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2002-P05667 - new septic
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3015 Watertown Road - 04-117-23-22-0030
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2002-P05667 - new septic
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Last modified
8/22/2023 5:09:42 PM
Creation date
7/24/2019 12:57:05 PM
Metadata
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Template:
x Address Old
House Number
3015
Street Name
Watertown
Street Type
Road
Address
3015 Watertown Rd
Document Type
Septic
PIN
0411723220030
Supplemental fields
ProcessedPID
Updated
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Mound Design Worksheet (For flows up to 1200 gpd) <br /> All boxed rectangles must be entered,the rest will be calculated. <br /> A. FLOW A l: Csknafed Sewage Ilam in Gdoru per Dav <br /> Estimated 750 gpd(see figure A-1) I Gass M bedrooms beds r m I <br /> Gass II Om III mass Ill <br /> 225 4 <br /> or measured x 1.5(safety factor)= 0 gpd I ; 8s <br /> 2 300 ; 80 6096 <br /> B, SEPTIC TANK LIQUID VOLUMES ! 3 450 ! 300 218 ctthe <br /> Septic tank capacity 2-1300 gallons(see figure C-1) 1 4 600 3 r 255 voWt <br /> 750 450 294 in the <br /> C. SOILS(Site evaluation data) 6 900 I' !25 332 <br /> 1. Depth to restricting layer=_ 1.83 feet 7 '� I l� 6011 370 1 .11 478 c1,u11I <br /> P 9 Y 1 l � �• <br /> 2. Depth of percolation tests- inches <br /> 3. Texture ILOAM <br /> 4. Soil loading rate(see Figure D-33) 0.6 gpd/ft <br /> Percolation rate I MPI <br /> 5. %Land Slope 6 % r�.�:: :11snrptluft��'IcHn�I t111g raltk <br /> f'errn4alron katr- � Londnlg It ate <br /> in Minute%pc'f 40,kNture ` 0Wkms Absllrgxml <br /> Inch 11CY Any`per k,lvn <br /> ('-L '+�c ptl(:1'�nl.e•a actties do allou5l <br /> f�Cltr{k11t15 t.` r } r. l�tnl(ll{ili�l.t5C:Idk)Ysi1a%lIfi�t, <br /> Iaer tcn.9 Cnrrsc•tind <br /> 131 1 00 <br /> V1mbur(if Millimul12 Lul"IS( Medium Sand <br /> AICl' i.oamst nd <br /> lift IISI(ic <br /> 7?; 1.ill <br /> 21 or Icss ^�) I I 1G to JU__ Loam II 6 1 y(NI <br /> I;yi .� _- -M to 4S I Sill l.. Im 1 ll it) _411 <br /> {)I'•4 1;};}',3 I +t 5115 <br /> db to GU SaruA-Chn'lxmi li 4S ?fcl <br /> 5 or 6 I`N ��•5l� if?f)I) swy f my Loam <br /> i`I'f) �I}lli 11)QQ Cary Lwm <br /> (•}to 120 14 K(1(1 <br /> tiancly C:la' <br /> Clan <br /> +ah'In•l':i•9ctlix 2!.(x x•:ii 5'.iJ'C'o9.c-. Iir.-i;.11:tC <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow(A)by 0.83 to obtain required area of rock layer:Item A x 0.83= <br /> 750 gpd x 0.83 ft"gpd= 630.0 fe <br /> 2. Determine rock layer width =0.83 fe/gpd x Linear Loading Rate(LLR)(see LLR chart) <br /> 0.83 ft?/gpd X 12 = 10.0 ft <br /> LLR Chart <br /> Perk Rate LLR <br /> <120 MPI <=12 <br /> >=120 MPI <=6 <br /> 3. Length of rock layer=area divided by width= <br /> 630 -ft' 10 feet= 63.0 feet <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 630 X 1 ft= 630.0 ft' <br /> 2. Divide ft3 by 27 ft/yd 3 to get cubic yards <br /> 630.0 -ft' / 27 = 23.3 yd' <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 23.3 yd X 1.4 ton/yd3 = 32.7 tons <br />
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