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;;��;��•;;���.y of�"inn�sc:u ^;���:���� DistribG�;c^ �ysterr ��s;�n - 10/25/04 <br /> All boxed rectangles must be entered,the 2st wil!be calculated. � <br /> �Ons�rc � <br /> Scwwoc ��� . <br /> 1. Seiect number of perforated laterals: 03 TwE�Ts.erat _;---�..= <br /> PROORAM ��� <br /> 2. Select perforation spacing= �ft <br /> �,�.. _.�,.., <br /> � _.. _ - -- <br /> - .<- - - __. <br /> 3. Since perforations should not be placed closer that 1 foot to j - -- <br /> the edge of the rodc layer(see diagram),subtract 2 feet from �"� "` ��� 'J ' <br /> I I �..,.n_k � <br /> the rock layer le th ,,�.,,,;,�,,, <br /> 25 -2ft= 23� ft ���`•-5�•a- � <br /> ��..� . ri <br /> .�....�.,K�.. <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 spaang= 23 ft/ 3 ft= 7 <br /> 5. Select perforation size 1!4 inch <br /> 6. Number of perforatans is aqual to one plus the number of perforation spaces(4). <br /> `Chedc figure E-4 to assure dhe number of per/orations per lateral guarantees <br /> <10%discha�ge varfetion. <br /> 7 spaces+1 = 8 perforations/lateral <br /> E�Maximum Number of 1/4 inch perforations E-5 Maximum Number of 3H6 inch perforations <br /> r lateral to uaranbee a10'/.discha e variation r lateral to uararKee<10X discha 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 /> 8 perfs/lat x 3 laterals= 24 perforations <br /> B.Calculate the square footage per perforation. <br /> Recommended value is 6-10 sqftlperf.Dces not apply to at-grades. <br /> 1. Rock bed area=rodc width(ft)x rock length(ft) <br /> 10 ft x 25 ft= 250 ft <br /> 2. uare foot per perforatan=Rodc Bed Area/number of perfs(6) <br /> 250.0 ft� / 24 perfs = 10.4 ft�/perf <br /> 8. Determine required flaw rate by multipying the total number <br /> of perforations(6A)by flow per perforations see figure E-&) <br /> 24 perfs x 0.74 gpm/perfs= 17.8 gpm <br /> E�Perforatlon Discha 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 Tor s�gle-famiy homes. <br /> b.Use 2.0 feet for ' else ` <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 selec:t minimum required lateral �w�•.E-,:M�,�����E�«�•»�.� <br /> diameter;eMer fgure E�l or E-5 with perforation spacing and <br /> number of perforations per lateral.Seled minimum diameter <br /> for perforated laterals= 1.5 inches <br /> - --- ----_ __ ... _ _ _ ._ <br /> B. Cerrter Manifold. If perforated lateral system is attached to �E��;,;�;y;,� ' , <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 ' i <br /> one half of that in step A. Using these values,seled , . ! <br /> minimum diameter for perforated lateral= 1.5 inches • ``�- ' <br /> , I <br /> ---------J <br /> I hereb certify that I have ed this work in accordance with all applicable ordinanoes,rules and laws. <br /> (signature) 810 (license#) 01127/08 (date) <br />