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OiNS�TE rockbed#2 Job# L4 62 <br /> SCWA6C �`;��, <br /> TREATMENT �'=�---�.=�-_�-- �'����` <br /> PROGRRM '.��_` ' � <br /> University of Minnesota Mound Design Worksheet <br /> Greater than 1%Slopes <br /> a F�ow <br /> Estimated 75Q gpd(see figure A-i) <br /> a���� x 1.5(safety factor)= 0 gpd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> Septic tank capaaty 3000 gallons(see figure G1) <br /> Number of tanks/compartments � <br /> Effluent Filter (yes/no) no <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 orless 750 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 evaluation data) <br /> 1. Depth to restricting layer- 2.0 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture loam <br /> 4. Soil loading rate(see Figure D-33) 0.60 9Pd�ft2 <br /> Percolation rate 8 MPI <br /> 5. %Land Slope 6.0 % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Mul�ply 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 ft2/gpd= 23 ftZ <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 ChaR <br /> Perk Rate L-�R <br /> <120 MPI <=12 <br /> >=12Q MPI <=6 <br /> 3. Length of rock layer=area divided by width= <br /> 23,0 {� / 10.0 feet= 23.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 23.0 X 1.0 ft= 23.0 ft3 <br /> 2. Divide ft3 by 27 ft3/yd3 to get cubic yards <br /> 23.0 ft3 / 27 = 9.0 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of roc:k in tons; <br /> g.p yd3 X 1.4 tontyd3 = 12.6 tons <br /> Page 1 of 5 <br />