|
` � ' .
<br /> i : . '
<br /> ' . PRESSURE DIST�tTB.UTIpN�SY�TEM
<br /> GeoEexti�e fabric
<br /> � I
<br /> 1• Select number of perforat�d laterals ���''��4e�:Y�ya�;�U�1,1��%'Urb�°��i"A�r�7�'c,°�.cS';;Ay;•vag.�,:;.•t.,�•t._a,,.
<br /> � . .. .Ma.n.4'R:1s q15.�.V P':C%o^�. l7da n
<br /> � arter lnch erforaHons s aced�3' ''••����3'%I"••Y�:
<br /> 2. Select OI'a I_ �''•��j��'��',,�!a?Gt4� :�-. . :.- � r,�"�e�`
<br /> erf ��n �e a .I' Nrbd;�JS:R�s'iC�'G aeCila�%4 r.:�Ji'���{�xw:enoi�+u�
<br /> P spacin --�._ �f,�C: (j.ri.�t .�Sv.f•�ep I1:%��.5� a• �. �c .5 r•(�'.so6"�'{.y::�1:
<br /> g I feet. �� .�� ;� � <.
<br /> � .4-'�5G 61�u' ly Q- ix'�i• kl_ •a�(�.7..5.1'ta•.�,
<br /> . ti'��.q?����°''t�rti.ci}f Viob a�+n`�e x: •R.�"�.C�.��,�AT�tp r...;:�:<
<br /> 3. ' � �. �v}�. J�',s" �...,��•..J ro !�'�.�rGe 1
<br /> °'d'a'�'s�- e:�.L'CT G'p�:a. . s�e��' h.��j��:i7�;��'u f%i,:�:0
<br /> � ' w ti' rl�-0Y.., . ..,,C•��5 it�c+:c
<br /> Since perforations should ot b e p l a c e d c l o s e r t h a n 1 ft. to Perf Spalcug►g 15��5 1��
<br /> the edge of the rock la yer e e d i a g r a m),su b t ract 2 ft. fronl�
<br /> the rock layer length, i
<br /> ' Perforatlon Dlscharges In gpm �
<br /> oc ayer enp, - I ft. = L
<br /> � ��—� � �_feet. porforatlon dlpmeter
<br /> ' • � • head� � (Inches
<br /> 4. Determine the number of spaces between perforations. (feet) �/$" s/�6 7/32 1/4
<br /> Divide the length above by�p�rforation spacing and round ���� �'�$ ��42 0.56 o.��t
<br /> dourz1 to nearest whole num�ber, 2.Ob 0.26. 0.59 0.80 1.04
<br /> Length perf. s acin lP I . 5.0 � 0.41 0.94 1.26 1.65
<br /> P g=_.L ft.T_�___ft. - �v �
<br /> , �3� �i r2` spaces a UsA 1.0 foot for sfnple-tamlly homes,
<br /> l l b Use 2.0 feet for anyfhine else.
<br /> 5. Number of perforations is e�qual to one plus.the number of ' Potentlalforplugeing
<br /> perforation spaces . � .
<br /> _ . �, . .
<br /> ; . .
<br /> �d spaces + 1 = � � � � � �
<br /> �_perforations/latexal M�m�number of quazter inch perforations pei
<br /> . , � . lateral to guaznantee<10%dischazge variaHon
<br /> 6• Multi 1 erforations eI'I�d ' PerEoraHon
<br /> P Y P p teral by number of laterals to � S�fe�g 1! y
<br /> get total number of.perforat�ons. . � � s 2
<br /> � . 2-5 14 . 18 28
<br /> at =a X pea a,er = �-�-�perforations: . � 3.0� 13
<br /> � 3.3 12 . 6, �
<br /> Calculate the square footage jper perforation(6-10 sqft/per� .4,Q 11 2'S
<br /> System area: lfl 3' x a o i �_ �,V� . . • �5 ,2,3
<br /> � 5.0 10 14 22
<br /> r �ea er o� o - �$_Sqfdperf - .
<br /> 7• De Le � M4NIfDlD{OGTED qT ENO pf�ppES.+NRE OISTIiIBUiION 1YSTEM
<br /> rnnine required flow rat�'by multiplying, � �
<br /> � nuznber of perforations by.fl�w per perforation � �
<br /> . �r.r�
<br /> i � .
<br /> l� ' /-� . � �-�-
<br /> � X gp7 per � �•K �Pm • "�
<br /> �..�'�.
<br /> �,,�a�'��'`� �f�Y�Py;..�
<br /> 8. Tf laterals are connected to h ader pipe as shown on upper
<br /> d""
<br /> example, to select ininimurn�'e uired lat • � �R
<br /> table with peTforation spacin�nd number ofperforationser
<br /> � per lateral. Selecf minimunl '�diameter for � � `"'°"r°°""'""_��"E�•TEM«�^
<br /> . mceauae o�nrrueunoN v�uouHo
<br /> perforated lateral = - in'Iches. �^^������
<br /> ,
<br /> .��,,,�,,:.a ,.• ��,�.� .
<br /> . 9- If perforated lateral system is attached to nZanifold pipe near. � ,.��,��,�,,,�,a
<br /> the center,lower dia arn r���
<br /> . .,.,s�.°r.
<br /> �' ,pe i forated lateral length ancl '° �.
<br /> number of perforations per Ia�eral will be approxunately.one .
<br /> half o£that .-�°�'''�•j�'""`��'�
<br /> in step 8. Using t�ese values,select minunum � ��� �� �,�o���^
<br /> diameter for perforated laterai -� J�o � .•���° �� ���.�.�
<br /> �--��inches. � � �„�
<br /> ; . .. .. .
<br /> , !
<br /> �' . : .
<br />
|