solutions3 - 4.54» Air enters an elbow with a uniform...

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Unformatted text preview: 4.54» Air enters an elbow with a uniform speed of 10 m/s as shown in Fig. P454. 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 t = 0.01 s and (b) the fluid that has entered and exited the control volume in that time period. From {=0 7‘0 f= 0.0/5 parfic/es fl, 3, Q D, and E Move 7% following dis/anew: Jfl= Vqu =(103’3’v){0.ms) =0./m = JD 68: MBJ£=(5 %)(0.0Is) =0.05m 6C: (/5331) (0.015) =0./5m J and 65 =0 TfiI/SJ fluid //./76J‘ dfld or‘I-glhal/y moves 7‘0 line‘s- w’ and B’E’C’ shown below. sysfem at {=0 ._ _ _ _ sysfem af {=0.0ls //// f/m'al {hat exi/ed com’ro/ volume fluid Marl em‘ered amt/“o/ Vol/m e 4—45 45$ A layer of oil flows down a vertical plate as shown in Fig. P455 with a velocity of v = (Vo/hz) (th — x2)j where V0 and h are constants. (a) Show that the fluid sticks to the plate and that the shear stress at the edge of the layer (x = h) is zero. (b) Determine the flowrate across surface AB. Assume the width of the plate is b. (Note: The velocity profile for laminar flow in a pipe has a similar shape. See Video V6.6.) 0 flame, {he {/U/d tell/ck: 7‘0 fhe p/m‘e and 7%8/‘6’ 115 I70 shear stress 0/ Me free surface. X=h h = 1% (2hx 42)de 4’46 4. 56 Water flows in the branching pipe shown in Fig. P45 6 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 t = 20.2 s, (b) the fluid that left the control volume during that 0.2-5 interval, and (c) the fluid that entered the control volume during that time interval. - - — Control voiu me FIGURE P4.5 6 Since V/s consfdm‘, f/Je f/m’d travels a distance Z: Vét I}; Hive At . 777%, I, = V, M = (2%") (20. v20)5 .—. 0,4)» If V2625 = (1 £1) (20. ~20): = 0.2,» ~ and fl3= V3 625 = (2.5%)(20. —zo)s = 0.50,» 777a system af #2023 and Me f/I/ia/ that has en/erea’ or exited the com’r‘o/ volume are indican in the figure below. — — -— control Volume \ (2) ------- - - sysfem 47‘ #2023 4-47 4.57 Two liquids with different densities and viscosities fill the gap betWeen parallel plates as shown in Fig. P457. 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 = 0.1 s and (b) the fluid that has entered and exited the control volume in that time period. #3 FiGURE P4.57 The flu/J 4% y = — 0.642”! ( {he [poi/om ,o/a/e) reMa/msfazf/Md/y [H y :0 {lie f/I/id spew/Ls L611“! :0 Mai af fime f=0./.r i/fla: Moved for #29 MW 4 d/Lr/ame X = V2! = A595 (0.1:) = 0J5”. In r’fle same fibre period {he fap ,b/a/P and Me f/w‘a’ SfVCIé {0 I31 #4: Moved a d/lr/ance X = 2:49 (Ms) ’03- ff. S/hce 77m Ve/oaifl/ prof/79 is pie“ wire Ifnear, #13 and: of {/79 epic»: (WV/Move .ra f/m’ linar AD and 86 remain straight. 72f: ir/M/km‘ed in fies/wick be/ow. 9.2.1“?l 0.sz *——r— — *EJDV'FBIH cH/n‘ - .r"——- ”--H— ‘ - (Ii 0 f @5612“ ’ \' ex/fed cam‘ml ain‘t/.0] ' VOW/he Volume x if; lTvJsff 'Z/JH 3:§I‘—“‘=""=’""="”'”c=c’ .. __ __ cam’r'a/ (lo/mu? -_--- - -- - - system mt f‘OJ—r [Mas 4.58 Cylinder 4.5 6 Water is squirted from a syringe with a speed of V = 5 m/s by pushing in the plunger with a speed of V1!, = 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 when t = 0.5 s. FIGURE P453 DWI/19 that =0.'55 hm interval H73 p/l/nqer "Miles I, = 14, 6£=0.015m. and Me wafer/b/fi‘a/fil m‘ {/79 ex/l‘ mm: [2: Vdi =2.5m. 77m correspond/I77 control surfaces and sysfems m‘ {:0 and #053 Show» in Me figure be/aul. — —— — com‘ral volume 41‘ #0453 .“ 1‘ .5‘ yS fam qf {=- 0,55 4-H 4.59 Water enters a 5-ft-wide, l-ft-deep channel as shown in Fig. P459. 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. Control surface a: HGBRE [34.59 During {be ‘1? =05: {I‘m Ihfc/l/a/ Meir/#1177742! W45 a/mq line BC af #1076 i=0 I14: Mal/ed 7’0 {/79 r/MM a/m‘a/ke l " V 1’ 5 2;? (0.55) = Hy. SIM/70°40 part/M: #7153 f/U/d 4/0179 hm flD fiat/e mat/ed Z=/#(a.ss) = 0.5261 and 1‘ 5-?! (0'53) ‘ 75/: arm/We: 7%? /§-{ afld 6 a? f/w’a’ dream: d0 2702‘ mi pr IbferM/Wy/e’ aim»? Me 0.53 We xh/em/afl See f/‘Wre below. Hm w /l' ‘5‘ enlered - .' - conira/ ' Volume " . ’. . a .. ’. a -‘l . . . f .| J 3 a . a ’ . «ex/{ed coml/v/ Volt/Ma _J . ' _ .. -_ _—-— -_.. _..-._I __=.=- =_-_._-.__ _-:.--_:_-=.__ .— D D’ C C __ _ _. Her confro/ Vo/the --.._------. :yn‘em ai‘ has: I 4'50 4.60 Water flows through the 2-m—wide rectangular channel shown in Fig. P4.6O with a uniform velocity of 3 m/s. (3) Di- rectly integrate Eq. 4.16 with b = 1 to determine the mass flowrate (kg/s) across section CD of the control volume. (b) Repeat part (a) with b = l / p, where p is the density. Explain the physical interpretation of the answer to part (b). - —- — Control surface . __ D a) BOW =5 M “on A I s A Wit/7 b =/ and = V c039 {/7119 became: Z n a 8.00,; = 591/6059 5M ‘QVCOSQ [am an 60 =9Vca~¢9 146,, , where 5’“, = .2 (2m) .— c 9:9 = I cos 9 Wills, w/M Van/s, B,” = (3 £1) 60:9 (~—’—— m’Mt/ry’ffi) = 3000? a 50:49 1:) Will; A = //p £7.1/Jbec0mes 8'0“. =jV~fidA =fVcost9 4/2: V0050 ,4” CD 60 3 =(3%)ca:9 (33%)”? = 3.00% WW) ’5 5 1/? = 27%;) x 11L i7! fa/hm Mai '19 =W/1/m” mass (‘19., b = 3%) $0 f/m‘ IV'I? 416’ = 8;”, represen/s Me tie/um erm’e (01%) from #19 can f/‘a/ 1/0/0079. 4'5/ 4.6} I I 4.6] The wind blows across a field with an approximate 18 velocin profile as shown in Fig. FILM. Use Eq. 4‘16 with the : parameter 5 equal to the velocity to determine the momentum I 8‘? fiowrate across the vertical surface Awfi. which is of unit depth II n ,. into the paper. I I Y L I I I I I x . .. _ .__ ..... .. . .. .. A I FIGURE P4.6l A y=20H 3,13 = 9b we at = jg v w; cm = (a [(ther‘] (HIM/v AB 8 20 AB y 0 1 = (32‘ V a’y D 3mg 1/ = éé-y iii to» 0: .1 IOH(¢.e We aim; vmti (nu/=10) and V = {51? For y3/0 H ThusJ to 20 ' 3 to 10 1" # A La 9- 1 _ A l — MW to» 4y}- ea [2 253 wt] 0 I0 0 o I 4 s: =0.00238 5%??[750 iii 4- 2250 i; F = 7.215499% 4'52. ...
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solutions3 - 4.54» Air enters an elbow with a uniform...

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