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. PUN1P,S'a::LECTION PROCEDURE Perforation Discharges in gpm <br /> �A. Deter�nine pump capacity: perforotion diameter <br /> - gravity distribution head inches <br /> .. Minimum required di�charge is 10 gpm (feet) 1/8* 3/16 7/3? 1/4 <br /> 2. Maximu�n su rested dischar e is 45 tn 1.Oa 0.18 0.42 0.56 0.74 <br /> ��-, g gP <br /> pressu re distribution 2•Ob 0.26 0.59 0.80 1.04 <br /> see pressur� desibn worksheeet 5.0 0.41 0.94 1.26 1.65 <br /> Selected pum p capacity: � gpm °Use 1.0 foot for single f�,m��y no�-�Ps <br /> b Use 2A feet for anythm���Ise. <br /> ' Potential for plugging <br /> B. Detea-r,�in� he��! r�qui���mnents: . <br /> 1. Elevatic�ti ��ift:���-�-�c��i�E:��1��'en }�u�����:nd point of diGcharge. <br /> S� fe�i � <br /> 2.Special h���l�� rc���uirem�nt: <br /> If pum���n�; t� a pressure distribution system,five feet for pressure <br /> requir��c� �it m��nifold. If gravity system,zero. `S feet <br /> SUd treoim�nt <br /> 3. Friction l,,tis sy�t�•m „_ �,� <br /> a. Enter �ricti�n loss table with m and i e diameter. total pipe <br /> �p p�p �ength <br /> Read fri,•tion loss in fcet er 100 feet from table. ; �.�. � � ela��otion <br /> Z ,. p h,et : �•_• r�. diffc�r��nce ��� <br /> � F.L. _�2-� `� ft./100 ft of pipe piPe �. .2 3 �� � � <br /> ; - - ;. r� <br /> b. Deterinine tohl pipe length from pump to discharge •` ' Z <br /> .. , <br /> . ;--------------------- ---- - ---- <br /> point. E�timatc by adding 25 percent to pipe len�th for fitting �• •--`- <br /> loss. F.��«ivalent pipe length times 1.25 = tota�pYpe length <br /> s <br /> `3.5 x 1.�5 = � � feet , <br /> c. Calcul.ite tc�t�,l friction loss by multiplying friction loss Friction Loss in Plastic Pipe <br /> in ft/1l)i i ft by eyuivalent pipe length. Per 100 feet <br /> Total fri��tion lc�tis = �_x z.� v =100= � Z feet pipe diameter <br /> 4.Total hr���i rec�uired is the sum of elevation difference, special head gpmrate 1.5" 2" 3" <br /> requircrncnts, ind t�t�l friction loss. 20 2.47 0.73 0.11 <br /> `� __ + . � + ��� (1) (2) (3c) 25 3.73 1.11 0.16 <br /> Total he�rc�i� __ /�% Z_ fee� so 5.23 1.55 o.2s <br /> r- � S�l� 2.06 0.3G <br /> ,� 4Q i c�.91 �64 �p.39 <br /> 45 11.G; 3.�si 0.48 <br /> C. �'ump selection . ° 5u ��3.4� 3.9v o.5s <br /> _ __- - 55 4.7u 0.70 <br /> 1. A punlp mu�t be selected to deliver at least�gpm 60 5.60 0.82 <br /> (Step f11 ���ith �it least �-Z feet of total head (Step B). 65 6.48 0.95 <br /> • 70 7.4-1 1.09 <br />