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` �Nu�� <br /> a�..d� Job#� <br /> TR...,�r.�rr <br /> P�ioo�ew� <br /> Universi of Minnesota Mound Design Worksheet <br /> Greaterthan 1%Slopes <br /> A FLOW <br /> Estimated � 750 gpd Qsee figum A-1) <br /> or measured ' x 1�.5(safety factor)_ � 9Pd <br /> B. SEPTIC TANK LIQUID VOIUMES <br /> Septic lank capacity 2250 gallons(see figure G1) <br /> Number of tanks/compaMients 0 <br /> Effluent Filter (yes/no) yes <br /> C-1 Sepfk Tank Capacity in Galbns <br /> Number of Minimum Capacity wfth Capacity with <br /> Bedrooms Capacity Garb.Disp. Disp.and Lrft <br /> 2 orless 7 1125 1500 <br /> 3 or4 1000 1500 1090 <br /> 5 or 6 ' 1500 2250 �Od <br /> 7,8 or 9 2000 3000 4000 <br /> c. sa�s�s►te e�aivana►d�ra� <br /> 1. Depth to resUiccting layer- 1.8 feet <br /> 2. Depth of percolation tests= 12 inches <br /> 3. Texture loam <br /> 4. Soil bading rate(see F'�gure D-33) 0.60 gpd/ftZ <br /> Percolation rate 6 MPI <br /> 5. %Land Slope 6.0 % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flaw(A)by 0.83 to obtain required area of rock layer:Item A x 0.83= <br /> 750 gpd x 0.83 ftZ/gpd= 630 ft <br /> 2. Detennine 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 /> Peric Rate LLR <br /> <120 MPI <=12 <br /> >=120 MPI <=6 <br /> 3. Length of rodc layer=area divided by width= <br /> 630.0 ft� / 10.0 feet= 63.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rock 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 ft3lyd3 to get cubic yards <br /> 630.0 ft3 / 27 = 23.3 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rodc in tons; <br /> 23.3 yd3 X 1.4 tontyd3 = 32.7 tons <br /> Page 1 of 5 <br />