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Unformatted text preview: 4.1 The velocity ﬁeld of a ﬂow is given by V =
(3y + 2)i + (x — 8)j + 52k ft/s wherex y, andzare 1n feet.
Detennine the ﬂuid speed at the origin (x = y = z = O) and
on they axis (x = z = 0). a:3yf2 ’ =X ",8 W =5
I . 1
”US, of X=y=z:0 V: W} 2‘+(—8) 3 25L and on Me line Xsyz= 0, V=.]/(3y+2)"+(—8)2 =./9y2 +/2y+68 g w/iere y~{{ 4.2 A flow can be visualized by plotting the velocity
ﬁeld as velocity vectors at representative locations in the
ﬂow as shown in Video V4.1 and Fig. E4.1. Consider the
velocity ﬁeld given in polar coordinates by v, = ~10/r
and v9 = lO/r. This flow approximates a ﬂuid swirling into
a sink as shown in Fig. P42. Plot the velocity ﬁeld at
locations given by r = l, 2, and 3 with 6 = 0, 30, 60, and
90 deg. WW1 N’?="/02/r'and Ar =10/r Men V= {new =,/(«/o/,~)z +(10/r)‘= ”" ’“
The any/e 0( AGZefu/een {he radial direoi/anr and
the ve/ocify vecfor is git/en by 7771/3 D(= 45°51" any r0 (5 e. {be ve/ocn’y V861l0f I: a/u/ay oer/Eden! 6‘5 re/m‘u/e form/14d ﬁnes) 4.3 4 3 The x and y components of velocity for a two dimen
sional flow are u = 3 ft/s and v = 9x ft/s. where x is in feet.
Determine the equation for the streamlines and graph repre
sentative streamlines in the upper half plane. a = 3 and/1r = 7x2 .90 Mai dream/flies are g/i/e/J by {7% : ,u/‘L's 7%: =3xz 0r fdy =f3xzc£r
777V: y= X3 +6 Were C /.r aca/mlaml. Represenfafxve dream/Mes correspana/My f0 0/1'7’1’8/‘90/ Lia/(la: of
'c are show» below. Streamlines y = x3 + C C: 4.4 A velocity ﬁeld is given by V = xi + x(x  1)(y + 1)}, where u and v are in ft/s and
x and y are in feet. Plot the streamline that passes
through x = 0 and y = 0. Compare this stream line with the streakline through the origin. a = X v = X (X “IN Y 1") where {he stream/Mes are obfaiﬂed 1’70”!) ='91'X "X +5 where C is'a consfa/If (I) For ‘H78 dream/[ﬁe Hia‘l passes “Wag/I H19 orig/II X=y60 7%? value of Ci: found from £910 as
In“) =6 1 or C=0 Thus, /n(y+/)I=fX2.'X or y=€
This sireamline is p/oHed below. (—ZL xz—x)_ I ”07% ‘ 756 Sff?0M/I'”¢ is Syramefr/‘ca/ 0500+ It: /ow pofm‘
of X=/ , y=— 0.373. ﬂ/ x=y=0 7%6 Vc/oc/{y is 0. For X<0, u<0 and for X>0j u>0, TM; 7% ma
Holds from #13 origin (x=y=0). Since fbe {/ow is steady, streak/mes are #23 same as stream/Mes. 4.5 Consider a ball thrown with initial speed V0 at an angle . .
of 0 as shown in Fig. P4.5a. As discussed in beginning physics, x,m ....................... y ’.....l..n.' ......
if friction is negligible the path that the ball takes is given by 0 0
, y = (tan 6)x — [g/(z Vozcosz 0)]x2 8:33 8 12
That is, y = clx + czxz, where c, and C; are constants. The path 075 013
is a parabola. The pathline for a stream of water leaving a small 10 000
nozzle is shown in Fig. P4.5b and Video V4.3. The coordinates 125 “020 (a)
for this water stream are given in the following table. (a) Use 150 "053
the given data to determine appropriate values for cl and c; in 1.75 ‘090
the above equation and, thus, show that these water particles 200 ‘143 also follow a parabolic pathline. (b) Use your values of c. and
c; to determine the speed of the water, V0, leaving the nozzle. (b)
H FIGURE P425 An E X 6E1 Program Was wed 7‘0 p/a/ Me X7 447% 4/7/
7’0 fill a second arc/er curve 7’0 7% d4 fa. Tie raw/Zr are warm {Se/ow. y vs x for Water Stream y = 8.4987x2 + 0.7 77m) mil/7 y = qx +6z x2 if fry/kw: iéaf
c,= 0.7/l5 mime or 6’ saw” and
,  .. _.__L_
C1="8.‘9‘7y7" Zvaz 19 606‘
01‘ 1
2.1 32. 2 _ 1%
V9 z 373.9727) cosﬁssﬂ“) 2‘95 5‘ 4.6 In addition to the customary horizontal velocity
components of the air in the atmosphere (the “wind”), there
often are vertical air currents (thermals) caused by buoyant
effects due to uneven heating of the air as indicated in Fig.
P4.6. Assume that the velocity ﬁeld in a certain region is ap
proximated by u = un, 12 = 110(1  y/h) for 0 < y < h,
and u = uo, v = O for y > h. Determine the equation for
the streamlines and plot the streamline that passes through the origin for values of uolvo = 0.5, 1, and 2. I FIGURE P4.6 “=55 , v=Vo(I—hL) {or 0<Y<h 50 Mai .sl‘ream/ioes
For y<h are given by y é’vl=——=‘*—— (if) 0’“ 7%)—
o Tbils, “h /h(/“‘},Z) =~Z€X 0 ”0’6: 7'59 /owe/~ Mir/'7‘: offni’eyraiion
(X=0, y=0) insure Mai ”If:
equaiian is for #79 siream/fne
”Wally/I Hie orig/I7. 77M: dream/file X :4) ("$32) In (I “ 77):) is p/oiied below. whlh — — — ratio = 2
~~~~~~ ratio = 1
ratio = 0.5 4.?" Repeat Problem 4.6 using the same in
fdrmation except that u = uoy/h for 0 S y E II
rather than u = u”. Use values of uo/vo = 0, 0.1,
0.2, 0.4, 0.6, 0.8, and 1.0. I FIGURE P4.6 a: uﬁy , V: Va (l'%) (or 0<y<h sci/701‘ streamlines for y<h are given by
or wh‘l) X=0 wheny=Q *( This stream/me is P/oHed below for 0s % s] I WM}; ratio= .911 and u ” nro
VI? = OJQ.IJ_0.2,0.1/,0.6, 0.8, and/.0. +ratio = 1.0
+ratio = 0.8
+ratio = 0.6
—)(—ratio = 0.4
+ratio = 0.2
+ratio = 0.1 A threedimensional velocity ﬁeld is given by u = x2,
v = ~2xy. and w = x + y. Determine the acceleration vector. M.
ax a w
”y‘ﬁ'rw‘g’lfwﬁ : X1(~2y) +(ZX,V)[.Zx) = szy
a; = 191% {iigag" 4711‘ 4J0 The velocity of air in the diverging pipe shown in Fig. P4..l0 is given by VI = 4r ft/s and V2 = 2! ft/s, where t is in seconds. (3) Determine the local acceleration at points (1) and (2). (b) Is the average convective acceleration between these V1= “ft/5
two points negative, zero, or positive? Explain. all All H
a) 3f = 4% and 5}— = 251
(I) (2) . . . J
b) conved/i/e accelerm‘mn along Me [71/08 = a 77% where u >0. ﬂ/ (my 79/776) 2’, V2 < V, . 751/5 bah/em (Ha/201(2) LIL” V’'V’<0 9X 1
Hence] a 7}; < 0 or Me ave/“479 amuse/{ye acce/eraf/an
is negaf/Va. 4.11 A ﬂuid particle ﬂowing along a stagnation stream line, as shown in Video V4.5 and Fig. P4.11, slows down Stagnation point, a. z 0 as it approaches the stagnation point. Measurements of the dye ﬂow in the video indicate that the location of a particle _ V _, F'Uid Pamc'e
starting on the stagnation streamline a distance 5 = 0.6 ft “‘ '
upstream of the stagnation point at t = 0 is given approxi
mately by s = 0.6 6‘”, where t is in seconds and s is in ft.
(3) Determine the speed of a ﬂuid particle as a function
of time, meidco), as it ﬂows along the streamline. (b) De
termine the speed of the ﬂuid as a function of position
along the streamline, V = V(s). (c) Determine the ﬂuid ac
celeration along the streamline as a function of position. as = all ) l FIGURE P4.11 «0.5:! (a) Wiih s=0.6 8 i/ follow: ‘l/Ia‘i
v . ' = ”'5‘ 1% P0 (b) From pam‘ (a),  0.5! 7 ms!
(—0.5)[a6e0 ] Where 3:0.58 0,‘ V: 4.53 {1% when: S~ﬁ (6) For sfead/ flow, 4, = V %
77711.5 Will/I V‘A‘0'53 and %= ’0'5J ' a; = (.0.55)(.o.s) = 0.255 H/s‘ where s~H ﬂ/ofei For S>OJ as is [Dari/file» 7% parf/c/e} accelera/fan L: 7‘0 f/Ie r/‘yéf
Since fhe pawl/ale is may/[77 7‘0 7%9 Mi a pas/Wile a: {or 7%: care
imp/ﬁes ﬁmi‘ {he panﬁc/e 1's Jase/6mm); (as i/ ”7167‘ be for Ml?
slay/mile!) poinf f/ow). 4J2 4.12 The ﬂuid velocity along the x axis shown W Va = 36 m/s in Fig. P4.l2. changes from 12 m/s at point A to VA = 12 ms 36 m/s at point B. It is also known that the ve
locity is a linear function of distance along the ——> A
streamline. Determine the acceleration at points
A, B, and C. Assume steady ﬂow. W C B I
0.05 m—J , FIGURE P4. l2 A =§¥+V'VV ,h/z'f/I qum J V''0, and w=0
+hl;$ becomes
Ez‘=—j—fr‘+u§§‘—)f=u—3x“f (I) Since (A is a //'/qur fund/on of X , a: C,X +62 w/wre Me
consfam‘s C, ,C‘2 are given as 9' “29"2 = (;z
“”51 1/3 =35 =o./c, +62 0" 6/:240 C 5/2
Thm, u=(24ox+/2)v’;;° will: x~m ’ 2 From Eq, {I}
51‘ = (1%? Z" =(21/0x +12) g1 (240%) (A 0!" .
For X,,=0 , 21:52 BBOZ‘fQ {or XB=0.Sm} 58:57”???
__ A ._s .— A,”
for x5 aim , “a — 8 540131 “'W'W_ 4J3 An incompressible ﬂuid flows past a turbine blade as
shown in Fig. P4J3 a and Video V4.5. Far upstream and down
stream of the blade the velocity is V0. Measurements show that
the velocity of the ﬂuid along streamline A—F near the blade is
as indicated in Fig. P4.13'b. Sketch the streamwise component
of acceleration, ax, as a function of distance, 5, along the stream
line. Discuss the important characteristics of your result. W FlGURE P4,!3
(a) as = V343! where from {he r’fyu/‘e of V = W!) the {ma 79m
5:", has {be fol/owing s/mloe. 7776 f/U/‘a’ dEGe/Cl‘a/es 79‘0”} ,4 7‘0 C) acce/grafes from C {a A, 404/
H79 niece/arm‘s: age/ﬂ from D I‘OF. The ”6/ aces/”4w” I’M/W
A {a F}: zero (is? 1/4 = V044»), 4’l0 4J4 AAﬂuid ﬂows along the x axis with a velocity given by
V = (x/t)i, where x is in feet and t in seconds. (a) Plot the
speed for 0 S x S 10 ft and t = 3 s. (b) Plot the speed forx =
7 ft and 2 S t S 4 s. (c) Determine the local and convective
acceleration. (d) Show that the acceleration of any ﬂuid particle
in the ﬂow is zero. (e) Explain physically how the velocity of
a particle in this unsteady ﬂow remains constant throughout its motion. (a) a=~zXﬁ so az‘ i=3s 5 1 (b) ForX=7HJ u: (0/) For any f/w'c/ parﬁc/e E=%¥+V'7V
Wh/c/I wf/ﬁ V'—'0 ’ w=0 becomes a=(§%+u%%)2=({1+7é)r so (6’) TAB [cart/dc: f/OW Ib/o areas at“ big/)6? 1/62/0039 (568 HyJ),
507‘ at‘ '0”; given boat/on 7% 1/6/06”) is decreasing /}7 21/7273
(see Fig.2). For Me yin/ex) yea/051.75! {is/61 Me /oca/ 4/10! awed/Va awe/gratin”: are e7ya/ and ale/0:178 ) y/V/IIy
,2 cm aces/erotic)” ﬁrm/M om‘. 4.15 A hydraulic jump is a rather sudden change in Hydraulic jump
depth of a liquid layer as it flows in an open channel as r—*—ﬁ
shown in Fig. P4.15 and Video V105 and 10.6. In a rela— F—4
tively short distance(thickness = d) the liquid depth changes from 21 to 22, with a corresponding change in velocity from V1 to V2. If V, = 5 m/s, V2 = 1 m/s, and E = 0.2 m, estimate the average deceleration of the liquid as it flows across the hydraulic jump. How many g’s deceleration does this rep— resent? FIGURE P4.l5 __x a bad V _;
a: T+VVV so Will/l V: ((0014 a =0xl =d5—xl W/f/Iout‘ knowmg the act‘aa/ ve/och‘y d/Lsfr/bohan a: Wx) Me acce/erafion can he approXI/nafea/ as M (V2. VI): Mil—1%
2 Z 0.2m 60m . . I 2
7—5/3 /: = 7—8721: = 6./2 .— 4.16 A nozzle is designed to accelerate the ﬂuid from VI to
V2 in a linear fashion. That is, V = ax + b, where a and b are
constants. If the ﬂow is constant with VI = 10 m/s at x, = 0
and V2 = 25 m/s at x2 = l m, determine the local acceleration,
the convective acceleration, and the acceleration of the ﬂuid at points (1) and (2). ‘
With u=ax +6 J V=0, ana’ w=0 Me acceleration 21’ can be wriHen as
all: Ef=0xf where ax=u7;,
Since u=V,='0'§'Z mlx=0 and 4:14:25én allX'J/ 14/30521th IO: 0 +b
25" a *1) so Hm" a=/5 and [3:10
7502‘ is, a=(/5X+/0){921 J Mere X~m , so H'Ial from Ezﬂ) ax .—. (l5x+l0)—én ([5 EL) =(zzsx+l.so)§’ﬂ /V0fe‘ The /oca/ accelerah'on is zero J £4 = con Vecf/i/e acceleration is ﬂlx=0 ,' E=I502%; afx=/m , a‘ #43 4. I 7 Assume the temperature of the exhaust
in an exhaust pipe can be approximated by T =
T0(1 + ae'b‘)[1 + c cos(wt)], where T0 =
100 °C, a = 3, b = 0.03 m”, c = 0.05, and a)
= 100 rad/s. If the exhaust speed is a constant 3
m/s, determine the time rate of change of temp
erature of the ﬂuid particles at x = O and x = 4mwhent=0.p Slhce u= 3 30!. 1 V=0 , and w=0 if fa//ows that 91,31". 2I 21 W aT_a‘r 3T
Dt’at+VVT’at+“ax*VW+WETr+u3y =0: _
%{ ::  abu7; (1 +6) e’bx 1 or my}; H78 git/8r) dafa’
%I= ”(3NO'037‘5M3 én‘)(lao°C)(l+o.os) e— °‘°3X
=28.’% e—O'OBX 3%— , w/zere X~m Thus, %TT=—28.43_E» afx=o i=0 J 4—1:? 4.18 The temperature distribution in a ﬂuid is given by T =
10x + 5y, where x and y are the horizontal and vertical coor
dinates in meters and T is in degrees centigrade. Determine the
time rate of change of temperature of a ﬂuid particle traveling (a) horizontally with u = 20 m/s, 1) = 0 or (b) vertically with
u = 0,1) = 20 m/s 0T: JT 3T ’6
77):: if 11:20? ant/V: 0 Many 4.19 Assume that the streamlines for the wingtip vor
tices from an airplane (see Fig. P4.19 and Video V4.2) can be approximated by circles of radius r and that the speed is
V = K/r, where K is a constant. Determine the streamline
acceleration, as, and the normal acceleration, a," for this
ﬂow. I FIGURE P4.19 where Slhce V‘é J 4.20 At the top of its trajectory, the stream of water shown
in Fig. P420 and Video V4.3 flows with a horizontal velocity
of 1.80 ft/s. The radius of curvature of its streamline at that point is approximately 0.10 ft. Determine the normal compo at Fl (3 U R E P420
nent of acceleration at that location. 2.
_ 1
_ Lﬁiil: 32 2%. Note' an sacccleraiion ’) 4. 2! Water ﬂows steadily through the funnel shown in Fig.
P421. Throughout most of the funnel the ﬂow 15 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 and 8.
FIGURE P4.2l 2
:0” n + 05W Wile/"e a,,=% =0 since W5” (4 3 H79 dream/me: 1‘ hi)
19/5905 ——§j\;/‘V V7; 1 where V= are S r‘mg 5/0619 V: 0 43’!) when r= 0. lm if {0/3/014/3 Mai _3 4x
c= v,» =(0‘f;")(0.lm) =4x/o £13, orV=——— ’° gram/13pm," 1/46 4. 22 Air ﬂows from a pipe into the region
between two parallel circular disks as shown in
Fig. P412. The ﬂuid velocity in the gap between the disks is closely approximated by V = V0R{r, [KI/11111111111111]!!!
where R is the radius of the d1sk, r rs the radial I I coordinate, and V0 is the ﬂuid velocity at the edge of the disk. Determine the acceleration for r =
1. 2. or 3 ft if Vo = 5 ft/s and R = 3 ft. FIGURE P4..22 . 2 . '
21’ = an [7‘ + as é‘ , where 0,, =7\'é‘ =0 since W5“ (Leo, the streamlines
by 31 y, R are straighi )
ﬂlsa, as = V75 = V 0,. , where V= *7.“ Since 14:5? and fir—3701’, V5 if; if, wbere rNﬁ Tﬁl/S, .,,c+ a. )2.
13 ”LB. ’_—ﬁ2='(5§)(3H :
a$=( :1 )( :2) "’ r3 ' [.3 H3 [qt/“3h”, gJ=2251f§’;~ A! r =2 H, as =' ~28./ fl A+r=3H, 05= ~8.33.§i 1M7 4.25% Water ﬂows through a duct of square cross section as
shown in Fig. P4.23 with a constant. uniform velocity of V =
20 m/s. Consider ﬂuid particles that lie along line A—B at time
t = 0. Determine the position of these particles, denoted by line
A'—B', when r = 0.20 5. Use the volume of ﬂuid in the region
between lines A—B and A’—B’ to determine the ﬂowrate in the
duct. Repeat the problem for ﬂuid particles originally along line
C—D; along line E—F. Compare your three answers. FIGURE P4..23 Since V is comic/77‘ if) time and space , a// [Darth/es on line ”3
move a disfance Z: l/m‘ ='(20;.’1)(o.23) = 4”, {NM {:0 /0 ,1: (223
Thus, #76 1/0/0076 oar/98142903 AAA’B’ = (0.5m)2(‘fm) =/.00 m3 50 ”mi
Q= JV/‘taxt’ia'__l.00m __50_m_
A2 ~ 0.25 _ ' ~5 S/hI/ar/y from i=0 fa 7.90.25 1%? f/w'd a/My Mes C D and E F
move 2‘0 C D'amd ETIrespecfit/ea/y. ﬂ/sa, ’VéDc/D/ slérs’r’ =Jqugﬂg/
50 ﬁnd we obtain 62: 3’:— =5.a{;" regard/m which line we comic/an l ~—— ... « ~~~~~ ‘L/Control surface
Slurce gate l 4. 2.4 In the region just downstream of a sluice gate, the wa
ter may develop a reverse ﬂow region as is indicated in I
Fig. P4.21fand Video V105. The velocity proﬁle is assumed to I
consist of two uniform regions, one with velocity V“ = 10 fps
and the other with V,, = 3 fps. Determine the net flowrate of I
water across the portion of the control surface at section (2) if , the channel is 20 ft wide. (1) (2)' FIGURE P4. 24 Q = 14 ,44 , Via/)1) = (/0 ﬂimzmmm —(33’1)(I,9f+)(20ril V
= /32 .24: we 4.2.5 Air enters an elbow with a uniform speed of 10 m/s
as shown in Fig. P415. At the exit of the elbow the velocity
proﬁle is not uniform. In fact. there is a region of separation or reverse flow. The ﬁxed control volume ABCD coincides with
the system at time t = 0. Make a sketch to indicate (a) the system at time I = 0.01 s and (b) the ﬂuid that has entered and
exited the control volume in that time period. I FIGURE P4.25 From 2‘ =0 7‘0 {= 0.0/3 par/ic/es 6’, B, Q D, and E Max/e Me
fol/oWI'ny ah'sfances: Cl”: “90‘! =(/0—?')(0.015) =0./m :60
63: 1435545 %)(0,013) =0.05m
(lo: Var”: (/5?) (0.015) =0./5m , and (55 =0
772ml f/l/I'a/ 0/0/79 ”/78: ,90 and BEE orig/rm”) Moves f0 ””65
ﬁ’D'and B’E'C’ shown [no/aw, sysfem at {=0 ._ _ _ _. sydem a1l [=0.0/s //// {/z/fal ‘l/7a7‘ exi/ed com’m/ vo/ume \\\\ fluid f/zm’ entered oom‘ro/
wwme 4—H 4. 2.6 Water ﬂows in the branching pipe shown in Fig. 134.26
with uniform velocity at each inlet and outlet. The ﬁxed control
volume indicated coincides with the system at time t = 20 5.
Make a sketch to indicate (a) the boundary of the system at time
I = 20.2 s, (b) the ﬂuid that left the control volume during that
0.2s interval, and (c) the ﬂuid that entered the control volume
during that time interval. — — — Control volume FIGURE P4.26 Since Vis COMM/77‘, 71/78 f/w’d iravels a d/sfcmce Z: Wz‘ /‘/2
ﬁrm At. ms, 1;: v, at = new 4% = 0,4177
lz= V2. (if = (l .522) (20. do): = 0,2,"
and 13: V3 62! = (2.5%)(20, —7_o)s = 0.50,» 7728 sysfem m‘ 71:20:23 and Ma f/w'a/ Ma/ has en/erec/ or
exf/ed Me control volume are ind/baled in ”)6 fly/Ire below. ’(mV/HW W0 cam’ra/ Vo/.
’K/ \Ab\( ilow oui of confroi vol: /' ,/ )0. ‘fm — ~— — COﬂ/f‘o/ volume \ 7}”)
 sysfem 47‘ #2023 4—20 4. 2.7 Two plates are pulled in opposite direc
tions with speeds of 1.0 ft/s as shown in Fig.
P4.27. The oil between the plates moves with a velocity given by V = 10 yi ft/s, where y is in Feet The ﬁxed control volume ABCD coincides
with the system at time t = 0. Make a sketch to indicate (a) the system at time t = 0.2 s and (b)
the ﬂuid that has entered and exited the control volume in that time period. ~
FIGURE P4127 Since V460)? = /0y(A if {Wows Him! 7%6‘ ﬂuid ﬂows [/7 7‘58
X'direcf/‘an a dis/once of 6X = a 61‘ = /0)/ (0.2) H = 2y {7’
from i=0 *0 {‘0'25 7776 //‘/76.5 ﬂ—B and 5’0 (Me ens/s of
Hie original system loaf/on) deform info line: 245F401 CLO/as
shown in #73 figure below. The portions of Me sysfem Hm‘ have
em‘ered and exh‘ed {/79 cam’ro/ volume all/ring 2%}: ﬁne are indicated . /X _  _ 11\;i_\:j\f ow m/o cam’ro/ V0/. D
— — — — com‘ro/ Vo/z/me ——— system all [=01 s 421 6628 2 ft/s
4.28 Two liquids with different densities and viscosities ﬁll '
the gap between parallel plates as shown in Fig. P4.28.The bot
tom plate is ﬁxed; the top plate moves with a speed of 2 ft/s.
The velocity proﬁle consists of two linear segements as indi
cated. The ﬁxed control volume ABCD coincides with the sys
tem at time t = 0. Make a sketch to indicate (a) the system at
time t = 0.] s and (b) the ﬂuid that has entered and exited the
control volume in that time period. ' Oft/s
i———————1.6ft——————————i iii FiGURE P428 The f/u/‘J 47’ y = , 0.642”! (Me bot/om p/az’e) fawn/m s/af/Md/y.
14/ y=0 {be f/I/id spew/115 ASH/r .ra Med 41‘ {five i=0,/s [Mm
mweal iv #29 [776/ a old/Mae X = W = /5 £15 (0.13) = 0./5f'/. In Me same //'/»9 period {he fop ,b/a/e and Me f/w‘d sill/6:4 2‘0 {7’ #4:
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__ __ _ con/ro/ Vo/mre _    4 system 47‘ {‘0./: L/wzz 4.29 Water enters a 5ftwide, lftdeep channel as shown in Fig. P4329. Across the inlet the water velocity is 6 ft/s in
the center portion of the channel and l ft/s in the remainder of it. Farther downstream the water flows at a uniform 2 ft/s
velocity across the entire channel. The ﬁxed control volume
ABCD coincides with the system at time t = 0. Make a sketch
to indicate (a) the system at time t = 0.5 s and (b) the ﬂuid
that has entered and exited the control volume in that time period. sag
 Control surface W FlGURE P429 During Me 1‘ ‘05 5 {I’m Ihfc/‘Va/ Me {/w'c/ Mm’ W4: 4/0/79
line 36 47‘ film i=0 hop Moved 7’0...
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 Fall '07
 Longmire

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