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by: Vidal Goyette


Vidal Goyette
GPA 3.99


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Class Notes
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This 16 page Class Notes was uploaded by Vidal Goyette on Friday October 23, 2015. The Class Notes belongs to ME 531 at University of Kentucky taught by Staff in Fall. Since its upload, it has received 30 views. For similar materials see /class/228235/me-531-university-of-kentucky in Mechanical Engineering at University of Kentucky.

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Date Created: 10/23/15
Strongly Viscous Flows These flows are ones where The viscous Term plays a dominanT role in The physics Flows ThaT will considered are PoiseuilleChannel flow HagenPoiseuillePipe flow Reynolds EquaTionLubricaTion flow Creeping flow STokes Ossen HeleShaw The sTrengTh of The viscous Term significanle reduces The complexiTy of The flow suppressing much of The fluid moTion lncompressible constant viscosity flow J7Du1invBulfi ij Dt p Bxi BxJij Bu Bv Bw Bx By BZ Bt x By BZ p Bxl 1 1 sz By2 BZ2 Bu Bu Bu Bu 1Bp Bzu Bzu Bzu 3 u v w v Bt Bx By Bz pr 3x2 By2 322 Bv Bv Bv Bv 1Bp Bzv Bzv Bzv u v w v Bt Bx By BZ pBy 6 i2 Bui Bui Bui Bui 1Bp Bzui Bzu 3 ua v w gi Bx 2 By2 BZ2 Simplifying Assumptions ZDflow 0 v constant 3y S ready flow 3 2 O at e 7 16 Fullydeveloped Bu 0 D 006ReD Turbulent flow ax m z 006ReD Laminar OIFDPoiseuille Channel Flow Eqns y OFD 02D 39Oseady u a V 82 82 amp 35 3 x2 y2 822 05139eady 0 2D av av gwp amp 82v i E u v w v a 8y az p 8y iaxz ayz azz J Poiseuille Channel Flow Eqns aw 0 Variables Li W Pa 82 a 1 8p 82 p cons ran r prescribed w 2 82 p ax aZ pgradien r prescribed 8w 18p 82w a Z T g 2 Z P Z 82 U1 29 h g x U2 Poiseuille Channel Flow Eqns Bw U 1 0 32 2 w3 l3PV3 I 29 h 32 p BX 2322 g IWLL Con rinui ry 0 con rinui ry a w 0 3 w const Boundary Z 0 U2 az Condi rions ZhU1 WZ0W0 wz00 wzhw0 wzh0 Zmomen rum conserva rion Poiseuille Channel Flow Eqns U 1 3 px z MOO pg reduced I h Define f p pgz pressure g 2 13 pfx pgzpgz pm pm 2 12100 ng Xmomen rum conserva rion 2 A 2 w3M2i3PV3 quot 3 M Liz constant 32 p 3x 322 3x 32 2 A A a 2iapoc 3 BizBJiiPL 322 u 3x 32 K ax 3 uz ABZrLEWZ2 K2u Bx Poiseuille Channel Flow Eqns U 1 ApplyBC u0U2A I 2 h uhU1 g 1 8A x U2 ABh ph2 2 8x 2 A 2 A 2 2 uzU2U1 U2 h a p 5h a pij 2 8x h 2 8x h 8132819 w 0 p f pgz constant 8x 8x Poiseuille Channel Flow Eqns Coue r rePoiseuille U1 U U2 2 0 Favorable 8p 0 8x Constant Unfavorable PlanePoiseuille U1 U2 2 0 aP max 8H ax Skin Friction Since w 0 3 0 then the only lg 11 z non zero stress tensor quantity is x Bu 3 39 Bu T172 3 13 a k xj xi xk ixz jxz Bu 3 sz sz H a 2 Define skin friction coefficient as follows c szlzzo h rm u U Couette f Plane pu pu max Poiseuille Channel Flow Eqns U1 8 w I 82 a h Bu 1 8p 82L g w V 82 3 8x 822 U2 w uz g U2 Symme fr ic uz g 2 U1 Boundary Condi rions wz 0 0 con rinui ry wz0 HagenPoiseuille Pipe Flow Eqns Assume 2D x Xx r s ready incompressible fullydeveloped cons ran r viscosi ry flow 1 8rur O r 8r ppgz BCs 1 urR0 r 8r pax arz r 8r urr00 u Bulaf7v azuula i uxrRgt0 r 8739 par arz r 8739 r2 8 x r20O 8r HagenPelsewlle Pipe Flow Eqns Con rinui ry 1 3ru IT urRO 2 u0 ur00 2 u0 Conservation of rmomen rum 7 2 i2i2ltxgt pialtxgt pgz 0 2 m c0nstant Conserva rion of xmomen rum A 2 A 02 iapv a 1 1811 t iapli rauquot p 3x arz r 3r p 3x r Br Br 1 3f 1 3 BuC r constant u 3x r Br Br d A d 2 2 r u drIL dr 2 r u Ar dr dr u dx dr 21 dx HagenPoiseuille Pipe Flow Eqns Conserva rion of xmomen rum duC A r d dr r d A quot r0oi AO dr 0 Zydx BC d 2 A u drIL dr 2uxrBr d p dr u dx 41dx 2 A 2 A BCuxrROBR 2 B R d p 41 dx 41 dx A d z dp u R 0 Z c0nstant r p p pg dx dx HagenPoiseuille Pipe Flow Eqns 3L 9r32 232r39 Bux 3 Trx Txr 2 W Twall TrxrR 2739 2Humax Humax T Z R rR R HagenPoiseuille Pipe Flow Eqns x J Jtxrr dr d0 75 2 R J O 1 2rdr 2amf1 2jd2um 1 lt gt2 gtd 1 u 2 1 2 max 2u ltL max 2 4f 0 2 2mm L Twau limrznaxR WmaxRIu ReR ReD 8 16 WW 2 2mmwx 31 151731 atme ReR ReD T DarcyFIiction Factor f W 1124Cf 64 6 RED Incompressible constant viscgsity flow Bu J7Du1iBpvBulfi ij Dt p Bxi BxJij Bu Bv Bw Bx By BZ P axi 1 i Bt Bx By 32 sz By2 322 g 2 Bu Bu Bu Bu 1Bp Bzu Bzu Bzu 3 u v w v Bui Bui Bui Bui 1Bp Bzui Bzu Bzu u v w v at ax ay FE egg 372322 Bv Bv Bv Bv 1 Bp Bzv Bzv Bzv u v w v Bt Bx By BZ p By 3x2 ay2 322 Bw Bw Bw Bw 1Bp Bzw Bzw Bzw u v w gv Bt Bx By BZ p BZ 3x2 ay2 322 Scaling Assumptions x yL Zhov W39g uvU wUhU p hl L Bu Bu Bu Bu pr B211 B2u B211 u v w v Bt Bx By Bz p Bx 3x2 By2 Bz2 U2 Buquot Buquot Buquot Buquot 3 u w L Btquot Bxquot Byquot Bzquot uU Bpquot UL1B2uLB2uLL2 B211quot hgp 395 L2pLBxquot2 Byquot2 kg B2quot2 u My Scaling Assumptions U2h Bu Bu Buquot Bu 3 u v w L Btquot Bxquot Byquot 32 uU BpLUuh 82uLB2uLL2 3211 p Bxquot L2p LBx2 By2 113 322 Given that lubrication films are very thin assume hU L gt 0 then 2 2 30 iU auiaP Bu V p Bxquot p 322 p 396 822 2 Similar analysis yields 0 la pv 8 p 0 gt p px y p 3y 322 32 Given that we can integrate the above equations with the following BC uz 0 U v0 w0 0 uz h vh 0 wh W Reynolds Equation yielding the following result u ia pz2 zh 1 5 U wig pd zh 2y Bx h 2 By Next integrate the continuity equation 11 h J 35 dz a Walzw0 whWa h 0 Bx By 0 32 at 3x ay ax 0 Plug in the equations for u v and integrate to get 1 3 3 BP 3 3 3p ah ah h h 6U 12 u3x Bx 3y 3y ax at Reynolds Eqn h h h By Leibnitz39 s Theorem J iJalz i J u dz 2 J v dz 0 y 0 Low vs High Re Flows Momentum equation for steady incompressible const viscosity ows 2 Bu l 1 8 u Bx Bx Bx 18x 1 Consider ow of freestream velocity U over a body of size L 9 Nondimensionalizeby x u ui p 17 1 L U pU 2 B a 1 a 2 UL 117 R357 Bx Bx Re Bx Bx v If Re gtgt 1 viscous term is negligible gtinviscid solution breaks down 8 U near surfaces boundary layer Where we can no longer assume that u I Opposite caseis if Re ltlt 1 In this case the inertial forces LHS aresmall compared to the viscous forces But if thereis no moving surface Couette the ow must be generated by pressure forces This implies that there are ows in Which pressure and viscous forces are in balance Poiseuille But by the above this means that Creeping Flows we cannot use the scaling 7 pU 2 Rather 82L U x u p p L I i 3 x u w p axjaxj L L UprL Bu39 Bp39 Bzu39 Reu397 I 1 8x3 Bx BxJBxg These types of ows are called creeping motions They occur With low velocity high viscosity andor small bodies A classic exampleis creeping ow around a sphere solved by Stokes in 1851 Re gt o 2 Vp uV2u 2 VgtltVp uVZu 2 W20 0 Axisymmetric ow Bldg O u O vorticity u lEBO u g Bauer 1 w e r r 1 811 1 811 Definea pheriml u ug r2 sing 849 rs1nl9 Br 2 jaw 1 ail 1 31 r s1nl9ar2 r2 849 s1nl9 849 Creeping Flow around a Sphere V2 1 a 280 1 a 980 O w VT r W r25m9 sin 86 jisin i 1 i 2 0 ar2 r2 849 sint9849 V The boundary conditions are wlta 9gto Ballta9gt0 VIlt ltgtOgtUr2sin29 739 The last is the streamfunction for uniform ow in spherical coordinates This last BC suggests a solution form of 11 fr sin 2 t9Sub this in 4 2 d f id f i if0gtfz474 Br2Cr2 dr4 r2 dr2 r3 W V4 df r gtltgtltgtfr gtUr2 OBU iaOC Ua fa0DUa3 3 3 uUcos91 31L u9 Usin 1 31 017 27 2V3 41 4V3 Creeping Flow around a Sphere Pressure Complicated math to integrate these equations to get 311 U 9 05 13 U 2r To get drag integrate the stress I over the sphere surface r a dFSI 039 njdA I75U le njdA pnl 1an dA We need the forcein the direction of the stream dFSX ldA 17 cos 91quot cos 49 1 sin 49a 3 3 3 3 1quot 2 u 2UCOSQ3J3L 359 r3 1 li 3Ua sing Br 2 2r4 Br r r 849 2r4 Creeping Flow around a Sphere 2 if 7p cos 49 1quot cos 0719 sin 49a MU cos2 0O MU 2a 2a Since this is location independent just multiply by surface area sin2 49 2a dF F D 47212 67rIaU oc U e Stokes39law of resistance One third of thisis pressure drag two thirds is skin friction drag cD 24 W5 pU m ReD The above result can be modified to consider the flow in the frame of reference of the freestream This creates the streamlines of iU 2 ean an21sin2 eelUr2 sin2 497 Ur2 sin2 0 73ii V 2 4 4 2 T 4 M3 This is a completely symmetric ow field indicating no wake Mllllkan s OII Drop Experiment It was Stokes drag that allowed Millikan39 s oil drop experiment 1910 to determine the charge on an electron For a non accelerating oil drop the force balanceis 67Fl Uamp m3gltpmz 7 pa neE Milliken repeatedly measured the veclocity of oil drops caught in an adjustable electricfield E between two charged plates 7 Photos from http chem chhuji acileugeniildhirtorymillikanhtml Ossen s Improvement For Stokes Flow the assumption was that inertial forces are dominated by viscous and pressure forces This is not valid far from the surface which is why we see a symmetrical streamline pattern when it should be asymmetrical For large r can approximate the viscous and intertial forces by Br i uUa Br 3 Ha r Bu 9 pUZa pm Br Hm quotM raw inertia pU 2a r3 pUa r r 2 2 7 7Re7 asr gtoltgt v1scous r uUa u a a So when r 11 Re the inertial forces are no longer neglibileStoke39s did not realize this in fact he concluded that Stokes ow Vp uVZu is not possible over a 2D cylinder because of an inability to mathematically match the boundary conditions when r gt oo Ossen proposed addressing this by partly accounting for inertia at large r Ossen s Improvement Ossen assumed u U u39 v v39 w w39 U gtgt u39 v39 w39for large r a xmomentum Uuaiv Biw ai ilVV2u Bx By Bz p Bx Bu39 1 Bp 2 Neglecting quadratic terms write as U 7 77 vV u B p Bx B 39 olt 0 Orgenerally Uiialvvlu 3C1 17 gt 3x P 3x u agtUVawa0 Ossen39s solution is L lsin2 49 i1cos 91 exp amp1l cos 49 Re E U11 2 2a 2 47 Re 4 a v To recoverStoke39s result assume 2 1 and expand exponential taking the lowest order term New drag coefficient C D D 24 3 7 1iRe Re lt5 iJUZMZ RED D D 16 HeleShaW Flow is a useful to visualize streamlines The arrangement is essentially Poiseuille ow With a cylinder blocking the channel Assume the geometry shown The ow is low Re Ub Re7 ltltL ltltL Reltlt 2ltltl v a 2 U Non dimensionalize 0 17 pin uUO a Cof Mass u 5 1 ar39uiamp 0 Bx r Br r 849 U0 u1 Uiop Cof Momentum Bu 5 u Bu u2 Bx39 r 8 r39 849 r39 Re tC Bp39282u 2 282u 21 Bu 1 Blu u 2 Bug 872 7 87 72 892 72 72 Br39 ax 2 HeIeShaw Flow 31 31 u Bu u u 6 6 6 6 6 r Reur 5 u 777 3 lar r 349 r x 32 32 a 32 a 713LJr u6 2 u61 us 1 us us i u 32 7 a r 2 3492 7772 349 7 736 ax Re u Bu 5 u Bu 14 Bu 73p 4rg2 3211 1311 1 3211 ax 3 r as ax 32 r Br r 3492 Drop primes for clarity First1et Re gt 0 then 9 gt O aa quot0s gt0 uxxr10 2 0 x 3211 3p 2 32116 13p 2 3p 2 7Os 777O 05 a0 6 3x2 3r gt 3x2 rag gt ax gt P 11r Integrate the two momentum equations with BC u x i1 us x i1 O a 1 2 18 1 2 3 ii 1 7777 1 u arlzm Ml us raglzm ml 31 a 1 2 3 77 7 1 201 ax axl 2Plt x gt HeIeShaw Flow If new argue thatu1C S 052 then 2 a r W 0 3x1 3x1 liltmrlal lig0g0 3 rar 349 3x1 BC 31010 r gteoRcoso917x22 r 17x2 1 2 rcoso91i r2 2 1 17x2 1 17x2 2 us 7s1n 01 u cos 417717 Note thatasr gt1u gtOu6 gt7251n4917x22 1 Ir A39P on cylinder Ilrfamn mall D 001 1 very thin BL


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