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Future N Job#� <br /> Twaw7�err <br /> P�eoawwr <br /> University of Minnesota Mound Design Worksheet <br /> Irsr"WItho 1%83140e8 <br /> A FLOW <br /> Est hated 900 9Pd(��factor) <br /> of measured x 1.5(safety faactor_ � 9Pd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> tank 3000 gallons(see rigiae G1) <br /> Number of tankskwVaMalts 0 <br /> Mot Filer (yeshmo) yes <br /> C-1 Sqft Tank In t3allons <br /> Number of MiNmum Capacity with capacity whh <br /> Bedrooms CapacRy Garb.Disp. Disp.and UR <br /> or less 1125 <br /> 3 or 1500 <br /> 5 or 6 2250 <br /> 7,8 or 9 3000 <br /> C. SONS(S#&evaluellm data) <br /> 1. Depth to restricting layer= 1.4 feet <br /> 2. Depth of parmlatim tests= 12 Inches <br /> 3. Texture loam <br /> 4. W loading rate(see Figure a33) 0.60 9pd/fe <br /> Percolation rete 10 MPI <br /> 5. %Land Slope 6.0 J% <br /> D. ROCK LAYER DNENSM <br /> 1. Multiply average design flow(A)by 0.83 to obtain required area of rock layer:Item A x 0.&3-- <br /> 900 <br /> .83=900 gpd x 0.83 fe/gpd= 750 fe <br /> 2. Determine rock layer width =0.83 600 x Linear Loading Rate(LLR)(see LLR chart <br /> 0.83 fe/gpd x 12.00 1 = 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 I 10.0 feet= 75.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 750.0 X 1.0 -ft= 750.0 ft3 <br /> 2. Divide ft3 by 27 ft3/yd3 to get cubic yards <br /> 750.0 ft3 I 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 ton/yd' = 38.9 tons <br /> Page 1 of 5 <br />