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' a� Job#� <br /> T.�...t.�.� <br /> Preoo�e...r <br /> Universit of Minnesota Mound Design Worksheet Site B <br /> Greater than 1%Slopes <br /> A FLOW <br /> Estimated 750 9Pd(see figure A-1) <br /> or rr�asu►ed x 1.5(safety factor)= U 9Pd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> ��k� �p gallons(see figure G1) <br /> Nurt�er of t�kslcompartments � <br /> Effluent Flter (yesJno) yes <br /> C-1 Septic Tank Capac�j►�Q�allons <br /> Number o� Minimum Capacity with Capacity with <br /> Bedrooms Capaaty Garb.Disp. Disp.and lift <br /> 2 a less � 11 5 :�� ;` <br /> 3or4 �,, �� Yiaz �k3 � <br /> s �� > ,, <br /> 5or6 ` ° `` <br /> # T <br /> �4 � <br /> 7,80f9 �:. ��.; i �� ��f s� .. <br /> �. .S�IL$�$it@ BV9�Uet1011 defe� <br /> 1. Dep4h to restric�ing laya= 1.5 feet <br /> 2. Deplh d percolation tests= 12 inches <br /> 3. Texture � <br /> 4. Sal bading rate(see Fgure Q33) 0.60 9P�� <br /> p�p�� 6 MPI <br /> 5. %Land Sbpe 11.0 J6 <br /> D. ROCK LAYER DIMENSIONS <br /> 1, t�uitiply average design flow(A)by 0.83 to obtain required�ea of rodc layer.Item A x 0.83= <br /> 750 gpd x 0.83 ft2/gpd= 630 ft2 <br /> 2. Detertnine rodc layer width =0.83 ft`lgpd x Linear Loading Rate(LLR)(see LlR chaR <br /> 0.83 ft�/�d x 12.00 = 10.0 ft <br /> LLR Chart <br /> Pertc Rate LlR <br /> <120 MPI <=12 <br /> >=120 MPI <� <br /> 3. Length of rock 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 rodc depth to get cubic fcet of rock <br /> 630.0 X 1.0 ft= 630.0 ft <br /> 2. Divide ft3 by 27 ft31yd3 to�t cubic yards <br /> 630.0 ft3 ! 27 = 23.3 yd3 <br /> 3. Muftiply 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 />