HW#4%20solution%20corrected - I 4 6 Show that the...

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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 'y-xf/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 x-m 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) == ’X-é‘xly" Thus, . , Zz’mxz‘myj‘mgfl . = XZ‘ +8x‘y’(1£Xz-yvl)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 (bf-YR: =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.‘f-3m-)(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 = Life-é": 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 AWN-2% 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, (St-10.0152»; 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 = 0-55 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" 46-51% 4.60 Water flows through the 2~m-wide 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 0.1-m 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—mm-diameter 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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HW#4%20solution%20corrected - I 4 6 Show that the...

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