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2003-P06867 - septic
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725 Ferndale Road North - 36-118-23-12-0002
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2003-P06867 - septic
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Last modified
8/22/2023 5:00:59 PM
Creation date
9/6/2016 12:09:06 PM
Metadata
Fields
Template:
x Address Old
House Number
725
Street Name
Ferndale
Street Type
Road
Street Direction
North
Address
725 Ferndale Road North
Document Type
Septic
PIN
3611823120002
Supplemental fields
ProcessedPID
Updated
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� �vIOUND DESIiGN WORK SHEET (For Flows up to 1200 gpd) <br /> ' A Average Design FLOW <br /> A-1: Estimated Sewage Flows In Gallons per Day <br /> num er o <br /> EShmated `��� gpd (see figure A-1) bedrooms Class I Class il Class III Class IV <br /> or measured x 1.5 (safety factor) = gpd 2 300 225 t so � <br /> 3 . 450 300 218 ofthe <br /> 4 600 375 256 values <br /> B. SEPTIC TANK Capacity 5 750 450 294 in the <br /> b 900 525 332 Ciass I, <br /> I -) � S� � �-��ovgallons (see figure G1) > >�� � 3�o u, or iu <br /> ��op 8 1200 675 408 columns. <br /> ��) ��.]�M'C' Lka raw���is�.. <br /> C. SOILS (refer to site evaluation) c-i: se ticTankCa acities(in allons <br /> Number of Minimum Liquid Liquid capacity with Liquid capacity <br /> 1. Depth to restricting layer = � . 3 ,a.��e3.I feet B���� Capxity garbage disposal W��disposal& <br /> lift iuside <br /> 2. Depth of percolation tests = /- o feet Zo�ies5 �so �i2s <br /> 3or4 �ppp 15� 1500 <br /> 3. Texture ��r�-� �o x�v� 5 w 6 isoo Zzso 200° <br /> Percolation rate mpi �,s o�9 a000 3000 300° <br /> 4. Soil loading rate � `-1 S �d/sqft(see figure D-33) <br /> 5. Percent land slope % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow (A) by 0.83 to obtain required rock layer area. <br /> `�c�v gpd x 0.83 sqft/gpd = � 4� sqft <br /> 2. Determine rock layer width = 0.83 sqft/gpd x linear Loading Rate (LLR <br /> 0.83 sqft/gpd x ) � gpd/sqft = /t� ft Mound LLR <br /> 3. Length of rock layer = area = width = <br /> � �L sqft (Dl) = /� ft (D2) _ '1� ft < � 2O M PI < � 2 <br /> E. ROCK VOLUME > 120 M PI < b <br /> 1. Multiply rock area (D1) by rock depth of 1 ft to get cubic feet of rock <br /> �y '� sqft x 1 ft =�cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> '��l ) cuft +27 cuyd/cuft =�_ cuyd <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons <br /> a.� cuyd x 1.4 ton/cuyd = 3g tons <br /> D-33: Absorptlon Width Slzing Table <br /> F. SEWAGE ABSORPTION WIDTH PercolationRale LoedingRate <br /> in Minules per Soil Tezture Gellons Absorption <br /> Inch pu day per Ratio <br /> MP� s uarc foo� <br /> Faster than 5 Coarse Send ],2p �,pp <br /> Medium Sand <br /> Absorption width equals absorption ratio (See Figure D-33) �gmYSend <br /> times rock layer width (D2) <br /> 16 to 30 Loem p.bp 2,pp <br /> /� 31 io 45 Silt Loam .50 2.40 <br /> a.�O' / x �� ft = a� ft <b�0 60 SandY Q.Y Q.45 2.6� <br /> Silty Clay Loam <br /> 6l to 120 Silty Clay 0.24 5.00 <br /> Sandy Clay <br /> 1 <br /> Slower than 120' <br /> •Sya�em duigned for�hue wils rtus�be qDer or per{orrrunce <br />
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