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4185 Bayside Road - 06-117-23-14-0012
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Last modified
8/22/2023 5:23:37 PM
Creation date
9/1/2015 2:25:58 PM
Metadata
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Template:
x Address Old
House Number
4185
Street Name
Bayside
Street Type
Road
Address
4185 Bayside Rd
Document Type
Septic
PIN
0611723140012
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�VIOUND DESIGN WORI< SHEET(For Flows u t 1200 d) <br /> A. Average Design FLOW � -1: Estimated Sewage Flows in Gallons per Day <br /> 3'��`�'+�-��'�^'\ � num er of <br /> Eshmatea ��� gpd (see figure A-1) � bedrooms Class I Class 11 Class III Class IV <br /> or measured --�' x 1.5 (safety factor) _ � gpd 2 300 225 �eo eo% <br /> 3 450 300 218 ofthe <br /> 4 600 375 256 values <br /> B. SEPTIC TANK Capacity 5 750 450 294 in the <br /> 6 900 525 332 Class I, <br /> `a�--1�Q� gallons (see figure G1) ' � �05o boo s�o ii, o�ui <br /> 8 1200 675 408 columns. <br /> /ocsc9 ��) �i��h� �-1�►�,n'l��'�:.. i - <br /> �. SOILS (refCY t0 Stte CYJAIUQt10tt� , C-]: Se tic Tank Ca acilies(iu allons <br /> �N�' 1�z..�-►S Liquid capacity <br /> Number of Minimum Liquid Liquid capacity with with disposal& <br /> 1. Depth to restricting layer = I-3 �1,� feet Bedrooms Capacity garbage disposal �if1 inside <br /> 2. Depth of percolation tests = /. U feet zo�ie55 �so >>zs �soo <br /> 3. Texture �-�-�`� �..c�►��►'1 3o�a i000 isoo Zo� <br /> S or 6 1500 2250 3000 <br /> Percolation rate I�• �` mpi �,s o�9 z000 3000 <br /> 4. Soil loading rate • �t a gpd/�qft (see figure -33) <br /> 5. Percent land slope `� % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. NYultiply average design flow (A) by 0.83 to obtain equired rock layer area. <br /> �-is� gpd x 0.83 sqft/gpd = �?� sqft-► ��=�t�Q`a! <br /> 2. Determine rock layer width = 0.�3 sqft/gpd x linea Loading Rate (]LLR � <br /> 0.83 sqft/gpd x 1�_. gpd/sqft = c� ft �ound LLR <br /> 3. Length of rock layer = area =width = __ _ <br /> '--!t� sqft (Dl) = � ft (D2) = �� ft < 120 M Pi < � 2 <br /> E. I�OCK VOLUME > 120 M PI < 6 <br /> 1. Multiply rock area (D1) by rock depth of 1 ft to get cubic feet of rock <br /> '-� ! t� sqft x 1 ft = '-�!o cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cuUic yards <br /> �, � cuft = 27 cuyd/cuft = ! ,; cuyd <br /> 3. Multiply cubic yards by 1.4 to get weight of rock i tons <br /> I� cuyd x 1.4 ton/cuyd = a � tons <br /> D-33: Absorption Width Sizing Table <br /> �. 5Eti0lAGE ABgOIZPTION WIDTH PercolationRate LoadingRete <br /> in Minutes per Soil Tezture Gallons Absorption <br /> Inch per day per Ra�io <br /> MPI s uarc foot <br /> Faster than�5 Coarse Sand 120 I.00 <br /> � Medium Sand <br /> Absorption width equals absorption ratio (See Figure D 33) 'Fi c Sannd <br /> times rock layer width (D2) ,b�o3o Loam o.�o �.o� <br /> i 31 to 45 Silt L.oam 0.50 2.40 <br /> �,�(s�� X �'0 ft — �f�i.� f t � a6�o bo Sandy Clay Lo o.as z.6� <br /> Silty Clay Loam <br /> 61 io 120 Silty Clay 0.24 5.00 <br /> � Sandy Clay <br /> Cla <br /> � Slowcr than 120• <br /> I °Sysiem dui@ned far these soils Irws�be ott�er or performence <br /> I <br />
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