RAYLEIGH-BÉNARD-, MARANGONI-, RAYLEIGH-BENARD DOUBLE DIFFUSIVE-, CONVECTION CONVECTION BENARD’S Convection EXPERIMENT Heat transfer by diffusion Solar panels T instability —> 1 Heat transport by convection Forced Convection (pump, fan, suction device) T0 or Natural convection (buoyancy force) Buildings & Environment industry mantel of the Earth, Atmospheres !3 mantel of the Sun... HEAT CONVECTION onset Optical deformation due to heat gradient using index refraction wavelength Cell formation stability onset t >> tc turbulence Ra >> Rac Cell formation RAYLEIGH BENARD CONVECTION HEAT CONVECTION T1 < T0 h Driven by density gradients due to gradients in concentration, temperature or composition T ρ STATIC STABILITY u=0 ΔT 0 ρ(z) = ρ (1+ βz) β = α T0 0 h EQUATIONS With u=(u,v,w) and g=(0,0,-g) STEADY SOLUTION ρ(z) ρ(z) ⌅u Momentum ⇥ + u. u = p ⇥gk + µ 2u ⌅t ⇥ ⇥ − ⇥ U=0 € ⇥ ⇤T Diffusion of heat + u. T = κ 2T ⇤t ∇ ∇ Hydrostatic balance dp Continuity .u =0 = ρ(z)g Stable Statically unstable ∇ dz − ⇥(T ) ⇥0 ⇥(z) ⇥0 α(T0 T1) Density ρ = ρ(T ) − = α(T T0) − = − z ⇥0 − − ⇥0 − h parameters g ν κ β Dimensions LT-2 L2T-1 L2T- L-1 Boussinesq approximation ∆ρ ≈ 1% , (ρ - ρ0)/ρ0 =α(T0 - T)<< 1 only density variation in z considered € * * 2 -1 * 2 2 * X=x H, t=t H /κ, u=u*H κ, p=p ρ0κ /Η T=T ΔΤ Note: Boussinesq approximation thus gives for density perturbations ρ = ρ +ρ’ 0 Substitution gives for the linearized equations: and related pressure perturbation p=p0 +p’ with hydrostatic equilibrium : 1 ⌅p0 + p 1 ⌅p0 + p g = g ∂u 2 −ρ + ρ ⌅z − −ρ (1 + ρ /ρ ) ⌅z − = p + Ra P r T k + Pr u 0 0 0 ∂t 1 ⌅p + p ρ 1 ⌅p 1 ⌅p ρ ⇥ ⇥ 0 0 + ... g = + g ⇥ρ ⌅z − ρ ρ ⌅z − −ρ ⌅z ρ ∂T 2 0 0 0 ⇥ 0 0 w = T ∂t − ⇥ Linearized perturbation equations around the basic state: .u =0 u = (u’ ,v’,w’) ; T = T0 + Τ’ and with ρ’/ρ0 = α T’ ∇ into the equations and keep linear terms in the perturbation Prandtl number Rayleigh number 2 2 2 ⇤u 1 ⇤p 2 2 ⇤ ⇤ ⇤ 3 4 = + ν u and = 2 + 2 + 2 ⇥ diffusivity of momentum gα∆Th g⇥h ⇤t −⇥0 ⇤x ⇥ ⇥ ⇤x ⇤y ⇤z Pr = = Ra = = κ diffusivity of heat ⌅⇤ ⌅⇤ ⇤v 1 ⇤p 2 ⇥u ⇥v ⇥w = + ν v .u = + + ⇤t −⇥0 ⇤y ⇥ ∇ ⇥x ⇥y ⇥z ⌅w 1 ⌅p 2 ⇥T T0 T1 2 = + gαT + ⇥ w w − = κ T Pr = 7 for heat in water and ~O(100) for salt in water ⌅t −⇤ ⌅z ⇥ ⇥t − h ⇥ 0 0.7 in air and has very small values in liquid metals !9 Five equations, five unknown --> eliminate the pressure with continuity to obtain a single equation by taking the curl of the velocity ω¯ = u¯ and using .u¯ =0 ⇥ ⇥ ⇥ω¯ 2 2 ω¯ = u¯ = Pr ω¯ + Ra P r T k¯ (A) ∂ 2 6 ∂ 4 2 ⇤ ⇥ ⇥t ⇤ ⇤ 2 w + Pr w (1 + Pr) w = Ra P r hw ⇥ ⇥T ∂t ⇥ ⇥ − ∂t⇥ ⇥ .u¯ =0 2u¯ = Ra P r 2T k¯ + Pr 4u¯ ⇤ ⇥ ⇥t⇤ ⇤ ⇤⇥z ⇤ ⇥ Normal modes w(x, y, z)=Re [W (z) exp i[(kxx + kyy)+st]] 2 2 ∂ 2 2 4 ∂ ∂ w = Ra P r hT + Pr w and h = + (B) 2 2 2 2 1 2 2 2 ∂t∇ ∇ ∇ ∇ ∂x2 ∂y2 (∂ k s)(∂ k Pr− s)(∂ k )W + k RaW =0 z − − z − − z − ∂T 2 (C) w = T 2 2 2 ∂t − ⇥ k = kx + ky Eliminate w or T from B, C to obtain a single equation in T or w (same eq.) ∂2 ∂ 2w + Pr 6w (1 + Pr) 4w = Ra P r 2 w ∂t2 ⇥ ⇥ − ∂t⇥ ⇥h Boundary conditions ---> Note that from (A) we can solve the z-component NOTE: FOR TAYLOR COUETTE FLOW, THE RELATION IS ≈ THE SAME ⇥v ⇥u ⇥ ω = z = Pr 2ω z ⇥x − ⇥y ⇥t ⇥ z T0 1 Stress free, perfectly conducting boundaries ( Rayleigh 1916) 2 2 2 2 1 2 2 2 T(z) ρ(z) (∂ k s)(∂ k Pr− s)(∂ k )W + k RaW =0 z − − z − − z − Marginal stability curve s=0 (Neutral stability curve) Boundary conditions 0 ρ0 2 2 3 2 [( @z k ) + Ra k ] W =0 In all cases W=0 and Tz=0 at the boundaries z=0 and h (0 and 1 scaled) − the temperature is maintained at a constant at z=0 and h 2 2 2 Boundaries: W=∂z W=(∂z - k ) W=0 (z=0,1) A) Bounding surfaces are rigid: no slip (zero tangential velocity) most realistic Solution W=sin nπz with (n2π2+k2)3=Ra k2 u=0, v=0 at z=0, 1. Since this is valid for all x and y, this is also valid for 3 4 ux=0 and vy=0 so that ⇤w αgΔTh βgh .u =0 = 0 and idem for vorticity ω =0 First unstable mode for n=1 Ra = = ⇥h → ⇤z z κν κν n=3 n=2 B) Bounding surfaces are free surfaces: free slip (zero tangential stress) (easiest!) Ra unstable n=1 w1(z) ⇧u ⇧v 2 2 3 2 τ = τ =0 = 0 and =0 Neutral curve Ra = (k +π ) /k xz yz ⇥ ⇧z ⇧z € ⇧ ⇧2w w2(z) .u¯ =0 = 0 at z =0, 1 2 4 ⇥ ⇧z ⇤ ⇥ ⇧z Ra Rac=27π /4 =658 ⇧ ⇧u ⇧v ⇧⇥ c stable = z = 0 at z =0, 1 (k=π/21/2, n=1) ⇧z ⇧y − ⇧x ⇧z k ⇥ k 0 z 0.5 ⇧4w c with equations = 0 at z =0, 1 Racrit does not depend on Pr but the growth rate does! ⇧z4 NOTES Perfectly conducting boundaries - critical value O(103) depends on boundary conditions Free and one rigid boundary (Benard problem) Rac≈ 1101 at kc= 2.68 Realistic temp differences result in Ra>>Rac Ra increases rapidly with h: deeper layers produce higher Ra numbers ! Two rigid boundaries Rac≈ 1708 at kc=3.12 3 4 at 200 αair≈ 0.003K-1 αgΔTh βgh Ra = = αwater≈0.0002 K-1 κν κν - Lowest mode n=1 has the least vertical structure. Vertical velocity has With fixed heat fluxes Tz=constant at boundaries (different T-curve) only one sign but changes sign for higher modes. € Ra (correlation between vertical motion and buoyant fluid results from the momentum generated by the buoyancy forces.) - linear analyses determines only the horizontal scale and not the structure ! Rac Two stress free boundaries Rac= 120 k Two rigid boundaries Rac= 720 0 Flows related to Rayleigh Benard ! How do we measure the transition to convection ? Heat Flux : The Nusselt number (⌘ = µ) Heat transfer rate F=λ(∆T)/H, λ is heat transfer coefficient FH Nu = λ ∆T total heat flux Nu = conducted heat flux In unsteady flows the Nusselt number is averaged in time no flow Nu=1 flow Nu>1 Guyon, Hulin et Petit, 1982 Boundary layer structure = Ra/Rac increasingly turbulent (mixed) in Ra/Rac≤1 the interior for higher Ra-numbers T – T0 |∆T| onset of Rayleigh Benard convection For larger Ra the boundary layer thickness decreases Interior ≈ homogenous Turbulence convection laminar boundary layer δ laminar boundary layer H T Turbulent interior H T Turbulent interior laminar boundary layer " laminar boundary layer δ u=w=0 the boundary layer thickness grows with δ~(κt)1/2 until it becomes heat flux in each boundary layer δ is Fδ = λ Tδ / δ unstable and releases its heat (see*) 3 Fδ ↵g(∆T/2)δ the Nusselt number for the boundary layer is then Nu = for the boundary layer " Raδ = = Raδ crit.=1708. λ∆T/H ⌫ ✓ ◆ T H Nu = δ δ 2Ra 1/3 ∆T δ ⟹ c = δ crit H Ra ✓ H ◆ From observations Tδ ≈ ∆T/2 so that Nu ≈ H/2δ the scaling of the Nusselt number was Nu ≈ H/2δ: H 1 1 Ra 1/3 Nu = = 0.077Ra1/3 2δ 2 2 Ra ⇡ ✓ c ◆ (*note: there are other scalings, see e.g. Lohse & Xia Ann. Rev Fluid Mech!22 2009, Ahlers et al (2009)) Turbulence convection *Plumes released from the boundary laminar boundary layer T=T0+∆T/2 h T turbulent interior h Turbulent interior laminar boundary layer thermal boundary layer δ(t) T=T0+∆T k∆T HEAT FLUX => F = Nu Howard 1966 H h 1/3 in the interior Nu ~ Ra The thermal BL grows by thermal diffusion Interior is turbulent and uniform : FH independent of h, ∆T z Then we have for a fixed Prandtl number P T = T0 + 1 + erfc δ = p⇤⇥t 1/3 2 2pκt 3 ✓ ◆ 1/ 3 ' αgΔTh * Nu = f (Ra,P) ∝ Ra = ) , this goes on until it is unstable and breaks down at tc ( κν + (i.e. at Ra_c) and δ breaks down to zero: and for the turbulent heat flux: αg∆T ⇥3 2⇤⌅ 1/3 3 1/3 c = Ra ⇥ = Ra1/3 ↵gk 4/3 4/3 2⇤⌅ c ) c αg∆T c F = c(P ) ∆T = C ∆T ✓ ◆ H ⌫ € ✓ ◆ the time cycle of the plume formation can be calculated as Also called 4/3 law. (Howard, Appl Mech.1964) h2 Ra 2/3 t = c !24 ⇥ Ra ✓ ◆ Turbulence convection Castaing et al, J. Fluid Mech. 1989: boundary layer relation with turbulent interior general comments These laws are also known as Malkus, Kraichnan, Howard, Ginsburg-Landau (GL) model. The general law is written as Nu~Ra2/7 Nu=C Raβ Prγ Nu~Ra1/3 allows to represent all models for infinite horizontal dimensions. For a finite horizontal size, D, the constant C=C(Γ)depends on the aspect ratio Γ=D/H.