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- ONS.�Tt' � <br /> s�w.a�c Sibe A lot 4 block 1 Job#� <br /> TREATME7VT <br /> PROGR/�M <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 LIQUID VOLUMES <br /> Septic tank capacity 3000 gallons(see figure G1 J <br /> Number of tanks/compartments 0 <br /> Effluent Filter (yes/no) yes <br /> C-1 Septic Tank Capac"rty in Gallons <br /> Number of Minimum Capaciry with Capacity with <br /> Bedrooms Capaaty Garb.Disp. Disp.and Lift <br /> 2 orless 75fl 1125 1500 <br /> 3 or 4 1000 1500 2000 <br /> 5 or 6 1500 2250 3000 <br /> 7,8 or 9 2000 3000 4000 <br /> C. SOILS(Site evalua6on data) <br /> 1. Depth to restncting layer- 2.5 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture loam <br /> 4. Soil loading rate(see Figure D-33) p.gp gpd/ft <br /> Percolation rate 9 Mp� <br /> 5. %Land Slope 7.0 % <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 /> 750 gpd x 0.83 ft/gpd= 630 ft <br /> 2. Determine rock layer width =0.83 ft`/gpd x Linear Loading Rate(LLR)(see LLR chart <br /> 0.83 ft/gpd x 12.00 = 10.0 ft <br /> LLR Chart <br /> Perk Rate LLR <br /> <120 MPI <=12 <br /> >=120 MPI <� <br /> 3. Length of rodc layer=area divided by width= <br /> 630.0 ft/ 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 ya�ds <br /> 630.0 ft3 / 27 = 23.3 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of ra�c in tons; <br /> 23.3 yd3 X 1.4 tontyd3 = 32.7 tons <br /> Page 1 of 5 <br />