Lecue 36 Pipe Flow and Low-eynolds numbe hydodynamics 36.1 eading fo Lecues 34-35: PKT Chape 12. Will y fo Monday?: new daa shee and daf fomula shee fo final exam. Ou saing poin fo hydodynamics ae wo equaions: Coninuiy equaion: " v = Navie-Sokes equaion: " v + " v $ ) v = & P + ' 2 v g is se o eo, see Lec. 34) Thee examples: 1. Sokes dag las lecue) 2. Seady-sae pipe flow a abiay!) eynolds numbe. full calculaion) Hagen-Poisseuille Fomula: Q = " 8 4 $ P, whee Q is volume/second hu pipe. L Geomey: Pipe adius, lengh L v) End pessues P 1 high) and P 2 low) P 1 independen of ) Pogam: a) find flow field v). L b) Calculae volume flow Q hough pipe. Assume cylindical symmey. Seady sae flow imples " v = All velociies along he axis ˆ : v ) = v,)ˆ Coninuiy eq: dv =, so v ) = v )ˆ and ineial em vanishes: " v $ ) v = "v v d =. We ae lef wih " 2 v = 1 " P. Bu, " P has o poin along!, so "P =, i.e., P is independen of. " Thus, " 2 v = 1 dp dp, which mus be independen of, so d d = P 2 " P 1 = " P L L, whee "P P 1 $ P 2 > is he posiive) pessue dop beween he ends of he ube. Finally, hen " 2 v ) = 1 d d dv & = ) 1 +P $ d ' * L. Inegaing: d " $ d dv ' = *P d & ) L dv d = 2 *P 2) L + A dv d = *P 2) L + A v ) = 2 *P 4) L + Aln + B Bounday condiions ae: dv d = = no singula behavio a he axis A=) v ) = no slip on walls fixes B) you ex ges he las sep f his wong, eq. 12.21) P 2
) $P Upsho: v noe paabolic velociy pofile in pipe. 36.2 4 L Finally, calculae he seady-sae ae of volume flow: ) = 2 " 2 ) Q = " dav = " d2v ) = 2 4$ &P " d 2 ' 2 = L 8 4 $ &P L This is Hagen-Poiseuille fomula fo lamina flow hough a pipe. Noe he high powe of, which makes blood flow vey sensiive o vessel dilaion/conacion. This fomula is valid a high eynolds numbe as well as low. I is believed ha his flow is linealy sable a all eynolds numbes. Howeve, in pacice i is exceedingly difficul o se up he seady sae a he enance o he ube. In pacice, wih sufficien cae, you can ge lamina i.e., ubulence-fee) flow up o e ~1 5 ; howeve, wihou such cae, ubulence onses in he sandad ange. 3. Wha s special abou Low-eynolds-numbe flow? I s had o mix. see below) Discussion of Couee expeimen. I s had o swim. see below) ecipocal moions wok fo you if he fowad soke is significanly slowe han he backwads soke. The eason is ha you can ansfe momenum o he backwads flowing packe of wae; you ge he coesponding fowad momenum. In viscous fluids, ha backwads-flowing packe does no ge away fom you, so i is eniely cancelled on he evese soke. In high- e fluids, i is he ineia em which caies he packe away. Avoidance and dag ac a long disance. This las feaue is a consequence of he absence of a naual lengh scale beyond in he low-eynolds limi. Enainmen falls off as powe law ahe han exponenial. Discussion of demixing expeimen: Couee Cell A low eynolds numbe " v = $ P + & 2 v wih " v = ) Equaions of ceeping moion foce em ficion em o, in seady sae, " 2 v = P. This is invaian unde v " v ; P " P moion evesal c.f., he ineial em!), so a) seady-sae flow lines evese b) fo a symmeical objec, upseam and downseam flow lines ae idenical c.f., ubulence). Befoe applying his o Couee expeimen, I wan o develop an analog, in which he mah is simple enough so you can follow i in deail: Damped paicle moion in he non-ineial limi wih a ime-dependen foce Conside a single paicle in damped moion subjec o a ime-dependen foce: m dv d = "v + f ). Suppose f)=f consan), hen i is easy enough o solve. ficion em foce em In he long-ime limi, v appoaches a consan dif velociy, v " = f /, a which he diving foce and he viscous damping cancel. The full soluion is v ) = v " + v v " )e $ m, i.e., any iniial deviaion fom v " elaxes wih a chaaceisic ime " elax = m. Now, suppose ha f) vaies wih ime, bu slowly on he scale of τ elax, so ha 1 df f d " 1 1 << " $ diving elax m.
Wha we expec o happen is ha, afe an iniial ansien of ode τ, he moion jus follows 36.3 he foce, accoding o v ) = 1 " f ), so ha x ) = x + 1 d f ). This moion is vey much like " he Couee cell in he sense ha he paicle euns o is saing poin wheneve " d f ) =. gaphical inepeaion and elevance o swimming and o Couee) The boxed equaion is an exac soluion in he sic non-ineial limi m=. Howeve, i is no an exac soluion fo m>, since m dv d ". Quesion: Can i be a good appoximae soluion when m is small, excep, of couse, fo he sho iniial ansien? Answe: Yes. Suppose we eliminae he iniial ansien by assuming v = 1 " f ). Then, wha happens as he sysem evolves in ime is ha he sysem evolves away fom he boxed soluion because he mdv/d em is no eo; howeve, i is coninuously elaxing back owads i due o he fas exponenial decay and neve ges fa away. I wan now o skech a sysemaic expansion in which he boxed non-ineial) equaion is he lowes appoximaion and hee is a sysemaic expansion abou i: m dv d = f ) " v subjec o bc f ) = "v ). i.e., if f)=, hen v)=) In his case, i s easy. We do no have o develop a peubaion heoy because hee is an exac soluion: I is easy o veify ha he following is an exac soluion of he equaion plus bounday condiions: v) = e " m f ) + 1 m d f )e m ' $ *. don woy fo now abou whee his comes fom!) ' & * ) Veify ha, when f is ime independen, his gives you back v)=v d. Thee is a nice ick fo expanding his soluion o see how i woks. " m f ) Simply, inegae by pas unde inegal sign: " e m & " df e m, so $ d ' v) = e " 2 + m f ) + 1 $ & m m f )e m ' ) " m d df / 5 4 -, & ) d e m - 7 4 * 7 = f ) " e" m d df * e m 4 - - 7 d 3. 16 Coninuing his pocess: v) = f ) m df " " 2 d + m df " 2 e " m + m " d = " 2 e m d df " $ e m. d The aio of second o fis em is m " 2 " 1 df f d ~ elax. When his aio is small, he expansion diving conveges apidly and he fis em is a good appoximaion. NOTE: Can also deive his expansion by a caeful ieaion of he oiginal equaion:
Take eoh ode appoximaion as v equaion of moion: m dv d = m " ) = 1 " f ) and subsiue ino he 36.4 df d =? f ) "v ) =. Fails o solve he equaion because of he ime deivaive on he lef. To make a fis coecion, we need o add a em o v ) o cancel he exa ime deivaive: You fis hink o add a em o cancel his ime deivaive on he lef: v 1 )=? 1 " f ) m df " 2. Check whehe his is soluion by subsiuion: d m df " d =? f ) "v ) = m dv 1 = m df d " d m2 d 2 f " 2 d 2 =? f ) "v 1 ) = + m df " d, which has, indeed, canceled he fis ime deivaive on he lef. Bu, hee is a poblem: The bounday condiion a = is no longe saisfied: v 1 ) = 1 " f ) m df " 2 $ v. d = Bu, we can always add an addiional soluion of he homogeneous equaion, so v 1 ) = 1 " f ) m df " 2 d + m df " 2 e " m, which now obeys he bc. d = noe ha his is wha came ou of exac equaion) Can ieae his pocess, hus building up he full exac soluion. Now, back o he Couee cell demo done in he las lecue. I will discuss he analog in linea geomey. X)=x +V V x Claim: x,) = x +V) L = X ) L When he uppe-plae velociy V is consan, he simple shea flow v = v ) x ˆ = V x L ˆ wih P= saisfies he equaions of seady-sae ceep AND, moe geneally, he full seady sae N-S equaion. Commen: This means ha an iniially compac blob ges sheaed ino a hin shee. Poof: As in pipe flow, he incompessibiliy condiion is saisfied auomaically. ) x Fuhemoe, " d2 v d 2 =, and i saisfies he bounday condiions a he lowe and uppe plaes. v ) = ; v L) = V. evesing he plae moion also a consan speed) eassembles he iniial blob. Bu, wha is less obvious is ha, a low eynolds numbe, ANY moion X) almos saisfies he ceep equaion a low eynolds numbe only). Claim: Povided acceleaions ae no oo lage, any moion which sas a X=)= and ends a
some lae ime a XT)= leaves he enie fluid almos in is iniial sae. 36.5 Poof: Ty he guess v,) = v,) x ˆ = V ) x L ˆ, whee V ) = dx d, i.e., x,;x ) = x + X ) L, so ha x,;x ) euns o x wheneve X) euns o eo. Noe ha his expession fo v,) saisfies he bounday condiions a he wo plaes. Bu, does i saisfy he Navie-Sokes equaion and incompessibiliy? All he ems of he ime-independen equaions including he incompessibiliy condiion) vanish, as befoe; bu, hee is an uncancelled ime deivaive on he lef: " v = " dv ) $. So, his is no a soluion. How big is he eo? Can i be fixed? d L We need o solve " v + " v $ ) v = & P + ' 2 v, which, fo he shea geomey, educes o " v = $ 2 v 2. Thus, we can cancel he ime deivaive on he lef by adding o he velociy field a coecion: v,) = V ) + "v,) wih "v,) = L2 3 dv ) ' *. This exa ems leaves he L 6$ d & L) incompessibiliy and ineial ems OK; howeve, i poduces wo new poblems: i) Unde he ime deivaive, i poduces a new, uncompensaed em of he lef of " $v = "L2 d 2 3 V ) & ) 6 d 2 +. ' L* Bu, his can be compensaed via a new em in 5, ec. convegence?) ii) This em spoils he ageemen wih he uppe plae bounday condiion: v L,) = V ) + "L2 dv ). 6 d Bu his can be compensaed by coecing he coefficien of he oiginal em, so v,) = V ) " L2 dv ) ' & 6$ d * ) L + L2 3 dv ) ' *. 6$ d & L) You can imagine ieaing o povide a fully coeced soluion in his way, POVIDED ha he coecion ems ae small. Bu, ae hey? Check: Suppose ha V is a chaaceisic speed of plae moion. Then he elaive sie of he coecion em is measued by 1 " L2 dv ) = 1 V 6$ d 6 " LV $ " L 1 dv T " ' * ~ V & V d e, whee T is he ime fo he uppe plae o move a ) + disance of he laye hickness and τ is he scale of vaiaion of V). Is his small? If so, he pue-shea soluion is vey good. If we ae in a vey-low-eynolds numbe egime, hen he T/τ aio does no have o be small. Ty ealisic numbes fo he Couee cell: " ~ 1 3 kg/m 3 ; L ~ cm =1 2 m;v ~ cm/s =1 2 m/s;$ glyceine ) =1Pas; ~ 1 s "LV 6 $ L 1 dv $ ' * ~ 13 $1 +2 $1 +2 $ 1+2 V & V d ) 6 1 +2 $ 1 1 = 1 6 wih some addiional facos of /L if you ae no a he uppe plae. Upsho: The pue shea appoximaion is pey good.