327examF-fall06 - 530.327 - Intro” to Fluid Mechanics u...

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Unformatted text preview: 530.327 - Intro” to Fluid Mechanics u 811 Final Exam Name gfiumws 180 points Short answers. (5 problems, total 40 pts) As usual, justify any assumptions, explain your answers, and be as brief as possible. Problem 1. (8 pts) One idea. for reducing flow separation from, say, a. cylinder is to punch small holes in the surface and generate suction through the holes“ If the zero degree point is the most upstream point on the cylinder and 180° is the most downstream point, what angular position (approximately) would be the best place to put these suction holes? What’s the physical reasoning about why they would w01k to reduce separation"? an» / $5.?AZAI’WN ‘PT (Foil ex AM? be) —> w...» on “‘00 3"‘\‘ ‘1". ° Sfiittmwr‘ 0W5 WW“ “WWW/s Hear. me Wilma. wt mmng luv/Meme seesaw (Mame magma emblem) o Sam’th Hate; LOW? 26800.6: Weave M we suemceva (D a To WW?- 5€?ARMLoH-. M sumo}: Hates NW Sfivmkfiom T? To ‘ V 1360606 A‘DWV—Sé F usuemiius-r DowsWM oF‘lO Problem 2. (8 pts) 011 a road trip you bring your bike along by attaching it to a rack on you: back bumper“ The wheels protrude a. little bit into the airstream on either side of the car. If you want to minimize the drag caused by the bike, should you lock the wheels in piece, or allow them to rotate freely? W“ / fie U Q5) as 11% mum’s“ homo mews : Wage/D M‘ECMAMCRL WW. Problem 3. (8 pts) Consider two models to be placed in a wind tunnel: a cylinder of diameter d, and a streamlined shape consisting of a half—cylinder (with diameter 03) followed by a long tail on the downstream side, as shown in the figure. When the airstream flows from left to right, the streamlined shape has significantly less drag than the cylinder. Now suppose that the flow is turned around so that it goes from right-to-Ieft and strikes the sharp end, not the blunt end, of the streamlined shape. How does the drag of the streamlined shape compare to that of the cylinder now? (The air is, of course, treated as viscous.) .3 CD site” 093912th occugzs towpgrm or “EB more“ tom -. 1F ‘foxx "were we STWWUP‘ZD SHAW More, Tet not:ng 'YAFT \5 iDfiNfiCM/To m6 at uWW— WBMSRME sweeten WLGE’Wl’rEgC-D 3 magma meme new 0 éteeamuneo same we Maw 5012mm»: am =0 may; warm: DWG’D D o o Ovepwl’me “amen—Aware) “emanate Same we We om. We at damage . C3) Problem 4. (8 pts) When your car is dusty, why can‘t you ciean it off just by driving it and relying on the wind to blow the dust away? Tam; A tamer wav— vm m Caz}, Seem was at»: umcmes Kim/Lame To The W) F??? W3 =3 ANT tom ere, meet; Tag 52:4,. whit Gar e,me AMT ® Problem 5. (8 pts) (Assume that standard atmospheric conditions pertain in this problem.) Suppose you have a. spherical weather baiioon, of diameter 2 111, whose skin is rigid (ie the balloon’s shape and volume are fixed), totally leakproof‘, and weightlessw in your sea—level garage you fill the balloon with helium at the same temperature and pressure as the atmosphere“ Helium has density 0. 138 times that of air at the same thermodynamic conditions” You attach an instrument package (with mass m = 605 kg) to the bafioon that transmits meteorological data back to you, then you go outside and let go of the balloons. If there is no wind, describe What happens to the balioon as time passes, and state as precisely as you can where the balloon will end up in infinite time. N61 d9vJA1LD were : .. 3 k a ignitth 1.225 3M. “1565; (0,5 K3 r 7‘ “r 7‘ Z? 3 "“ V‘émon ‘ “35 {Jmtfl 4' \IWW emfifl" m3 WNW—.4 L (ms to mom: G) EFOMW mun “Farms Q) @ " \JBA'LLOON ‘ (“€752”) Fm " M3 k .i a (axwfiosumnsfi4.62; {0.8%.43’531 @ ”‘ “2-0.5 N we Baum?» 1317—0175 To 1% .WM Giacomo M30 SW5 “Fri-9 PUFF/“EP— - P’roblem 6. Lab-related questions. (Total 40 pts) 3) (15 pts) In Lab 2 (the pipe flow lab), when you made pressure drop measurements on the ‘cleen’ pipes (the ones without the valves), you may have found that you got values for the friction factor, f, that led to nonsensical values for the relative roughness, e/ D (that is, the intersection of the f value and the Reynolds number, Re, was below the ‘smooth pipe’ curve on the Moody diagram). Does it make sense to blame this result on buildup in the pipes, which would reduce the effective diameter of the pipes? Be as mathematically precise as you can. b) (25 pts) [11 Lab 3, suppose that instead of a. cylinder, we put a triangular wedge into the wind tunnel, where the wedge engie is 60° and the wedge height is h. Suppose the free-stream tunnel speed is U , the tunnel height is H m 10h, and its width is w in the out~oprlane direction. Now suppose that the boundary layer on the wedge is infinitely thin, so that upstream of the back of the wedge (region 1 in the figure), the velocity is uniform on tunnel cross—sections; also suppose that the flow separates right at the back corner of the wedge, and that the pressure is uniform everywhere downstream of the wedge (region 2), With these assumptions, what’s the drag coefficient of the wedge? , Tin If/r ////// (Den-ms W! ‘iw MMSUEED be , Q (sow we), Ann L (m more] . ‘iou «swan PM 1 pm J Am) D (viva Dominates.) "a THAT 15/ you obs-iv Meme “Wee mum, W64 mamas . .— * W n \ItD END , 50-??096 too Fm: x 5/ We: mg?“ Eff“ "'5" “4 as; (9 Fe = W l m‘q Yaw- Dm (hm-r A I”? D vitae. MD :3: 9593\idiéii2" “ms oEe‘mficb D =- 1" ~01- (amt D -=. 1A '11- ~ w M" C?) W a so 1 \f D 15 119130669 t1 WWW} lZQ (pesos as "‘5 J went“?me As D5 So You; DATA ymm’ mw MINE: \ (9 if is M W f]? rom W“? was A wmew D R5 '5 mu, m Youz pm 4 been im‘o 13W» MUD twice as 0‘5 we set mm mm sane to @ amigos. mix on scum? (Extra workspace for problem 6) Swag E “f A W swam Mao weewgzas on waves suEFA-cg 1‘“ T’s 0" m M1 . kampso" ngwm: “(6. +' E. g. .6)... 4- M2,, *3. outlaw CW mss'runna L {W1} fin» 30° We?“ Ppsmwm (Wawtif 6mm "= a“ “was: Mmop') 2. “fig” flwnMLWfi-j 2. ° Fag. \Jtsz): was Covswvmm/ \IM My.) : GA = W \ohwu O a 1- *4: Acx); w. (H—lxrgnso) = Mich—Ex) =2.) PM a y“ gitz” 6110?]: {SEEP mesh“ 11_ (39 c 1‘ 2- .1 Kapowfms A 1‘ (wk fix) \1 E Ram“? V0239 FROM 611M m -—~'> m L L Q) Vfi 15 ?Cx\~5m30" NE???“ 7* lw'wwoj ERMA" To“! ’90 Human.“ a? 0 WM 0,; V05.sz mEDGE mm L- l L— 7. 9;; Reg“ (29} W 133W 2:2. eg‘x 3,2553 Wk“ {‘3 o 2. 2- (IoL-fix} ‘13 2‘ to [towing] a —.-. 4- 50904141“) (— J— 4r L) = 9(5sz =‘0.055bf\51wlfi lob. fih @ 2. 1‘ ‘ :— mo: F2 ~= ... Pump. = «39. [b- %]wk :- o.m» 907‘th @ Problem 7. (Total 35 pts) Consider a large water tank that contains water at 15° C, The bottom of the tank opens into a, horizontal Spfilwey; the spiliway has a. square cross—section that measures 2h tall by 2k in the out—oprlane direction. A massless gate, which tums on a horizontal hinge at its midpoint, sepmetes the water in the tank from the air in the spillwey. In the closed position, the gate makes an angle of 6 from the horizontal. Let d = 5m and h = 1 In, and use a. value for the gravitational acceleration of g "2 9313/52" pmmh'll ’V a) (15 pts) What’s the tOIque (magnitude and direction) you need to apply at the hinge to open the gate when 6 = 90°? 1)) (20 pts) What’s the torque (magnitude and diz‘ection) you need to apply at the hinge to open the gate when 6‘ = 60°? I (.160 can sowe met N Flnbme Fog-we mm was 6? km“ gamma : \u» {Wham Tee Mm.st beam ewe 5mm w; emweefi A to E . .. l X W" X W T“? WWW Goof-mare TM“? :0 M” Tee. Hmee. l“ “a Tee MOMfiNT o? met-item Aeou-r “ma {tube \9 mm “a h =AA @ k m 3 (>0 x-olA = B MCho—x) x (it?) xwh J, \J W -14 mswae Mi“ G = (Moxdfl'hb‘ C3) m” W“ (.08an was “In “WWW” M: MOMENTPO _ h ‘ 1 ‘ 3 h J! QCWOKWIQE) "' 2‘ [Ethic la apfio'aich‘3‘x) ' 1 5 L; H 2 W5; @ “Ml”? lhel‘fl a pm (MM/{MM Kg/m3/3sfifiW/S1.) Se ‘ggg M36010 AWN 1 11505-10" N-m (memes) To OWN “meow. JCH’“, e“f%' (Extra workspace for problem 7) New var-m commam )4 as o Arr-ma amuse, mm BE mm: M’ be" Fpm veg.er (sa rré mast mm “me em) @69 1‘ \n 313' " SW10 WM wear Hmbé = 5 . di ( >0: momma] .32. ""‘ E“ T web 2 S PfiChw‘éflx Zkokx = M1933 (M x—‘B. £13436 é“ 3w '2: «- w ‘1 Lt “L 33 he most *m + %h“ffl: H‘t0«lo"' H-m (momma)! To OPEN “me (3M6 (“Wt ms Ffimurm W612— TRAN m E 91:06 You Web 12: We 06mm mass m: m mm mm Anuser Problem 8" (30 pts) Consider a tank with squaxe crosswsection that measm'es L m 03:11 on each side“ The tank sits on a very accurate scale“ Water (at 20°C) flows into the tank, at volumetric flow rate Q = 02 fiteIs per second, from a faucet with diameter d m 1.5cm, whose outlet is located H = 1m above the bottom of the tank" You went to put 50 liters of water in the tank, which (using 9 = 9.8 1321/52 and p = 998 kg/ 1113) weighs 489 N; you decide to do this by running the faucet until the dial on the scale indicates 489 N. What volume of water is actuaily in the tank at that point? What weight W113 the scale indicate when there are in fact. 50 liters in the tank? I» mg on“ g Tue some uses»? MD in; Neither 0? Tee wkféfifionct) ‘Becwse 1116 TAMKJMbD Aim-$5 Tue “MAEN‘WM 0? we were: swam _ Mo” “34 ‘ may “I «new? a ' 5 CD a 0. Q i q i w m ®= w- »?wm teem H ( @ a I! K@ a (2): \I1 comes womeemuuu: V. +$fl+32igza+§flaz 1“ t “F 2- 2- "iuzo :5 a) v2. '7 J‘Wiflizu‘ifl 1* Vim-r3 (vi—h} mkmon $09.06 or scene (mum we smut films but} C29 @{i .MOMENTUM 53w: ‘-‘- 4' 7‘“ CM. .- ..... ..,. 35 T‘DMT J 5“ ; (emcee W E'Mwsufi i i (win???) 2’ wmp. m TAAK ' - "41 " M” s \r u .a J2. CV = __ W Hzo + s as. O 4‘ (no WW mews: whey. m (mew) kmflnzpm GW'A‘WZI-e') M :5 CD " ® ’1‘ swxmfi ‘ = w - Puzo.3 L ‘ g u 30 “3—7) 3:6 5 e “a”? UstoF m" wean.) (NW) (Extra workspace for problem 8) 4' Naenfim Ape 5!) L w ‘rfleTANK: V-‘ODS M3: L1M:Co.3m\1h -. = 4661 .c, w m Wow wan» 'V- r- 50L _. C9 69 ., TE]: H-is‘i N = fiQ%%Y_(o.3m\z-‘t$§‘1.kado-*1“; (man‘2«1.$§.z(\m—h\fl ° “Amine 5m mos Rama: MAM V (M’s week: 4“ = 33011,. .» o.m%~ [L117 + \fi-HPMI 1 “£333? Oiluf'ffifld v 11" 7' “a” \ -..-.> (Hm-33am} = (o.xwY‘[x.z-n MW Wm] ‘L ‘K Saws Tm; Quaopknom. 1146 MM! Fm; Mossm m *> if: [fin = (oétflzo‘SBI-fl 4191.611? L l wmavwm wk?» 5mm ® mus W19. Problem 9. Random short problems. (35 pts) a) (10 pts) Consider an incompressible velocity field of the form V = u(:1:, + v(:t, . Suppose u, 2 2x3; + y in meters per second when m and y are in meters, and you’re given that v = 4m/s when :1: = 1 m and y m 2111“ Specify a form of v 2 v(m, y) (doesn’t have to be the simplest) that satisfies the given conditions“ b) (10 pts) Consider a turbulent pipe flow, in a round pipe with radius R, where the average velocity profile happens to be given by V0“) 2 Vgcos (1 —~ What would the kinetic energy coefficient, 05 (which shows up in the head loss equation), be for this case? c) (15 pts) In HW 9, we analyzed the flow of viscous fluid between palatial plates, Where the top plate moved with speed U1 and the bottom plate moved with speed U2“ For the case where dp/da: = D, the velocity fieid came out to be My) = U2 + (U1 W U2)%« (1) Write the Navier—Stokes equation for the wcomponest of velocity Which term describes the viscous forces? Show why, even though we know the fluid is viscous, the viscous terms go to zero for the solution in Eq. 1, and explain what’s going on" Q9 .3 @ Ianfmflw => V V: O = ifs? \N Tans case a ‘ 69 So sigma m “21' , \M‘Eb‘f—ATE: \I': .Ym-r'pch3 (in-€— 1F XIV In) “A J a)»: W C9 Ar x=\w\,i=2-M; \f= W5 = " *% W100. C33 coo ¥m(xl=§% were, a) W(~‘(z+3\m“:};l 03%» ‘t 15%» lM E (SW: W: “W ‘rr M6 Mesa: To me \lm : \IO (Hi?) :3 stngs em memee.’ (oven) 10 (Extra workspace for probiem 9) u,va @ Nos 661., meoml FofiAmw 9.1mm: Names 69 a“ g». a“ wéfi-..__\__3? 6.3% 6‘ 313*“9xWSfJ' ea" PE'SEHJL3%*WJ?%’ O \2 O 92 (33 So 1; "333 \s WM 116sz momma \nscosm'. L a“ M 3 2 W {3012. meme» v.00, Aha .-: 0 =3 \nstous “raw 3“ (ma—1b 2.61:0 g : W-fi XNW‘M VT 1%") (A M Am: 73me men 1/ amp Ykilflcw vs m4 Raw: (m mans Fem): -3 To "THE 7.11m B? THE stTW- CED wab wove “"3 To me m N we stow JwaD 8mm @ b a c, flew \nscous vopmwTwe mwwznan so . 11 ...
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327examF-fall06 - 530.327 - Intro” to Fluid Mechanics u...

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