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septic design-2005
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Last modified
8/22/2023 4:42:02 PM
Creation date
4/12/2017 9:09:53 AM
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x Address Old
House Number
140
Street Name
Kintyre
Street Type
Lane
Address
140 Kintyre Lane
Document Type
Septic
PIN
3211823430019
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� �"'•"'� Job#� <br /> a..�� <br /> -r....►„»� <br /> P�eos�ewM <br /> Universi of Minnesota Mound Design Worksheet Site A <br /> Greaterthan 1X Siopes <br /> A. FLOW <br /> � 750 9Pd(see figure A-1) <br /> or measured x 1.5(sa�ety facto�= 0 9pd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> ��k�� �p gallons(see figu►e G1) <br /> Numbe�d tanks/oort�tments 0 <br /> Eifluent Fdter (Yes/no) f� <br /> C-1 Sepdc Tank Capadly in C�Ilons <br /> Number of Minimum Capacity with Capacity witl► <br /> gedroorr�s Capaiaty Gar�.Disp. Disp.and Lift <br /> a� :: � 1 25 �� h <br /> 3or4 � �. hr° �� 1500 �,�����' <br /> �`�'' i J '�'�". Jf. <br /> 5 or 6 � 2250 �� �,���k' <br /> 7,8or9 � <br /> �, � ����r <br /> C. SOILS(Site evalu�ion d�a) <br /> 1. Deqh to restricting layer= 1.8 feet <br /> 2. Oeqh d perodation tests= 12 inches <br /> 3. Texlure � <br /> 4. Soil loa�ng ra�e(see Figure D-33) 0.60 9P�� <br /> p�ap��; 6 MPI <br /> 5. 96 Land Slope 9.0 96 <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Muldply average design flow(A)by 0.83 to obtain required�ea of rodc layer:Item A x 0.83= <br /> 750 �d x 0.83 ftZlgpd= � � <br /> 2. Detertrrine rock layer width =0.83 ft`/gpd x Linear Laeding Rate(LLR)(see LLR chart <br /> 0.83 fl�/gpd x 12.00 = 10.0 ft <br /> LLR q�art <br /> Pe�k Rate LLR <br /> <120 MPI <=12 <br /> >=120 MPI <� <br /> 3. Length of rock layer=area divided by width= <br /> 63p,p ft/ 10.0 feet= 63.0 ft <br /> E. ROCK VOLUME <br /> 1. Mu�iply rodc area by rodc depth to get cubic feet of rodc <br /> 630.0 X 1.0 ft= 630.0 ft3 <br /> 2. Divide ft3 by 27 ft3tyd3 to get cubic yards <br /> 630.0 ft 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 />
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