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Pressure Distribution Design Worksheet : 1. Select Number of Perforated Laterals: 3 Geotextile w sacro Minimum►..�yf ;6►...firm§ 2. Select Perforation Spacing: 3.0 ft '/.'perforations spaced 3'apart.7,..r.., 2.of rock:•-•.2-.-.-, ,.:1,9:;....2,0e0.1- 3. Select Perforation Diameter 7/32 inc1.,,...!...!..L--; ..ter v .-. /, y_., - __ 12" ��i�a.r i�t+iraa� i i .'�.iia-� �jJ,g D' . . r.- . i 11:�, 9 of narks..jam`.it ►.j�D�.:r6( 4 Length of Laterals=Media Bed Length-2 Feet ! ., { { > '♦ 1.F'41:0--4-4. ri`:0--4?.ss-.. rl . i= a r� n nPr, 63 - 2ft = 61 ft Perforation sizing:'/.'to 4." Perforation spacing:2'to 3' 5. Determine the Number of Perforation Spaces. Divide the Length of Laterals(Line 4)by the Perforation Spacing(line 2)and round down to the nearest whole number. Number of Perforation Spaces= 61 ft 3 ft = 20 Spaces 6. Number of Perforations per Lateral is equal to 1.0 plus the Number of Perforation Spaces(line 5). Perforations Per Lateral= 20 Spaces . 1 21 Check Table I to ensure that the number of perforations per lateral guarantees less than a 10%discharge variation. Table I Maximum Number of Perforations Per Lateral to Guarantee c10%Discharge Variation '/.Inch Holes s/„Inch Holes '4 Inch Holes Perforation Pipe Diameter(Inches) Perforation Pipe Diameter(Inches) Perforation Pipe Diameter(Inches) Spacing(Feet) 1 114 114 2 3 Spacing(Feet) 1 114 115 2 3 Spacing(Feet) 1 114 114 2 3 2 10 13 18 30 60 2 12 18 26 46 87 2 21 33 44 74 149 214 8 12 16 28 54 214 12 17 24 40 80 214 20 30 41 69 135 3 8 12 16 25 52 3 12 16 22 37 75 3 20 29 38 64 128 7. Total Number of Perforations equals the Number of Pet/mations per Lateral(Line 6)multiplied by the Number of Perforated Laterals(Line 1). 21 Perforations Per Lateral X 3 Number of Perforated Laterals = 63 Total Number of Perforations Calculate the Square Feet per Perforation Table IL Table Ill Recommended value is 6-10 ft2 per perforation. Does not apply to At-Grades Volume of Liquid in Pipe Perforation Discharge(GPM) 8. Bed Area = Bed Width(ft)X Bed Length(ft) Head Perforation Diameter 10 ft X 63 ft 630 Pipe Diameter Liquid Per Foot (ft) v/a /ha r/rz V4fe (inches) (Gallons) 9. Square Foot per Perforation=Bed Area(Line 8)divided by the Total Number of Perforations(Line 7). i 0.045 1'� 0.18 0.41 0.56 0.74 630 ftt 63 perforations - 10.00 ft2/perforations 1.25 ovals 2.0° 0.26 0.59 0.80 1.04 1.5 0.110 5.0' 0.41 0.93 1.26 1.65 10. Select Minimum Average Head: 1.0 ft 2 0.170 a:Use 1.0 for dwellings using 1/4 inch or 3 0.380 3/16 inch holes. 11. Select Perforation Discharge(GPM)based on Table III: 0.56 GPM per Perforation 4 0.661 b:Use 2.0 for dwellings using 1/8 inch holes,or,for other establishments using 12. Determine required Flow Rate by multiplying the Total Number of Perforations(Line 7)by the 1/4 inch or 3/16 inch holes. Perforation Discharge(line 11). c:Use 5.0 for other establishments using 63 Perforations x 0.56 GPM per Perforation= 36 GPM 1/8 inch perforations. manifold pipe 1 13. Select Type of Manifold Connection(End or Center): Center I 14. Select Minimum Diameter of laterals based on Table I: 1.50 in ` pipe from pump Determine Volume of Distribution Piping L.► 15. Pipe Diameter of Distribution Pipe 1.50 dean outs - t in 16. Volume of Liquid Per Foot of Distribution Piping: 0.110 Gallons L.- aftegate of pipe fropump 17. Length of Laterals=[Number of Perforations per Lateral-i)X Perforation Spacing __Ckvaam --- ------ 20 - 1 )( 3.0 ft = 57 ft J manifold pipes 1 18. Volume of Distribution Piping= .,e. ' J -[Number of Perforated Laterals(Line 1)X Length of Laterals(Line 17)X ' -...'s'ac- ----- (Volume of Liquid Per Foot of Distribution Piping(line 16)] �� \= ....----_ Aiternate location 3 X r 57 I X 0.110 = 18.81 Gallons N......- of pipe from pump '�" ripe from pump I hereby certify that I have completed this work in accordance with all applicable ordinances,rules and laws. Joseph J Olson810 112912008 (Oesigner) (:---a) (License ti) (Date)