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Unformatted text preview: — — in: ‘ —. q—qv‘ 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)
Vw: ii< 3°31 + Lawm + 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 34103): =  1033 + constant —)  th = constant ‘
5." “q gum11mm 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 11031
= 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 EranIlse 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 dGr 51mg, 1'
0 f; D I pr2 [Hr pl‘zh r1 1
=——h2n' —=———1——_—r.
241:2 ,1 r 4n_"r. 2” Q '5 W Ecerciw 5.8 a) qu=w=23=const. b) [hi(1A: . uds=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 UniversityWest Lafayette.
 Spring '11
 Wereley

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