058:0160 Chapter 6-part1Professor Fred Stern Fall 2009 1Chapter 6: Viscous Flow in Ducts 6.1 Laminar Flow SolutionsEntrance, developing, and fully developed flowLe = f (D, V, , μ)(Re)fDLtheoremei f(Re) from AFD and EFDLaminar Flow: Recrit ~ 2000 Re < RecritlaminarRe > Recritunstable Re > RetransturbulentRe06./ DLeDDLcrite138~Re06.maxMax Le for laminar flow058:0160 Chapter 6-part1Professor Fred Stern Fall 2009 2Turbulent flow:6/1Re4.4~/ DLe(Relatively shorter than forlaminar flow)Laminar vs. Turbulent FlowRe Le/D4000 181042010530106441076510895Hagen 1839 noted difference in Δp=Δp(u) but could not explain two regimesLaminar Turbulent Spark photoReynolds 1883 showed that the difference depends on Re = VD/ν058:0160 Chapter 6-part1Professor Fred Stern Fall 2009 3Laminar pipe flow:1. CV AnalysisContinuity:.021constQQdAVCS1 2 1 2. . sin , .,avei e V V ce A A const and V Vr= = = =Momentum:{2 21 2 2 2 1 1( ) 2 sin ( )/0x wF p p R RL R L m V Vz LWpp t p gp J b b�= - - + = -D=D�1 2 3142 43 1 442 4 430222 zRRLRpwRLzpw2058:0160 Chapter 6-part1Professor Fred Stern Fall 2009 41 22( / )wLh h h p zRtggD = - =D + =or)(222zpdxdRdxdhRLhRwFor fluid particle control volume:)(2zpdxdri.e. shear stress varies linearly in r across pipe for either laminar or turbulent flowEnergy:LhzVgpzVgp2222111122RLhhwL2 once τw is known, we can determine pressure dropIn general,),,,,(DVwwΠi Theorem)/,(Re82DffactorfrictionfVDwwhereVDDReroughness058:0160 Chapter 6-part1Professor Fred Stern Fall 2009 5gVDLfhhL22Darcy-Weisbach Equationf (ReD, ε/D) still needs to be determined. For laminar flow, there is an exact solution for f since laminar pipe flow has an exact solution. For turbulent flow, approximate solution for f using log-law as per Moody diagram and discussed late.2. Differential AnalysisContinuity:0 VUse cylindrical coordinates (r, θ, z) where z replaces x in previous CV analysis0)(1)(1zrrrrzrwhere ^^^zzrreeeVAssume = 0 i.e. no swirl and fully developed flow0zz, which shows r = constant = 0 since )(Rr =0^^)(z zzerueV 058:0160 Chapter 6-part1Professor Fred Stern Fall 2009 6Momentum:VzpDtVD2)( z equation:uzpzuVtu2)( )()(1)(0rfzfrurrrzpzboth terms must be constantppBrArzpuArzpruArzprurzprurrrˆlnˆ41ˆ21ˆ21ˆ)(22)0( rufiniteA = 0u (r=R) = 0dzpdRB^42dzpdRrru^4)(22dzpdRuu^4)0(2max058:0160 Chapter 6-part1Professor Fred Stern Fall 2009 7zpRruyuzprstressshearfluidruruzRrRrwrˆ2ˆ2y=R-r,42max0^1( )28 2RR d pQ u r r dr u Rdzpp pm-= = =�2max2^12 8aveQ R d pV uR dzp m-= = = Substituting V = Vave28VfwDVRVRVRaveavew84822DDVfRe6464or DwffVCRe164212VgDLVgVDLDVgVDLfhhL222322642As per CV analysis058:0160 Chapter 6-part1Professor Fred Stern Fall 2009 8for Vpz 0Both f and Cf based on V2 normalization, which is appropriate for turbulent but not laminar flow. The more appropriate case for laminar flow is:64Re16Re)(#000fPCPPPoiseuilleffcffor pipe flowCompare with previous solution for flow between parallelplates with ^xp2max1hyuuxphu^22maxxphhuq^32343max058:0160 Chapter 6-part1Professor Fred Stern Fall 2009 9max23232^uphhqvxhVw3hDhhVhfRe42Re96Re4824hDhhffVhCfCRe42Re24Re1264/96Re24Re)(#000hhfDfDfcfPCPPPoiseuilleSame as pipe other than constants!32966424160000hffhffDonbasedchannelpipeDonbasedchannelcpipecPPPP058:0160 Chapter 6-part1Professor Fred Stern Fall 2009 10Exact laminar solutions are available for any “arbitrary” cross section for laminar steady fully developed duct flowBVP0)()(00^huuupuZZYYxxhyy /*hzz /*Uuu /*)(^2xphU 12 uPoisson equationu(1) = 0 Dirichlet boundary conditionCan be solved by many methods such as complex variables and conformed mapping, transformation into Laplace equation by redefinition of dependent variables, and numerical methods.Related umaxRe only entersthrough stability and transition058:0160 Chapter 6-part1Professor Fred Stern Fall 2009 11058:0160 Chapter 6-part1Professor Fred Stern Fall 2009 12058:0160 Chapter 6-part1Professor Fred Stern Fall 2009 13058:0160 Chapter 6-part1Professor Fred Stern Fall 2009
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