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QNSinrE •1.%' <br /> SlWA6! Job# <br /> TREATMENT <br /> PROGRAOM <br /> University of Minnesota Mound Design Worksheet <br /> Greater than 1%Slopes <br /> A. FLOW <br /> Estimated 450 gpd(see figure A-1) <br /> or measured x 1.5(safety factor)= 0 gpd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> Septic tank capacity 2000 gallons(see figure C-1) <br /> Number of tanks/compartments 0 <br /> Effluent Filter (yes/no) yes <br /> C-1 Septic Tank Capacity in Gallons <br /> Number of Minimum Capacity with Capacity with <br /> Bedrooms Capacity Garb.Disp. Disp.and Lift <br /> 2 or less 750 1125 1500 <br /> 3 or 4 1000 1500 2000 <br /> 5 orb 1500 2250 3000 <br /> 7,8 or 9 2000 3000 4000 <br /> C. SOILS(Site evaluation data) <br /> 1. Depth to restricting layer= 1.3 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture loam <br /> 4. Soil loading rate(see Figure D-33) 0.60 gpd/ft2 <br /> Percolation rate 4 MPI <br /> 5. %Land Slope 11.0 <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 /> 450 gpd x 0.83 ft2/gpd= 380 ft2 <br /> 2. Determine rock layer width =0.83 ft`/gpd x Linear Loading Rate(LLR)(see LLR chart) <br /> 0.83 ft2/gpd x 12.00 = 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 /> 380.0 ft2 / 10.0 feet= 38.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 380.0 X 1.0 ft= 380.0 ft3 <br /> 2. Divide ft3 by 27 ft3/yd3 to get cubic yards <br /> 380.0 ft3 / 27 = 14.1 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 14.1 yd3 X 1.4 ton/yd3 = 19.7 tons <br /> Page 1 of 5 <br />