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� PRESSURE DISTRIBUTION SYSTEM Geotextile fabric <br /> 1. Select number of erforated laterals 3 iz�� <br /> P Quarter inch erforaHons s aced Q 3' <br /> 2. Select perforation spacing = ;�,Ca ft 9"of rock <br /> Perf Sizing 3/16"-1/4" <br /> 3. Since perforations should not be placed closer than 1 foot to Perf Spacing 1.5'-s' <br /> the edge of the rock layer (see diagram),subtract 2 feet from <br /> the rock layer length. E-4: Moximum allowabie number of 1/4-inch perforations <br /> per lateral to guaranfee<10%discharge variation <br /> '���� -2 ft = c�v ft <br /> Rock layer englh perforotion <br /> spacing <br /> 4. Determine the numUer of spaces between perforations. <br /> Divide the length (3)by perforation spacing (2) and round �Qet 1 inch 1.25 inch 1.5 inch 2.0 inch <br /> down to nearest wh.ole number. <br /> 2.5 8 14 18 28 <br /> Perforation spacing = (PO ft= 3 ft= a0 spaces 3.0 8 13 ,.....iZ.., ��-.26 � <br /> ---__ __ <br /> 5. Number of perforations is equal to one plus the number of 3.3 � �2 �6 25 <br /> perforation spaces(4). Check figure E-4 to assure the nu»iber of Q�� � >> 15 23 <br /> perforations per lateral guarantees <10% discharge variation. 5.0 6 10 14 22 <br /> �?O spaces + 1 =�_perforations/lateral E-6: Per(orotion Discharg� in gpm <br /> 6. A. Total numUer of perforations = perforations per lateral (5) perforation diameter <br /> times number of laterals (1) head inches <br /> (feet) 3/16 7/32 1/4 <br /> �21 perfs/latx :� lat= l�'`, perforations <br /> 1.Oa 0.42 0.56 ''0.74 � <br /> B. Calculate the square footage per perforation. <br /> Should be 6-10 sqft/perf. Does not apply to at-grades. 2•0b 0.59 0.80 1.04 <br /> Rock bed area = rock width (ft) x rock length (ft) 5.0 0.94 1.26 1.65 <br /> 1� ft X (.�'�� ft= (ad.c� sqft � Use 1.0 foot for single-family homes. <br /> Square foot per perforation = Rock bed area =numUer of perfs (6) b Use 2.0 feet for an n�� else. <br /> ��',t;) sqft= �"r, perfs = r�� �?, sqft/perf <br /> MANIFOLD LOCATEO AT ENO OF PqE55URE DISTRIBUTION SYSTEM <br /> 7. Deterinine required flow rate by multiplying the total number of <br /> perforations (6A) by flow per perforation (see figure E-6) w�;,a„ <br /> ,) ,�. <br /> �� �,�. <br /> (r=� perfs x , `)'� g-pm/perfs = � ' gpm �` <br /> 8. If laterals are connected to header pipe as shown on upper ����'� <br /> �ViN� u�E.r...[ <br /> example, to select minimum required lateral diameter;enter d,���`" "�"'w° " <br /> figure E-4 with perforation spacing (2) and number of perforations ��``�M <br /> per lateral (5) Select m.inimum diameter for <br /> 1Y1^� 4YOUT Of PERfOMTEO PIPE LCTERnLS fDN <br /> perforated lateral = , yy��jl�S. PAESSWIE DiS�R�BUIION w ta0uu0 <br /> �o n.snc.�n <br /> 9. If perforated lateral system is attached to manifold pipe near �,�,�,�,,,o„,,,� u. y.�� <br /> V Hp ...�.ro�'°�o.� , ^Kp�pNrra^' <br /> the center,lower diagram,perforated lateral length (3) and �Ew �""`.""°` �� <br /> Y�n;>�io <br /> number of perforations per lateral (5)will be approximately one _.�a.;;:;o�,,�,w„�,o, <br /> half of that in step 8. Using these values, select min.imum '°\ � 1- <br /> � � •-�a��K:��.^�.�:::, <br /> diameter for perforated lateral = inches. � <br /> � �,t[D tIr[N� -n <br /> � �nrc <br /> �i�NtM <br /> N <br /> I hereby certify that I hav .completed this work in accordance with applicable ordinances, rules and laws. <br /> -�,,..�'(_-- v�', � �- � � < < <br /> '` � (signature) 3 1 (license�) � �'� � `'C? �, (date) <br />