ch04 - 4.1 The velocity field of a flow is given by V...

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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% ‘ 7-56 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 0-13 is a parabola. The pathline for a stream of water leaving a small 1-0 0-00 nozzle is shown in Fig. P4.5b and Video V4.3. The coordinates 1-25 “0-20 (a) for this water stream are given in the following table. (a) Use 150 "0-53 the given data to determine appropriate values for cl and c; in 1.75 ‘0-90 the above equation and, thus, show that these water particles 2-00 ‘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’v-l-=——=-‘*—— (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 three-dimensional 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% {ii-gag" 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) 3-f- = 4% and 5}— = 2-51 (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—3-x“-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‘f-Q {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=~zX-fi 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+WE-Tr+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..2-3 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(‘f-m) =/.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. P4.21f-and 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 0.2-s 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 4-21 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: moved a d/lr/Mce X 5 2 £3 (Ms) 50.?- 79‘. 5/054? 7% Ve/ocxi’y prof/79 is [Gibson/ire ”flea/'1 #19 8/751: 0f My cycle/n w/V/fllal/e so Md line: All and 86 remain simiqbi’. 72/: is ind/baled in flesh/ah be/ow. Y 0.2.H _‘ 0.2. H I" — lDNI-B/ {I J hi , “""—"“ “—""=‘"— " y” 7‘ :ltiéiefim‘i , / K\'*_, ex/{fed Zuni/VI o’c’miml I yo/yme Volume . X ”1/4. , l/I alsff J. Lei/5207’ 3:2;u - ———_-_-=_=-;_.=_=’.=_r--c = CI __ __ _ con/ro/ Vo/mre ---_- - -- - 4 system 47‘ {‘0./: L/wzz 4.29 Water enters a 5-ft-wide, l-ft-deep 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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