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septic info including 1995 design
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1680 North Farm Road - 27-118-23-44-0010
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septic info including 1995 design
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Last modified
8/22/2023 4:23:20 PM
Creation date
9/29/2017 12:51:26 PM
Metadata
Fields
Template:
x Address Old
House Number
1680
Street Name
North Farm
Street Type
Road
Address
1680 North Farm Road
Document Type
Septic
PIN
2711823440010
Supplemental fields
ProcessedPID
Updated
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' ' MOUND DESIGN WORKSHEET <br /> (For Flows up to I200 gpd) <br /> . , <br /> A. �.OW Estimued Seavage Floa in Gallons pet Day(gpd) ,_.� <br /> Estimated (�o o gpd <br /> N�nba <br /> or measured - x 1.5 = -- gpd. B�°�,,,, �"�t �II �'�rti ry�rv � <br /> s 30o Zu �so � <br /> 3 450 300 218 °� <br /> B. SEPTTC TANK LIQUID VOLUMES � 4 s� 3�5 �w .� <br /> s �so aso ZQa � <br /> 6 900 525 332 � <br /> �-/oo u g ons � iaao boo s�o '°�' <br /> �w�u�'-� $ izoo b�s aos <br /> �a.�,. <br /> C. SOIIS (refer to site evaluation) �'�"'�"° �K`' <br /> � x,�� �� �,�. <br /> 1. Depth to restricting layer = ����T� a�" inches �� �,,. �, <br /> 2. Depth of percolation tes�s = 1 a inches `"'°m" `""m`' <br /> 3. Permlation rate `%- �' mpi Z3 a 4� i'.oso°o i;'so�o <br /> 5 a 6 1�00 2,150 <br /> 4. Land slope 3 % ,a g z000 3,000 <br /> wc 9 See fig.C-6 (z 1S) <br /> D. ROCK LAYER DIlvIIIVSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer:A x 0.83 = , <br /> (<oo . gpd x 0.83 sq. ft./gpd = L�� sq. ft.-� ��`��=Sy�II <br /> 2. Select width of rock layer (10 feet or less) _ �o ft. <br /> 3. Length of rock layer = area=width = R�g� <br /> <4-� sq. ft. i �o ft. = SS ft. ti.1•ti•�•�•ti.ti•�H•L•\•��•�•�•ti•ti <br /> J•J•t•/•1N•t•!•t•1•r•1•f•r•r•f• <br /> {•�•ti•ti•1•tiK•'L�t•ti•ti•ti•1•ti•ti•'L•ti <br /> �•r•r•r•�•r•r.r•f r•r•r•r•f•r•r• idth 510 ft. <br /> ti.ti•ti•ti•ti•ti•ti•ti•ti•ti.ti•ti•ti•ti•ti•ti•ti <br /> . �f�f;��f1��:yf���`r��tf��'�f�f1f� <br /> �r�f�f'f�r�f�r�r�r�f�r�r�f�/��'r�f� - <br /> E. ROCK VOLUME �- �'g�' � ` <br /> 1. Multiply rock azea by rock depth to get cubic feet of rock; �� � <br /> :y� sq. ft. x l,o_� ft. _�cu. ft. <br /> 2. Divide cu. ft.by 27 cu. ft./cu. yd. to get cubic yards; <br /> ��u cu. ft. -:-27= m � 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. ADSORI'TION WIDTH L�`� l-o'�'''� ,�,o e�,wa�,s�, Tible <br /> 1. Percolation rate in top 12 inches of soil is�_mpi r�a,e,�,�,� �°� a�°°d <br /> Soil Texture '°d`°ed <br /> lNi�,veh pe+i:+ct, p�p,,.. .;ad,� <br /> (mpil r.a ,�a <br /> M <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 0�sand 1.20 l.00 <br /> 0.1 to� Sand 1.20 1.00 <br /> ,y " gpd/ft� 0.1 to� Fne Sand*' 0.60 2.00 <br /> 6 to 15 ndy Loam 0.79 1 S2 <br /> 3. Calculate adso Hon width ratio b dividin rock la er 16 to�o [.oam o.60 2.00 <br /> � Y g }' 31 to 45 Silt Loam OSO 2.40 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; �-to bo c�a L,oan, C0.45i 2.6� <br /> 1.20 gPd/ft2-'.- ��I gpd/ft� = a.la� Slowerthan 120 Clay 0.24 �.00 <br /> "Shc having 509G or maee of fu�e or vcy fiae saad <br /> 4. Multiply adsorption width ratio by rock layer width to get <br /> required adsorption width; <br /> .:�,�� x lo ft= �L•� ft <br /> .J <br />
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