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<br /> P11MI'�E1.i.LT1�N PI�C)C:EDUItC
<br /> FIO PfR�nRAfl(MI (Y A P(R1(N7AlFn 1.41FHp�
<br /> A. [)ctcrmine pump capacity: ���
<br /> <:ravity Distribution ` -
<br /> 1_ Minimum su��stcd is 1�00 gallons Ex�r h<iur(10 f,pm) Ic� stay ahcad c�t �� ���—
<br /> w a Mr u tic ra te. ,.-.+s..e� •�"b::.'"•':;"°,K'>,a„
<br /> .��.�.�..,,;^-�......
<br /> 2 Maximum su�,esMd for delivcry lo a drop Lx�x o(a home system is 2,7(x) 'T -M���a��..a�,.a,,,.,ti
<br /> .,»� ,.... �
<br /> gallons per how(45 gpm)to prevent build-up ol pressure in drop box. �.,•pw � ��,,, 17•,.�.a
<br /> �{go rie � ei a«��,.,
<br /> wwe�w�.�o.�.a.i
<br /> Presswe Distibution �„�,�„� 8.,,.,,,«�„„„
<br /> 3.a. Select numberof perforated laterals __
<br /> b. Select perforation spacing= feet_ �«�.��.,� ►�ro«�r sm��ae
<br /> �����e�
<br /> c. Subtract 2 R.(rom the rock layer length.
<br /> -2 ft.= feet
<br /> d. Deter�mine the number of spaces bet�vicen perforations.
<br /> Length perf.spacing= ft. + ft. = spaces iteq�urd Pet�oraeon D�scha�e
<br /> e_ _spaces+ 1 = perforations/lateral v,gauon5 p�Q,;,,�ce f•
<br /> f. Multiply perforations per lateral by number of laterals to ��x��e
<br /> get total numberof perforations. x - H�d ���� �.,aa.,,
<br /> �„m,. -�,,,7�Q,�- perforations. �f�)
<br /> £ _� x�r�r--gPm.
<br /> I.Oa 0.56 0_74
<br /> SELEC'TF_-D PUMP CAPACTIY� $pm 2.0� 0.80 1.�4
<br /> B_Uetermine head requirements:
<br /> 1. Elevadon difference between ump and point of discha e, a. Use for single family homes
<br /> teet
<br /> � b_Use for all other applications
<br /> 2. If pumping to a pressure distribution system,five feet for pmssure
<br /> mquired at manifold if gravi system,zero.
<br /> t feet
<br /> 3. Friction loss
<br /> a. Enter friction loss table with gpm and pipe diameter. PRpe I.rngct�
<br /> Read friction loss in feet per 100 feet from table. ' Point of Discharge
<br /> F-L-= ft/100 ft of pipe
<br /> b. Determine total pipe length from pump to discharge n����D���T
<br /> point_ Add 25 perc�nt to pipe length for fitting � 1
<br /> loss,or use a fithing loss chart. Equivalent pipe p
<br /> length- 1_25 times pipe length=
<br /> x 1.25 = (�t F-186
<br /> c. CalCuldte tOtal friction loss by multiplying 1.S inch 2.0 inch 3.0 inch
<br /> �1 Friaim Im pc]00 h d Pipc
<br /> friction loss in ft/100 ft by equivalent pipe length.
<br /> Total friction loss= x +7pp-�_feet 10 0_69 0.20
<br /> 4. Total head required is the sum of elevation difference, 12 0-� 0.28
<br /> 14 ].28 0.3R
<br /> s�xrial head mquirements,and total fnction loss. 1G 1.63 0.48 j
<br /> 18 2.03 O.C�
<br /> 20 2_47 OJ3 0 t]
<br /> + � � 7_S �.7� 1.11 i 0 16
<br /> --°1--- --_ ---- Z
<br /> (1? (2) (3c) j �0 � S.�� I SS � 02 i �
<br /> � �� � ;_y(I 'Oh I 0 iU
<br /> L, � -itl ;1-137 Z.W j O3`i �
<br /> TOTA[_}�ilAll -�V-- f����t a�, I�� 7: 3-28 o.ar;
<br /> �0 3 S�) � O.SR �
<br /> SS 4.7(, � 0 7(i �
<br /> � . �'illtl}) SPICCilOII 6U 5 6(I U}+2 '
<br /> j �
<br /> ---- — --------- - �
<br /> �-
<br /> � 1- A pump must b�> sc�lc��- �>d to deliver at lc�ast
<br /> � � t;�tn (5ic�� A) �ti�ith af li�.��i � (rct of (otal hcad (�t��E, It)
<br /> � � �--
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