2006 - P10462 - new septic system

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Oct 19 06 09: 09a Josh Swedlund (952) 873-3292 p. 8 • PRESSURE DISTRIBUTION SYSTEM - Trenches Ll, r_u_lu_If1 Y\I 172.ll l!>,l�\,iitt<1 P17' _� 1.2 All boxed rectangles must be entered,the rest will be calculated. 1 k 1',rr slot.$;i f 1.T'-'1/4" 1'a:rt!+F.:ia-ing;1.5'- 1. Select number of perforated laterals: 3 2. Select perforation spacing = 3 ft E4: Maximum dlonableryntbetaf1/4-Inchperforations pet lotted lo guarantee<10%discharge vadalion 3. Since perforations should not be placed closer that 1 foot to "specienerati"' spacing the edge of the rock layer(see diagram), subtract 2 feet from cher? _i Inch 1.25 IncIL, 1.5 Inch 2.0 Inch the rock la er len•th 2s e 14 18 28 50 -2ft= 48 ft 30 a 13 17 26 rock layer length 4.0 7 11 I 15 23 L 5.0 1 6 10 _v 14 22 4 Determine the number of spaces between perforations. Divide the length (3) by perforation spacing(2) and round down to nearest whole number. Perforation spacing= 48 ft/ 3 ft= 16 spaces 5. Number of perforations is equal to one plus the number of perforation spaces (4). *Check figure E-4 to assure the number of perforations per lateral guarantees < 10% discharge variation. 16 spaces+ 1 = 17 perforations/lateral 6. A. Total number of perforations = perforations per lateral (5)times number of laterals(1). 17 perfs/ lat x 3 laterals = 51 perforations E-6: Perforolion D:lischorcjo in upm B. Calculate the square footage per perforation. Should be 6-10 s ft/ etf. Does not I to at-grades. per fora lion diameter q P apply 9 head Qr';ches) • 1. Rock bed area= rock width (ft) x rock length (ft) (feet) 1/8 3/16 7?32 114 10 ft x 50 ft= 500 ft2 "Loa 0.18 0.42 0.56 0.74 2. Square foot per perforation = Rock Bed Area/number of perfs (6) 500.0 ft2 / 51 perfs = 9.8 ft2/ pert 2'0b 0.26 0.59 ' 0.80 1.04 5.0 0.41 0.94 1.26 1.65 7. Determine required flow rate by multiplying the total number 1i1,_1.0 r,_._,11,f^.11,,;:..1:,,-III,: 1CcTF1'8. of perforations(6A) by flow per perforations(see figure E-6) 0s.;, t 1l r,.�l'-1rn-irena.Av.,. 51 perfs x , 0.56 gpm /perfs : 28.6 gpm 8. If laterals are connected to header pipe as shown ." in Figure E-1, to select minimum required lateral ' � diameter; enter figure E-4 with perforation spacing (2) and -"' . number of perforations per lateral (5). "�:'" I Figure E-1:Manifold Located al End of System Select minimum diameter for perforated laterals = 2 inches 9. If perforated lateral system is attached to manifold pipe FlgureE--2,Man1oldLocaled , ' In the Center of the System near the center, like Figure E-2, perforated lateral length (3) and number of perforations per lateral (5)will be approximately one half of that in step 8. Using these values, select - minimum diameter for perforated lateral = inches. �-- I her certify th hay ompleted this work in accordance with all applicable ordina ces, rules and laws. (signature) ,�`,S-Og (license#) /Q /3 C) 6 (date)