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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, , μ) L theorem f (Re) f(Re) from AFD and EFD e i D Laminar Flow: Recrit ~ 2000 Re < Recrit laminar Re > Recrit turbulent L / D .06 Re e L e max .06 Re D ~ 138D crit Max Le for laminar flow 058:0160 Chapters 6 Professor Fred Stern Fall 2007 2 Turbulent flow: Re Le/D 4000 18 L / D ~ 4.4 Re 1/ 6 104 e 20 105 30 106 44 (Relatively shorter than for 107 65 laminar flow) 108 95 Laminar vs. Turbulent Flow 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/ν 058:0160 Chapters 6 Professor Fred Stern Fall 2007 3 Laminar pipe flow: 1. CV Analysis Continuity: 0 V dA Q1 Q2 const. CS i.e. V1 V2 sin ce A1 A2 , const., and V Vave Momentum: Fx ( p1 p2 ) R2 w 2 RL R2 L sin m(2V2 1V1 ) p W z / L 0 pR 2RL R z 0 2 w 2 2 L p z w R 058:0160 Chapters 6 Professor Fred Stern Fall 2007 4 2 w L h h1 h2 ( p / z ) R or R h R dh w 2 L 2 dx R d ( p z ) 2 dx r d ( p z ) 2 dx i.e. shear stress varies linearly in r across pipe for either laminar or turbulent flow Energy: p1 1 p2 2 V1 z1 V2 z 2 hL 2g 2g 2 L h h w L R once τw is known, we can determine pressure drop In general, roughness ( ,V , , D, ) w w Πi Theorem 058:0160 Chapters 6 Professor Fred Stern Fall 2007 5 8 w f friction factor f (Re D , / D) V 2 VD where Re D LV2 h hL f Darcy-Weisbach Equation D 2g 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: V 0 Use cylindrical coordinates (r, θ, z) where z replaces x in previous CV analysis 1 1 (r r ) ( ) z 0 r r r z 058:0160 Chapters 6 Professor Fred Stern Fall 2007 6 ^ ^ ^ where V r er e z e z Assume = 0 i.e. no swirl and fully developed z flow 0 , which shows r = constant = 0 since z r (R ) =0 ^ ^ V z e z u ( r ) e z Momentum: DV ( p z ) 2 V Dt z equation: u V u ( p z ) 2u t z 1 u 0 ( p z ) r z r r r f ( z) f (r ) both terms must be constant ^ 2 r dp ^ u A ln r B p p z 4 dz u (r 0) finite A=0 058:0160 Chapters 6 Professor Fred Stern Fall 2007 7 ^ 2 R dp u (r=R) = 0 B 4 dz ^ ^ r R d p 2 2 2 R dp u (r ) u u (0) 4 dz 4 dz m ax u u r z r r As per CV analysis ^ rp 2 z ^ u u Rp w y r R r r R 2 z y=R-r, ^ R 4 d p 1 R Q u (r )2 r dr umax R 2 0 8 dz 2 ^ Q 1 R2 d p Vave u R2 2 max 8 dz Substituting V = Vave 8 w f V 2 058:0160 Chapters 6 Professor Fred Stern Fall 2007 8 R 8Vave 4Vave 8V w 2 R 2 R D 64 64 f DV Re D L V 2 64 L V 2 32 LV h hL f V D 2 g DV D 2 g gD 2 for z 0 p V w 16 or Cf 1 Re D V 2 2 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: Poiseuille # P C Re 16 0 f for pipe flow Compare with previous solution for flow between parallel ^ plates with p x 058:0160 Chapters 6 Professor Fred Stern Fall 2007 9 y h ^ 2 2 u u 1 h u p 2 max m ax x ^p 3 4 2h q hu 3 m ax x 3 q h ^ 2 2 v p u 2h 3 3 x max 24 w 3V h 48 f Vh Re 2h C 12 Po =12 f Re 2h Same as pipe other than constants! Exact laminar solutions are available for any “arbitrary” cross section for laminar steady fully developed duct flow 058:0160 Chapters 6 Professor Fred Stern Fall 2007 10 BVP u 0 x ^ 0 p (u u ) x YY ZZ u ( h) 0 2 Re only h ^ enters y y/h * z z/h * u u /U * U ( p ) through x stability and Related umax transition u 1 2 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. 058:0160 Chapters 6 Professor Fred Stern Fall 2007 11 058:0160 Chapters 6 Professor Fred Stern Fall 2007 12 058:0160 Chapters 6 Professor Fred Stern Fall 2007 13 058:0160 Chapters 6 Professor Fred Stern Fall 2007 14 058:0160 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 gd gd 4bouyancy force Raleigh #: Ra w / d 2 k Racr viscous force 1 α = coefficient of thermal expansion = T P T / d dT dz 1 T 0 d = depth of layer k, ν =thermal, viscous diffusivities w=velocity scale: convection (w) = diffusion (k/d) from energy equation, i.e., w=k/d 058:0160 Chapters 6 Professor Fred Stern Fall 2007 16 Solution for two rigid plates: Racr = 1708 for progressive wave disturbance ^ i ( x ct ) αcr/d = 3.12 w we ct e i [cos( x ct ) sin( x ct )] ^ i ( x ct ) λcr = 2π/α = 2d T T e α = αr c=cr + ici For temporal stability αr = 2π/λ=wavenumber cr = wave speed Ra > 5 x 104 transition ci : > 0 unstable to turbulent flow =0 neutral <0 stable Thumb curve: stable for low Ra < 1708 and very long or short λ. 058:0160 Chapters 6 Professor Fred Stern Fall 2007 17 (b) finger/oscillatory instability: hot/salty over cold/fresh water and vise versa. Rs gd ds dz /k 4 (Rs – Ra)cr = 657 S (1 T S ) 0 058:0160 Chapters 6 Professor Fred Stern Fall 2007 18 (c) Centrifugal instability: Taylor Problem Bernard Instability: buoyant force > viscous force Taylor Instability: Couette flow between two rotating cylinders where centrifugal force (outward from center opposed to centripetal force) > viscous force. r c 2 2 Ta i i o c r r r 2 0 i i centrifugal force / viscous force Tacr = 1708 αcrc = 3.12 λcr = 2c Tatrans = 160,000 Square counter rotating vortex pairs with helix streamlines 058:0160 Chapters 6 Professor Fred Stern Fall 2007 19 058:0160 Chapters 6 Professor Fred Stern Fall 2007 20 (d) Gortler Vortices Longitudinal vortices in concave curved wall boundary layer induced by centrifugal force and related to swirling flow in curved pipe or channel induced by radial pressure gradient and discussed later with regard to minor losses. For δ/R > .02~.1 and Reδ = Uδ/υ > 5 058:0160 Chapters 6 Professor Fred Stern Fall 2007 21 (e) Kelvin-Helmholtz instability Instability at interface between two horizontal parallel streams of different density and velocity with heavier fluid on bottom, or more generally ρ=constant and U = continuous (i.e. shear layer instability e.g. as per flow separation). Former case, viscous force overcomes stabilizing density stratification. g 2 12 12 U1 U 2 2 2 ci 0 (unstable) U U 1 2 large α i.e. short λ always unstable Vortex Sheet 1 1 2 ci (U1 U 2 ) 2 >0 Therefore always unstable 058:0160 Chapters 6 Professor Fred Stern Fall 2007 22 058:0160 Chapters 6 Professor Fred Stern Fall 2007 23 (2) Transition from laminar to turbulent flow Not all laminar flows have different equilibrium states, but all laminar flows for sufficiently large Re become unstable and undergo transition to turbulence. Transition: change over space and time and Re range of laminar flow into a turbulent flow. U Re ~ 1000 δ = transverse viscous thickness cr Retrans > Recr with xtrans ~ 10-20 xcr Small-disturbance (linear) stability theory can predict Recr with some success for parallel viscous flow such as plane Couette flow, plane or pipe Poiseuille flow, boundary layers without or with pressure gradient, and free shear flows (jets, wakes, and mixing layers). Note: No theory for transition, but recent DNS helpful. 058:0160 Chapters 6 Professor Fred Stern Fall 2007 24 Outline linearized stability theory for parallel viscous flows: select basic solution of interest; add disturbance; derive disturbance equation; linearize and simplify; solve for eigenvalues; interpret stability conditions and draw thumb curves. ^ u u u u, v mean flow, which is solution steady NS ^ v vv ^ ^ ^ p p p u,v small 2D oscillating in time disturbance is solution unsteady NS ^ ^ ^ ^ ^ 1^ ^ u t u u x u u x v u y v u y p x 2 u ^ ^ ^ ^ ^ 1^ ^ v t u v x u v x v v y v v y p x 2 u ^ ^ ux v x 0 ^ ^ ^ Linear PDE for u, v, p for ( u , v , p ) known. Assume disturbance is sinusoidal waves propagating in x direction at speed c: Tollmien-Schlicting waves. ^ i ( xct ) ( x, y , t ) ( y ) e ^ Stream function ^ i ( xct ) u e y =distance across y shear layer ^ ^ i ( xct ) v i e x 058:0160 Chapters 6 Professor Fred Stern Fall 2007 25 ^ ^ u v 0 x y Identically! i = wave number = 2 r i c c ic = wave speed = r i Where λ = wave length and ω =wave frequency Temporal stability: Disturbance (α = αr only and cr real) ci >0 unstable =0 neutral <0 stable Spatial stability: Disturbance (cα = real only) αi <0 unstable =0 neutral >0 stable 058:0160 Chapters 6 Professor Fred Stern Fall 2007 26 ^ ^ Inserting u , v into small disturbance equations and ^ eliminating p results in Orr-Sommerfeld equation: inviscid Raleigh equation i (u c)( '' 2 ) u '' ( IV 2 2 '' 4 ) Re u u /U Re=UL/υ y=y/L 4th order linear homogeneous equation with homogenous boundary conditions (not discussed here) i.e. eigen-value problem, which can be solved albeit not easily for specified geometry and ( u , v , p ) solution to steady NS. 058:0160 Chapters 6 Professor Fred Stern Fall 2007 27 Although difficult, methods are now available for the solution of the O-S equation. Typical results as follows (1) Flat Plate BL: U Re 520 crit (2) αδ* = 0.35 λmin = 18 δ* = 6 δ (smallest unstable λ) unstable T-S waves are quite large (3) ci = constant represent constant rates of damping (ci < 0) or amplification (ci > 0). ci max = .0196 is small compared with inviscid rates indicating a gradual evolution of transition. 058:0160 Chapters 6 Professor Fred Stern Fall 2007 28 (4) (cr/U0)max = 0.4 unstable wave travel at average velocity. (5) Reδ*crit = 520 Rex crit ~ 91,000 Exp: Rex crit ~ 2.8 x 106 (Reδ*crit = 2,400) if care taken, i.e., low free stream turbulence 058:0160 Chapters 6 Professor Fred Stern Fall 2007 29 Falkner-Skan Profiles: Reδ*crit : 67 sep bl (1) strong influence of 520 fp bl 12,490 stag point bl Recrit > 0 fpg Recrit < 0 apg 058:0160 Chapters 6 Professor Fred Stern Fall 2007 30 058:0160 Chapters 6 Professor Fred Stern Fall 2007 31 058:0160 Chapters 6 Professor Fred Stern Fall 2007 32 Extent and details of these processes depends on Re and many other factors (geometry, pg, free-stream, turbulence, roughness, etc). 058:0160 Chapters 6 Professor Fred Stern Fall 2007 33 Rapid development of span- wise flow, and initiation of nonlinear processes - stretched vortices disintigrate - cascading breakdown into families of smaller and smaller vortices - onset of turbulence Note: apg may undergo much more abrupt transition. However, in general, pg effects less on transition than on stability 058:0160 Chapters 6 Professor Fred Stern Fall 2007 34 058:0160 Chapters 6 Professor Fred Stern Fall 2007 35 Some recent work concerns recovery distance: 058:0160 Chapters 6 Professor Fred Stern Fall 2007 36 Turbulent Flow Most flows in engineering are turbulent: flows over vehicles (airplane, ship, train, car), internal flows (heating and ventilation, turbo-machinery), and geophysical flows (atmosphere, ocean). V (x, t) and p(x, t) are random functions of space and time, but statistically stationary flows such as steady and forced or dominant frequency unsteady flows display coherent features and are amendable to statistical analysis, i.e. time and place (conditional) averaging. RMS and other low-order statistical quantities can be modeled and used in conjunction with averaged equations for solving practical engineering problems. Turbulent motions range in size from the width in the flow δ to much smaller scales, which become progressively smaller as the Re = Uδ/υ increases. 058:0160 Chapters 6 Professor Fred Stern Fall 2007 37 058:0160 Chapters 6 Professor Fred Stern Fall 2007 38 Physical description: (1) Randomness and fluctuations: Turbulence is irregular, chaotic, and unpredictable. However, for statistically stationary flows, such as steady flows, can be analyzed using Reynolds decomposition. 1 t0 T 1t 0 T u u u' u u dT u' 0 u ' u ' dT 2 2 etc. T t0 T t0 u = mean motion u ' = superimposed random fluctuation u ' = Reynolds stresses; RMS = u ' 2 2 Triple decomposition is used for forced or dominant frequency flows u u u' 'u' Where u ' ' = organized oscillation (2) Nonlinearity Reynolds stresses and 3D vortex stretching are direct results of nonlinear nature of turbulence. In fact, Reynolds stresses arise from nonlinear convection term after substitution of Reynolds decomposition into NS equations and time averaging. 058:0160 Chapters 6 Professor Fred Stern Fall 2007 39 (3) Diffusion Large scale mixing of fluid particles greatly enhances diffusion of momentum (and heat), i.e., viscous stress Reynolds Stresses: u ' u ' i j ij ij 2 Isotropic eddy viscosity: u 'i u ' j t ij ij k 3 (4) Vorticity/eddies/energy cascade Turbulence is characterized by flow visualization as eddies, which vary in size from the largest Lδ (width of flow) to the smallest. The largest eddies have velocity scale U and time scale Lδ/U. The orders of magnitude of the smallest eddies (Kolmogorov scale or micro scale) are: 1 3 4 LK = Kolmogorov micro-scale = 3 U LK = O(mm) >> Lmean free path = 6 x 10-8 m Velocity scale = (νε)1/4= O(10-2m/s) Time scale = (ν/ε)1/2= O(10-2s) Largest eddies contain most of energy, which break up into successively smaller eddies with energy transfer to 058:0160 Chapters 6 Professor Fred Stern Fall 2007 40 yet smaller eddies until LK is reached and energy is dissipated by molecular viscosity (i.e. viscous diffusion). Richardson (1922): Lδ Big whorls have little whorls Which feed on their velocity; And little whorls have lesser whorls, LK And so on to viscosity (in the molecular sense). (5) Dissipation 0 L u0 k k u '2 v '2 w '2 Energy comes from 0 (U ) largest scales and Re u0 0 / l arg e fed by mean motion ε = rate of dissipation = energy/time 2 Dissipation u0 0 o o u0 occurs at smallest scales 1 3 u0 3 4 = independent υ LK l0 058:0160 Chapters 6 Professor Fred Stern Fall 2007 41 Fig. below shows measurements of turbulence for Rex=107. Note the following mean-flow features: (1) Fluctuations are large ~ 11% U∞ (2) Presence of wall cause anisotropy, i.e., the fluctuations differ in magnitude due to geometric and physical reasons. u ' is largest, v ' is smallest and reaches 2 2 its maximum much further out than u ' or w' . w' is 2 2 2 intermediate in value. (3) u ' v' 0 and, as will be discussed, plays a very important role in the analysis of turbulent shear flows. (4) Although u u 0 at the wall, it maintains large values i j right up to the wall (5) Turbulence extends to y > δ due to intermittency. The interface at the edge of the boundary layer is called the superlayer. This interface undulates randomly 058:0160 Chapters 6 Professor Fred Stern Fall 2007 42 between fully turbulent and non-turbulent flow regions. The mean position is at y ~ 0.78 δ. (6) Near wall turbulent wave number spectra have more energy, i.e. small λ, whereas near δ large eddies dominate. 058:0160 Chapters 6 Professor Fred Stern Fall 2007 43 058:0160 Chapters 6 Professor Fred Stern Fall 2007 44 Averages: For turbulent flow V (x, t), p(x, t) are random functions of time and must be evaluated statistically using averaging techniques: time, ensemble, phase, or conditional. Time Averaging For stationary flow, the mean is not a function of time and we can use time averaging. 1 t0 t u u (t ) dt T > any significant period of u ' u u T t0 (e.g. 1 sec. for wind tunnel and 20 min. for ocean) Ensemble Averaging For non-stationary flow, the mean is a function of time and ensemble averaging is used 1 N i u (t ) u (t ) N is large enough that u independent N i 1 ui(t) = collection of experiments performed under identical conditions ( phase aligned for same t=tref). 058:0160 Chapters 6 Professor Fred Stern Fall 2007 45 058:0160 Chapters 6 Professor Fred Stern Fall 2007 46 Phase and Conditional Averaging Similar to ensemble averaging, but for flows with dominant frequency content or other condition, which is used to align time series for some phase/condition. In this case triple velocity decomposition is used: u u u' 'u' where u΄΄ is called organized oscillation. Phase/conditional averaging extracts all three components. Averaging Rules: f f f' g g g' s = x or t f ' 0 f f fg fg f 'g 0 f f f g f g fg f g f ' g ' s s f ds f ds 058:0160 Chapters 6 Professor Fred Stern Fall 2007 47 Reynolds-Averaged Navier-Stokes Equations For convenience of notation use uppercase for mean and lowercase for fluctuation in Reynolds decomposition. ~ u i U i ui ~ p P p ~ ui 0 xi ~ ~ ~ ~ NS ui ~ ui 1 p 2 u i ui g i 3 equation t xi xi x j x j Mean Continuity Equation U u U (U u ) 0 i i i x x x x i i i i i i ~ u U u u 0 i i 0 i x i x x i x i i Both mean and fluctuation satisfy divergence = 0 condition. 058:0160 Chapters 6 Professor Fred Stern Fall 2007 48 Mean Momentum Equation 1 U i ui (U j u j ) (U i ui ) ( P p) t x j xi 2 (U i ui ) g i 3 x j x j U u U (U u ) i i i t t t t i i U u U u (U u ) (U u ) U U u u i i i i x x x x x j j i i j j j j j j j j j U U uu i x x j i j j j u u u Since uu u u u j i i x x x x i j i j j j j j j P p P ( P p) x i x x x i i i g gi3 i3 058:0160 Chapters 6 Professor Fred Stern Fall 2007 49 2 2U i 2 ui 2U i (U i ui ) x j 2 x 2 j x j 2 x 2 j U i U i (ui u j ) 1 P 2 U j 2 U i g i 3 t x j x j xi x j DU i 1 P U i Or g i 3 ui u j Dt xi x j x j DU i 1 Or g i3 ij Dt xi RANS U U Equations P x x u u ij i j i j j i U with 0 i x i The difference between the NS and RANS equations is the Reynolds stresses u u , which acts like additional i j stress: (1) This leads to the closure problem: 4 equations 10 unknowns for RANS versus 4 equations 4 unknowns for NS, (2) NS and RANS paradox: NS deterministic with non-deterministic solutions while RANS non- deterministic with deterministic solutions. 058:0160 Chapters 6 Professor Fred Stern Fall 2007 50 u u = u u i j j i (i.e. Reynolds stresses are symmetric) u 2 uv uw uv v 2 vw uw vw w 2 u are normal stresses 2 i uu i j i j are shear stresses For isotropic turbulence u u i j = 0 and u v w i j 2 2 2 constant; however, turbulence is generally non-isotropic. dU For example, consider shear flow with 0, dy V (U u, v, w) : x-momentum tends towards decreasing y as v0 u0 turbulence diffuses uv 0 v0 u0 gradients and decreases dU dy x-momentum transport across y = constant AA per unit area (U u )v U v uv uv i.e u u i j = average flux of j-momentum in 058:0160 Chapters 6 Professor Fred Stern Fall 2007 51 i-direction = average flux of i-momentum in j-direction Turbulent Kinetic Energy Equation k u u v w = turbulent kinetic energy 1 1 2 2 2 2 i 2 2 ~ Subtracting NS equation for u i and RANS equation for Ui results in equation for ui: ui ui U i ui 1 p 2 ui U j uj uj (ui u j ) 2 t x j x j x j x j xi x j Multiply by ui and average Dk 1 1 2 U i pu j ui u j 2 ui eij uiu j 2 eij e ji Dt x j 2 x j x j x j PT T D P Dk k k 1 ui u j U j e Dt t x j and ij 2 x j xi Where C 058:0160 Chapters 6 Professor Fred Stern Fall 2007 52 T= turbulent transport illustrates the rate at which TKE is transported through the fluid by turbulent fluctuations. PT =pressure transport is another form of turbulent transport resulting from correlation of pressure and velocity fluctuations. D=viscous diffusion represents the diffusion of TKE caused by the fluid’s natural molecular transport process. P = shear production (usually > 0) represents loss of mean kinetic energy and gain of turbulent kinetic energy due to U interactions of u u and . i i j x j ε = viscous dissipation represents the rate at which TKE is converted into thermal internal energy. 2 ui u j xi x j 0 where is called pseudo-dissipation and defined by ui ui x j x j C= turbulent convection represents the rate of TKE convected by the mean flow 058:0160 Chapters 6 Professor Fred Stern Fall 2007 53 Recall previous discussions of energy cascade and dissipation: Energy fed from mean flow to largest eddies and cascades to smallest eddies where dissipation takes place. (1) Kolmogorov Hypothesis 1: local isotropy: for large Re, micro-scale l l0 and turbulence is isotropic. Where eddies in the largest size range are characterized by the length scale l0 , velocity scale u 0 , and turn-over time scale 0 . The Reynolds number of these eddies is Re0 u0l0 All of the experimental evidence confirms that most eddies break-up on a timescale of 0 . So the rate at which energy per unit mass is passed down the energy cascade from the largest eddies with size l0 is: 2 2 3 u0 u0 u0 0 l0 u0 l0 For statistically steady conditions, this must match exactly the rate of dissipation of energy at the smallest scale Sij Sij (from TKE equation), where the turbulent rate 1 u u j Sij i Since Sij u , so: of strain 2 x j xi . 058:0160 Chapters 6 Professor Fred Stern Fall 2007 54 u 2 2 Apply 3 u 2 u0 2 l0 It is also known that Re based on u and is of order unity: u 1 we get: 3 / 1/ 4 length / l0 Re0 3/4 u 1 / 4 velocity u / u0 Re01/4 u / 1/2 / 0 Re01/2 time Micro-scale<<large scale As Re increases, the range of scales increases. It is also known as (2) Kolmogorov’s first similarity hypothesis: for large Re, micro-scale has universal form uniquely determined by υ and ε. (3) Kolmogorov’s second similarity hypothesis: for large Re, intermediate scale ( l0 l ) has a universal form for u l and l that are uniquely determined by and l and independent of υ, i.e., dimensional reasoning results in: ul l u l u0 l l0 13 13 13 058:0160 Chapters 6 Professor Fred Stern Fall 2007 55 l l l 0 l l0 2 13 23 23 Show that the velocity scales and time scales decrease as l decreases. (2) and (3) are called universal equilibrium range in distinction from non-isotropic energy-containing range. (2) is the dissipation range and (3) is the inertial subrange. Spectrum of turbulence in the Inertial subrange u S (k ) dk 2 k = wave number in inertial subrange. 0 S A 2 / 3k 5 / 3 For l01 k 1 (based on dimensional analysis) A ~ 1.5 Called Kolmogorov k-5/3 law 058:0160 Chapters 6 Professor Fred Stern Fall 2007 56 058:0160 Chapters 6 Professor Fred Stern Fall 2007 57 Velocity Profiles: Inner, Outer, and Overlap Layers Detailed examination of turbulent-flow velocity profiles indicates the existence of a three-layer structure: (1) a thin inner layer close to the wall ( y / .02) where turbulence is inhibited by the presence of the solid boundary, i.e. u u are negligibly small (= 0 at the i j wall) and the flow is controlled by molecular viscosity. (2) An outer layer (.2 y / 1) where turbulent shear dominates. (3) An overlap layer (.02 y / .2) where both types of shear are important. Considerably more information is obtained from dimensional analysis and confirmed by experiment. Inner law: u f ( , , , y) w yu* u u /u f * u* w / Wall shear velocity f ( y ) 058:0160 Chapters 6 Professor Fred Stern Fall 2007 58 u+, y+ are called inner-wall variables Note that the inner law is independent of δ or r0, for boundary layer and pipe flow, respectively. Outer Law: Ue u g ( w , , y, ) for px = 0 velocity defect Ue u * g ( ) where y / u Note that the outer wall is independent of μ. Overlap law: both laws are valid It is not that difficult to show (also see AMF chapter VII) that for both laws to overlap, f and g are logarithmic functions: Inner region: 2 dU u df dy dy Outer region: dU u dg dy d 058:0160 Chapters 6 Professor Fred Stern Fall 2007 59 2 y u y u dg df ; valid at large y+ and small η. u dy u d f(y+) g(η) Therefore, both sides must equal universal constant, 1 1 f ( y ) ln y B u / u (inner variables) 1 U u g ( ) ln A e (outer variables) u , A, and B are pure dimensionless constants = 0.41 Von Karman constant Values vary somewhat depending on B = 5.5 (or 5.0) different exp. arrangements A = 2.35 BL flow The difference is due to = 0.65 pipe flow loss of intermittency in duct flow. A = 0 means small outer layer The validity of these laws has been established experimentally as shown in Fig. 6-9, which shows the profiles of Fig 6-8 in inner-law variable format. All the profiles, with the exception of the one for separated flow, 058:0160 Chapters 6 Professor Fred Stern Fall 2007 60 are seen to follow the expected behavior. In the case of separated flow, scaling the profile with u* is inappropriate since u* ~ 0. Details of Inner Law u Very near the wall, ~ constant y w w u y i.e. varies linearly Or u+ = y+ y+ ≤ 5 This is the linear sub-layer which must merge smoothly with the logarithmic overlap law. This takes place in the blending zone/buffer layer (5 y 30) . Evaluating the RANS equation near the wall using an eddy-viscosity μt turbulence model shows that: μt ~ y3 y 0 Several expressions that satisfy this requirement have been derived and are commonly used in turbulent-flow analysis. That is: 058:0160 Chapters 6 Professor Fred Stern Fall 2007 61 t e B eu 1 u u 2 2 Assuming the total shear is constant very near the wall the equation for μt can be integrated to obtain a composite formula which is valid in the sub-layer, blending layer, and logarithmic-overlap regions. y u e B eu 1 u u 2 3 u 2 6 Fig. 6-11 shows a comparison of this equation with experimental data obtained very close to the wall. The agreement is excellent. It should be recognized that obtaining data this close to the wall is very difficult. Details of the Outer Law With pressure gradient included, the outer law becomes (Fig 6-10): Ue u g ( , ) (*) * u * dpc Clauser’s equilibrium y / w dx = parameter 058:0160 Chapters 6 Professor Fred Stern Fall 2007 62 Clauser (1954, 1956): BL’s with different px but constant are in equilibrium, i.e., can be scaled with a single parameter: Ue u * vs. y / u u U defect thickness e * dy * 0 u 2/C f Also, G = Clauser Shape parameter 1 U u 2 e * dy 6.1 1.81 1.7 0 u Curve fit by Nash Which is related to the usual shape parameter by H 1 G / 1 const. due to ( x) Finally, Clauser showed that the outer layer has a wake- like structure such that 058:0160 Chapters 6 Professor Fred Stern Fall 2007 63 t 0.016 U e * (**) Mellor and Gibson (1966) combined (*) and (**) into a theory for equilibrium outer law profiles with excellent agreement with experimental data: Fig. 6-12 Coles (1956): A weakness of the Clauser approach is that the equilibrium profiles do not have any recognizable shape. This was resolved by Coles who showed that: Deviations above log-overlap layer u 2.5 ln y 5.5 1 W(y / ) (***) U e 2.5 ln 5.5 2 Max deviation at δ Single wake-like function of y/δ y W wake function 2 sin 3 2 , y / 2 2 3 2 curve fit 058:0160 Chapters 6 Professor Fred Stern Fall 2007 64 Thus, from (***), it is possible to derive a composite function, which covers both the overlap and outer layers, as shown in Fig 6-13. 1 u ln y B W ( y / ) wake parameter = π(β) 0.8( 0.5) 0.75 (curve fit to data) Note that the agreement of Coles’ wake law even for β constant Bl’s is quite good. We see that the behavior in the outer layer is more complex that that of the inner layer due to pressure gradient effects. In general, the above velocity profile correlations are extremely valuable both in providing physical insight and, as we shall see, in providing approximate solutions for simple wall bounded geometries: pipe, channel, and flat plate boundary layer. Furthermore, such correlations have been extended through the use of additional parameters to provide velocity formulas for use with integral methods for solving the b.l. equations for arbitrary px. Also used for CFD turbulence modeling. 058:0160 Chapters 6 Professor Fred Stern Fall 2007 65 6-4 Summary of Inner, Outer, and Overlap Layers: Mean velocity correlations Inner layer: U U / u y y / u u* w / * Sub-layer: 0 y 5 U+ = y+ Buffer layer: 5 y 30 region where sub-layer merges smoothly with log law Outer Layer: Ue U * u* g ( , ) y / , px w Overlap layer (log region): 1 U ln y B inner variables Ue U 1 ln A outer variables u * Composite Inner/Overlap layer correlation (U )2 (U )3 U y e B eU 1 U 2 6 Composite Overlap/Outer layer correlation 058:0160 Chapters 6 Professor Fred Stern Fall 2007 66 1 2 U ln y B W ( ) W sin 2 3 2 2 3 2 0.8( 0.5) 0.75 Equilibrium Boundary Layers ( = constant) Ue U * vs. y / , ,* 2/C f u (shows similarity) Reynolds Number Dependence of Mean-Velocity correlations and Reynolds stresses 1. inner/overlap U+ scaling shows similarity; extent of overlap region (i.e. similarity) increases with Reθ. Fig 10, 11, 12 2. outer layer may asymptotically approach similarity for large Re as shown by U ( 2 / ) vs. Reθ, but controversial due to lack of data for Reθ > 5 x 104. Fig 13, 14 2 3. streamwise turbulence intensity u u vs. y+ u* shows similarity for 0 y 15 (i.e., just beyond the point of kmax, y+ = 12). u+max vs. Reθ increases with Reθ. Fig 18, 27 058:0160 Chapters 6 Professor Fred Stern Fall 2007 67 4. v+ and uv do not show similarity: peak values increase with Rand move away from the wall. Fig 25, 31, 34 058:0160 Chapters 6 Professor Fred Stern Fall 2007 68 058:0160 Chapters 6 Professor Fred Stern Fall 2007 69 058:0160 Chapters 6 Professor Fred Stern Fall 2007 70 058:0160 Chapters 6 Professor Fred Stern Fall 2007 71 058:0160 Chapters 6 Professor Fred Stern Fall 2007 72 058:0160 Chapters 6 Professor Fred Stern Fall 2007 73 058:0160 Chapters 6 Professor Fred Stern Fall 2007 74 6-5 Turbulent Flow in Pipes and Channels using mean- velocity correlations Circular pipe Recall laminar flow exact solution 8 w f 64 / Re D Re D uave D 2000 uave 2 A turbulent-flow “approximate” solution can be obtained simply by computing uave based on log law. 1 yu* u u ( y ) .41, u ln B B 5 u* w / u * y Rr dy dr 1 R * 1 ( R r )u * V uave Q / A 2 u ln B 2r dr R 0 1 * 2 Ru* ln 3 u 2R 2 V Ru * 2.44 ln 1.34 Or u * f 1/ 2 1.99 log(Re D f 1/ 2 ) 1.02 EFD adjusted f 1 / 2 2 log(ReD f 1 / 2 ) 0.8 constants f only drops by a factor of 5 over 104 < Re < 108 058:0160 Chapters 6 Professor Fred Stern Fall 2007 75 Turbulent Flow in Rough Pipes U f ( y , k ) f f (Re d , k / d ) 1 U ln y B B(k ) Log law shifts downward which leads to three roughness regimes: 1. k+ < 4 hydraulically smooth 2. 4 < k+ < 60 transitional roughness (Re dependence) 3. k+ > 60 full rough (no Re dependence) 1 / 2 k / d 2.51 Moody diagram f 2 log 3.7 Re d f 1 / 2 approximate explicit formula 6.9 k / d 1.11 ~ 1.8 log Re d 3.7 There are basically three types of problems involved with uniform flow in a single pipe: 1. Determine the head loss, given the kind and size of pipe along with the flow rate, Q = A*V 2. Determine the flow rate, given the head, kind, and size of pipe 3. Determine the pipe diameter, given the type of pipe, head, and flow rate 058:0160 Chapters 6 Professor Fred Stern Fall 2007 76 1. Determine the head loss The first problem of head loss is solved readily by obtaining f from the Moody diagram, using values of Re and ks/D computed from the given data. The head loss hf is then computed from the Darcy-Weisbach equation. f = f(Red, ks/D) L V2 p p hf f h h 2 z 2 1 z1 D 2g p = z Red = Red(V, D) 2. Determine the flow rate The second problem of flow rate is solved by trial, using a successive approximation procedure. This is because both Re and f(Re) depend on the unknown velocity, V. The solution is as follows: 1) solve for V using an assumed value for f and the Darcy- Weisbach equation 1/ 2 2gh f V f 1/ 2 L/D known from note sign given data 2) using V compute Re 058:0160 Chapters 6 Professor Fred Stern Fall 2007 77 3) obtain a new value for f = f(Re, ks/D) and reapeat as above until convergence 1/ 2 D 3 / 2 2gh f Or can use Re f 1/ 2 L scale on Moody Diagram 1) compute Re f 1/ 2 and ks/D 2) read f L V2 3) solve V from h f f D 2g 4) Q = VA 3. Determine the size of the pipe The third problem of pipe size is solved by trial, using a successive approximation procedure. This is because hf, f, and Q all depend on the unknown diameter D. The solution procedure is as follows 1) solve for D using an assumed value for f and the Darcy- Weisbach equation along with the definition of Q 1/ 5 8LQ 2 D 2 f 1/ 5 gh f known from given data 2) using D compute Re and ks/D 3) obtain a new value of f = f(Re, ks/D) and reapeat as above until convergence 058:0160 Chapters 6 Professor Fred Stern Fall 2007 78 058:0160 Chapters 6 Professor Fred Stern Fall 2007 79 Concept of hydraulic diameter for noncircular ducts For noncircular ducts, τw= f(perimeter); thus, new 8 definitions of f w and C f 2 w are required. V 2 V 2 Define average wall shear stress 1P w w ds ds = arc length, P = P0 perimeter of duct Momentum: z pA w PL AL 0 L W wL h p / z A/P=Rh=hydraulic radius A/ P circle: A/P = R2/2R=R/2=D/4 Energy: wL h hL A/ P 058:0160 Chapters 6 Professor Fred Stern Fall 2007 80 ^ A h A dh A d ( p z ) A d p w P L P dx P dx P dx non-circular duct A/P = length= Dh/4 Dh = 4A/P hydraulic diameter For multiple surfaces such as concentric annulus P and A based on wetted perimeter and area 8 w VDh f f (Re Dh , / Dh ) Re Dh V 2 L V 2 f L V2 L V2 h hL f f Rh 8 4Rh 2 g Dh 2 g However, accuracy not good for laminar flow ( 40 %) and marginal turbulent flow ( 15 %) 8 w fVDh / VD / 64 using 8V / Dh i.e. exact solution V 2 h w f Cf 64 / Re Dh 16 / Re D 4 P 0 C f Re Dh 64 16 Circular pipe 058:0160 Chapters 6 Professor Fred Stern Fall 2007 81 For laminar flow, P 0 varies greatly, therefore it is better to use the exact solution For turbulent flow, Dh works much better especially if combined with “effective diameter” concept based on ratio of exact laminar circular and noncircular duct P0 numbers, i.e. 16 / P = exact laminar solution for 0 noncircular duct. First recall turbulent circular pipe solution and compare with turbulent channel flow solution using log-law in both cases 058:0160 Chapters 6 Professor Fred Stern Fall 2007 82 Circular pipe: as already shown 1 yu* u u ( y ) .41, u ln B B 5 u* w / u * y Rr dy dr 1 R * 1 ( R r )u * V uave Q / A 2 u ln B 2r dr R 0 1 * 2 Ru* ln 3 u 2R 2 V Ru * 2.44 ln 1.34 Or u * f 1/ 2 1.99 log(Re D f 1/ 2 ) 1.02 EFD adjusted f 1 / 2 2 log(ReD f 1 / 2 ) 0.8 constants f only drops by a factor of 5 over 104 < Re < 108 058:0160 Chapters 6 Professor Fred Stern Fall 2007 83 Since f equation is implicit, it is not easy to see dependency on ρ, μ, V, and D f ( pipe) 0.316 Re1 / 4 D 4000 < ReD < 105 Blasius (1911) power law curve fit to data p L V2 hf f D 2g p 0.158 L 3 / 4 1 / 4 D 5 / 4v 7 / 4 Near quadratic (as expected) Drops weakly with Only slightly pipe size Nearly linear with μ 0.241L 3 / 4 1/ 4 D 4.75Q1.75 laminar flow: p 8LQ / R 4 p (turbulent) increases more sharply than p (laminar) for same Q; therefore, increase D for smaller p . 2D decreases p by 27 for same Q. Ru* u (r 0) * ln umax 1 B u * u V 2.44 ln Ru * Combine with 1.34 u* 058:0160 Chapters 6 Professor Fred Stern Fall 2007 84 V u max 1 1.3 f 1 Recall laminar flow: V / u max 0.5 Channel Flow Laminar Solution C f Re h 24 Re D P0 = 6 based h= 6 h half width (=24 based on Dh=4h) 1 h * 1 (h y )u* V u ln B dY Y=h-y wall coordinate h 0 * 1 1 hu* u ln B VDh A=2h*w Define Re Dh Dh 4A lim 4(2hw) 4h P=4h+2w P w 2 w 4h f 1/ 2 2log Re Dh f 1/ 2 1.19 Very nearly the same as circular pipe 7% to large at Re = 105 4% to large at Re = 108 058:0160 Chapters 6 Professor Fred Stern Fall 2007 85 Therefore error in Dh concept relatively smaller for turbulent flow. Note f 1 / 2 (channel ) 2 log0.64 Re f Dh 1/ 2 0.8 P0 (circle) 16 Define Deffective 0.64 Dh ~ Dh P0 (channel) 24 Laminar solution (therefore, improvement on Dh is) VDeff 16 Re Deff Deff Dh P0 (lam) From exact laminar solution 058:0160 Chapters 6 Professor Fred Stern Fall 2007 86 058:0160 Chapters 6 Professor Fred Stern Fall 2007 87 058:0160 Chapters 6 Professor Fred Stern Fall 2007 88 058:0160 Chapters 6 Professor Fred Stern Fall 2007 89 058:0160 Chapters 6 Professor Fred Stern Fall 2007 90 058:0160 Chapters 6 Professor Fred Stern Fall 2007 91 058:0160 Chapters 6 Professor Fred Stern Fall 2007 92 058:0160 Chapters 6 Professor Fred Stern Fall 2007 93 058:0160 Chapters 6 Professor Fred Stern Fall 2007 94 058:0160 Chapters 6 Professor Fred Stern Fall 2007 95 058:0160 Chapters 6 Professor Fred Stern Fall 2007 96 058:0160 Chapters 6 Professor Fred Stern Fall 2007 97 058:0160 Chapters 6 Professor Fred Stern Fall 2007 98 058:0160 Chapters 6 Professor Fred Stern Fall 2007 99 058:0160 Chapters 6 Professor Fred Stern Fall 2007 100 058:0160 Chapters 6 Professor Fred Stern Fall 2007 101 058:0160 Chapters 6 Professor Fred Stern Fall 2007 102 058:0160 Chapters 6 Professor Fred Stern Fall 2007 103

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