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, I � � ' � . <br /> PRESSTJRE DISTRIBUTION SYSTEM Geotextile fabr�c � � <br /> � :.;�: . :,,.�. .: -;;� ��;:•; _�... .. ...: . _; _ <br /> 1. Select number of perforated laterals 3 � arter irch •e:foraMorv. ,aced @.3' •. 'S 1�'". . � <br /> '. �'� :: •r.:,,;^.:'•j i �. :.�� ..�; ';� . , . <br /> nr•.. . <br /> ..J�:,;,.�.���r'�• '�r:�.tfif�G�C' <br /> 2. Select perforation spacing= 3.0 ft .� ' . � .".' �{�:t ..`,� . . . .. : . <br /> .:r .r, . <br /> PerE Sizin 3/16"-1/4" <br /> 3. Since perforations should not be�placed closer than 1 foot to P�spa�►s 1.5'-5' <br /> the edge of the rock layer(see diagram),subtract 2 feet from <br /> the rock layer length. E-4: Maxirrwm allowoble number of 1/4-(nch pedoration <br /> ' s v -2 ft . p�r laterd to guarcmtee<107L dtscharge variatton <br /> o t �yez eng �f t . , pAf�OfC}�0� <br /> .sp��9 <br /> 4. Determine the number of spaces between perforatioris. tee 1 Inch � :1.25 Inch .T:s tnch� 2.o inch <br /> � Divide the length(3)by perforation spacing(2)and�� <br /> �.p�to nearest whole number.. � <br /> 2.5 � � 8 •14 18 28 <br /> Perforation spacing= � ft+,,,�ft=1�a spaces 3,0 � 8 13 17 2b <br /> •3.3 7� 12 16 ' 25 <br /> 5. Number of perforations is equal to one plus the number of 4 p � �� 15 23 <br /> perforation spaces(4)..Check figure E-4 to assure the number of <br /> perforations per'lateral guarantees<10%discharge variation. 5.0 6 10 14 22� <br /> �1 ,�spaces+1 =-1 !L p�rforations/lateral E-6: Partoratton�ts e t gpm <br /> 6. A. Total number of perforations= perforations per l�teral{5) perforatlon d(ameter <br /> times number of lateraLs.(1) � �� heod inches <br /> �`� a��• (feet) 3/16 7/32 1/4 <br /> �'�perfs/lat x,�_lat= �1 perforations �.�0 0.42 0.56 0.74 <br /> B. Calculate the square foatage per perforation. 2,pb 0.59 0.80 1.04 <br /> Should be 6-10 sqft/perf.Does not apply to at grades. <br /> Rock bed azea= rock width(ft)x rock length(ft) 5.0 0.94 1.26 1.65 <br /> �o ft x .tu ft= $O� sqft , � a u�i.a roor i«��i�-ro�h�no�:. <br /> Square foot per perforation=Rock�bed area+number of perfs (6) b u�2.0�aer ro�o� i►, e�3a. <br /> So o sqft+�-Perfs= �•t� sqft/perf .W„� �,�,,,�.T �a.,��„E �a�.�,�, �, <br /> 7. Deternvne required flow rate by multiplying the total number of . <br /> perforations(6A) by flow per perforation(see figure E-6) � �,. <br /> S 1 " perfs x�m�pper�s= 3�_gpm �� � <br /> 8. If laterals are conriected to heacier pipe as shown on upper � <br /> ►�'"'t ya"�'R'.' <br /> example,to select aL��mum ret�uired lateral diameter;enter ,,�••'° <br /> figure E-4 with perforation spacing(2)and number of perforations `""� � <br /> per lateral(5) Selectminimum diameter for \� <br /> u�a�w cMa iiiinnw�n MTi�a�v�wo� <br /> perforated lateral= inches. <br /> .�..�....�..,K..� <br /> 9. If perforated lateral system is attached to manifold pipe near � � „� <br /> the center,lower diagram,.perforated lateral length{3) and � �ti""`�`` ��� M"'�� <br /> r.r�e.• <br /> number of perforations per lateral(5)will be approximately one K.,�;�.M MfqM M � <br /> half of that in step 8. Using�these values,select minimum � _ � <br /> diameter for perforated lateral= �I 1 Z inches. �*�� <br /> a w �. � <br /> . . . . . ►�.�'"" ,,,w. . <br /> . . �� <br /> r <br /> I hereby certify that I have completed this work in accordance with applicable ordinances, rules and laws. <br /> , <br /> �� �� �� (signature) 3�� (license#) 11 - �o-O Z (date) <br />