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a , <br /> PRESSURE DISTRIBUTION SYSTEM <br /> znata„xr%%;, srt:<':.aai�!•:.;«�•aaaa,.sr�%;:arcrs�::t;.nns nt„aaax�.!ss•.. <br /> All boxed rectangles must be entered,the rest wig be calculated. " <br /> .....:::: .....:.::: <br /> 1. Select number of perforated laterals: 0 <br /> 2. Select perforation spacing= 2.5 ft <br /> M is OO 104L%001tea9<IVA<60tUW W,0fWAWj F <br /> 3. Since perforations should not be placed closer that 1 foot to <br /> sgatawr�g � F <br /> the edge of the rock layer(see diagram),subtract 2 feet from .!�, •r <br /> the rock layer length <br /> zu- a 14 1A 98 E <br /> 75 -2.5 ft= 72.5 ft st, 1:1 1; 7:6 <br /> rock layer length ? la 16 2 <br /> 1 f is av F <br /> �-0 d 10 # 14 <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= 72.5 ft/ 2.5 ft= 29 spaces <br /> 5. Number of perforations is equal to one plus the number¢f perforation spaces(4). <br /> "Check figure E-4 to assure the number of perforations per lateral guarantees <br /> < 10'0 discharge variation. <br /> 29 spaces+ 1 = 30 perforations/lateral <br /> 6. A.Total number of perforations=perforations.per lateral(5)times number of laterals(1). <br /> 30 perfs/lat x 3 laterals= 90 perforations <br /> E•fs; 1�h$'tU1`�s•`1Q`1�1•f<,i'f:llf�ls�k31�43t�'n <br /> B. Calculate the square footage per perforation. •»w»»»• ...-« <br /> Should be 6-10 sgft/perf. Does not appiyto at grades'. PWf tx-00ry cfbMator <br /> 1. Rock bed area=rock width(ft)x rock length(ft) 116 ,,1l� 7;r2 t? <br /> fff� <br /> 10 ftx 75 ft= 750 ftZ 1.00 £3.18 0,42 CMM Q74 <br /> 2. Square foot per perforation=Rock Bed Area/number of perfs(6) <br /> 750.0 fe / 90 perfs = 8.3 fel pert 2•Db 0.246 0,5559 0-80 ;•04 <br /> 510 A?,4•I 0,94 1.26 1.65 <br /> 7. Determine required flow rate by multiplying the total,number <br /> of perforations(6A)by flow per perforations(see figure E-6) <br /> 90 perfs x F 0.74 gpm/perfs= 66.6 gpm <br /> .............................................................................................................. <br /> 8. If laterals are connected to header pipe as shown <br /> in Figure E-1,to select minimum required lateral <br /> diameter;enter figure E-4 with perforation spacing(2)and �! `="' -°K! =3,:.� <br /> number of perforations per lateral(5). <br /> i ftwe a-a:%tar4ftid Loand•<ad at End of S wt9m <br /> Select minimum diameter for perforated laterals= FN-A-1FN—A-1 inches <br /> ...................... .. ..............................•----.......................................; <br /> 9. If perforated lateral system is attached to manifold pipe r5gu-•e E .7YSanab y Losatad •:*1-'-<�>_ <br /> 6:the coosw of ft,e ayntam <br /> near the center,like Figure E-2,perforated lateral length(3) -5• <br /> and number of perforations per lateral(5)will be approximately <br /> one half of that in step 8. Using these values,select <br /> k 'y <br /> minimum diameter for perforated lateral= 1.5 inches. <br /> I hereby certify that I have completed this work in accordance with all applicable ordinances,rules and laws. <br /> (signature) <br /> (license#) 3Y39 (date) <br />