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'a�w s � rodcbed#1 Job#� <br /> Tw�wTMe�rt <br /> r�� <br /> Universi of Minnesota Mound Design Worksheet <br /> Greaterthan 1%Slopes <br /> a F�ow <br /> Estimated ; 750 9Pd(see figwe A-1� <br /> or measured I x 1.5(safety factqr)= 0 gpd <br /> B. SEP'TIC TANK LIQUID VOLUMES <br /> Se�tank�paary 225p gallons(see figwe G1) <br /> Nurnber of t�ksloompartrnents 0 <br /> Effluent Filter (yeslno) yes <br /> G1 Saplic Tank Capadty in Galbns <br /> Number ofi Minimum Capadty with Capacity with <br /> Beckooms Capaary G�b.Disp. Disp.and Lift <br /> 2 or less 1125 <br /> 3 or 4 1500 <br /> 5 or 6 2250 <br /> 7,8 or 9 3000 <br /> C. SO�S(Site evaluation data) <br /> 1. De�h to restricting layer= 1.5 feet <br /> 2. DepRh d per�daUon tests= 12 inches <br /> 3. Texlure day lo�n <br /> 4. Soil loading ra6e(see F'gure D-33) 0.45 9P�� <br /> Perodabon rate 6 MPI <br /> 5. %Land Sbpe 3.0 % <br /> D. ROCK LAYER DqYffNS10NS <br /> 1. Multiply average design flo�r(A)by 0.83 to obt�n required area of rodc layer.Item A x 0.83= <br /> 750 gpd x 0.83 ft/gpd= 630 ftZ <br /> 2. Determine rock layer width =0.83 ft`lgpd x Linear Loading Rate(LLR)(see LLR chart <br /> 0.83 ftZ/gpd x 12.00 = 10.0 ft <br /> LLR Chart <br /> Pe�lc Rate LLR <br /> <120 MPt <=12 <br /> >=120 MPI <=6 <br /> 3. Length of rodc layer=area divided by width= <br /> 630.0 ft/ 10.0 feet= 32.0 ft <br /> E. ROCK VOLUME <br /> 1. Mul6ply rodc area by rock depth to get cubic feet of rock <br /> 630.0 X 1.0 ft= 630.0 ft3 <br /> 2. Divide ft3 by 27 ft3/yd3 to get cubic yards <br /> 630.0 ft3 / 27 = 23.3 yd3 <br /> 3. Mufliply cubic yards by 1.4 to get weight of rodc in tons; <br /> 23.3 yd3 X 1.4 ton/yd3 = 32.7 tons <br /> Page 1 of 5 <br />