip,a rat UMW iu&ad a ua.t a jwfuaa.a.. <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow(A)by 0.83 to obtain required area of rock layer:Item A x 0.83= <br /> 750 gpd x 0,83 ft21gpd= 622.5 ft2 <br /> 2. Determine rock layer width =0.83 ft2/gpd x Linear Loading Rate(LLR)(see LLR chart) <br /> 0.83 ft2/gpd X 12 ! = 10.0 ft <br /> LLR Chart <br /> Perk Rate LLR <br /> <120 MPI <=12 <br /> >=120 MPI <=6 <br /> 3. Length of rock layer=area divided by width= <br /> 622.5 ft2 / 10 feet= 62.5 feet <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 622.5 X 1 ft= 622.5 ft3 <br /> 2. Divide ft3 by 27 ft3/yd3 to get cubic yards <br /> 622.5 ft3 / 27 = 23.1 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock In tons; <br /> 23.1 yd3 X 1.4 ton/yd3 = 32.3 tons <br /> F. ABSORPTION WIDTH <br /> 1. Abso I tion width e.uals absorption ratio(see Figure D-33)times rock layer width <br /> 2.4 x 10.0 ft = 24.0 ft <br />