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sem:,as Job# <br /> Tslue^rMlNr <br /> P�eoo�eww� <br /> University of Minnesota Mound Design Worksheet <br /> Greater than 1%Slopes Lot 6 Site A <br /> A. FLOW <br /> Estimated 1050 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 C-1) <br /> Number of tanks/compartments 0 <br /> Effluent Filter (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 lessa 112514:C-4.1%V <br /> 3or4 ,11r,ir , 1500 ' ' aIi <br /> 5 or 6 .1504 2250 `� '' <br /> 7,8 or 9 d t 3000 r i 0 - <br /> C. SOILS(Site evaluation data) <br /> 1. Depth to restricting layer= 1.5 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/ft2 <br /> Percolation rate 12 MPI <br /> 5. %Land Slope 8.0 % <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 /> 1050 gpd x 0.83 ft2/gpd= 880 ft2 <br /> 2. Determine rock layer width =0.83 ft`/gpd 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 /> 880.0 <br /> 880.0 ft2 / 10.0 feet= 88.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 880.0 X 1.0 ft= 880.0 ft3 <br /> 2. Divide ft3 by 27 ft3/yd3to get cubic yards <br /> 880.0 ft3 / 27 = 32.6 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 32.6 yd3 X 1.4 ton/yd3 = 45.6 tons <br /> Page 1 of 5 <br />