327examF-fall04 - 530.327 - Intro. to Fluid Mechanics - Su...

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Unformatted text preview: 530.327 - Intro. to Fluid Mechanics - Su Final Exam Name Semicaug 180 points Short answers. 5 unblems, mml 45 its AS usual, éustify any assum )ttimis, and be brie l 1 I Problem 1. (9 pins) Considel two laminar bouncimy layers that giow at the same 1am (i re. 6(a:) is the same for both) and have {the same flee-stimuli velocity value, U, but £01 which the pmfile shapes (liffeiz Which inofile will have 111010. viscous ding: the solid pzofile, whose veiucity values am given by m, 01' the dashed ptofile, whose veiocity values am. given by :12? Explain matiienmtically ' \lxswus DIM, is magnum EN 1% \hscou‘: mm mm Arraewkw,‘l=o; Le. ’tjwping‘i GD :0 ° “MM Tue Mow: i9? 79% / so Tm 50MB Yum/e / NM Hm Maine meow; M. @w CD 55 Problem 2.. (9 pins) FlOlll a fluid nmclmuicul standpoint, why is it bad fcn yum hezut'. if you have, say, Cholesteml buildup in yam blood vessels"? (Aftel all, it seems like a blockage should just i1'1c19ase the iecal blood Velocity; the flow late will be the same ) oggxzwnow, Luce moveswjao» Dcfasfls wfizoww MIMOF. WAD CS) urges, www panama "me “emf 10 woman may WAD] fomap. To magma ewe» mu) fifw® Problem 3. (9 ptzs) Consic‘lm the slant-tabla mmxonmte: pictmecl Assume the density of ni1 is negligible and let p be the density of watt-21 As shown, 1); > pg Suppose the pIRSSUIE ciiifmeuce A}; m m m pg incmases; then I); {leueusos in! some amount (1h, hg increases by the same amount, and the change in A}; is given by pg(2 (Hit) Howevel, in the lab, we moasmerl 1319551119 changes using; [the iiquid level 111 only; that is, changes in Ag) wele given by pg (1h How does the lab nutl'louletel diffen £10111 the one shown? lu w m mummy ?WUF6 wwws mega team-rep m Lamas. w M MM 86st in Ms Ix. eonsv Ms ms Time G) 7» all / 60 lag was wen-mew Rams-a me Mama's; boom: MW We: wperkm Wm As l4, V‘A‘FWD. Problem 4. (0 131:5) Looking, at the (can; on the mad, it seems that Inflaclll‘mcks, station wagons and SUV’S, whose 19211 windows am often close to \relsical, generally have mm window wipms, while ‘Ieguim’ ems, whose 19m windows am 030.991 to liOllZOl'lEEIL almost nevm have such wipels Explain whyr this might im the. case ‘E I A W MW mm u): AlfigmuM emw' Ml“: ‘5'" fl \Ewws’ gman ow® 5 N15.) cm. ~. dim? 4—m— ¥wuo swamnom a» max mnww .- C5) Ame.va wasn’t ow gmnpgoiz on: / Ntme flow 56 kmfiw@ a) 10 Problem 5. (9 pts} A}: inviscid, incompiessilfie fluid flows tinough a section of ciicnhu- pipe that. initially has ciiamote: (11, the]: 3111111ng to a section with diametei (lg <. (11, then expands to a section with diameter (13 > (El, then ietiuzzs to its initial diameter (11 befme exiting into the air-11105331191 e Ignom g1; vity in this pioblmn imagine that: you are :1 iiiiscmam who wants ino (hill ;1 small hole to make the fluid leak out ol £2119, pipe bofme The exit. Will you be successful if you (hill the hole at point. A? Point B? Point C"? Explain ii: C19 0‘3 l i: M Jim “we mwa (Aifiiosfriwe'l ’T T l (?= Vm 60:5me 9 © Ms'lUE'TUEe) . g? wise sweepinme -. Jib < \lA New (\lg @ Fm 6D M - homo/so Saginaw may, ,awmu vavfis'fiefifi flaw > R5 ® WW. '3 ° Fwib wwb m but mx 1'? Weséviae msme 1% We Pn’ me: WM: no cinema: Wm we move amine (mm mes W pm To m» m w...) So FWD WW5 Fir—0M 'fmm’CcD; owe. NOT w’kK MM ®°i®i GD Problem 6. (Toiual 30 ptS) You am interested in iclei‘ltif3-'i1'1g the paianietei's that clelzeimine the teiinénal velocity, UT, of spinning, Smooth, aluminum spheres (Recall that £21m touninai velocity is the velocity at which the dou’nwmcl acceleiation of giavity is balanced by the drag, foace.) You identify the set oi possible pamn‘aeteis as the initial spin late, a), the fluid density and dynamic Vlfi— cosiiy, p and ,u, iesgiecctiveiir, and the 5131mm. diauietei, D. (You’m only (:oncoined about. aluminum spheies, so the. sphele donnii‘v is not: a. paiainelei ) Then, you coiioct inoasnmmenlis for dill'eient sized spiiezes, difl'kéieni‘. fluids and (hill-2mm Spin IEItE‘S, which am tabulated below: or (an/s) w p (kg/m3) ,u. (kg/1n s) D (in) 0 5 i500 1200 1 10-3 0 25 {J 62 500 .500 2 10"“1 0 1 4.22 2000 500 .5 10-3 0 3 9.9 ":50 5 n 510"“? 0 15 6417 1400 1 4 3 10-5 0 05 Afti-u you have collected these nmamnmnenls, a blend suggests to you that, in fact, the fluid viscosity isn’t important; then ai‘iotliei hit-ind suggests alias: the spin late isn’t inipoi mm: You decide to appioinfii the pi'ohieni logically using; you: lu'iowledgo ol dimensional analysis (a) {U pits) Using dimensional analysis. dotennine iiho non—dinmnsimial pamniatem that govmi‘i {the problem i§ all of the pmamotois ale iinpmtzai‘it. (i3) (6 pts) Deterniino the nonwdiinensional paullneteis {that govein the in ohlmn if the fluid viscosity isn’i. impmtant. (c) (6 pm} Deteunine the non—dinmnsional paiainetem $ha‘r. gown 1'1 the piolilein if the spin lat-e isn’t l1]1§)01 taint (d) (12 Iii-S) Based on yum paniculai data. which pan anion-218 are inipmtant? (Hint: one. oi your fziends i5 conm't ) Explain. E 5 VaslAME'fextsz 3 DlMCNSIwQ H: 7. nowbmmémnau WM’EWJV-é (D C5) Smog?» we fch (A, p, D as mamas imam .m ll1 “-‘- VT.ma€b Dc > W11. Var/(004)) .,. x“ 100001 1T2: M’ 00an Do < N016 2 Aug \F {BIND kite aftan 1061: If}: V11? [A T— ..j “7 Muff/A. A516 fieféfirfimn YWMETE’ESIQET'. i ., (Extra workspace for problem 6) E] \F [a 1001’ wroemm: L\- kas, 3 Dummslm9, \ Mora-Dmmsmm, vmmmz G) View ['9] , me flown kawrep. Momma w) vale}; VS ‘33,: \[TE({,J (9 El Luce [E]; 'w m: wmwm uo «mu m MWD Manama \nxzowmc, VT) ML D I “mp: E \F [A us new” \Mfomrm/mmYfl Tm; us'mm’ \IT/(vo D):'.C.0H$T. WW4 ‘1: M Kim MYOWNY, 'W \IS'MM’ @ QWOK w» ( Fog. THE W; 9MP: f0 ms w Tue Have) "-3 ‘5 £3340 L5'10 TTL «a Dmmm Nerf £05351. T10 w) Guns-a To CszT. "3-? DNM 5066631" “MT (:9 \11 i6 WTWDEN‘T on {II/Alma I, "ma. 5sz $411 00 19 NOT 1Mfo1zTAMF. (NOTE: We Womm is cowl/flew Me U?) 5 Problem 7. (Total 30 pts) Let‘s do some math In the following, 11, v and m the the :1:—, y" and zwcomponents of velocity Justify all conclusions mathenmtieally ( a.) (b) [0:] what, \I= argue? LE] \nmmmsme: (10 ptg) Cansidel a Steady flow field wheie you know u and 1;, but. not w Suppose 1: = 2:1: + 33} and 1.: = 2:1: w 1:; Is it possible ft)! the flow field to he ineolnpzessible and t.\\r(J~cli111eiisi(miii? (10 pts) Consiclei the steady. inecnnlnessihle flow of a. viscous fluid between two fixed, parallel, infinite fiat plates located at y n 0 and g; a: 1, ie the :r:- and z—Lliiections me paiallel to the plates Once again, you know u and U, but not '21). A19. u m (’33; -— 1}2 -— l and if at My m 1) possible values of the '21 and n components? (10 pts) F01 steady, inviscid flow, Eulei’s equation 1101 mind to a stiemnline is 10]) u (a: _ f." {1012 an m R l wheie “n. is the eomtlinate 1101111211 tn the stiemnline, points v01 tically upwach and H is; the iatlius of cm venue of the stu-amnline 1:; the lmzizoutal plane. we pieviouely derived that them is 110 piessuie vmiation 1101 11ml to $121 eight stiemnlines IS them. any 1)). (aseilie valiatémi 1101 11m] to stmight stmamlines in the ver tim.’ plane? It so, rlezive the piesmue vmiatinn mat.izeazmticrailv and explain it physically leMrt'ZcfilSfil‘i'aW'. gill-tig— AJLJ =0. éé‘M—52)%¥'=“m\ [60 art! =‘l ll: Ax a1 :32. x bl tml lncom'fwssieue got: a? #0 Mess m \6 5; Humor: 6?@, WW “W45 “W 9W9“ Ftew Wfénvs 0:5 )ch WC, 2 3-) Wwa 3 mweanL = > W <9 (9 \lxscous FWWJSO mm} to M t: o,\ (235 ‘ll’t clv 6w M ____. av M 1 W l 91 32 O Baunbapt WNDETION; M50 (datum) Wtqu vows: \(zo M tau/i 0 (It mm \1 sweet m we)ka muvmopsi 0 FtELD Wat Be maritimle 59.9 24>. elm s : flzufimdhzowlmx at» l u was muff a; M H ( m ) 2i lutebweéopmfl, ( So N: 2.0"?” "' “$68)” (D (we) (Extra workspace for problem 7) 0 N660 To Know: 9669 mm U0 6mm Tow Eouvbhfl commons? LOOK Prf $0“? u) “ Influx-'0) cAN’1%=o Fogéj sz.’ *0 \TM am“ a man \! Manes «Fe gal 70561313] (7-D W E] , meg—e 1; E’Wééu‘ae \M-mnom MiiMkb To arms!“ afiemun‘es \F TM MM Memo» M$ q Mariam Wch (swamp-r . \1 e. . fl Euufiz: in i?- «1—3'32: ; E}: f) &V\ W K A A \A "7‘ sqvmwr STFEMun-a: 12 =. Va ":- A"). ;, 3 Pg "flsx £2 _ A: 3.; G) (Mm €39“ 7&0 n: 334:0 émng we: \r’lmwwm momma «MY—e» \/ 3"":‘5. , 13.21. = --\ W a) if. 2 up 51mm 46 vafiosmm Wessopa a 3 I V‘ Vkflkfion/ (No «gnaw RWWFK’EGN, so Fww cm Be “Imago K4, ‘srmc,’ u: TM mama, 121mm») Problem 8‘ (~10 pts) Consiciel in: enounous, sealed tank, initially filled with iiquiri, that empties into the ntmosphem though :1. hmizontsl. cast hon pipe, shown: (D (Cowman To It cousmm- mug», m “I?” 5W”) A 4) GD '3 "‘> Q / T lbw-'91 SQUAFGEDhE‘D ‘Q INLET Because the tank is sealed. we can anhitlzuily the aii p1 essum at the liquid Sillftlcf’. (location 1) The entrance into the pipe from the tank is squale edged, and these is a gate valve in the pipe that allows the flow to he iegninted Assmne the following: the liquid in the tank is wntm at (58°F; the depth 12 is 43 ft ; the messumtl innel {lizunntm of the pipe is d = E in ; the guts valve is fully open; and the pipe has length i m ‘2 it (As usual: instihr all assumptions ) (a) (8 pts) Suppose the piessme at incutiun E is atnmsphelic Tientil'lg the “rate: in the tank as inviscid, what is the whims flow 1am, Q, of write: out of the. pipe? (13) (24 pts) Now heat the warm as viscous What does the ‘rlii pressum at location 1 need to be if we want the flow late to he the. same as fol pent (a)? (c) (8 pts) Imagine now that you stick a. C01 k in the pipe. outiet so that no watt-n gets out... What‘s the total {'01 (‘8 applied on the Cfliii in; the watts {'01 the inviscid case of part (it), and the viscous case (if part (b)? If the Estes are different, expiain how they can he (lifimeut even thmigh the (unsalted) flow rates me, the sums .3 [9—] \Nwsup: mat safsouui mes-o 6) sim@ (Nu wss Tag/us} O 21+ 3E¢Ygi+fi(0) Picvz-T'PA’YN © 7‘ in V‘ = 0 (mm is (nervous) =3 v1: zflt 25%;) Q: Awful: giiflgg @twwm amqtvn t ‘i’mweeu @169 : ‘63" ‘31" ‘5 Q 611bt=0 \1 1.45 \I 8 : ~32 . --~- ,1: 0 L if“? D 2 vi- . 6mm»: ca mm: .5 MW Mmoil tosses (Extra workspace for problem 8) (mm cm.) (rkape‘fi J M \N @1 plrwknawfi '92 u IPMM; K6“ =05; £5.13; ~5- wz-lf ; V1=Bikfg {WE} . JP .14 Toast-$3 Kc: will. 7.5% 25b; “Lop” CD "0 1.0% IO'S‘FVS (Tits. A41) L m 6.». AW 3 , 630004;“ O m Meow mmm; D VIZ. F u “ f ‘F: -; ALSO". €81 mou 50 v(,_ =1 15 vansomguECD —-'2. 6o@ sesame. Prpzc 463th .\- 10%.. [no.5 +o.osss(g+24fl faztmafif (“9:”) ® (An) 1. c) . _ My '1“... 55.111. 5M5") Elaflo- MR {3 51132.5.L H’H‘ 4.1.0111“ 15 .11”, = Heoolgi a‘m as; =9)“ G) G) E \$ New maven-519mm Qeeowas Sta—“n , Fem“ = 31. Ame W “2° M5 balsam 'PMM “NW/“9 “5693‘ ?z= Wm + {963% a VAT,“ + 7.0% 15%; *- 430013;: a») $.ka = $72“ .3... (LL11 Fame, mums Wm FMM (samba, = WC: Hp) . ‘1'?ka ? TM"? «600% “((96 Vm= vbfi+fflk $21 =3 : A «f (m me = mm 51.5 TM“, 16 005199695121 MOW Fogwlvwvv'eluem w me \hscous case ~: me how was Maw magma Bewse ma mews; mm: \s us-ao To 9 cova \Mwus Hem; Leases. Problem 9. (Total 35 pts) Ah flows tlnongh a. Cll‘Culal tube with diametc-u (lg, at a. Winnie flow rate Q A ennui-r11 disk with (intuit-alter (12 is pinned at the exit of the: pipe, :15 shown: LAW“) FLOW L}; (l gig 05.4"“? (fi’vawl The (link is centemtl in the pipe Assmne thth the ah that flows past the disk moves in the (If—{liletttiOi‘s {with The ah iizunediateiy behind the dink is stationzuy Suppose that (11 m {l 4 m, (12 mt O 35 In, the density of an is p = 1 23 kg/in”, and Q m (l 5 1113/5 Twat ah as; inconlgn‘essihle (21} {5 13213) "heating ail as invist'iti, what is the inesmue, 1);, well upfiheall‘l of the disk? (NOTE: this will not he the 311 (2551110, on the npstmmn smlace Of the disk ) (h) {17' gits) ‘Nlmt is the inltwnul [0109. F. neat-essan’ to keep the disk in plat‘t-a‘? Define yam mntml volume elem 1y (c) (5 pm) 'Wimt is the (hng (:(mffirient, CD, of the disk in the pi obit-2:11? Pick appi-opiiate values of V and A £01 the tiefinitiml of CD (d) (8 pts) Compam V0111 CD with the value given in Table 9.3 in the text. \Vhat me some plausible masons why they are (lifl‘hmnt'? Explain [El ' its“: la‘l ms (109591.kach , AND Assxmwc. mam new .9 me 19mg mm m“ we am (geksom‘aw \F trauma») “2) Q—"VlAi :VZA’mr “la. (‘G‘l‘l’lzl L—bmsk Pct-2A m 0.5“3/5 > \ll=~ 2 fiwmzalmsg- Nil CD l _.. - \I A at“ m 3 Vz- NAA=V\'J';§“.L-L+1’I \Il- Hog-N1. It" '2. l 2. m“ a .- _..—. a T 2- f’ 2, We. = ?MM\ 2 - 1. =5 l" ,PMM : fig-(V2241?) = 111.3 “fl/Ma (213 “4/513 1 10 N "' WSW: ‘“ PIILAGE 0 ° law: 81 BUY-Mum Pt ‘11?" l"kill; “will?” C’FAU‘“/“°V"‘°m“q Ea CWWCM‘E ®i2;"“:{]7€>emv (Awor@~ma ,_ ._ __ ._ fist-"D Q 7 7 ’ 7 ’ ’ ’ f1 whom or Mona/mm @ a g / 516M WFT we“ 1m: Meme,»qu m. WAN- ?2, A“ “F = f?" A 3 - PQV‘ + PQV2® (.x-themon) srém‘f .1, mm @ mam: => F: ($912M; +PQ(V\N23@ "‘" Pwoe 'Al 4' 1”an “’23 = lb‘fia'gkkm" + Lisa—0.599. £6.02) ‘ m3 5 .%2 I: {\SJ N a: F ' (La?rwpqzv\@ G) ¢F F m N E cpu‘ a;le :- ‘vLs‘SSL W” "a :6 L 2' “3&3;qung max GD 053V“ A?“ ‘anmwswm‘t Bl $110M “(#:8126432 CD Fog. A 7.0mm DISK 00?. CD 15 MUCM matey. owe momma éxmnm'wn \5 mm 0051 “Pg-when “If-ms m van-me 2%on fienmb "rm: mm: M A STkTmMp‘f udmm Woe;ka WGGEFATEE’ Tug Q26 0? “m wkx’e {Lemma AND Trws wages-cw Mes “OPAL. C9 sefmfion Wotan, 11 ...
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327examF-fall04 - 530.327 - Intro. to Fluid Mechanics - Su...

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