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masse Job# <br /> . "Temakinwesonry <br /> Preosrewwa <br /> University of Minnesota Mound Design Worksheet <br /> Greater than 1%Slopes <br /> a FLOW <br /> Estimated 750 gpd(see figure A-1) <br /> or measured x 1.5(safety factor)= 0 gpd <br /> B. SEPTIC TANK UQUID VOLUMES <br /> Septic tank cepacity 2250 gallons(see figure C-1) <br /> Number of tankh/compartments 0 <br /> Effluent Filter (Os/no) yes <br /> C-1 Septc T k Capacity In Gallons <br /> Nu of Minimum Capaklty with Capacity with <br /> Bedrooms Capacity Garb.Disp. Disp.and Lift <br /> 2 or less r.1 1125 <br /> 3or41500 <br /> rfi <br /> 5 or 6 '` 2250 <br /> Z�dih. <br /> 7,8m9 Avalakt 3000 a. <br /> C. SOILS(Site evaluation data) <br /> 1. Depth to restricting layer= 1.8 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture loam <br /> 4. Soil loading rate(see Figure D-33) 0.60 9th ft2 <br /> Percolation rate 7 MPI <br /> 5. %Land Slope 4.0 96 <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow(A)by 0.83 to obtain tequired area of rock layer.Item A x 0.83= <br /> 750 gpd x 0.83 ft2/gpd= 630 ft2 <br /> 2. Determine rock layer width =0.83 ft`/gpd x Lined Loading Rate(LIR)(see LLR chart) <br /> 0.83 ft2/gpd x I 12.00 = 10.0 ft <br /> LLR Chart <br /> Pe'k Rate LLR <br /> <120 MPI <=12 <br /> >=120 MPI <=6 <br /> 3. Length of rod layer=area divided by width= <br /> 630.0 ft2 / 10.0 feet= 63.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 630.0 X 1.0 ft= 630.0 ft3 <br /> 2. Divide ft3 by 21 ft3/yd3 to get cubic yards <br /> 630.0 ft3 / 27 = 23.3 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 23.3 yd3 X 1.4 ton/yd3 = 32.7 tons <br /> Page 1 of 5 <br />