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Slee A Job#� <br /> TwtwTMwrr <br /> rwos�ewr <br /> University of Minnesota Mound Design Worksheet <br /> joraW-than 1%Slopes <br /> A. FLOW <br /> Estimated 750 9Pd(see figure A-1) <br /> or measured I x 1.5(safely factor)= 0 9pd <br /> & SEPTIC TANK LIQUID VOLUMES <br /> Septic tank qty 3000 gallons(see figure G1) <br /> Nuniber of tanks/oompartrnents 0 <br /> Eff m t Filter (yeslno) yea <br /> C-1 Sgft Tank Cqmxty In Gdbm <br /> Neer of Minimum Capacity with lCapacity with <br /> Bedroorns Capacity Garb.Disp. Disp.and Lift <br /> or less 1125 <br /> 3 or 4 1500 <br /> 5 or 6 2250 <br /> 7,8 or 9 3000 <br /> C. SOILS(Slfe evaluation data) <br /> 1. Depth to restricting layer= 1.1 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture loam <br /> 4. Sol loading rate(see Figure D-33) 0.60 9pd/ft' <br /> Percolation rate 14 MPI <br /> 5. %Land Slope 4.0 % <br /> 0. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow(A)by 0.83 to obtain required area of rock layer.hem A x OM-- <br /> 750 <br /> .83=750 gpd x 0.83 fe/gpd= 630 ft' <br /> 2. Determine rock layer width =0.83 ft`/gpd x Linear Loading Rhe(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 /> 630.0 ft2 I 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 27 ft3/yd3 to get cubic yards <br /> 630.0 ft3 / 27 = 23.3 yd' <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 ,: ,`y'° <br />