3.185 Problem Set 7. Fluid Dynamics. Solutions CORRECTED

Transcription

1 3.185 Problem Set 7 Fluid yamic Solutio CORRECTE 1. Power-law o-newtoia fluid behavior (a) If the fluid i peudoplatic, the a we icreae the hear trai rate, the ratio of trai rate to tre ux y /τ yx decreae, it hear-thiig. Thi happe if the power law expoet i le tha oe. Aother way to look at thi i that peudoplatic i qualitatively imilar to the Bigham platic hape. Sice trai rate look more like that tha (trai rate), hould be le tha oe. (b) Start with a hell balace, with flow i the poitive x-directio ad y the vertical directio ormal to the (horizotal) plae: 0 = [A τ yx ] y [A τ yx ] y+δy + V legth ( i the differece betwee ilet ad outlet preure). Agai divide by V (= AΔy) ad let Δy go to zero: 0 = dτ yx + dy legth That wa pretty eay. (Next year we ll ue L for legth ad omethig ele for thicke, to avoid cofuio.) (c) I ll firt itegrate to give τ yx v. y, the ubtitute the hear tre expreio. Splittig the term to oppoite ide ad itegratig oce with repect to y give: τ yx = legth y + C 1 Now we ubtitute the power law cotitutive equatio, ad thig tart to get ugly: dux η 0 = y + C 1 dy legth We have to divide by η 0 ad take the th root of both ide: du x = dy η0 legth y + C 1 where C 1 = C1 η 0. Note that becaue of the upredictability of ig of a egative umber to the power (poitive for =, egative for = 3, two imagiary root for =, etc.), thi i oly trictly valid for o-egative value of η0 legth y + C 1. Now we ll itegrate by ubtitutio, lettig f = η0 legth y + C 1, o df = η0 legth dy, ad we have: 1 η 0 legth du x = f df 1

2 η 0 legth +1 u x = f C Subtitutig f back i give the geeral olutio: +1 η 0 legth u x = + 1 η0 legth y + C 1 + C That pretty ugly! Let put y = 0 halfway betwee the plate, o the boudary coditio are u x = 0 at y = ±L/. Becaue we have ymmetric boudary coditio ad cotat (ad therefore ymmetric) drivig force ad cotat propertie, we ca aume the velocity i ymmetric about y = 0. For thi reao, we ca ay that at y = 0, dux = 0, ad ee how thi help u. We ca get dux dy dy from above: du x = dy η0 legth y + C 1 ad et it to zero at y = 0: 0 = 0 + C 1 η 0 legth 0 = C 1 So, C 1 = 0. What a relief! Thi implifie the geeral olutio to: +1 η 0 legth u x = + 1 η0 legth y + C ad we ca go a tep further to: +1 u x = y + 1 η0 legth + C 1 Agai, thi i oly trictly valid for o-egative η0 legth y ad therefore for o-poitive y. We ca get the poitive y velocitie by ymmetry. (d) Now ue the boudary coditio y = L u x = 0 ad olve for C : +1 L 0 = + C + 1 η0 legth +1 L C = + 1 η0 legth which ca alo be writte L L C = + 1 η0 legth So the velocity profile for o-poitive y i give by: [ +1 ] L +1 u x = y + 1 η0 legth Note that the maximum velocity at y = 0, which i give by C, i proportioal to the thicke to the power. For a Newtoia fluid, thi will be quadratic i thicke (i.e. proportioal to thicke quared); for dilatat fluid ( > 1), thi will be betwee quadratic ad liear; for peudoplatic fluid ( < 1), fater tha quadratic. Maximum velocity ad flow rate are alo oliear fuctio of preure differece, peudo-vicoity, ad legth!

3 . Czochralki crytal growth (1) (a) Carteia coordiate clearly do t make ee here, ad the complexity of the pherical equatio oly jutify their ue if the boudary coditio would be greatly implified, e.g. if the crucible i exactly hemipherical ad the top urface of the liquid i flat. So we ue cylidrical coordiate. The liquid ilico i icompreible, ad it homogeeity (the ilico i very pure) ugget cotat deity. A a liquid metalloid, it i mot likely Newtoia, ad ule there are very trog temperature gradiet, it ha cotat vicoity. Uder thee coditio with lamiar flow, we ca ue the upper form of the Navier-Stoke equatio give out i cla o page 4 of the Solvig Fluid yamic Problem hadout, which i alo the -F form i Poirier ad Geiger p. 59. Now let look at the other aumptio: Steady-tate: the crytal ad crucible are rotatig at cotat peed, o the oly ource of time-depedece i removal of the liquid a the crytal grow ad pull out. Becaue thi pullout i order of magitude lower tha the rotatio, we ca aume a quai-teady-tate where at ay give momet the flow i at teady-tate, but that teady-tate i chagig very lowly with time a there i progreively le liquid i the crucible. So we ca cacel the time derivative. Fully-developed flow: becaue thi i rotatig, it hard to defie a fully-developed, ice etrace legth i eetially zero ad the overall legth eetially ifiite. Itead, it better to thik of it a havig axial ymmetry, that i, we ca loe the velocity derivative i the θ-directio. However, the cetrifugal force varie greatly with both r ad z, o there i likely to be igificat r-velocity variatio, ad the cotiuity equatio with the time ad theta derivative cacelled tell u that thi will produce flow i the z-directio. Edge-effect: it uclear whether the thicke or width directio i the r- or z-directio, becaue the ize i thoe two directio are o imilar, o we really ca ot eglect edge effect. The reultig equatio, tartig with cotiuity: 1 (rvr ) + v z = 0 r r z Motio r-compoet: [ ] v v p 1 r θ v r v r ρ v r + v z = + η (rvr ) + r r z r r r r z Motio θ-compoet: [ ] v 1 θ v r v θ v θ v θ ρ v r + + v z = η (rvθ ) + r r z r r r z Motio z-compoet: [ ] v p 1 z v z v z v z ρ v r + v z = + η r + + ρg z r z z r z r r I t that much better? We till eed a computer to olve the ytem though. (b) Thi i a bit of a trick quetio, a there are ozero r-, θ- ad z-compoet to flow, makig it eem three-dimeioal. However, becaue velocity ad preure are ot fuctio of θ but oly of r ad z, by the defiitio give i the problem, the flow i coidered two-dimeioal. (Likewie, all of the problem we actually olve i thi coure ca be coidered oe-dimeioal, eve if the flow directio i differet from the directio i which it varyig, e.g. x-velocity a a fuctio of y.) Note that becaue of the o-zero θ velocity, ome people ue the expreio.5-dimeioal... 3

4 3. Torioal vicometer (40) (a) It pretty clear that the cylidrical form i relevat. We re told to aume Newtoia flow with uiform deity ad vicoity, o we ue the upper form of the Navier-Stoke equatio give out i cla o page 4 of the Solvig Fluid yamic Problem hadout, which i alo the -F form i Poirier ad Geiger p. 59. (b) Simplifyig aumptio: Lamiar flow: do t eed to ue Reyold tree. Steady-tate: the vicometer ru util it reache teady-tate, the a meauremet i take, o it eem o. Time derivative ca be elimiated. Fully-developed: a with problem a, it i hard to defie fully-developed for rotatig flow. Itead, we ll ay it axially ymmetric, o we ca elimiate derivative of velocity i the theta directio. Edge effect: Flow i i the θ-directio, o we have to chooe thicke ad width from the r- ad z-directio. I the r-directio, the ditace betwee the rod ad cup i 3 cm, but the width of the flow i about 15 cm, o we ll chooe r a the thicke directio where the velocity gradiet are teep, ad i the z-directio, the edge i the bottom of the vicometer, which we ll eglect for thi aalyi. So we ca et the velocity derivative i the z-directio to zero, ad becaue cetrifugal force i a fuctio oly of r, we ca elimiate r ad z-compoet of velocity. I eece, wherea problem a i oe big edge effect, o it circulatig throughout; i thi cae, there a mall amout of r ad z circulatio i the bottom, but we re callig that the edge ad eglectig it. Gravity i oly i the z-directio, o we ca igore g r ad g θ, ad et g z imply to the gravitatioal acceleratio g. The equatio with thee elimiatio implify to the followig, tartig with cotiuity: Motio r-compoet: 0 = 0 (everythig cacel) v ρ θ = p r r Motio θ-compoet: ) 0 = p + η ( 1 (rvθ ) θ r r r Motio z-compoet: 0 = p + ρg z That much better tha what we tarted with. (c) Let tart by takig the derivative of the θ mometum equatio with repect to θ to get the variatio of preure with repect to θ (a i cla with the tube flow example we took the derivative of the z mometum equatio with repect to z): )] 0 = p [ + η ( 1 (rvθ ) θ θ r r r We ve aumed cotat vicoity, ad ice r doe t vary with θ ad the mixed partial are equal, we ca move the θ derivative all the way ito the vicou term: 0 = p ( 1 ( + η r v )) θ θ r r r θ 4

5 Now we kow that θ derivative of velocity are zero, o that whole ecod term drop out ad we re left with 0 = p θ whoe olutio i p = A(r, z)θ + B(r, z) Sice the preure i the ame at θ = 0 ad θ = π, the A(r, z) mut be zero, o p = B(r, z), ad p θ = 0. So we ca get rid of that i the θ motio equatio. Now that equatio i reduced to: 1 0 = η (rv θ ) r r r We ca divide through by η ad itegrate oce to give 1 C(θ, z) = (rv θ ) r r The multiply by r, itegrate agai ad divide by r: C (θ, z)r + (θ, z) = v θ r where C = C/. Now becaue we ve aumed fully-developed/ymmetric flow without edge effect, v θ i a fuctio of r aloe, o C ad are cotat rather tha fuctio of r ad z: v θ = C r + r Thi i all we ca do without boudary coditio. [You could top here, or cotiue if you liked.] Next we ca go to the z-directio: 0 = p + ρg z We move the preure to the left ide ad itegrate to give p = ρgz + E(r) where E i ot a fuctio of θ becaue we howed above that preure doe t deped o θ. Ad to the r-directio: v ρ θ = p r r Subtitute our v θ from above: ρ ρ C r + = p r r r Multiply it out: ρ C r + C p + = r r 3 r Ad itegrate: C r p = ρ + C l r + F (z) r where F i agai ot a fuctio of θ. Now we equate the two preure expreio: C r ρgz + E(r) = ρ + C l r + F (z) r 5

6 C Thee ca be equal if F (z) = ρgz + G ad E(r) = ρ r + C l r r + G, (G i a cotat), o our preure i give by C r p = ρ gz C l r + + G r Thi give u the hape of the liquid meicu a a preure ioquat. Pretty cool! (d) The θ-compoet of velocity i give by v θ = C r + r At r = R 1, v θ = ωr 1 (ω i the rotatioal peed i radia/ecod; ome jut ued V a the boudary coditio at R 1 ad that fie too), ad at r = R, v θ = 0. Subtractig the two equatio would t help ice there o cotat to cacel, o we ll multiply each by it radiu the ubtract: R 1 ωr 1 = R 1 C R 1 + R1 R 0 = R C R + R ωr1 = C (R1 R) ωr1 C = R 1 R Now we get from the ecod boudary coditio: So the velocity profile i give by ωr = C R = 1R R R 1 v θ = ωr1 R r R 1 r I uppoe we could ubtitute thi C ad back ito the preure at the ed of part 3c... Nah! The torque i give by the itegral of theta force time radiu. I thi cae, we itegrate the hear tre-radiu product r(τ rθ + τ zθ ) over the iide urface of the outer cup. The problem tate that we ca eglect the hear tre τ zθ o the the bottom of the cup, ad both τ rz ad r are cotat o the cylidrical ide, o the torque i imply the product rτ rz A, o we ue the torque from p. 61 of P&G: ( v ) θ torque = r ηr πrl r r Subtitute i the velocity from above: ] [ 1 ωr R torque = πr 3 1 Lη r r R r R 1 r r Pull the cotat out of the derivative, ad ote that the derivative of r i zero: πr 3 LηωR1 R torque = R R 1 r r R torque = 6 πr 3 LηωR1 R R R 1 r 3

7 The r 3 cacel, o the torque i idepedet of r: 4πLηωR1R torque = R R 1 That the torque i idepedet of r i igificat becaue it mea the torque i equal the torque out, o there i o agular mometum accumulatio i the fluid, which cofirm teadytate. Thi i aalogou to the flux-area product idepedece of r i diffuio. (e) Solve the above torque expreio for vicoity: Subtitute i give data: R R1 η = torque 4πLηωR 1 R η = N m (0.04m) (0.01m) 4π 0.m 10 rev π radia 60 1 mi (0.04m) (0.01m) mi rev ec m η = N m m 5 η = 0.59 N m (f) Accordig to the torque expreio above, torque hould be proportioal to rotatioal peed. Sice the torque double ad the rotatioal peed icreae by a factor of.5, thi doe ot look like a Newtoia fluid. So the vicoity calculated above i oly approximate. The effective vicoity for thi ytem i maller at the higher trai rate, o the fluid i peudoplatic, which make ee for a molte polymer. 4. Fluid flow i blood veel (5) (a) A a rough ballpark etimate, ay about 6 ml of blood i pumped with every heartbeat. At a 3 pule of 80 beat/miute, the flow rate i 8 cm. (b) Average velocity i flow rate divided by cro-ectio area, which i 8 cm [π(0.35cm) ], or about cm 1. Reyold umber i u ν, u = 1 cm, = 0.7cm, ν = 0.01 cm for water, o it i about Flow i therefore lamiar. (c) For lamiar flow, we ca ue the Häge-Poieuille equatio for flow rate i a tube, derived i cla: P 1 P π Q = L 8η R4. We kow the flow rate, vicoity ad radiu, o we ca olve for the preure gradiet: 3 ad the uig the force balace P 1 P 8Qη = L πr 4, πr (P 1 P ) = πrlτ rz τ rz = P 1 P R L, we plug i the above equatio for (P 1 P )/L to give: 8Qη R 4Qη τ rz = = πr 4 πr 3. 7

8 3 Pluggig i our value Q = 8 cm = m N, η = 10 3 m, R = 0.035m give m 10 m τ rz = π (0.0035m) N = 0.4Pa. You could alo tart with the frictio factor (though that wa t required, ice we had t covered 16 it i time), which i Re = Becaue we do t have the legth, we ca oly approximate hear tre, which i give o p. 76 of P&G a ( u = kg 0.1 m ) τ0 = fk = f = 0.4Pa ρ m 3 If you ued a larger vetricle volume or higher pule rate it part 4a, the flow would be traitioal or turbulet. I that cae, you could ue the frictio factor aalyi, though a metioed i cla, you were ot repoible for kowig how to do that for thi problem et, o you could leave the awer a Ca t do thi yet. (d) If flow i lamiar, we ca exted the equatio for the preure drop P 1 P derived above: P 1 P 8Qη 8ūη = =. L πr 4 R With the parameter of thi part of the problem u = 10 3 m, R = m, η = N m, thi give u a preure gradiet of m N P 1 P = m = 1800 Pa. L ( m) m Multiplyig by the legth (1mm) give 1.8 Pa for the preure differece. To ue the frictio factor, firt calculate the Reyold umber which i u ν = 0.05, o flow i defiitely lamiar, ad f = 30. Thi give a hear tre of 0.16 Pa, ad total drag force i thi time the urface area of m, o F d = N, ad = 1.8 Pa. (e) There are everal importat reao why the above aalyi will be iaccurate: The mot importat problem i that it aume teady-tate flow, which i obviouly far from true. The tube flow aalyi aume rigid wall, wherea blood veel are ofteough to give lightly durig each urge of preure. The wall of blood veel are ot mooth, though thi i much more of a factor for mall blood veel tha large oe. For capillarie where blood cell ize i a igificat fractio of the diameter, the Newtoia flow aumptio break dow. For the aorta i particular, it i ot a traight tube, but ha a large bed, ad dyamic preure i that bed ca be quite high. It i alo hort relative to the diameter, o fullydeveloped flow i quetioable. Ad there are blood veel brachig off very early. For thoe itereted i further readig o fluid dyamic of blood flow, Jewell Wu (who took i Fall 000) foud a very iteretig webite: 8