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2010-00600 - mound septic system
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2675 Fox Street - 04-117-23-43-0003
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2010-00600 - mound septic system
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Last modified
8/22/2023 5:14:32 PM
Creation date
11/10/2016 12:36:41 PM
Metadata
Fields
Template:
x Address Old
House Number
2675
Street Name
Fox
Street Type
Street
Address
2675 Fox St
Document Type
Septic
PIN
0411723430003
Supplemental fields
ProcessedPID
Updated
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MOUND DESIGN WORK SHEET(For Flows u to 1200 d) <br /> 1�. •Av�rage Design FLOW <br /> A-1: Estimated Sewage F ows in Gallons per Day <br /> num er o <br /> Estirnated �_gpd (see figure A-1) bedrooms Closs I Closs II Class III Closs IV <br /> or measured �--- x 1.5 (safety factor) _�- gpa 2 300 225 t8o bo% <br /> 3 450 300 218 ofthe <br /> B. SEI'TIC TANK Capacity 4 600 375 256 voiues <br /> 5 750 450 294 in the <br /> 6 900 525 332 Class I, <br /> � -1 a�� gallons (see figure G1) `ou� � �05o boo s�o u, or iii <br /> lar�t� e�,.,) 'r'►tvtt'�'" C.N►avti1���... 'Fa�',�E=��_5�n-'��1`�5 8 1200 675 408 columns. <br /> C. SO:[L5 (refer to site evaluation) c-�: se tic Tank Ca cities(in allons <br /> � Number of Minimu Liquid Liquid capacity wilh Liquid capacity <br /> 1. Depth to restricting layer = 1� 'PO 1.7 {eet ' Bedrooms Cap city garbage disposal With disposal& <br /> lifl inside <br /> 2. Depth of percolation tests = /•t� _ feet 2 0�iess So �i25 <br /> 3. Texture ��.��( t.p�e�� 3or4 � 15� �soo <br /> s o�6 z000 <br /> Percolation rate �.�' mpi �,g o�9 �0 32� 3000 <br /> 4. Soil loading rate , �i.� gpd/sqft (see figure D-33) <br /> 5. Percent land slope � % <br /> D. ROCK LAYER DIMENSIONS <br /> l. Multiply average design flow (A) by 0.83 to obtain required rock layer rea. <br /> _�-k S� gpd x 0.83 sqft/gpd = ��'�� sqft-+ 1�a"':�� r�tc��'� <br /> 2. D�termine rock layer width = 0.83 sqft/gpd x linear Loading Rate (LLR <br /> 0.33 sqft/gpd x_ )2�., gpd/sqft = l� ft <br /> 3. Length of rock layer = area = width = OUCICI LLR <br /> �� I c� sqft (D1) = � ft (D2) = 4J ft < 120 M PI < � 2 <br /> E. ROCK VOLUME > 120 M PI < 6 <br /> 1. Multiply rock area (D1) by rock depth of 1 ft to get cubic feet of rock <br /> ,v sqft x 1 ft = � 0 cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> �- � cuft = 27 cuyd/cuft = _ I„S' cuyd <br /> 3. Mu— ltiply cubic yards by 1.4 to get weight of rock in tons <br /> � cuyd x 1.4 ton/cuyd – _ a I tons <br /> D-33: Absorp(ion Width Sizing Table <br /> F. SEWAGE ABSORPTION WIDTH Percolalion Ralc Loading Ra�e <br /> . in Minutos per oil Tex�ure Gallons Absorption <br /> Inch per dey per Ralio <br /> � MP� s uarc(ool <br /> � Fastcr than 5 oarsc Send L20 �_pp <br /> Absorption width equals absorption ratio (See Figure D-33) �diumSand <br /> amy Send <br /> Fine Sand <br /> times rock layer width (D2) 16 io 30 Lo,,,, o.� Z� <br /> It 31 to 45 Silt Loam 0.50 2.40 <br /> ���� X !Q i t = ��Q�� f t 46 to 60 S y Qay Lo 0.45 Z.67 <br /> Si ty Clay Loem <br /> 61 l0 120 Silty Clay p.zq <br /> 5.00 <br /> Sandy Clay <br /> Cle <br /> slowcr than 120• <br /> •Sys�em designed for Nue ils musi be ot6er or perTarmence <br />
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