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2002-P05491 - repair septic system
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2002-P05491 - repair septic system
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Last modified
8/22/2023 4:16:38 PM
Creation date
4/10/2017 3:12:21 PM
Metadata
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Template:
x Address Old
House Number
1100
Street Name
Knoll Manor
Street Type
Road
Address
1100 Knoll Manor Road
Document Type
Septic
PIN
2611823310012
Supplemental fields
ProcessedPID
Updated
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'� 1VlO�J1�iD DESIGN WORK SHEET(For Flows u to 1200 d) <br /> A. . Average Design FLOW A-1: Eriimated Sewage Fiows In Gallons per Day <br /> nu r o <br /> Estimated 7sC� gpd(see,figure A-1) bearooms c�ass� c�ass u c.�.ass�n c�au�v <br /> 2 300 225 180 60% <br /> or measured — x 1.5 (safety factor) _ — gpd 3 �p 3pp 2t 8 of tne <br /> 4 600 375 256 values <br /> B. SEP'TIC TANK Capacity 5 �50 450 294 in the <br /> 6 900 525 332 Class I, � <br /> �'���u� 7 1050 600 370 II,o�Iil <br /> a - /v�c� gallons(see figure G1) lvvo g.,.� 8 �� 675 408 co�umns. <br /> ��5�5-C��V -���5 �v�'"�? <br /> �1��A-w�Si� <br /> C. SOILS (refer to site evaluation) G�: s «Teorc� �tn� ib� <br /> Number of Mialnnm I.iqnid liqnid capacity wit� ���� <br /> 1. Depth to restricting layer= /.S�►- a.� feet B� �'�' �� urc��a� <br /> 2. Depth of percolation tests = feet • � �. s a� �so uas �� <br /> 3. Texture L�A-�i L-v�-v� 3«a �000 �soo � <br /> S or 6 1500 2?SO �p <br /> Percolation rate 3c9•v mpi �ss���. ��8°f 9 � � <br /> 4. Soil loading rate .�.S gpd/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 /> � S� gpd x 0.83 sqft/gpd = c�a a sqft <br /> 2. Determine rock layer width= 0.83 sqft/gpd x linear Loading Rate (LLR <br /> 0.83 sqft/gpd x / �, gpd/sqft= /o ft Mound LLR <br /> 3. Length of rock layer = area+width= <br /> �aa Sq�(Dl� j /o � �2� -�ft < 120 M PI <12 <br /> E. ROCK VOLUME � � 2O M PI < 6 <br /> 1. Multiply rock area (Dl)by rock depth of 1 ft to get cubic feet of rock <br /> �az sqft x 1 ft= (oa�.cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> �_cuft +27 cuyd/cuft= _ a'�_cuyd <br /> 3. Muliiply cubic yards by 1.4 to get weight of rock in tons <br /> a3 cuyd x 1.4 ton/cuyd = 3� tons <br /> D-33: Absorptlon Width SfziaglLble <br /> F. SEWAGE ABSORPTTON WIDTH �"`°''�°"�"` `'oid�'`�"` <br /> in Miwtes per Sal Textwe Gdlais Absmpuo� <br /> Ineh pa day RaGo <br /> MPI wre oot <br /> Fasur th�n S Cwrse Su�d 1.20 1.00 <br /> Mediwo Sand <br /> Absorpdon width equals absorption ratio (See Figure D-33) `AiiySind <br /> times rock layer width (D2) � <br /> ,�o as ut <br /> ,(p� �( 1� ti= o[� -/� tt 46 to 60 Si1tdYQay Lwm 0.45 2.67 <br /> ll 1 <br /> 61 l0 120 Silty C9�y 0.24 5.00 <br /> S�nd Qq <br /> lowa ihan 1 <br /> •Sraam dadywA for Wae�oik mw M ahr a pafam�ebe <br />
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