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8.93 Water is circulated from a large tank, through a filler," d
hack to the tank as shown in Fig. P893. The power added to the water by the pump is 260 ft “3/5. BelemLinc the ﬂowlpile
I through the ﬁlter. \200 f1.nHJ,1ﬂdin. ‘ pipe with em = 0.01
E FIGUHE P8311 ;
k
i v‘ V1
%+z,+ia—q'+hp=4f£+z +5; D
wbere ﬁzp, J V, =V1=o, and fir‘32. Also, L54: ramp or h .—. 422$: m
P 52.#ﬁ%{,§m/HJ‘)Vi v
77mg Eq, (I), became: . 2 409 2001»? ! V ._v._ = (mf +(0.8+53(l.5) +12+6 +0) M31132) "3 _ 13.13 i v z ———— §
(f+0.01355) 1 7M1! and error salt/I‘M)“ ﬁssume f= 0.04 .. Ham} Ey. (2), V= 6.26 5 from ff. ('3),
Re = 5,2ox/a“. Tﬁm; Frag 5'9. 9.25; f: o. 039 '# 0.04: 193502179 PIMP?) or V: 6i29'3ﬂ and We =522/X/04aﬂd 10:19.03?
(Checks) 7%ka [Q =I9V== £(0.IHJ1(F.29§5)= 0.04:9??? HHernaﬁt/eéi, #75 Ca/gerok much“ (5?, 835) could be wed
rafher Man {he Moody chin/"f. 77mg (aim ’1‘)
‘3—8? M where from 139;,(2;J (4:) f=(/3.13//3)—0.01365' ' (5)
77505; by cambinfng 13$. {3; [ﬂiand(SJ We obfain 1776* following Eyym‘z‘m
for V3 l
l /[( 13.13/v3) ~001365]4‘=—2. Ia Egg2.51/[ezfoanaJa/v3yo.anﬁfﬁ} Using a smut/fer mﬁ—fthdmy pro mm gills: {he solﬁian 1‘0 EfJJ) as m __ H 
V— 6.295— , Me same a: 0571mm 1 by Hm 4501/9 {rig} and eﬂw mko
i 8.97 Air. assumed incompressible, flows through the: was
pipes shown in Fig. P837. Determine lhe ﬂowratc if r‘n'mor'
losses are neglected and um friction factor in each pipe is 01.016.
Determine the Howrah: if the 0.5in.diume1r:r pipe waft re placed by a MILdiameter pipe. Comment on Lhe assumlpliunl
of incumpmssibiiily. I FIGURE P837 +23} where %=0120=z;1ﬂ3501 (I)
‘ 1 ‘ 2 .  9 %,and4=14%3=m%)= v2
777st £?_(/) becomes =6L2.‘51{;l
_ 0.25V." 17 W a

or 2,0,, = £91611}; {2; (0.25324 {2% + I] (2) L511 ' {71.1
WI 100%“? or Po: 0 75 *(17/6 rug"? )(ISomoﬁR
and ﬁ= )9: 0.015 Eq. (2) gives 5 I” R = 0.0020? Elf‘3'? (0.5%,).(IWgif) =19 (0.002 9%)fﬂo05XﬁgN0Ef ffﬂ D” 1] 2+9
or 1/1: 970.4 gt 77m: Q: 1 t4 = f(i%'f{)2(9a#§)=o.123 g
If 507% pipes Were / 1}). Jib/Dyer; 2%” V, =Vz and 0.} beam/71.9:
gag gel/3R +76%: +1] 011" vii/1 Hz, 44:} m: 4:9; 2192) ‘
2% = 2 P WE (*5; “1 #801381 _J 4 ﬁt 11. .111. =J_ 5:" 1 . 0
(0.537304%) 26200209 )Ié[0°’5( TWP]
0r V1=9L7 gt ms, Q=ﬂzm%%(ﬁ—H)2(w.7§t)= 0.500%;
5/3059 69=FRT f7! Ital/aw £11611! 1
{Jo—a) If We «same 73 =75 (I'l’ﬂNéqﬁ/y “HY/#01156",
RTn ‘
503' BMW/J be a remand/e oppkxm‘m’im) M917 / . ‘ .
'6; x j = W} =0.§’67 _ 730 Mavis gawk Mchresslﬁ/e. 
Bull?!) The ﬂuwmte between tank A and itank
3 Shawn in Fig. 138.100 is to be increased by 30% _ 5indiamﬁﬂzn 6in diameter:
(i.e.. from Q to 1.3(3Q) by the addition of 'sec.  . . 5°” " "’"B 5"“ " “‘"E
and pipe (indicated by the dotted lines) ru IHing, =
from node C to tank B. If the elevation of the
free surface in tank A is 25 ft above that in tank
3, determine the diamcter. D, of this new ipa.
Neglect minor tosses and assume that the fri lion Nathan‘1? 509 It l'unB
factor for each pipe is 0.02. a] FIGURE P8109 Mil/J ﬁle sing/e Pipe = where ﬂfﬂgﬂaJ M}:
and V, = K. (szhce D,‘ ). TM, 39 = 2‘, (11¢ it“; , or 25340.on (daemon) H V,"
' (£85 2(32.2 1:1,) r i 3
" v, = 6.05 ti 0w =£G2Hfrm$ = we ii WM 4% Second pipe 0;} 1.300.193 ) = [5.55 £15 x 3
77m, QI=L54¢~§E=QZ+Q$ or 1kg} =i£§$ =Z'ggzgt
4 75 For fluid ﬂowing from 24 {0:8 threw/J pipe: land 2"  _ 1 1 g =
2,9 =51, 1%; 157$ 2% “‘92 £3: 3% {593 Eff”)
0!" a. H: . 600G rumba _ nary v,L
25 (092) GER, 432.2%) (0 02‘) (fiftj Haggai) Heme: V2: 2.60% I Grid :1
0;: ﬂ; t4 = {56% It); (2606?) =05H £5
77”" 623 = 0. d 01" 1.5410,”? = I. as? ' t
For 1%?! flowing from H {eff threw/2 prpes I mam, H
zﬂ=bir+b£3 = 1&5? 2;,wéere lé=£f=40317 "3'
Tilly's: I ( 2 £0: IF
= man (299?): M _&a__‘_‘
25” I’D“2’ (#9275312 1) “0'0” 03 292.2%) or
DJ :7 O. 6621‘} Mule ' MW) 1%? parameters aim1:, ﬁe :a/m’rbn :1: gaffe semifim 2’0
rounder? arran: 1:1) #15 blew/MM; Elevation c: 60 m 8.102 The three waterﬁlled tanks shown in Fig.
P8.102 are connected by pipes as indicatecg. If
minor losses are neglected. determine the ﬂow
rate in each pipe. 1 _ 7 FIGURE P8492
ASSUME Hie fluid ﬂows Win? )9 2‘03 and/41596. Mus, Q, =Qz+93
gyfmm)?‘ V, = —§(o.oam) y; + £(mamfwg
’ 14 = 0,64 V2 +0.54: 14 m
For ﬂuid ﬂowing From ﬁfe? wH/J ﬂq=ﬂ3=0 and Hq=lé=0J .2 2
39:38 +££LJ +££§ V1 or D, 2‘9 D: a?
_ __ 200/): I V1" 20cm; V:
60m 20m  (0'0153( mm awafg) +(0.029)(m)m
Hence, 5
5:0 : /.529 Vf + 2.55 v: i (2) Sim/afﬁx, for f/Iéid ﬂowing from x? 7‘06 .WIWJ £970; =0 and K942 =19J
24:26 + ‘23 £1 60m '= (0.0I5)(203;m 2m 1 f1) +(o.ozo)(3%3§£) zmﬁﬁf Hence, I 50:].529 1/,2 +5Jovaz (3)
Solve 59.5. (04(2), and (3) for V“ Va, Ma'Lé, Frw» @311) and (a):
50=L529(0.H)1(I/2+b3)2+449 [all or 95.8 =(V2H/3)‘+ 611‘sz (49 Subzlr‘acf £911) From 5?, (3) ;‘= 5040 = 5.10 I42+ 2.551411'0r 1/2 =W (5)
. 1 2 7720.53. from 59¢. (We/761(5). EBJIJCIGIW f2V3z_7.5¥' +16) #4158 =0 t =
777/19 can be Sim/1'10}?! 1‘01 2 waif—284* = 10336 ﬁlm/31 Square bail: side: and m
rearrange 1‘0 give ‘4“ #9163 W2 + 425 =0 w/Jich am he SOIL/ea]
by #79 quadrafic fanny/é fa nil/e 2 .SBiJ—fﬁ
H? = M 1953 40:25) :H'Wor 7.86 Thus V3 2. 
or V3=2.80gl I _ (gen?) (can’t) M’fe ‘ The Value lé=3"*$5ﬂ [9 1'70?l (1 so/Ufian of H18 original Eyum'r’aﬂs,
Efs. (I), (2), and (3). i Wﬂﬁ Mg Valve H73 rig/51' banal 5:21.? a)“ Efféj
is ﬂeqm'iua (:25. [0.3% —I!./’/ M32": 103.641.}? (3.413325‘2495)‘ ,9;
Seen {ram Me [229’ Mm! side of £1716), “W: can/mt be. 7771': exfra
r004 was t'nl'r‘aduced Squaring £716). Tm, a; = as 1/3 = 7202.0 mﬂmogl) = amt/£13
Also, from Eq. (331 '
60:1.529 Mz+5i0 (2 so): a, 14: 3.52
or 3
and From 557.0):
3.52 = 0.54 14 +0.6Iﬁ(2.ea) or V2: 2.85% 01" l 3
Q1 =14; V: = §(o.aam)2(2.36 %) = 0. ants—g“ 8103 ...
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 Spring '06
 wark
 Tank, pipe, ﬁle sing/e Pipe, Mavis gawk Mchresslﬁ/e., um friction factor, 5indiamﬁﬂzn 6in diameter

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