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University of Minnesota Pressure Distribution System Design - 10/25/04 <br /> . Al boxed rectangles must be entered,the rest WM be calculated. <br /> Orserrit <br /> ate■ <br /> 1. Select number of perforated laterals: 3 Tfraaa d'* <br /> r.o.n.w <br /> 2. Select perforation spacing= 3 ft <br /> Cialextik,(b... <br /> 3. Since perforations should not be placed closer that 1 foot to I -......••••••••........• fF ,Z- <br /> the edge of the rock layer(see diagram),subtract 2 feet from 9'„r mck <br /> the rock layer lengthPert Sasal <br /> 1 <br /> I 63 1-2ft= 61 ft ^R ''$=-"• <br /> 4. Determine the number of spaces between perforations. <br /> Divide the length(3)by perforation spacing(2)and round down to nearest whole number. <br /> Perforation spacing= 61 ft/ 3 ft= 20 <br /> 5. Select perforation size 1/4 inch <br /> 6. Number of perforations is equal to one plus the number of perforation spaces(4). <br /> Check figure E-4 bp assure the number of perfoatons per lateral guarantees <br /> <10%discharge variation. <br /> 20 spaces+1= 21 perforationsAateral <br /> E-4 Maximum Number of 1/4 inch perforations E-5 Maximum Number of 3116 inch perforations <br /> .: lateral to . ,: : <10%discha . : variation lateral to .uarantee<10X discha . = variation <br /> Perforation Perforation <br /> Spacing Pipe Diameter Spacing Pipe Diameter <br /> ft 1 inch 1.25 inch 1.5 inch 2.01 ch feet 1 inch 1.25 inch 1.5 inch 2.0 inch <br /> 2.5 MUNI 14 18M 2.5 12 19 25 39 <br /> 3.3 12 16 3.310 17 23 36 ,.: <br /> ,e'"�! ��&�° - r i:. ..re's,!' ��,-','.7,4k.,... 111�., . �. ,y,.. _ .�-a-- ,. <br /> 5.0 10 14 y 5 9 15 r 20 31 <br /> 7. A.Total number of perforations=perforations per lateral(5)times number of laterals(1). <br /> 21 perfs/lab( 3 laterals= 63 perforations <br /> B.Calculate the square footage per perforation. <br /> Recommended value is 6-10 sgft/perf.Does not app)r to at-grades. <br /> 1. Rock bed area=rock width(ft)x rock length(ft) <br /> 10 ftx 63 ft= 6 ft2 <br /> 2. Square foot per perforation=Rock Bed Area/numh r of perfs(6) <br /> 630.0 ft2 / 63 perfs = 104) ft2/perf <br /> 8. Determine required flow rate by multiplying the total number <br /> of perforations(6A)by flow per perforations(see figure E-6) <br /> 63 perfs x ( 0.74 Igpm/perfs= 46.6 gpm <br /> E-6 Perforation Discharge in GPM <br /> Head Perforations diameter <br /> (feet) (inches) <br /> 3/16 7/32 1/4 <br /> 1' 0.42 0.56 0.74 <br /> 5 0.94 1.26 1.65 <br /> a. Use 1.0 foot for single-family homes. <br /> b.Use 2.0 feet for anything else <br /> 9. Determine Minimum Pipe Size "; <br /> A. Manifold on End. If laterals are connected to header pipe , <br /> as shown in Figure 6-1,to select minimum required lateral Noun E-1 Piardlokl L000led a1 End of drawn <br /> diameter;enter figure E-4 or E-5 with perforation spacing and <br /> number of perforations per lateral.Select minimum diameter <br /> for perforated laterals= 2.0 inches <br /> B. Center Manifold. If perforated lateral system is attached toEM of Om 11,01wn i <br /> manifold pipe near the center,like Figure E-2,perforated lateral length(3) <br /> and number of perforations per lateral(5)will be approximately • <br /> one half of that in step A. Using these values select - <br /> minimum diameter for perforated lateral= 1.5 I inches L_ <br /> I hereby certify that I have completed this work in accordance with all applicable ordinances,rules and laws. <br /> "‘"- - (signature) 810 (license#) 9/11/2005 (date) <br />