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University of Minnesota Pressure Distribution System Design -10/25104 i���L <br /> Al/bwced iecta�ks must be enteied,fAe�est wiH be calw/ated. S �� <br /> ONSIT! � <br /> $1lWp6C <br /> 1. Select number of perforated laterais 03 T'u"T"'"'T <br /> Preoaww�w _„ _, <br /> 2. Select pertoration spacing= 03 ft <br /> �.k f�,.,. <br /> 3. Since perforations should not be placed Goser that 1 foot to ��� <br /> the edge of the rock layer(see diagram},subtrad 2 feet ftom c` '�" <br /> � �._� I .�...n_� <br /> the rock layer len th <br /> 75 -2 ft= 73 ft ,��.r�>„ K.,.,.:--��.� <br /> ['cel+�aa�'��xK I.V-S <br /> 4. Determine the number of spaces befween perforations_ <br /> Divide the�ngth(3)by perforation spacing(2)and round down to nearest whole number. <br /> Perforation spacing= 73 ft/ 3 ft= 24 <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 /> `Chedc figme E�to assure the number of peiforatans per lateral guarantees <br /> <1096 discharge variation. <br /> 24 spaoes+1 = 25 perforations/lateral <br /> E-4 Maximum Number of 1/4 inch perforations E-5 Maximum Number of 3/16 inch perforations <br /> r lateral to uarantee<10%dischar e variation r laterel to uararrtee<10%dischar e variation <br /> Perforation Perforation <br /> Spacing Pipe Diameter Spacing Pipe Diameter <br /> ft 1 inch 125 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.Totat number of perforations=perforations per lateral(5)times number of laterals(1). <br /> 25 perfs/lat x 3 laterals= 75 pertorations <br /> B.Calculate the square footage per perforation. <br /> Recommended value is 6-10 sqft/perf.Dces not appy to at-grades. <br /> 1. Rodc bed area=rock width(ft)x rodc length(ft) <br /> 10 ft x 75 ft= 750 ft <br /> 2. Square foot per perforation=Rock Bed Area/number of perfs(6) <br /> 750.0 ft� / 75 perfs = 10.0 ft/perf <br /> 8. Determine required ftow rate by muftipfying the total number <br /> of perfora6ons(6A)by flow per perforations see figure E-6) <br /> 75 perfs x 0.74 gpm/perFs= 55.5 gpm <br /> E-6 Pertoration Dischar e 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-fam�y►romes. - <br /> b.Use 2A feet fw 'ng else --_`--' - , <br /> _ - <br /> . .....,. <br /> _-::-.-._=-_'-'_' �-'\% <br /> 9. Determine Minimum Pipe Size �" , -- � <br /> A. Manifold on End. If laterals are connected to header pipe , <br /> _ . ..__ ... <br /> as shown in Figure E-1,to select minimum required lateral F+c�••E-,:N��a�a�«�.�...+ <br /> diameter,enter figure E-4 or E-5 with perforation spacing and <br /> _-- _. - --- ---- - - - <br /> number of pertorations per lateral.Select minimum diameter <br /> for pertorated laterals= 2.0 inches <br /> B. Center Manifold. If perforated lateral system is attached to ���;��� ' <br /> man"rfold 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 inches - `L_ �"�`'�` <br /> _ _ , <br /> I hereby certify that I have completed this work in accordance with all applicable ordinances,rules and laws. <br /> (signature) 810 (license#) 02/11/08 (date} <br />