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, - , C E'd ,�'`,�,�',1.l���� <br /> . � r4"F.�%'e<'�;::��' S` �-19 <br /> MOUI�,'D DESIGN PROCEDURE <br /> (For Flows up to 1200 gpd) <br /> A. Sewage Flow Rate I'. Pressure Distribution System <br /> See D-7 or I-3, 4, or 5, or tise <br /> metered alue; Flow a = 1• Select number of perforated � <br /> D gpo �7 laterals {� <br /> l � 2. Select perforation spacing / <br /> B. Septic Tank Liquid Volume / _ .3 ft ✓ <br /> (see C-3 or C-5) �oo a gallons ✓ <br /> 3. Select perforated lateral <br /> C. Soi1 Characteristics length; Note if manifold is <br /> at end of rocic layer, lateral <br /> 1. Depth to restricting layer length is rock layer length <br /> such as seasonally saturated less half a perforation <br /> soil, bedrock, coarse soil, / spacing. If manifold is in <br /> etc. ; i,�6 inches `� center of rock layer, lateral <br /> 2. Depth of percolation tests; length is one-half rock l�er <br /> �S inches � length less half a perforati�i <br /> spacing. Perforat d' latera� � <br /> 3. Number of percolation test / length = ��, ,�_ f� � �p ,,� <br /> holes; _� holes <br /> �/ � 2 <br /> 4. Divide lateral lengt��y perf�or- <br /> 4. Ave. p�rcolation rate; / ation spacing to get number of <br /> -_32 mPl `� perforations per lateral (� <br /> 5. Landslope = �_% ✓ ,� � � :J feet : ,3 feet = perfs <br /> � Note: last perforation must be <br /> `_ D. Rock Layer Dimensions in end cap, (see page E-14) <br /> l. Tlultiply gpd by 0.83 to 5. Multiply perforations per <br /> obtain required area,d� ��``� lateral by number of laterals <br /> rock layer; to get total number of <br /> � �23 -; <br /> ` ��'(� c�O�� gpd x 0.83 =,>��c�-ft perforations; n./ � <br /> � ' � �perfs/lat x � lats = � �D <br /> ' 2. Select width of rock layer /''' 6. Determine required flow rate <br /> (10 feet or less) _ /O feet " <br /> by multiplyin� number of <br /> 3. Lengt�2q�f rock layer = Area perforations by flo:a per � f'�;�- -.;�, <br /> - t,Tid f D s9 f.t -:�..1/n f t � erfor�tion : �,) y �'� <br /> p (see page E-17) <br /> -. �, <br /> - c� f t/ � �, � � �perfs x .7Y�PmIPerf =.��Pn���� <br /> � �OZ• .� ! — �__ <br /> E. Rock Volume �` --•'" � � 7. Select minimurn required lateral <br /> ,"""y�P diameter from table on Pa�;e E-17; <br /> 1. Multiply rock area by rock deptl enter table witli �erforation <br /> to get cubic feet of rock; (of1r3 spacing, perforation diameter, <br /> �22� � sq ft x / ft =„s�y�cu ft '� and number of perforations per <br /> j� lateral. Select min ' <br /> 2. Divide cu ft by 27 cu ft/cu,.� diameter for per � ated lat al <br /> to get cubic yards; /� �� _ /�� " inctles UIf / �/� '� <br /> 3. Multiply cubic yards by 1:4_� <br /> get weight of rock in tons; G. Basal Width <br /> ��;�cu yds x l.4 = Z , 9 t�,As��~ ' / <br /> ,,L 1. Percolation rate in top 12 � <br /> 32 � inches of soil is �.3,� mpi <br /> 2. Select allowable soil loading <br /> '- rate from table on pa�e E-16; / <br /> C.�J'� �. �� �P d/f t� ✓ <br />