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University of Minnesota Pressure Distribution System Design - 10/25/04 <br /> Al boxed rectangles must be entered,the lest will be calculated. <br /> ONs,rc <br /> SEWAGE <br /> 1. Select number of perforated laterals: 3 Tw�wr -- <br /> 2. Select perforation spacing= I 3 ft <br /> 3. Since perforations should not be placed closer that 1 foot to I ..mcly per.� a3 n <br /> 12_ <br /> the edge of the rock layer(see diagram),subtract 2 feet from <br /> the rock layer len.th ! - <br /> 38 -2 ft= 36 ft ".� .^Ka_5• <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= 36 ft/ 3 ft= 12 <br /> 5. Select perforation size 1/4 finch <br /> 6. Number of perforations is equal to one plus the number of perforation spaces(4). <br /> "Check figure E-4 to assure the number of perforations per lateral guarantees <br /> <10%discharge variation. <br /> 12 spaces+1 = 13 perforations/lateral <br /> E-4 Maximum Number of 1/4 inch perforations E-5 Maximum Number of 3/16 inch perforations <br /> per lateral to guarantee<10%discharge variation per lateral to guarantee<10%discharge variation <br /> Perforation Perforation <br /> Spacing Pipe Diameter Spacing Pipe Diameter <br /> ft 1 inch 1.25 inch 1.5 inch 2.0 inch feet 1 inch 1.25 inch 1.5 inch 2.0 inch <br /> 2.5 8 14 18 28 2.5 12 19 25 39 <br /> 3.0 8 13 17 26 3 11 18 24 37 <br /> 3.3 7 12 16 25 3.3 10 17 23 36 <br /> 4.0 7 11 15 23 4 10 16 21 33 <br /> 5.0 6 10 14 22 5 9 15 20 31 <br /> 7. A.Total number of perforations=perforations per lateral(5)times number of laterals(1). <br /> 13 perfs/lat x 3 laterals= 39 perforations <br /> B.Calculate the square footage per perforation. <br /> Recommended value is 6-10 sgft/perf.Does not apply to at-grades. <br /> 1. Rock bed area=rock width(ft)x rock length(ft) <br /> 10 ft x 38 ft= 380 ft2 <br /> 2. Square foot per perforation=Rock Bed Area/number of perfs(6) <br /> 380.0 ft2 / 39 perfs = 9.7 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 /> 39 perfs x 1 0.74 lgpm/perfs= 28.9 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 /> 2" 0.59 0.80 1.04 <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 y __,__,---,"---''''":',7n'' T-m L.,n, <br /> a-. <br /> 9. Determine Minimum Pipe Size I __ <br /> A. Manifold on End. If laterals are connected to header pipe _ �- _ <br /> as shown in Figure E-1,to select minimum required lateral FM <br /> rogue E-1: d <br /> odalEndSodom .1 „c <br /> J <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 J inches <br /> B. Center Manifold. If perforated lateral system is attached to D`oz..*:its=,----; ,;-s " _= <br /> manifold pipe near the center,like Figure E-2,perforated lateral length(3) ley, <br /> and number of perforations per lateral(5)will be approximately =one half of that in step A. Using these values,select <br /> minimum diameter for perforated lateral= 1.5 inches ,,---- <br /> I hereby certify that I have completed this work in accordance with all applicable ordinances,rules and laws. <br /> ",....... "-m------- (signature) 810 (license#) 01/07/08 (date) <br />