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Aug 04 07 07: 48a FYLE 'S 7632955422 p. 5 <br /> Job# <br /> Prooawaur <br /> University of Minnesota Mound Design Worksheet <br /> Greater than 1%Slopes <br /> A. FLOW <br /> Estimated 600 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 Fitter (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 - '4`r 1125 -r <br /> 3 or 4 K< i ;' 1500 ' $t <br /> • <br /> 5or6 �, t'At:' 2250 <br /> 7,8or9 ga �,.y 3000 @<< <br /> C. SOILS(Site evaluation data) <br /> 1. Depth to restricting layer= 2.1 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture loam <br /> 4. Soil loading rate(see Figure D-33) 0.60 9Pdl ft2 <br /> Percolation rate 10 MPI <br /> 5. %land Slope 9.0 % <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 /> 600 gpd x 0.83 ft2/gpd= 500 f <br /> 2. Determine rock layer width =0.83 ft`lgpd 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 /> 500.0 ft2 / 10.0 feet= 50.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 500.0 X 1.0 ft= 500.0 ft3 <br /> 2. Divide ft3 by 27 ft3lyd3 to get cubic yards <br /> 500.0 ft3 / 27 = 18.5 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 18.5 yd3 X 1.4 ton/yd3 = 25.9 tons <br /> Page 1 of 5 <br />