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aOMWA::. Future Site 8 Job#� <br /> Tw�w�r�rr <br /> P�eos�ewrr <br /> University of Minnesota Mound Design Worksheet <br /> Greater than 1%Slopes <br /> A. Flow <br /> Estimated 900 gpd(see figure A-1) <br /> or measured l x 1.5(safety factor= i 0 gPd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> Soptic tank qty 9000 gallons(see figure C-1) <br /> Number of tanks/compartments 0 <br /> Effluent Filter (yes/no) Yes <br /> C-1 Septic Tank Capady►In Gallona <br /> Number of I W&-T--1 Capacity with lCapacity with <br /> Bedrooms CapaJ,,tfty _,Garb.Disp. Disp.and Lift <br /> 2 or less 1125 <br /> 3 or 4 1500 <br /> 5 or 6 2250 <br /> 7,8 or 9 3000 <br /> C. SOILS(Site evaluation data) <br /> 1. Depth to restricting laye- 1.4 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/ft' <br /> Percolation rate 9 MPI <br /> 5. %Land Slope 8.0 % <br /> D. ROCK LAYER DIMENSIM <br /> 1. Multiply average design flow(A)by 0.83 to obtain required area of rock layer.Item A x 0.83= <br /> 900 gpd x 0.83 f elgpd= 750 f? <br /> 2. Determine rock layer width =0.83 ft`/gpel x Linear Loading Rate(LLR)(see LLR chart <br /> 0.83 fe/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 /> 750.0 fe 1 10.0 feet= 75.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply nock area by rock depth to get cubic feet of rock <br /> 750.0 X 1.0 -ft= 750.0 ft' <br /> 2. Divide ft3 by 27 ft3/yd3 t0 get Cubic yards <br /> 750.0 ft3 / 27 = 27.8 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 27.8 yd X 1.4 tonlyd3 = 38.9 tans <br /> Page 1 of 5 <br />