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Oct 19 06 09: 08a Josh Swedlund (952) 873-3292 p. 5 <br /> Mound Design Worksheet (For flows up to 1200 gpd) <br /> All boxed rectangles must be entered,the rest will be calculated. A-1: Estimated Some ROM in Gallons per Cloy M <br /> A. FLOW <br /> Estimated 600 gpd(see figure A-1) ' ter a <br /> iV <br /> or measured x 1.5(safety factor)= 0 gpd bedrooms GG0IIl Gas <br /> 300 <br /> 2 uu 2 2255 180 00% <br /> � <br /> B. SEPTIC TANK LIQUID VOLUMES 3 450 300 218 of the <br /> 4 600 375 250 values <br /> Septic tank capacity 2000 gallons(see figure C-1) 5 150 450 294 in the <br /> C. SOILS(Site evaluation data) b 900 525 332 Cess I- <br /> 1. Depth to restrictinglayer= 1.8 feet 8 1250 605 400 oII,of III <br /> p Y 8 1200 b75 408 column, <br /> 2. Depth of percolation tests= 12 inches -�-�- <br /> 3. Texture loam <br /> 4. Soil loading rate(see Figure D-33) 0.6 gpd)ft2 <br /> Percolation rate 20 MPI <br /> 5. %Land Slope f 4 % D-33: Absorption W law Wing Table <br /> rerculatton Race I.000ing RAW <br /> In Minutes.per Sod Texture Gallons Absolplton <br /> C.-.1; 1'Sept Tank a!actffes c In :anus's) Inch per d per keUo <br /> iwwas. w . ....._-ti Still=1001 1 <br /> Li quid capacity Feslerthan 5 Coarse Sand 1.20 1 0 <br /> Number of . Minimum Liquid Liquid capacity With with disposal& Mechu Sand <br /> Bedrooms Capacity garbilge disposal lift inside --------... .----- Ntltcy Sand..... <br /> .Gto.LS__..---_Su>d>. 'Im--.___-. OA. ..... -TOO.-O <br /> 2()Hess I 1 25IG 1u 30 _ -.._...)own 0.60 <br /> .. I_ _.._._ <br /> 1�.y ;1lw45 MIL Loam U71) 2.40 <br /> 3 or 4 10,a) 1500 2000 -aG to till ;min s .{lk ..... 1)45- 7.67......... <br /> S or{i 1500 2250 ; Silty Loam <br /> ?;8 t)C 9 2000 3000 000 -io1 in 1`20 7 1.vales.--- 2....----.__....00_..-- <br /> uuru+arun�nae�a�i Salty t.7ay I)?4 S QU <br /> Sandy Clyy. <br /> - Clay... <br /> Slower than 12t) <br /> .y.urn dcsiawd foribem wilt must 4.too ue meioses+un10 <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 /> 600 gpd x 0.83 fegpd= 498.0 ft2 <br /> 2. Determine rock layer width =0.83 ft2/gpd x Linear Loadin' 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 /> 498 ft2 I 10 feet= 50.0 feet <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 498 X 1 ft= 498.0 ft3 <br /> 2. Divide ft3 by 27 ft3lyd3 to get cubic yards <br /> 498.0 ft3 / 27 = 18.4 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 18.4 yd3 X 1.4 ton/yd3 = 25.8 tons <br /> F. ABSORPTION WIDTH <br /> 1. Abso tion width uals absorption ratio(see Figure D-33)times rock layer width <br /> 2 x 10.0 ft = 20.0 ft <br />