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058:0160 Chapters 6 Professor Fred Stern Fall 2007 1 Chapter 6: Viscous Flow in Ducts Entrance, developing, and fully developed flow Le = f (D, V, ρ, μ) (Re)fDLtheoremei=→Π f(Re) from AFD and EFD Laminar Flow: Recrit ~ 2000 Re < Recrit laminar Re > Recrit turbulent Re06./ ≅DLe DDLcrite138~Re06.max= Max Le for laminar flow058:0160 Chapters 6 Professor Fred Stern Fall 2007 2 Turbulent flow: 6/1Re4.4~/ DLe (Relatively shorter than for laminar flow) Laminar vs. Turbulent Flow Re Le/D 4000 18 104 20 105 30 106 44 107 65 108 95 Hagen 1839 noted difference in Δp=Δp(u) but could not explain two regimes Laminar Turbulent Spark photo Reynolds 1883 showed that the difference depends on Re = VD/ν058:0160 Chapters 6 Professor Fred Stern Fall 2007 3 Laminar pipe flow: 1. CV Analysis Continuity: .021constQQdAVCS==→⋅=∫ρρρ 12 1 2.. sin , .,avei e V V ce A A const and V Vρ==== Momentum: 2212 2211() 2 sin( )/0xwFppR RL RL mV VzLWpπτπ γπ ϑ β β•=− − + = −Δ=Δ∑ 0222=Δ+−Δ zRRLRpwγππτπ RLzpwτγ2=Δ+Δ058:0160 Chapters 6 Professor Fred Stern Fall 2007 4 122(/ )wLhhh p zRτγγΔ= − =Δ + = or )(222zpdxdRdxdhRLhRwγγγτ+−=−=Δ= )(2zpdxdrγτ+−= i.e. shear stress varies linearly in r across pipe for either laminar or turbulent flow Energy: LhzVgpzVgp+++=++2222111122αγαγ RLhhwLγτ2==Δ ∴ once τw is known, we can determine pressure drop In general, ),,,,(εμρττDVww= Πi Theorem roughness058:0160 Chapters 6 Professor Fred Stern Fall 2007 5 )/,(Re82DffactorfrictionfVDwερτ=== where υVDD=Re gVDLfhhL22==Δ Darcy-Weisbach Equation f (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 Analysis Continuity: 0=⋅∇ V Use cylindrical coordinates (r, θ, z) where z replaces x in previous CV analysis 0)(1)(1=∂∂+∂∂+∂∂zrrrrzrϑϑθϑθ058:0160 Chapters 6 Professor Fred Stern Fall 2007 6 where ^^^zzrreeeVϑϑϑθθ++= Assume θϑ = 0 i.e. no swirl and fully developed flow 0=∂∂zzϑ, which shows rϑ = constant = 0 since )(Rrϑ =0 ^^)(z zzerueV ==∴ϑ Momentum: VzpDtVD2)( ∇++−∇=μγρ z equation: uzpzuVtu2)( ∇++∂∂−=⎥⎦⎤⎢⎣⎡∇⋅+∂∂μγρ )()(1)(0rfzfrurrrzpz⎟⎠⎞⎜⎝⎛∂∂∂∂++∂∂−=μγ ∴ both terms must be constant BrAdzpdru ++= ln4^2μ zppγ+=^ )0( =ru finite Æ A = 0058:0160 Chapters 6 Professor Fred Stern Fall 2007 7 u (r=R) = 0 Æ dzpdRB^42μ−= dzpdRrru^4)(22μ−= dzpdRuu^4)0(2maxμ−== zpRruyuzprruruzRrRrwr∂∂−=∂∂−=∂∂−=∂∂=∂∂=⎥⎦⎤⎢⎣⎡∂∂+∂∂===^^22μμτμϑμτ y=R-r, 42max0^1()282RRdpQurrdr u Rdzπππμ−===∫ 2max2^128aveQRdpVuRdzπμ−== = Substituting V = Vave 28Vfwρτ= As per CV analysis058:0160 Chapters 6 Professor Fred Stern Fall 2007 8 DVRVRVRaveavewμμμτ84822==−×−= DDVfRe6464==ρμ VgDLVgVDLDVgVDLfhhL∝=××===Δ222322642ρμρμ for Vpz ∝Δ→=Δ 0 or DwfVCRe16212==ρτ Both f and Cf based on V2 normalization, which is appropriate for turbulent but not laminar flow. The more appropriate case for laminar flow is: 16Re#0===fCPPoiseuille for pipe flow Compare with previous solution for flow between parallel plates with ^xp058:0160 Chapters 6 Professor Fred Stern Fall 2007 9 ⎟⎟⎠⎞⎜⎜⎝⎛⎟⎠⎞⎜⎝⎛−=2max1hyuu xphu^22maxμ−= ⎟⎠⎞⎜⎝⎛−==xphhuq^32343maxμ max23232^uphhqvx=⎟⎠⎞⎜⎝⎛−==μ hVwμτ3= hVhf2Re4824==ρμ hfC2Re12= Po =12 Same as pipe other than constants! Exact laminar solutions are available for any “arbitrary” cross section for laminar steady fully developed duct flow058:0160 Chapters 6 Professor Fred Stern Fall 2007 10 BVP 0)()(00^=++−==huuupuZZYYxxμ hyy /*= hzz/*= Uuu /*= )(^2xphU −=μ 12−=∇ u Poisson equation u(1) = 0 Dirichlet boundary condition Can 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 umax Re only enters through stability and transition058:0160 Chapters 6 Professor Fred Stern Fall 2007 11058:0160 Chapters 6 Professor Fred Stern Fall 2007 12058:0160 Chapters 6 Professor Fred Stern Fall 2007 13058:0160 Chapters 6 Professor Fred Stern Fall 2007 14058:0160 Chapters 6 Professor Fred Stern Fall 2007 15 Stability and Transition Stability: can a physical state withstand a disturbance and still return to its original state. In fluid mechanics, there are two problems of particular interest: change in flow conditions resulting in (1) transition from one to another laminar flow; and (2) transition from laminar to turbulent flow. (1) Transition from one to another laminar flow (a) Thermal instability: Bernard Problem A layer of fluid heated from below is top heavy, but only undergoes convective “cellular” motion for Raleigh #: 42/crgd gdRaRawd kααυνΓΓ==> forceviscousforcebouyancy α = coefficient of thermal expansion =PT⎟⎠⎞⎜⎝⎛∂∂−ρρ1 /dTTddzΓ=Δ =− ()TΔ−=αρρ10 d = depth of layer k, ν =thermal, viscous diffusivities w=velocity scale: convection (wΓ) = diffusion (kΓ/d) from energy equation, i.e., w=k/d058:0160 Chapters 6 Professor Fred Stern Fall 2007 16 Solution for two rigid plates: Racr = 1708 for progressive wave disturbance αcr/d = 3.12 ^()[cos( ) sin( )]ctixctiwwe e xct xctα−==−+− λcr = 2π/α = 2d ^()ixctTTeα−= α = αr c=cr + ici αr = 2π/λ=wavenumber cr = wave speed ci : > 0 unstable = 0 neutral < 0 stable Thumb curve: stable for low Ra < 1708 and very long or short λ. For temporal stability Ra > 5 x 104 transition to turbulent flow058:0160 Chapters 6 Professor Fred Stern Fall 2007 17 (b) finger/oscillatory instability: hot/salty over cold/fresh water and vise versa. (Rs – Ra)cr = 657 ())1(/04STkdzdsdgRsSΔ+Δ−==βαρρνβ058:0160 Chapters 6 Professor Fred Stern Fall 2007 18 (c) Centrifugal instability:
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