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2006-P10221 - new septic system
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2006-P10221 - new septic system
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Last modified
8/22/2023 4:30:00 PM
Creation date
6/16/2016 12:30:47 PM
Metadata
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Template:
x Address Old
House Number
510
Street Name
Deborah
Street Type
Drive
Address
510 Deborah Drive
Document Type
Septic
PIN
3111823230006
Supplemental fields
ProcessedPID
Updated
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, MOUND DESIGTV WORK SHEET(For Flows u to 1200 d) <br /> A. Average Design FLOW <br /> A-i: Estimated Sewage Flows in Gallons per Day <br /> num er o <br /> EstimateCl `� C� gpd (see figure A-1) bedrooms Class I Closs II Class III Class IV <br /> or measured — x 1.5 (safety factor) _ — gpd 2 300 225 �8o bo% <br /> 3 450 300 218 ofthe <br /> B. SEPTIC TANK Ca acit 4 �0 375 256 values <br /> p �' S 750 �150 294 in the <br /> ���c�s�t�r-�v -f�-r��-s vn� w�_ ���,f�,�; �f w� �t"iu�a-5. <br /> ���a�T w v�-s{-�_ -s��.��a-�s- �+�5-�r���. a.-i�S��,.,��-q-. 6 900 525 332 Class I, <br /> �-1 ���"� /-1���7 gallons (see fZgL{Ye C-1� L��m``n' � 1050 600 370 II, or III <br /> 8 1200 675 408 columns. <br /> C. SOILS (refer to site evaluation) C-1: Se ticTankCa acities(in allons <br /> Number of Minimum Liquid Liquid capacity with w���disposal& <br /> 1. Depth to restricting layer = a.u�a,� U a.3 feet Bedrooms Capacity garbage disposal ��ft inside <br /> 2. Depth of percolation tests = 1.0 feet 2orless �So ��z5 <br /> 3. Texture��,�A�( l.����1 3or4 �� 15� �soo <br /> 5 or 6 1500 2250 20� <br /> Percolation rate 1 5.O mPl 7,8 or 9 2000 gppp 3000 <br /> 4. Soil loading rate .�.S gpd/sqft (see figure D-33) <br /> 5. Percent land slope � % <br /> 1��• �'-.J1"D'��- �f-�fV1�� -l9G' .�'1.V '�C7JJ P � <br /> uSL. -.. �„�c� __ V L.n /o .��, <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow (A) by 0.83 to obtain required rock layer area. <br /> �So gpd x 0.83 sqft/gpd = (�a,�, sqft-� <br /> 2. Determine rock layer width = 0.83 sqft/gpd x linear Loading Rate (LLR <br /> o.s3 Sqft��a X _ �a gpd/sqft = lo ft Mound LLR <br /> 3. Length of rock layer = area = width = <br /> �_ sqft (Dl) = /o ft (D2) _ ��ft < 1 �0 M�I < � 2 <br /> E. ROCK VOLUME � 1 �O M PI G � <br /> l. Multiply rock area (D1) by rock depth of 1 ft to get cubic feet of rock <br /> ��,�� sqft x 1 ft = t���, cuft <br /> 2. Divide cuft by 27 cuft/cuyd to get cubic yards <br /> �����-- cuft = 27 cuyd/cuft = a-3 cuyd <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons <br /> �� cuyd x 1.4 ton/cuyd = 3 a tons <br /> D-33: Absorption Wid[h Siaing Table <br /> F. SEWAGE ABSORPTION WIDTH PercolationRate LoadingRate <br /> in Minutcs per Soil Tex�ure Gallons Absorption <br /> Inch per day per Ralio <br /> Mp� s uare foot <br /> Fester than 5 Coarse Sand l,yp �.pp <br /> Medium Sand <br /> Absorption width equals absorption ratio (See Figure D-33) �-oamySend <br /> ine S nd <br /> times rock layer width (D2) 16to30 �o�,,, o.�o Z,o� <br /> l 31 to 45 Silt L.oem 0.50 2.40 <br /> d�'�� X �V 1 t = ��. / f t 46 to 60 Sandy Qay Lo OAS 2.67 <br /> Silty Clay Loam <br /> 61 to 120 Silty Clay 024 S.pp <br /> Sandy Clay <br /> Ch� <br /> Slowerthan 120• <br /> •Sysicm designed(or Wese soils rtmst be oWer or pedomwnce <br />
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