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2017-00696 - septic new/replacement
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1860 West Farm Road - 27-118-23-43-0018
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2017-00696 - septic new/replacement
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Last modified
8/22/2023 4:22:39 PM
Creation date
5/7/2020 12:59:38 PM
Metadata
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Template:
x Address Old
House Number
1860
Street Name
West Farm
Street Type
Road
Address
1860 West Farm Road
Document Type
Septic
PIN
2711823430018
Supplemental fields
ProcessedPID
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' MOUND DESIGN WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. FLOW Estimated Sewage Flow in Gallons per Day(gpd) <br /> Estimated l✓:�0 gpd Number <br /> of Type I Type II 'Type III Type IV <br /> or measured - x 1.5 = gpd. Bedrooms <br /> 2 3(X) 225 180 strs <br /> 3 450 300 218 �ti <br /> B. SEPTIC TANK LIQUID VOLUMES - 4 600 375 256 <br /> 5 750 450 <br /> -"/o 0 G gallons 6 900 525 33292 <br /> 7 1050 600 370 oe"�� <br /> C. SOILS (refer to site evaluation) ��'�'�� 1 a <br /> ii 8 1200 675 408 � <br /> I <br /> m <br /> Ser <br /> 1. Depth to restricting layer = 1P, 1 i t,, 1 ` inches Number <br /> of <br /> y <br /> DIcpwal <br /> 2. Depth of percolation tests = I D . inches `"8"' (pikes) <br /> 3. Percolation rate /e, . mpi 23c,c417.000 1.lso <br /> 4. Land slope - % 'j - - - lz S ;as 21000 3000 <br /> over 9 See fig.C-6 (z 1.5) <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: A x 0.83 = , <br /> ( cD , . gpd x 0.83 sq. ft./gpd = Lic)c sq. ft.-t-lo�Sys° <br /> � <br /> 2. Select width of rock layer (10 feet or less) = 10 ft. <br /> 3. Length of rock layer = area_width= Rock Bed <br /> S'-)cl sq. ft. -- = •......-....r......-. .. <br /> i� ft. SS ft. tti.ti.�.,.ti.ti.ti.ti.,...ti.ti.,.�.,.ti.ti. <br /> :.l•l•l•I•l•l•l•l•l•r•f•1•l•f•f•l•f I <br /> ti.%.%.,"ti....%.%....ti•ti.%.%.%.%•%•%r ---h <_10 ft. <br /> S•S•ti•MT:1L•t• :4,74:ti•t•ti• 7414.1 <br /> r_l.1.l�J 1.l�l�t�l;l�t_l:l•l�f•l�1 <br /> E. ROCK VOLUME 1 --- Length �; <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; <br /> 5-n sq. ft. x )..o - ft. = s')y cu. ft. <br /> 2. Divide cu.ft.by 27 cu. ft./cu. yd. to get cubic yards; <br /> 50,1 cu. ft. =27= a 1 cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> a ) cu. yd. x 1.4 ton/cu. yd. = a./ tons. <br /> F. ADSORPTION WIDTH L OcosiAbsorption Width siting Table <br /> 1. Percolation rate in top 12 inches of soil is /o. mpi Percolation Rate Cams I.e.d <br /> Mims per <br /> inch Soil Texture 4.::::, m"`d <br /> fee ab=an <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 Coarse Sand 1.20 1.00 <br /> 0.1 to 5 Sand 1.20 1.00 <br /> GO gpd/ftp 0.1 toy Fine Sand"` 0.60 2.00 <br /> 6 to 15 Sandy Loam 0.79 132 <br /> 3. Calculate adsorption width ratio bydividingrock layer -36_to 45__._ Loam ��0 2.00 <br /> � y 31 to 45 Silt Loam 0.50 2.40 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; 46 to 60 Clay Loam 0.45 2.67 <br /> 61 tO 120 Clay 0.24 5.00 <br /> 1.20 gpd/ft2y ,(..o gpd/ft2 = Q.0 (7 . Slower than 120 Clay -- - - <br /> "Soil having 50%or more of fine or very fine sand. <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> 7.o x /o ft= a,o.D ft - <br />
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