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OSTP Pressure Distri ution <br /> MinnesotaPoilvtion UI�IVERSITY �� � �� ; <br /> Desi n Workshe t ��,, --��=a <br /> ControlA enc � OF MINNESOTA �.��-_,�,�,�.,�� <br /> ProjettlD: v 11.09.22 <br /> 1. Select Number of Perforated Laterals in system/zone: C 3� <br /> (2 feet is minimum and 3 feet is maximum spacing) "'`'""" <br /> :..,,��-. <br /> �� •`\ ,' ' <br /> 2. Select Perforation Spacing: 3.0 ft "'= ' " �; �� - - <br /> � , <br /> ..:.,,,�a„ <br /> �,-�w,�.,,w„�...M�•:�.� ,--z-«,� ,�- <br /> 3. Setect Perforation Diameter Size 7/32 a� _ _ <br /> 6"��t�at <br /> 4. Length of Laterats =Media Bed Length-2 Feet. a�,+.,�.«�_�-,^., �-�n:_- �«.,=�n�:x;�:����3 <br /> 53 - 2ft = 61 ft Perforotion can t be closer then 1 foot from edge. <br /> 5• Determine the Number of Perforation Spaces. Divide the Ler�th of Late ls (line 4)by the Perjorotion Spocing {tine 2)and <br /> round down to the nearest whole number. <br /> Number of Perforation Spnces = 61 ft - 3 ft = 20 Spaces <br /> b. Number of Perforotiorls per Lnterat is equat to 1,0 ptus the Number of P forntion Spaces (Line 5). <br /> Perforations Per tateral = 20 Spaces + 1 = 21 Perfs. Per l.ateral <br /> Check tabte below to verify the number of perforations per tntera!guara tees less thvn u 10%discharge variotion. The value is <br /> doUble if the a center manifotd is used. <br /> Maximum Nc�r c,#Pesforati�zu Per laterat ta tee<1�6 Discharge Variatian <br /> '��inch P ora�ans 7/32 i�ch f?rforatians <br /> P2!#ora�i Spacmg Ifeet) Fipe Diarneter!i!xhesi Perfor don Spating PiPc'L}?"aR�#�f(IRC�1P5) <br /> i tt� 1}: 2 3 FeEtl i i'f is: [ 3 <br /> � 10 13 i8 30 b0 2 1t 1h 2( 34 68 <br /> z�� 8 12 f5 28 54 2� 90 i4 2Q 3Z 6�1 <br /> � & f2 i5 25 51 3 9 19 19 30 b0 <br /> �'161r,ch Perforations 9/8 Inci�Fe�faratioru <br /> Pipe Oiameief{inches? Perfor tion Spacinq Pipe Diameter�Inchesy <br /> Fer�orat�,n Spac:n,�iF�Tj - <br /> t t� 1�: 2 3 Feett 1 1t: ti; 2 3 <br /> t 12 !8 26 45 87 2 21 33 44 74 i49 <br /> 2�= 12 17 2� 4U �� 2'�? 2a 30 41 69 135 <br /> � 92 16 22 37 75 3 Zp ?9 38 54 128 <br /> �• Total Number of Perforations equals the Number of Perforatiorts per Lat ot (Line 6)multiplied by the Number of <br /> Perforeted Loterals (Line 1}. <br /> 21 Perf.Per i..ateral X ��Number of Perf. terals = 63 Total Number of Perf. <br /> _ <br /> 8. Catculate the Square Feet per Perforation. Recommended value is 4-10 ft per perforation. "�'•�°^°'��"���'�� <br /> Does not apply to At-Grades ""r°""°"D5i1etr <br /> HeW t�t3� . - <br /> ,�s '�. .�ss ,�. <br /> Sed Mea = Bed Width(ft)X Bed length(ft) �,a o.�a a.+ osa o.» <br /> f.5 ( 0.2? 0.51 tJ.i9 0.9 <br /> '!0 ft x 63 ft = 630 ftz Zw o.� 0.59 0.� ,.o, <br /> 7-5 0.29 0-65 D.89 1.f7 <br /> l.0 0.32 � O.]2 0.98 1_2C <br /> Square Foot per Perforation =Sed Area divided by the Tota(Number of P rforations {Line 7)_ �.o ; o.» o_e; ,.,3 j ,..� <br /> SA` O.af 0.93 /.36 1.65 <br /> 2 �Pa'tlli jz+nih 3)t65ncT W t:�i�r.h <br /> b30 #t .- 63 perforations = 10.0 ft2/perforations '#°°` ���� <br /> pv.tU��NIt1 1 J6 iKh nn[orM[Ns <br /> 3(ee[ Octrrp.iayishmMf[sandMSt5w2h3f:6 <br /> 9. Select Minimum Averoge Head: 1.0 ft tl6h 10➢JJ itKA pClSO�dt10n5 <br /> 5([CR ��GidFdi591f11P_�lY anG NST$vKT 1/8�nCh <br /> pF�de[iOrv3 <br /> 10. Select Aerforation Disrharge (GPM)based on Table IIi: 0.5b GPM per Perforation - <br /> 11• Determine required Fiow Rate by multiplying the Total Number of Perforo ions (Line 7)by the Perforation Oisclwrge {line 10}. <br />