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' University of Minnesota Pressure Distribution System Design - 10/25/04 <br /> an ooxsd re�xen�.s m�sr ee em�ed,a,e ne��r ae caroua�eo. <br /> o»s..: <br /> � <br /> 1. Select number of perforated laterals: 03 TP�� <br /> 2. Select perforation spacing= Oft <br /> <:.. .,.���,.... <br /> 3. Sinoe perforationa should not be plaoed c�oser that 1 foot to .....=W.W."`_______...... ..___. <br /> the edge of tl�e rodc lay�er(sae diagram),subtrad 2 feet from r_y.��M*..�..�,..�P= - <br /> v�r n.4 <br /> �11@ fOCI(�8y'@f� <br /> 63 -2 ft= 61 ft ^;`*^":^*�;'"'��4- <br /> 4. Determine the number of spaces between perforations. <br /> Divide the length(3)by pertoration spadng(2)and round down to nearest whole number. <br /> Perforation spacing= 61 ft/ 3 ft= 20 <br /> 5. Sel�d pe►foration size 1/4 inch <br /> 6. Number of perforations is equal to one plus d�e number of perforation spaces(4). <br /> `Ci►eck figwe E-4 to assure tl�e number of peArNeOions per late�aJ guarantees <br /> <10%discharye variation. <br /> 20 spaoes+1= 21 perforationsflateral <br /> � um umber 1/4 inch E-6 Maximum Number of 3H6 inch perforaUons <br /> isberal to ran�se<10X disc variaffon laberal to uara�se<10X�scha variation <br /> rForation Perforation <br /> Spacing Pips Diameter Spacing Pfpe Diameter <br /> R 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 /> �3A ' $,.�:�: ,£ �'�3;• :�.'!7 2g. i 3 i: 11.�;_`� 1�$- : 24 S? > ; <br /> 3.3 7 12 16 25 3.3 10 17 23 36 <br /> .i�i.0 �::; � ;- ,41':' �6 +�. :: 4 , 10;_ 7B'; 21 33.` <br /> 5.0 6 10 14 22 5 9 15 20 31 <br /> 7. A.Total number of perforations=perforations per la�ral(5)times number of laterals(1). <br /> 21 peAs!lat x 3 laterals= 63 perforations <br /> B.Cakx�late the square footage per perforaHon. <br /> Recommended value is 6-10 sqR/pe�f.Does not apply to at-grades. <br /> 1. Rock bed er�ea=rock width(ft)x rodc lengtl�(ft) <br /> 10 ft x 63 R= 630 R� <br /> 2. Square foot per perforation=Rodc Bed Area/number of perfs(6) <br /> 630.0 R2 / 63 peAs = 10.0 ft/perf <br /> 8. Detertnine required fbw ra�e by mulbplying tl�e total number <br /> of perforations(6A)by flow per perforaGons see figure E�) <br /> 63 perfs x 0.74 gpm/perfs= 46.6 gpm <br /> n <br /> Head Perforations diameter <br /> feet ind�es <br /> 3/16 7/32 1/4 <br /> 1 0.42 0.56 0.74 <br /> z° o.s� o:� t.a <br /> 5 0.94 1.26 1.65 <br /> a. uae�.o roa�ar anpe-ram�y no�s. <br /> b.Use 2.0 feet for else , <br /> ._�/ <br /> 9. Detertnine Minimum Pipe Size " , . <br /> A Manifold on End. If laterals are connected to header pipe .., . <br /> as shown in Figure E-1,M select minanum required lateral �..E•�:M����aE„�aw.�«., <br /> diameter,enter figure E-4 or E-5 with perforation spacing and <br /> number of perforations per lateral.Selec:t minimum diameter <br /> for perforated laterals= 2.0 inches <br /> ___ -----.__ _ _ __ <br /> B. CerMer ManifoW. If perforated lateral system is attac�ed to ^�E�;;,,,�,� . <br /> manifold pipe near the oenter,like Figure E-2,perforated lateral length(3) � ,.,, _ ' <br /> and number of pe�forations per lateral(5)will be approbmately ' <br /> one half of that i�atep A. Using these values,selec;t - <br /> minimum diameter for perforated lateral= 1.5 inches , 'Z�-�',, <br /> I hereby certity that I have completed this wo�ic in accordanoe with ali applicable ordinanoes,rules and laws. <br /> (signature) 810 (license#) 11/03/OS (date) <br />