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2012-00802 - septic
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2012-00802 - septic
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Last modified
8/22/2023 4:30:06 PM
Creation date
6/15/2016 10:44:01 AM
Metadata
Fields
Template:
x Address Old
House Number
420
Street Name
Deborah
Street Type
Drive
Address
420 Deborah Drive
Document Type
Septic
PIN
3111823230009
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORK SHEET (For Flo���s u to 1200 d) <br /> A. Average Design FLOW A-1: Estimated Sewoge Flows in Gallons per DQy <br /> ! <br /> num er of <br /> Estimated t� � � g�d (sec figure A-T) bedrooms ciass i ciass n cias5 ui ciass iv <br /> or measured x 1.� (safety factor) _ �� 2 300 225 �so 60°� <br /> 3 450 30C � 218 o�tne <br /> 4 600 375 256 values I <br /> B. SEPTIC TANK Capacity 5 `7�"�6' 450 � 294 in the <br /> b 900 525 332 Ciass i. <br /> J 4� t�,J�2.SC allons (see fi icre C-1) � toso boo 3�� !�, o� �,� <br /> � — � . � 8 1200 675 408 columns. <br /> C. S�IL,S �YEfPI' t0 SItE' �'i,�fl�2lllft0il� C-1: tie tic"i'unkCa�acilies(in�aliuns) <br /> " LiquiJ tap�cin� <br /> \umber of �1linimum L:quid Liq�id�apa:it��uid� �s-ith disposal�A <br /> 1. Depth to restricting layer = ,� %2- feet 6e�nwms Capacin� �:ul�age disposal ������iside <br /> 2. Depth of percolation tests = � feet '���i�SS %�o i�'-� �i;oo <br /> 3. Texture .�:c,E( "„a i000 '_ ��oo �°00 <br /> �or 6 I�00 "-�� 3000 <br /> Percolation rate �_ mpi %�s��� '-� �000 � „�, <br /> 4. Soil loading rate �S�l gpd/sqEt (�ee figtrrE� D-33) <br /> 5. Percent land slope C,� % <br /> D. ROCK LAYEIZ DIMENSIONS <br /> 1. Multiply average design flow (A) by 0.�3 to obtain required rock layer area. <br /> �, oa gpd x 0.83 sqft/gpd = �_ sqft d �a�a - .5,,"� <br /> 2. Determine rock layer �vidth = 0.83 sqft/g-�.�d x linear Loading Rate LLR) <br /> o.s3 sqft�g��a X /-� �d�Sqft = ro Et Mound LLR <br /> 3. Len�th of rock layer = area = width = <br /> J�._sqft (D1) = �_� ft (D2} _ .5�� ft � 120 M Pl <� 2 <br /> E. ROCK VOLUME > 120 M P! < b <br /> 1. Multiply rock area (D1) by rock depth of 1 ft to get cubic feet of rock <br /> .��� sqft x 1 ft = �.�D cuft <br /> 2. Divide cuft by 27 cuft/cttyd to get cubic yards <br /> ,.�5� cuft = 27 cuyd/cuft = �J cuyd <br /> 3. Muitiply cubic yards by 1.4 to get weight of rock in tons <br /> [%'T cuyd x 1.4 ton/cuyd = �� tons <br /> D-33: AbSor�Hion Wid[h Siriiig"CHble <br /> F. SEWAGE ABSORPTION WIDTH Pcrculauun F2aic L�����„b k�«� <br /> in hLnulcs�x:r Suii�1'cs!urc Gallons AbsurPlion <br /> hich per duy par Ratiu <br /> (MPIi �4uarc liwt <br /> Fasicr Oimi 5 Cuarsc SanJ L20 1.00 <br /> Mcdiwn Sund <br /> Absorption width equals absorption ratio (See Figure D-33) ��:"»>sy«s <br /> times rock 1 a.yer ��idth (D2) -- _ F�"�.°°� __ <br /> � <br /> 16 Iu�D Lcwm 0.60 ?.DO <br /> . � [ 2' — 31 to 45 Silt Lu:mi ---- 0,50 ?.40 .. <br /> _ � � 1 t = �/ `�.� _ .t J6(u 60 S:uxlv Cl:iy Luan U.45 '_'.67 <br /> Siltv Clay Lt�m <br /> ru <br /> 61 to t20 Sliy C��y 0.23 5.(� <br /> Sandy Clay. <br /> Clav ---- <br /> Siuwcr�han 120* <br /> •$��.i�ii�J.�icn.J lur�I�e.c.���I�nw.i Ik��il�.•r�x ry�rl��mvrw:i <br />
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