homework5solutions - -— -— in: ‘ -—. q—q-v‘...

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Unformatted text preview: -— -— in: ‘ -—. q—q-v‘ C311!\I?1EIEI% £5 Exggciee 5.1 Angular velocity u = 40 rad/s h = marl/2g h, “firf/zg = (40i‘rf/2(9.31) 31. h 2 a 4 + h| = gari/Zg = 81,50: Subtracting (2) - (1). we get n 2 2r(2) {4 - 3) or I 2 where we have used (1) and (2) to replace hI and hi. Using (3!. the above becomes 0? + rL = 2/{81.5}(r: - r?) = 2/(81.5i(0.0491) = 0.: (4) Subtracting (4; — (3), we get 2:} = 0,5 7 0.0491 = 0.45. Thus ‘rI = 0.475 m so that Area uncovered =11rf = 0.71 ml Exe '52 5.2 p = - 2000 um1 ' . = _ u L pz 000 N/m 0 = 1.13 kg/m? at 25°C v = 25 m/s (8) Applying Bernoulli equation between infinity 'and edge core, we get U. = J 2(p“, - g)/P = J 2(2000)/1.18 = 58.2 n/a I. F = Zflr’U = 23115)(5a.2) = 5485 n‘/s I (b) Apply Bernoulli equation between points 2 and 1: P. *4 if".1= P1 +‘éf’U: q: = J 2(pl - gfiq3+ u? = q’2(--2ooo + 500)/1.18-:MEE:;E = 29.0? m/s I Since circulation outside core is conetnnt, rIUl ='rlU2. So r; = rlu' IIJz = 30.0 III Tine required = (r:L w a )/V = (30 * 15)]25 = 0.5 s \ cine fl Given ua = O u? = aRx 3 x K ‘1 . t a W I E = __.__(Ru,) « i- = -—.__ (an x) = 23x 0)" KER 4’ R" Rbg (b) V-w: ii< 3°31 + Law-m + em” \ ~~ RBR an. “3‘? “Bx ’-/0 of a, vv“—""‘—'-‘4“'V'—"V‘vavv I a 2 '3 =——-{—aR)+O+—(2ax)=-23+Za= R 3R ‘ 9x (c) Vortex lines are given by :1 d9. ' d l £01 a 1 ax -a~ Integrating R 34-103): = - 1033 + constant -—-) - th = constant ‘ 5-." “q gum-11mm m mw—Jnuamjhju In IRMA—citra- figggige 5“ v r— 1.- 4W./X1—"+3L ,EXA The components of the net velocity at point (any) due to the three image vortices are I‘ ' I— y u:ZV=V3--\u"_.s:I.ne=_1 - _-——--—-_.___ ‘— W 41T\/ X1153?- V 11-0-31 = fl; _ 3 = L x" 411' 3 11+32' ‘HT 3 (x1. 37) v = 2v! =-V + Vlcose = - ‘— + r— _.__i“__ i I" I X. ‘_ r' y?“ -—- ___ .. _ — —— —----—.—___ln 1 1 11+ 51L . Path lines are given by dx/dt = u(t) Therefore ‘ i Eran-Ilse 5. 7 For an irrotational vertical flow with circulation P, “9 = F121". The volumetric flow rate is " '2 " M“ r; i = drf dzv = —In -, \ Q ' [n 0 9 2:: r. h r: 2:! l . Kinetic energy =f dzf drf dG-r- 51mg, 1' 0 f; D I pr2 [Hr pl‘zh r1 1 =——-h-2n' -—-=—-——1——_-—r. 241:2 ,1 r 4n_"r. 2” Q '5 W Ecerciw 5.8 a) qu=w=23=const. b) [hi-(1A: . u-ds=f A =34 0 since ua (r = R) = 0 by no slip. J w «M = 2:23:22, A as - R610 =0, 73%....- 75f: (1 ...
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This note was uploaded on 10/01/2011 for the course ME 509 taught by Professor Wereley during the Spring '11 term at Purdue University-West Lafayette.

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