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I� 17
<br /> CUMI'SELi.L�I1C?N1'I�UC:EI1UJZl:
<br /> (IO PFRfOliAT10N (lf q PfRF'(Xt47ED l_4iFRAl
<br /> 11. I)ctcrmine pump capacity: �„��
<br /> �_ . . _� _�
<br /> Gravity Distribution �- ��
<br /> 1 Minimum su�,*rstcd is 6(xl gallons Fx�n c�iur(10};pm) l�i stay ah��ad ol _ .
<br /> � l d G««.�w labr Ir�s.
<br /> wat�r u�c rate. • �--�+s�,..�.« �`b..• b,..,,,.�..,,,,
<br /> .��.�.r„,�,..�
<br /> 2 Maximum suF,gc�strd for dclivcry tci a dro��tx�x o( a homc•systcm is 2,7(x) 'r -�...«.�a a,��„�.,.d.,,,,,
<br /> .. . �.a cao w., �
<br /> �;allons per hour(45�m)to prevent bwld-up o(pressure in drop box. ;.,•R,„ -., �„„ ,:-„E.�,
<br /> b1A of Hoct�q�
<br /> MxO1Mr lnOe1M al
<br /> Pressure Distibution �,�,,,5a,,,� a.��..«,.�«,�
<br /> 3.a Select numberof perforated laterals
<br /> b. Select perforation spacing= feet. �o^«W�.�� h..-�,�,��
<br /> B�/�n Pledp S�M le�+r
<br /> c. Subtract 2 fL from the rock layer length.
<br /> ����-2 ft.= feet
<br /> d. Determine the number of spaces bet�veen perforations.
<br /> I-ength perf.spacing= ft. + ft. = spaces u,�d p�to�atio„ e
<br /> e. spaces+ 1 = perforations/lateral ��,uo�pet�,u,�c ��gp■,�
<br /> f. Multiply perforations per lateral by number of laterals to D�t,�ge
<br /> get total number of perforations. ��� x - erforations. H�d �z��'' �a�,°"
<br /> p�7G�aa— p cfeec)
<br /> �' �a�[ X�pei $P�-
<br /> I.Oa 0.56 0.74
<br /> SELECTED PUMP CAPACTIY�_gP� 2,�b �.g� 1.�4
<br /> B. Determine head requirements: a_Use for single family homes
<br /> 1. Elevation difference between pump and point of discharge. b.Use for all other applications
<br /> _�Z feet
<br /> 2. If pumping to a pressure distribution rystem,five feet for pressure
<br /> requimd at manifold if gravity s tem,zero.
<br /> _�feet
<br /> 3. Friction loss
<br /> a_ Enter friction loss table with gpm and pipe diameter. Pipe Lcngch �
<br /> Read friction loss in feet per 100 feet from tablE�. � Poinc of Discharg<
<br /> F.L=��./100 ft of pipe
<br /> b. Determine total pipe length from pump to discharge p`���,D,��,�T
<br /> point Add 25 percent to pipe length for fitting � 1
<br /> loss,or use a fitting loss chart. Equivalent pipe p
<br /> length-125 times pi length= F-ISb
<br /> 2�x 1.25 =_��feet
<br /> c Calculate total friction loss by multiplying iS�c�h ��_0 inch 3.0 inch
<br /> g� pc 100 h d Pipc
<br /> friction loss in ft/100 h by equivalent ipe length.
<br /> Total friction loss= �_x � • 2� +7 pp=�,Q _fee t 10 0_69 0.20
<br /> 12 0.96 0.28
<br /> 4. Total head required is the sum of elevation difference, 14 1.28 0.38
<br /> special head requirements,and tatal fnction l�ss. 16 1.63 0.48
<br /> 18 2.03 0_60
<br /> � 20 2.47 0 73 0.11
<br /> i � � � ?5 3.73 l.l l f O l b �
<br /> ------- — ---
<br /> �11 (21 (3c) �0 � 5.2_� LSS i O�i j
<br /> 3ti ` 7_9{1 ?Ob � 0.3U j
<br /> �1t1 11 (17 Zt.4 i 0 ��� �
<br /> T(�"CAI_NEAI) �.�_ feet -�5 14 7� �,:'.�- i 0��t, �
<br /> SU 3� � 0 St� !
<br /> 5S 4J6 , U 71� '
<br /> � . �'UI71}) SE'ItCt1011 (��) S.GO � U1i� '
<br /> I
<br /> �--- -- --�-- ---
<br /> � 1. A pump must be selr>�-t�>d to deliver at least
<br /> �� l;f rn (St�p A) t,�ith at 1��<�til �F _ fc�>l o} total hc�ad (�t���, fi)
<br /> �
<br /> � - - ----- --- -- ----J
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