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� ' � � <br /> s� �0�#1�� <br /> Tw�wtr�vT <br /> PreoswwM " <br /> Universi of Minnesota Mound Design Worksheet <br /> Groatarthan 1X Slopes <br /> a �ow <br /> Esti�ed 600 9Pd(see figu�e A-1) <br /> or measured x 1.5(safety factor)_ � 9Pd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> ��k�ty 225p gallais(see figvre G1) <br /> Nurtiber d tanksloanpartrnents 0 <br /> Etfluent Filter (yesino) YeS <br /> �1 Tank Ca�pacity in Gallons <br /> Nurtiber of Minimum Capacity with Capacity wiN <br /> 8edrooms Capacity Garb.Disp. Oisp.and Lift <br /> 2 or less 1125 . �. <br /> 3 or 4 �' 1500 <br /> 5 or 6 160� 2250 �� <br /> 7,8 or 9 2000 3000 <br /> C. SOILB(S�e eveluation deta) <br /> 1. Depth to restric�ing layer= 0.5 feet <br /> 2. Deplh d perooladon tests= 12 ir� <br /> 3. Te�xe bam <br /> 4. SoN loa�ng rate(see tigure D-33) 0.60 9P�� <br /> Peroolatlon rate 5 MPI <br /> 5. %L.and Sbpe 3.0 96 <br /> D. ROCK LAYER DIMENSIONS <br /> 1. M�lipiy average design flow(A)by 0.83 to obtain required area of rodc layer:Item A x 0.83= <br /> 600 gpd x 0.83 ft/gpd= 5(l0 _ft� <br /> 2. Deterrtdne rodc layer width =0.83 ft`/gpd x lir�ear 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 MPI <=12 <br /> >=120 MPI <� <br /> 3. Length of rock layer=area divided by width= <br /> 500.0 ft! 10.0 feet= 50.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rodc area by rock depth to get cubic feet of rock <br /> 500.0 X 1.0 ft= 500A ft3 <br /> 2. Divide ft3 by 27 ft3lyd'to get cubic ya�ds <br /> 500.0 ft3 ! 27 = 18.5 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 18.5 yd3 X 1.4 ton/yd3 = 25.9 tons <br /> Page 1 of 5 <br />