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2005 - P09111 - new septic
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1101 Willowbrook Dr - 26-118-23-41-0003
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2005 - P09111 - new septic
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Last modified
8/22/2023 4:18:29 PM
Creation date
2/19/2020 10:06:17 AM
Metadata
Fields
Template:
x Address Old
House Number
1101
Street Name
Willowbrook
Street Type
Drive
Address
1101 Willowbrook Drive
Document Type
Septic
PIN
2611823410003
Supplemental fields
ProcessedPID
Updated
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PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> 1. Select number of perforated laterals 3 Quarter inch perforations^apaced®3' <br /> U / ' 'i 9'':ofrock <br /> 2. Select perforation spacing= `3 • ft :' . 'j - <br /> Perf Sizing 3/16"-1/4" <br /> 3. Since perforations should not be placed closer than 1 foot to Perf Spacing 1.5'-5' <br /> the edge of the rock layer(see diagram),subtract 2 feet from <br /> the rock layer length. E-4: Maximum allowable number of 1/4-inch perforation <br /> 5per lateral to guarantee<10%discharge variation <br /> Rock layer length -2 ft = ^z3 ft perforation <br /> 4. Determine the number of spaces between perforations. spacing <br /> • Divide the length(3)by perforation spacing(2) and round (feet) 1 Inch 1.25 Inch 1.5 inch 2.0 inch <br /> down to nearest whole number.. <br /> 2.5 8 14 18 23 <br /> Perforation spacing= S 3 ft+ 3 ft= 1 ') spaces 3,0 ` 8 13 17 26 <br /> 3.3 7 12 16 ' 25 <br /> 5. Number of perforations is equal to one plus the number of <br /> perforation spaces(4)..Check figure E-4 to assure the number of 4.0 7 11 15 23 <br /> perforations per lateral guarantees <10%discharge variation. 5.0 6 10 14 22 <br /> l') spaces+1 = 15' perforations/lateral E-6: Perforation Discharge In gpm <br /> 6. A. Total number of perforations= perforations per lateral (5) perforation diameter <br /> times number of laterals (1) l� head (inches) <br /> 1)�1 't ) . (feet) 3/16 7/32 1/4 <br /> 1 CS perfs/lat x ' • lat= scLi, perforations 1.0° 0.42 0.56 0.74 <br /> B. Calculate the square footage per perforation. 2.Ob 0.59 0.80 1.04 <br /> Should be 6-10 sqft/perf. Does not apply to at grades. <br /> Rock bed area= rock width (ft) x rock length(ft) 5.0 0.94 1.26 1.65 <br /> /c) ft x cS ft= SS0 sqft • a use 1.0 foot for single-family homes. <br /> Square foot per perforation=Rock bed area +number of perfs (6) b Use 2.0 feet for anything else. <br /> Sca sqft+ .cel perfs= 10."2--• sqft/perf <br /> MANIFOLD LOCATED AT ENO Of PRESSURE DISTRIRUTNOT, •YS*ri <br /> 7. Determine required flow rate by multiplying the total number of <br /> perforations (6A) by flow per perforation(see figure E-6) M.;,; <br /> I)t•i,,I <br /> EL perfs x I'7 9 gpm/perfs= 4 gpm " <br /> mom <br /> r IM <br /> 8. If laterals are connected to header pipe as shown on upper _ /�„ <br /> example,to select minimum required lateral diameter;enter °,,,(r°"""`o �`"'" <br /> figure E-4 with perforation spacing (2) and number of perforations \,. - <br /> per lateral (5) Select minimum diameter for <br /> L•TOVT&KRTDIATED IHC LATERALS TOP <br /> PRESSURE OISTRIEVTION W MOUND <br /> perforated lateral= inches. <br /> KNOWN PLASTIC PRE <br /> 9. If perforated lateral system is attached to manifold pipe near �KATw.IeNR „0,,.��° <br /> the center,lower diagram,perforated lateral length(3) and `"l ms`s'""` '''�"! """� <br /> number of perforations per lateral (5)will be approximately one K.-maC{IIMCD-T,W I1. - <br /> half of that in step 8. Using these values,select minimum ..\ <br /> •-I�1f�I�MT[lK <br /> diameter for perforated lateral= ) )Z inches. <br /> ,./-.... .Rw <br /> from <br /> I hereby certifyy that have c mpleted this work in accordance with/ applicable ordinances, rules and laws. <br /> / ., 4 ,_:__".-....\ `2,7 1../ /1:.,.....n.,44\ 1 1 - 7 -/`1'7- /d-L-N <br />
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