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set9-answers - E 01o E06157(qua-ho"(W50 U09 Secixon 064...

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Unformatted text preview: E! 01o) E06157 (qua-ho" (W50) : U09 Secixon 064 P) “I V; Luz) (;f~¥:+5")“ (77* "r “‘5‘“ h“ R .4 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 " w’ V ,__'—_'3,4 ,V/ 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 cu-kr and t MEMMMM a'r 1h; wan of +1“ Fir-C 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 ° ‘ form-f)“ 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 (Ma-43””, 90" At) t p1,;p1 +04??? iiffi (Lp‘kz) F Eli-331+- 0.2:3 wot-r 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 y-commnem‘ 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-(I2-ff/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: fia-I'Zz Ali-7%..) 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 8-in. -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 (as-eh ’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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