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Unformatted text preview: I 4. 6 Show that the streamlines for a ﬂow whose ‘ velocity components are u = C(Jt2  yz) and v =
2€xy, where c is a constant, are given by the;
equation xzy  y‘l3 a constant. At which point; ~
(points) is the ﬂow parallel to the y axis? At which point (points) is the ﬂuid stationary? V W . : M=C(X"Yz) , v=426xy 5+reamlines given by y=f(X) 45's sac/2 Mai, =7:— ‘Consiz/er #78 {wad/Ton X‘y " 4% =60mﬁ I Wok: If is no} easy f0 til/"He 7%[5‘ exp/ioiﬂy as 'yxf/X) ' ' ” However, We can differs/7790716 5?, [0 7‘0 git/8
zxydx _{+x1a/y avid), so? I or
(Xi—WM), +2xydx=0 Thug, Me [Was I}; 7% x—y p/ane ' give/7 by 571/) 1541/3 a s/ope d ‘ 42x  " * 4“ x
I 7% s Ol‘ {Ur (my C0”Jfa”7l 6} —2§‘HHC(X:_>11) 3 ale. ﬁe 10005190” X 2y "3):— =com7l4~ represem’s Me stream/mas
' Of Me gEil/en f/ow. L ' ,l V (l) 7779 now ,3: Pam/M lo Hm xm We” %=o, or we.
‘ 771/5 066m? w/Jc/l 63/‘7’56’P’X1'0 or y=0 32:82,:71/16’ X~axxir or 7%6’ yea/Ls
or d :0. 7729 How parallel {9 H19 fax/‘5 ‘ when :00
777/: Occur: when X = i X I The flu/c1 has zero 1/6/0613} 07‘ X z y =0 4:15 4.15 Determine the acceleration ﬁeld for a three—dimensional
ﬂow with velocity cOmponents u = ~15, 1) = 4x23) 2, and
w = x — y. u= ~X, xv: #x’y‘l aﬂddﬂ‘x’750 “W r. 0 +(—x)(~/) +4X‘y‘(o) +(x*y)(0) = X ..&M Mr M Mr
ay*X?—4uﬁ “WWW turf; —= 0 + (—x) (8xy’) +(4‘X‘y3) {8x1y) +(x—y)(o) : waxzyz X4y3 : 8X2Y1(4X1y _l)
and
air as =7;—+u%’+nr%’£ur%r
= 0 + (x)(l)+(4x’y’)('I)+(x»y)(0)
== ’Xé‘xly"
Thus, . ,
Zz’mxz‘myj‘mgﬂ .
= XZ‘ +8x‘y’(1£Xzyvl)j e(X+6‘x’y?)k 4.45 Water ﬂows steadily through the funnel shown in Fig.
P4.45. Throughout most of the funnel the ﬂow is approximately
radial (along rays from 0) with a velocity of V = c/rz, where
r is the: radial coordinate and c is a constant. If the velocity is 0.4 m/s when r = 0.1 m, determine the acceleration at pOints
A and15’. FIGURE P4.45 2' (‘1‘ = an F3 + a‘S 3, where (bfYR: =0 sﬂace W600 (ale, ﬁve. streamlines
are draught?) V
A150, a5 = v37! =—v 53,: , where v = 7% 5/1709 V= 0.4—? when r=0.lm if fol/0m; “ml 2 _ 3 4x10“?
c= Vrz=lb.‘f3m)(0.Im) =‘HK/0 3% , or V= r2 3’55 where r~~m 7771/3, 2
“.9... _2._§__ , 26,
a$”(r7)( r3)" i~
/H~ mini/4: a '
P ,_ (six/04% )‘Z 320 m
a e“?  ~ ‘52
’5 (00,173) :=:::=_._— l9} poirnl B2 3
2(4):;633’9— 2
(cu/Hm)5 = Lifeé": 05: 4.5% 4.48 Air flows into a pipe from the region between a circular
disk and a cone as shown in Fig. P448. The fluid velocity in the
gap between the disk and the cone is closely approximated by
v = VoRz/rz, where R is the radius of the disk, r is the radial Q
coordinate, and V0 is the ﬂuid velocity at the edge of the disk.
Determine the acceleration for r = 0.5 and 2 ft if V0 = 5 ft/s
and R = 2 ft. IFIQURE P4.48 z
a? s an [7‘ +05 3 J biz/Jere a” 5 5 0 Shea Wﬁb (1'. e. {/19 dream/1.1794: a _ , . are sire; ln‘)
ﬁlm a: = Val = —'V;,‘’,¥ 51/706 1‘ 0M3 are pol/lied m qofosii‘e gored/om. ﬁll/SJ WWI V= VaRZ/I‘2 ht fol/0W3 {heat
as (v., 19%") M ii Iii/r3) = mid/r” = 2 (5fiA)’(2f+)“/r5 = goo/r5 f1” w/zere r~ if
Mr = 0.5%" q: = 9001.5.){éé = 256mg; 5
AWN2% a: :: goo/(2.0) ,ffg gggg “mm 4.5 8 Water is squirted from a syringe with a of V ==
5 m/s by pushing in the plunger with a speed of V " p 0.03 m/s as shown in Fig. P458. The surface of the deforming
control volume consists of the sides and end of the cylinder and
the end of the plunger. The system consists of the water in the
syringe at t = 0 when the plunger is at section (1) as shown. Make a sketch to indicate the control surface and the system
whent = 0.5 s. FIGURE P458 During {beef $553 'fime [ﬁfe/"Val H73 p/mqer‘ mot/es ﬂ, = l4, (St10.0152»;
and {/16 Wafer/b/Wa/ﬂl af 11/109 ex/f Mal/6L5 [2: 52.507. 7763 correspond/[I79 antral surfaces 4/701 sysfe/m m‘ {:0 405/ If = 055
show in Me 71290]? bee/ow. — ~=—— —— com‘ro/ volume 42‘ #053 ~‘.*.‘ «‘13! if 6‘ y S fem af 1‘ = 0.5 s u \l 3" 4651% 4.60 Water ﬂows through the 2~mwide rectangular channel
shown in Fig. P4.60 with a uniform velocity of 3 m/s. (a) Di
rectly integrate Eq. 4.16 with b = 1 to determine the mass
ﬂowrate (kg/s) across section CD of the control volume.
tb) Repeat part (a) with b = l/p, where p is the density. Explain
the physical interpretation of the answer to part (b). — — — —  Control surface IFIGURE 94.60 6') Bow: =5 (>1; 17”? 01/! f: (I) “our 2 ‘ V
.A A 9 A Will/7 b =/ and V'n = V 5039 21/7/19 becomes Z n ' c 800* ‘(Egchasa =QV60396£JI9 =9er9 14“, , where ﬂap 3,6310!) [56% 0‘5," =(ﬁfemﬂzm)
= 60:19 )mz 7/3113, MM V=3m/s,
8001‘ z (3 6056’ (50:9)m21997'513) = 3000.? b) WI'7’/7 b = //p (/1 becomes B'Wf 5] 61/) =fl/cm9 M = V6039 40
an op i 3
=(3%)co:9 (335%)”? = 3.00%" WA b : ’/p = = M H mm m; 39 = Viz/m ” Val mass (4199 b = 5%) so fﬂaf I 1751/9 = 8;”, Feynman/s Me Ila/(1m
flawmz’e (01%) from ﬁle can f/‘a/ 1/0/0070. 5.4 Air ﬂows steadily between two cross sections in a long.
straight section of 0.1m inside—diameter pipe. The static tem— Section (1) sects‘cm (2)
perature and pressure at each section are indicated in Fig. P54. If the average air velocity at section (I) is 205 m/s, determine P1: 77 “Pa (3‘35) p2 = 45 “:3 (abs) . . . T = 268 K T = 240 K
the average at veloc1ty at section (2). v1: 205 m/s 2 I FSGURE P5.4 7/7}: anaéw/x i: JIM/Var 75 #1: one a?“ éi’dmp/e 5’. 2 .
For sfead’y f/ow Ae/Weeh 'Jccr‘l'ans (I) ana’{2]
m . n
S .5: l  lee. Asfum/nj uﬂm; under ﬂie Mimi; 19m; 0/ f4}; fmé/em/ m} Aehmxe; a s an idea/ 945 we: use #16 z‘dea/ ya;
cguaﬁ'on 0f sfa/e #3) 79 967‘ = (2)
“ ’2 ’3 7, (Oman/Jay €55. / man 2. and abserw/la; Ma)! 14, =41
We. gef Vi = P; 72— V, = [feﬂms](240/<) “<5
H 5.7 Water ﬂows along the centerline of a SO—mmdiameter pipe
with an average velocity of 10 m/s and out radially between two
large circular disks as shown in Fig. P57. The disks are paraliel
and Spaced 10 mm apart. Determine the average velocity of the
water at a radius of 300 mm in the space between the disks. For sfeddy I}? compress/6!: flew ' (4)277 '71 _ mzdwoommom) 0” Q, g 82 fﬁ=10mls
AV, z 4sz FIGURE P5.7
Thus 1 ._.. ‘2.
\7=A.V, 779 V, _ (50mm)(10m/s?
z __
dz v1 = 1.04 g: 5.1% I ' 5.14 Oil having a speciﬁc gravity of 0.9 is pumped as Section (1)
illustrated in Fig. P5.i4 with a water jet pump (see Video '\"3.6).
The water volume flowrate is 2 m3/s. The water and oil mix
ture has an average speciﬁc gravity of 0.95. Calculate the rate. Section (3) in m3/s, at which the pump moves oil. _>_
water Water
Q} = an?
« "Lm3/s “‘3'”
Section (2) (8G = 0'95)
Oil (80 = 0,9)
For “eddy {my FIGURE P514
. . z ml + ﬂ’l2~ 3
Of“
ﬂqﬂk/fQLz/gqa (I)
14/50 Slhce, #12 Waller and 0// may be. cow/'45er ,,‘, Com/Wx/é/ﬂ
/
Q, + €22 = a} (2) COMé/n/nj 5?; /ama’z we. gei‘
pIQ/ + 542 = gamma) 0f” , Q; + SGZQz = 563(6)! + Q2)
and Q n a, (1— 563) z _
5G3  SGZ Thus 3 \ Q = (1%)(/”01%_/ .—, 2001": 0.625 r 0.70 5 ...
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