Laserfiche WebLink
lJni�versity of�AAinne�sota �rr�essure Distribution System Design - 10i25/04 <br /> a►aoX�ed.ecr�n�es musr ae enrered,me�esr w�n ne cercu�ared. <br /> o.,s� <br /> sewaae <br /> 1. Se�lect number of perforated laterals: [�3 P�� � J <br /> 2 Select perforation spacing= �ft <br /> ��ex..��,:,�..K <br /> 3. Since perforations shouid not be ptaoed doser that 1 foot fo �r <br /> the edge of fhe ro���...c���...k{{{,,,,,,layer{see diagram),subtract 2 feet from <br /> .a.. J M..�.<.�k <br /> the rock la�rer le� ParfSi•s�w}�5/E6'-1/A' <br /> 63 -2 ft= 61 ft t������-5•-5• <br /> 4. Determine fhe number of spaces between perForations. <br /> Divide the iength(3)by perforation spacing(2)and round down to nearest whole number_ <br /> Perfora6on spacing= fi1 ft/ 3 ft= 20 <br /> 5. Seled perfora�Fion size 1!4 inch <br /> 6. Number of�ertordtions is�{ual to one plus the number of perforation spaces(4). <br /> `Check fnju�E-4 to assuie the rximberof perlo�ations per�atera/guarantees <br /> <10%drscharge va►iad�on. <br /> 20 spaces+1= 21 pertorations/lateral <br /> E-4 Maximum Number ofi 1/4 inch perforations E-5 Maximum Number of 3/i6 inch perforafions <br /> r lateral to uaraMee<10°/.d@scha e variation lat�ral to uararrtee<10'k discha e varia�on <br /> Perforafion Perfaration <br /> 5pacing Pipe DiameEer Spacing Pipe Diameter <br /> fF 1 inch 1.25 inch 1.5 inch 2.0 inch feet 1 inc� 125 inch 1.5 inch 2.0 inch <br /> 2.5 8 14 18 28 2.5 12 19 25 39 <br /> 3A 8<` ,13 9T 26! 3'; '`.1'i 18 ' z4 37 <br /> 3.3 7 12 16 25 3.3 10 17 23 36 <br /> 4.Q , 7 it 95 ti 23; 4; *10 i6 2fi 33 <br /> 5 0 s 10 .. 1� �, : � _:,.:,.9 ` 15 _ 2a 31 <br /> 7. A.Totat numl�r of perforations=perforations per laterat(�times namber of laterals(1}. <br /> 21 perfs/tat x 3 Oaterals= 63 perforations <br /> 3.Caiculate the square fnotage per perioration_ <br /> Recvmmended value is 6-10 sqNpe�f.Does not apply to at-gcades. <br /> 1. Rodc bed area=rodc width(R)x rock tengt6 fff) <br /> 10 ft x 63 ft= 630 fl <br /> 2. Square foot per perforatiar-Rock Bed Area/number of AerfsE6) <br /> 630.0 ft/ 63 perfs = 1Q.0 ft�/perf <br /> S_ Determine required tiow rate by multipiying the total number <br /> of perforallons(6A)by flow per perForations see figure E�) <br /> 63 petfs x 0.74 gpm/perfs= 46.6 gpm <br /> E-6 Perforation Disc6�a in GPM <br /> Head Perf�raGons diameter <br /> (feet inches <br /> 3/16 7/32 1/4 <br /> 1 0.42 0.55 0.74 <br /> 2° 0_59 Q.Bt� a 1.04 <br /> 5 0.94 1.26 1.65 <br /> a. Use 1.0 foot for singtafartdly homes. i <br /> b.Use 2_0 feat ftx else ' i .>,f�,�.H„ t <br /> �:i��'\�z.. � <br /> ��� �,���c•�, . <br /> iyi-� Y ,,l_ <br /> 9. Detertnine Minimum Pipe Size i c•���° _=�� �- I <br /> A Manifold o�End. If laterals are connected to header pipe i ` _-��-'"���� �p� i <br /> as shovm in Figure E-1,to select minimum requi�ed{aterai I -'�� c ,A" ; <br /> siia�meter,e�ter figure E-4 or E-5 wiEh perforation spacing and ��El:Mantloltl Load�tl ot ErW oTSysisrt� �� � <br /> nuenber of perforations per lateral.Selecf minimum diameter <br /> for pertorated laterals= 2.0 inches <br /> B. Cen�r Manifald. If perForated lafaral system is attached to ��G,p„�� _= .��� I <br /> manifold pipe near fhe center,like Rgure E-2,pertorated lateral length(3) "� -- � <br /> and number of perforations per fatecal(5)wi{I be approxirnately _ " -- __> I <br /> on�:half of that in step A. Using these vafues,select - =�``� ---= --� <br /> minimum diameter for perforated lateral= 1.5 inches -=-� ��-'"��.=;T`""" � <br /> �_- �.....n«,_ � <br /> B heref�yAertify that i have wmpieted�is woric in acxordance with aii appiicable ordinances,rule,s and laws. <br /> /� <br /> (signaturej 810 (license#) 05/11/08 (dafe) <br />