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a� Future Site Job#� <br /> ,Tw�wri��rr <br /> Pw�oorsw�r <br /> Universi of Minnesota Mound Design Worksheet <br /> Greater than 1%Siopes_� <br /> a F�ow <br /> Estimate� 900 ! 9Pd(see figwe A-1) <br /> or meas�red i x 1.5(safety factor)= 0 9Pd <br /> B. SEP'TIC TANK LIQUID VOLUMES <br /> S�C tank c�aaty 715p gallons(see figure G1) <br /> Number of t�kslcompartments 0 <br /> Effluent Fifter (yes/no) Y� <br /> G1 Seplic Tank Capadly in�allons <br /> Number of Minimum Capacity with Capaaty with <br /> Bedroorr�s Capaary �arb.Disp. Disp.and Lift <br /> 2 or less 11 5 <br /> 3 or 4 1500 <br /> 5 or 6 2250 <br /> 7,8 or 9 3000 <br /> C. SOILS(Site ev�u�ian data) <br /> 1. Deplh to resUicting layer= 1.3 feet <br /> 2. Deplh of percoi�ion tests= 12 inches <br /> 3. Te�me bam <br /> 4. Sa�bading r�e(see Figure D-33) 0.60 9Pd�� <br /> peroda6on rate 19 MPI <br /> 5. 96 Lar�d Slope 6.0 % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multi�y average design flow(A)by 0.83 to obtain required area of rock layer:Item A x 0.83= <br /> 900 gpd x 0.83 ft�/gpd= 750 ft� <br /> 2. D��mine rodc 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 <� <br /> 3. Length of roc:k layer=area divided by width= <br /> 750.0 ft2 / 10.0 feet= 75.0 ft <br /> E. ROCK VOLUME <br /> 1. Muldply rodc arna 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 ft3/yd3 to get cubic y�ds <br /> 750.0 ft3 / 27 = 27.8 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 27.8 yd3 X 1.4 toNyd3 = 38.9 tons <br /> Page 1 of 5 <br />