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2005-P09396 - new septic
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2920 Fox Street - 04-117-23-31-0018
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2005-P09396 - new septic
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Last modified
8/22/2023 5:11:29 PM
Creation date
11/16/2016 11:05:29 AM
Metadata
Fields
Template:
x Address Old
House Number
2920
Street Name
Fox
Street Type
Street
Address
2920 Fox St
Document Type
Septic
PIN
0411723310018
Supplemental fields
ProcessedPID
Updated
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� MOUND DESIGN WORK SHEET (For Flows u to 1200 d) <br /> A. Average Design FLOW q-1: Estimated Sewage Flows In Gallons per Day <br /> num er o <br /> Estimated �5� gpd (see figure A-1) bedrooms Closs I Class II Closs III Class IV <br /> or measured � x 1.5 (safety factor) _ — gpd 2 300 225 �so � <br /> 3 . 450 300 218 ,pf the <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 /> I -�a-�d � )—JooO allons (see ure G1) � �05o boo 37o u, or�t� <br /> g fi$ 8 1200 675 408 columns. <br /> C. SOILS (refer to site evaluation) 1��'4�✓��►�'� C-l: Se ticTaokCa acities(in allons <br /> S �-s�• 1..3�4 1•� Liquid capacity <br /> a z,) Number of Minimum Liquid Liquid capacity with W��disposal& <br /> 1. Depth to restricting layer = l.S, i.�s �a.o feet <br /> Bedrooms Capaciry garbage disposal lift inside <br /> 2. Depth of percolation tests = I.� feet 2o�ies5 �so 1125 15� <br /> 3. Texture ��-,a^�t �� 3 or 4 �000 �soo 2� <br /> 5 or 6 1500 2250 3� <br /> Percolation rate I� - 1 mpi �,8°�9 z°°° 3°°° <br /> 4. Soil loading rate -H� �pd/sqft(see figure D-33) <br /> 5. Percent land slope 3 % ��s-s. <br /> c� � <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow (A) by 0.83 to obtain required rock layer area. <br /> '��o gpd x 0.83 sqft/gpd = �a sqft t►���=l�y�', <br /> 2. Determine rock layer width = 0.83 sqft/gpd x linear Loading Rate (LLR <br /> 0.83 sqft/gpd x 1�— �pd/sqft = /o ft Mound LLR <br /> 3. Length of rock layer = area =width = <br /> �� sqft (D1) = ►o ft �D2� - �� ft < 120 M PI < 12 <br /> E. Rocx voLUME > 120 MPI < 6 <br /> 1. Multiply rock area (D1) by rock depth of 1 ft to get cubic feet of rock <br /> l��_sqft x 1 ft = ���cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> lo�s � cuft T 27 cuyd/cuft = aa� cuyd <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons <br /> �� cuyd x 1.4 ton/cuyd = 35� tons <br /> D-33: Absorption Width Sizing Table <br /> F. SEWAGE ABSORPTION WIDTH Percolotion Rale Loeding Rate <br /> in Minutes per Soil Teznuc Gallons Absorption <br /> lnch per day per Ralio <br /> MPI s uerc(oot <br /> Fastcr than 5 Coersc Send 1.20 1.00 <br /> Medium Sand <br /> Absorption width equals absorption ratio (See Figure D-33) �`my S°"d <br /> times rock layer width (D2) 16 io 30 ��, o.� �.� <br /> 31�0 45 Silt l,oam 0.50 2.a0 <br /> �-�X �� f t = ac�•� f t 46 l0 60 $.ndy C7■y 0.45 2.67 <br /> s�u ci.y tA.m <br /> 6��o i2o s�i�y Ciey o.za s.00 <br /> Sandy Gay <br /> la <br /> Slowcr ihan 120' <br /> 'Synem desiFned for Nue wlls rt.�q be aber or perform.xe <br />
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