Laserfiche WebLink
SEWAGC �'�. Future Site Job# 11 <br /> TREATMlNT '�� <br /> PROaRAM \"� <br /> University of Minnesota Mound Design Worksheet <br /> Greaterthan 1°�Slopes <br /> A FLOW <br /> Estimated 900 gpd(see figure A-1) <br /> or measured x 1.5(safety facta}= 0 gpd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> Septic tank capacity 3000 gallons(see figu►e G1) <br /> Number of tanks/compartments 0 <br /> Effluent Filter (yes/no) yes <br /> C-1 Septic Tank Capacity in Gailons <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 or6 1500 2250 3000 <br /> 7,8 or 9 2000 3000 4000 <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. Sal loading rate(see Figure D-33) 0.60 gpd�ft� <br /> Percdation rate 11 MPI <br /> 5. %Land Slope 3.0 °lo <br /> D. ROCK LAYER DIMEN510NS <br /> 1. Multiply average design flow(A)by 0.83 to obtain required area of rodc layer:Item A x 0.83= <br /> 900 gpd x 0.83 ft/gpd= 750 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 <=6 <br /> 3. Length of rodc layer=area divided by width= <br /> 750.0 flZ / 10.0 feet= 75.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rodc depth to get cubic feet of rock <br /> 750.0 X 1.0 ft= 750.0 ft3 <br /> 2. Divide ft3 by 27 ft3tyd3 to get cubic yards <br /> 750.0 ft3 / 27 = 27.8 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rodc in tons; <br /> 27,g yd3 X 1.4 taUyd3 = 38.9 tons <br /> Page 1 of 5 <br />