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Unformatted text preview: E! 01o) E06157 (quaho" (W50) : U09 Secixon
064
P) “I V; Luz)
(;f~¥:+5")“ (77* "r “‘5‘“ h“
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MR4$ COASCFV“+\'O" a“ (2: AIM: ALVL Hoe/ow dﬁ r annular
ﬂ‘:A2 2:) 12:5; ‘ ‘2‘”? Jr (regim)
Uniform How 0d (pt) 0% C
A+®: V0) 'Z'SZ’WCI )I’7‘ 55,3: V
617,: %(§)f(¥'md4 K LORP€~2nrolr
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I m V‘/Z ‘ PAZVZ/Z ‘
“heed V1 ‘f‘b S‘O’ve for 043 ‘) Solve for 71 Miinj inky“! fo (95*qu
Howrah M’@ from Hm if ﬁllows Hm‘ V2  i I
~‘~ )
V1: ‘ {Di/AHA :~ 1. R w L” ”7 W __ 2‘" We, '2 u
A: 2( A2 30 c(7) MUN olr \ 1122. (ll/7‘ So r(,g,.) 7 01’,
2 2,740,; i“, if) R {27 f ‘« Inﬁafnhm by fariir:
“ newt", [~§r(ﬁ~/) ,0 +1, E 02") I, 01,] {If} : r f’(/) s I 9’0) = 92" )"7 50) 3 ’ $(‘V7ygh .‘ ch w 1 '57) a ‘ \ ‘7‘“ 4. ‘ w 3N3“. The murky: velocv‘ﬁ if ‘9“ Hum “\c eeAHr/iu {Igloo}?
one? bemwse 'H‘a ve’vcih‘ prov“: o~+ @ (yo/thins «’r #he cukr and t MEMMMM a'r 1h; wan of +1“ FirC
So)“ for 012 ‘4 ”ml/imam ‘ _ zTY ' _
6‘2 \ 12! SR Vzr ”(I : R r ’/7 60 k 3/7
’MV ° I?“ (SW) S; M17?” rd” :Z(‘I"I )3, '2‘”? So NIH} Jr 2: i ' 60 3 ‘1 lo I? Q 7 1/ ’nkﬁﬂﬂim A», par’j"
R'"? («75) 70"“) / ~+J15m~{)""o¢r] ’ rma» {m _,
0 ° ‘ formf)“ 9m =~2 (mm Z
~ wk) [(9‘0)+,3‘,%(12~r)"hlo I :3. (93,3.2‘1 pm. q” R‘nl? 9°) ,0 "7
2 ea .. 
s: l a) ' SMESH‘MQ M‘H merglz e7um‘han’ in. '5 PPPZ “f (Ma43””, 90" At) t p1,;p1 +04??? iifﬁ (Lp‘kz) F Eli331+ 0.2:3 wotr 50min) 5.103 Wake: ﬂows steadily in a pipe and exits as a; free jet
through an and cap that contains a ﬁlter as shown in Fig.” . 103.
"PM ﬂow is £13 a hm‘immal plane. The axial compancm, R,” of
the: anchoring force needed to keep the end cap stationary is
60113. Determine the head less far the flow through the and cap. Area = 0.12 ftz
v = 10 M H F I G 'U H E P511133 The ycommnem‘ of {be momeﬂh/m equ/mi
In)“ 6’ V4) (119 =2 F), J far #16 caan/ va/ume
5 05/10Wni3
6’) WP (“t/1);]: +(ﬁ‘é51b300)@‘/2.A1 zﬂﬂ/ ’Ry
where V, ":10 H/s 4M 1
V2: 742V! ‘(g’lijfiNmH/S) =12 H/s TIM/s, since 0/4/14 se/Iz V,” Eq.(l) gives
m. = Ry ~evm—ewsmao‘ﬂz :Ryveavlm +vmaon
= 60/5 {W ‘ﬁ;‘)(o.12H‘)g(Io iii/o? +12 £19,730? = 22.9/5 Hence, _ ‘
f), = 22.9/A/A" :22.8Ib/(o./§2Hz) = Mow/f!“ From Me energy equah‘m for Hm Hon; 1 1
1,4,2 _' 90 15ml (I0H/5)1(I2ff/s) = .
“€75 ' 15% + usual"Wl 2 36 5:101! I ; 5.109 A pump is to move water from a lake into a large, pres
smized tank as shown in Fig. P1109 at a rate of 1000 gal in 10
min or less. Will a pump that adds 23 hp to the water work for this
purpose? Support your answer with appropriate calculations.
Repeat the problem if the tank were pressurized to 3, rather than
2, atmospheres. %+zl+wv‘l;.+l7p~/41 r: ﬁaI'Zz Ali7%..) where F1
TAU: w h,— quiz+22, Also, 3 2
Himawd)/~mm>] align/w £453) = 0.223 1::
so Maia H 26/: (a) I f f, = 2 aim = 204715 )("f‘r‘MVrﬂz 45236": Men from 57 (/2 hp h+ 57723—4333;on =16 +82%?
7711/3/71. ILS “hp 97 r? fi= NW? HWY‘ 3/ 2 H f/Ie give!) lot/mp w;// war/f for/7f 2479)). ___ (b) If ,6; 3 an= 6, 356%, Men hp: [2+(g—f—fo—‘EES—Jr +20%: )7 Huff This ,‘i 2%,}: pump/s fa work Hm =6, +/2»2HJ or h, . ~3~H
Since if is nm‘ pasty/He %0 have thJ {/13 pump til/”1707‘ u/W‘k far 702 :Bozlm, rig—[2.5 5.11:; I "" 5.119 The turbine shown in Fig. P5.119 develops 2500 kW
when the water ﬂowratc is 20 Ins/s. The head loss acrossgthc tur—
bine fmm ( l) to (2‘) is negligible, but the head loss for this Entire
ﬂow is 2.5 m. (a) DG'CBI‘I’niﬂB the pressure diffemnce, m ~ p2,
across the turbine. (b) Determine the elevation h. ‘ IFIGURE P5.119 (a) 7776 every y egwn‘im across #749 farb/ha ix Tﬁw; ;
1% +65 =% or W
ﬁvfz , «(y/z.J J where Q. ‘3 "3
HemeJ  ' 3AM»
W: W; : ~2500X40 1;" _ 31! z .
,9,~p,_=—a‘ Y6)" «2 ""( 20m; ) “25x’0m1 £55. 1 (Mia: M4 <0 because a {kirk/n?) removes 9:617}! from #13 HUM“ _ ’ 2 .
£423 +2; ”LIZ: ”643.4 7: & 4.244.1/1 (”bf/’6‘ W3 5/7; ‘0) 254“?) V3"? 6? ¢ﬁl
7:7  and 2'34!
u: z .
1715/; —b = ‘4 a
5 ‘34 371:; J r
2
h5g4. ‘34" “A: om3
“(3 " 21/ 1". Fiﬁ .: 5'
14/50) 14—2; (3’:me 6 75 ’53? 2 in, 4/71
A"!!! I Therefore, from Eq. (I) h :: {6.3712531
2(7.8*/g';) + 2.5m '('/i.76m) = [7.3 m :5~m¢ . 5 [,2] Section (2) 5.12! Water is to be moved from one large
reservoir to another at a higher elevation as in 3;mnnside
dicated in Fig P5 121. The loss in available en diameter pipe
ergy associated with 2.5 ft3/s being pumped from
sections (1) to (2) is 61V2/2 where V is the av—
erage velocity of water in the 8in. inside diam
eter piping involved. Determine the amount of
shaft power required. Em» Hae [/ow 14pm 5&79'011 (I) 7% Sec/fan (2) 53. 51192 lead/1
7‘0 «z
=/242[7 (aseh ’0SS]=/aée[5/z,2,>+ 61%] (I) WsAaH
nef in —— i 0.5 ﬂy
17' *1 51’1: ’
9‘ /2/5;.
771a!) 757»: £25. /
w : (11.911 5/1115) )[2 s :11) (22 2 ﬂ)/50{f)
7641C!" ff? 1
ncf m +._(é£2(7'/62?2// ,1 a )I z 2/“, .13: {50 {L /4
‘W 5 151,12
= 21? l:
MC/mff :8
nef H1 5436 ...
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 Spring '09
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