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1997-008955 - replace septic sys
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2245 French Lake Road - 10-117-23-22-0004
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1997-008955 - replace septic sys
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Last modified
8/22/2023 3:20:54 PM
Creation date
12/6/2016 1:42:26 PM
Metadata
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Template:
x Address Old
House Number
2245
Street Name
French Lake
Street Type
Road
Address
2245 French Lake Road
Document Type
Septic
PIN
1011723220004
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WOItKSHEL'"T <br /> � � ,� . (For Flows up to 1200 gpd) <br /> _ �.� FLO`,N Eslimatcd Scwege Fbwx in(iallons�xx day <br /> Estimated 7,s-o gpa (see pages U-7 or I-3, 4, 5) �g'"'� <br /> um r <br /> or n�ieasured gpa x 1.5 = ►��a°�ms .�,r�� ��r`�n ,�r�'u� Tiv <br /> B. SEPTIC TANK LIQUID VOLUMES 3 4 0 3a, iis o«�� <br /> 4 600 375 25G <br /> Q't-(�So gallons (see pages C-3 or C-5) s ,so oso z94 '��' <br /> �_fas� �u���--A.nK 6 9W 525 332 Trat. <br /> 7 1050 G(X) 370 �°f <br /> C. SOILS (refer to site evalualioil) g t2� ��5 aoa �d',;;,n <br /> 1. Depth to restricting layer = _ � 111C�12S Seplk7'inkC�pacilic;h�gallom <br /> Numbcr of Minimum Liquid l.iquid cepr;i�y wiih <br /> 2. Depth of percolation tests = /Z inches ����m. � <br /> qncity gnbugc dis�s�l <br /> 3. Percolation rate 11 Tll�l Z��r.� �� ��u <br /> 4. Land slope �„-� % 4 0�6 '�x' �5`", <br /> �s��u Zuo <br /> 7,Bw9 �p � <br /> '�'�Ke�s s..SWa��l.`� �w�l.f'a'� /�7� �`�fe,f:�it��i�.d. ova9 ...... <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: Daily Flow x 0.83 = <br /> `7� gpd x 0.83 sc]. ft./gpd = loz3 sc]. ft.-����Io Pl�,res�/o <br /> 2. Select width of rock layer (10 feet or less) _ _�_D It. <br /> 3. Length of rock layer = Area + Widtli = <br /> ��D sq. fl. � �� ft. _ �_ ft. Rock F3ed <br /> Sf.f.f.f:r:f:f.�f:--f:�-f:f:f T <br /> ••,••,.ti.ti.ti•ti.ti,.. .. •. •. 'L.ti.•,. <br /> f•t•1.l•1'•f l•1':j�.f.f.f�,•.r�� <br /> ti•ti•ti ti•ti•ti•ti•ti••.••.••.••. •. . 4Vidth S10 ft. <br /> •f•�•�•r•f•r•f•f•j•r.r.f.j.r•f 1 <br /> . ti.•, ti.•..ti.•..ti.. <br /> E. ROCK VOLUME 'f'r'f'''''f'''f`''"''�'"�' <br /> � Length --1 <br /> 1. Multiply rock area by rock depth to get cul�ic feet of rock; <br /> .�Q sq. ft. x�ft. _ � cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. t� get cubic yarss; <br /> �Q,_cu. ft. + 27 = d3.3 cu. ya. <br /> 3. Multiply cubic yards by 1.4 lo get weiglll uf rock iil toiis; <br /> � cu. yd. x 1.4 ton/cu. yd. _ � toils. <br /> F. ADSORPTTON WIDTH <br /> 1. Percolation rate in top 12 inches af soil is �_ mpi ^hs°�•i������wtdu�s�����K�r��i� <br /> 2. Select allowable soil loading rate from table on a � g_• PcrcoledonRa�c ca����„ Re�;oor <br /> p g � in Minules per Soii Texwre ��cr clny per A�sorpiion widih <br /> � �/rtZ Inch(MYl) syuorc fooi io itock Lxyer <br /> g� i w��i�n <br /> 3. Calculate adsorption width ratio by dividing rock layer Facterihnn0.1• co:�uso�,a .._._ __... <br /> 0.1 W S SanJ 1.20 �.pp <br /> loading rate of 1.20 gpd/ft�by allowable soil loading rate; o_��os�• PincSand•• o.�o z.�,� <br /> 1.20 gpd/f t2+-(� gPd/f tz= Z.° 6`o'.o sA��a�A�,e�n o:�o z��; <br /> i 1 .oam 0.50 <br /> Check this value on �qge E-16. 46to6� ClayLoam o.45 Z:fi, <br /> 4. Multiply adsorption width ratio by rock layer widtll lo get � `°'Z" ��°y °=24 5_`"' <br /> Slower thrn L�e <br /> 120��• <br /> required adsorption width; <br /> �x 2. ft = 0�0 ft <br /> ' � � <br />
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