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s�:w,4 Job#� <br /> Twswrr�s�rT <br /> Pieos�ewM <br /> Universit of Minnesota Mound Design Worksheet S�e B <br /> Greater than 1 X Slopes <br /> A. FLOW <br /> Estimated 750 gpd(see 6gure A-1J <br /> or measured x 1.5(safety fact�)= 0 gpd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> Septic tank r�pacity 225p gallons(see figur�e G1) <br /> Nurt�er d tanks/oanparUnents 0 <br /> Eifluent Flter (Yes/no) Y� <br /> C-1 Septic Tank C�adly M Galloos <br /> Number of Minimum Capacity with Capaaty with <br /> Bed�oort�s Garb.Disp. Disp.and LiR <br /> or less .��v:, 1 �;, <br /> =:r . <br /> 3 or 4 � 1500 �� <br /> ,'. � <br /> 5 or 6 �� 2250 �4_ . <br /> 7,8 or9 .��,�a 3000 �� <br /> C. SOILS(Sife eveluatron dafa) <br /> 1. Depth to restric�i�k�yer= 2.0 fee,t <br /> 2. Depth af pe�dabon tests= 12 inches <br /> 3. Texture loam <br /> 4. Soil loa�ng r�e(sse F'gum D,?3) 0.60 �� <br /> Peroda6on raae 6 MPI <br /> 5. %Land Sbpe 4.0 % <br /> D. ROCK LAYER D�ENSIONS <br /> 1. MulGply average design flow(A)by 0.83 to ot�ain required area a�rock layer:Item A x 0.83= <br /> 750 gpd x 0.83 il�/gpd= 630 ft� <br /> 2. Determine rodc layer widih =0.83 ft`Igpd x Line�L.oading Rate(LLR)(sae LLR chart <br /> 0.83 it�/gpd x 12.00 = 10.0 ft <br /> LLR Ch�t -- <br /> Perk R�e 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. Muttiply rodc area by rock depth to get cubic feet of rodc <br /> 630.0 X 1.0 ft= 630.0 ft3 <br /> 2. Divide fC�by 27 ft3lyd3 to get cubic yards <br /> 630.0 ft3 I 27 = 23.3 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 23.3 yd3 X 1.4 toNyd3 = 32.7 tons <br /> Page 1 of 5 <br />