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� � Job#� <br /> Tw�tw�rr��r+T <br /> P�eoo�ewM <br /> Univers of Minnesota Mound Design Worksheet Site A <br /> Greate�than 1�6 Slopes <br /> A FLOW <br /> E.stimated 750 9pd(see figure A-1) <br /> or measu�ed x 1.5(safety factor)= 0 gpd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> Sepbc taMc c�adty 225p gallons(see figure G1) <br /> Numbe�of tanks/oompartments 0 <br /> Effluent F�ter (Yeshro) Yes <br /> C-18epfk Tank C�pacily in GWons <br /> Nurt�er a� Minimum Capacity with Capacity with <br /> Bedrooms Capaaty Garb.Disp. Disp.2nd Lift <br /> or less i f�: 125 x' <br /> 3 or 4 �� 1500 <br /> 5 or 6 �'`;�� �� 2250 1�" � <br /> 7,8 or 9 ��,arr � � ���"= 3000 <br /> C. SOILS(S�e evaluetion data) <br /> 1. Depth to reshic�ng layer- 2.0 feet <br /> 2. Deptlt d per�d2�ion tests= 12 inches <br /> 3. Texd�re bam <br /> 4. So�loec�ng rate(sae F'rgure D-33) 0.60 gpd�1t� <br /> Peroolatlon rate 6 MPI <br /> 5. 96 t.and Sbpe 3.0 % <br /> D. ROCK LAYER D�IIENSIONS <br /> 1. Muitiply average desiyn flow(A)by 0.83 to obtain required azea of rodc iay�:Item A x 0.83= <br /> 750 gpd x 0.83 ft�lgpd= 630 ft <br /> 2. Detem�ne rodc lay�width =0.83 ft%gpd x Line�Loading Rate(LLR)(see LLR c��t <br /> 0.83 ft�/gpd x 12.00 = 10.0 ft <br /> LLR Chut <br /> Perk Rate LLR <br /> <120 MPI <=12 <br /> >=1Z0 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. Mul6ply rodc area by rock depth to get cubic feet of rodc <br /> 630.0 X 1.0 ft= 630.0 ft3 <br /> 2. Divide ft3 by 27 ft31yd3 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 tonlyd3 = 32.7 tons <br /> Page 1 of 5 <br />