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University of Minnesota Pressure Distribution System Design - 10125/04 <br /> All boxed rectanQ;es must be ente�ed,�he resf will be cafculaled. r <br /> o.as,rn _�„ <br /> $CWA6G �..fj�� <br /> 1. Select number oi perforated Iaterals: 03 TRCATMGNT _,_' � <br /> PROGRAM <br /> 2. Seiect perforation spacing= ��ft <br /> ,�, <br /> 3. Since perforations should not be placed cioser that 1 foot to ; . , ��,__,;„ _ <br /> the edge of the rock layer(see diagram),subtract 2 feet from • , _, „ <br /> the rock fayer len th ----. �_ .____.___ <br /> 63 -2ft= 61 ft . ' .. ^ <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= 61 ft/ 3 ft= 20 <br /> 5. Setect perforation size 1/4 inch <br /> S. Number of perforations is equal to one plus the number of perforation spaces(4). <br /> 'Check f,gure E-4 to assure the number of perforations psr lateral gusrantees <br /> <10%discharge variation. <br /> 20 spaces+1 = 21 pertorations/laterat <br /> E-4 Maximum Number of 114 inch perforatians E-5 Maximum Number of 3116 inch perforations <br /> er laterai to uarantee<10°/a dischar e variation er latoral to uarantea<10"/a dischar e variation <br /> Perforation � Perforation <br /> Soacing 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 i 12 16 25 3.3 10 17 23 36 <br /> 4.Q 7 11 15 23 4 10 16 ?_1 33 <br /> 5.0 6 10 14 22 5 9 15 20 31 <br /> ', 7. A.Total number of perforations=perforations per lalerai(5j times number of laierals(1). <br /> 21 perfsl lat x 3 laterais= 63 pertorations <br /> B.Ca;culate the square footage per perforation. <br /> , Recommended value is 6-1�sqft/peri.Does not apply to at-grades. <br /> 1. Rock bed area=rock width(ft)x rock Isngth{ft) <br /> ' 10 ft x 63 ft= 630 ftZ <br /> 2. Square foot per perforation=Rock Bed Area/number of perfs(6) ' <br /> 630.0 ft2 / 63 perfs = 10.0 ft�/perf <br /> 8. Determine required flow rate by muitipiying the tatal numbar <br /> of perforations(6A)by flow per perforations see figure E-6) <br /> 63 perfs x 0.74 gpm/pesfs= 46.6 gpm <br /> E-6 Perforation Dischar e in GPM <br /> Head Perforations diameter <br /> feet inches <br /> 3/16 7l32 1/4 <br /> 1' 0.42 0.56 074 <br /> 2° 0.59 0.80 1.04 <br /> 5 0.94 1,26 1.65 <br /> a. Use 7 A foot for single-family homes. <br /> b.Use 2A feel tor anyltwn else . <br /> �/ <br /> 9. Detertnine Minimum Pipe Size , <br /> A. Manifoid on End. If laterals are connected to header pipe <br /> as shown in Figure E-1,to select minimum required laterai f�qa�ok 1 Mo�itokflacoroCa�ItnAnliys�om <br /> diameter;enier figure E-4 or E-5 with perforaiion spacing and <br /> number of perforations per lateral.Select minimum diameter <br /> for perforated laterals= 2.0 inches <br /> B. Center Manifoid. If perforated lateral system is attached to �w�rco4�M.,^��M�^���--� - <br /> . ��mo c..ro�a m�sn»,,, , <br /> manifold pipe near the center,like Figure E-2,perforated lateral length(3) <br /> and number of perforations per iateral(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 • Z� <br /> �I herebj�certify that I have compfeted this work in accordance with all applicable ordinances,rules and laws. <br /> ��!.�•�'' ��� (signature) 810 (license#) 12/31/08 (date) <br /> � ,. <br />