6.6 Horton-Rogers-Lapwood instability in a porous layer

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1 1 November, Notes on 1.63 Advanced Environmental Fluid Mechanics Instructor: C. C. Mei, Horton-Rogers-Lapwood instability in a porous layer Nield & Bejan, Chapter 6 Convection in Porous Media Related: Rayleigh-Bernard Problem (Chandrsekhar, Chapter II, Hydrodynamic and Hydromagnetic Stability) Ifalayerofviscousfluidisheatedfrombelow,instabilitycanoccurandleadstoconvection cells important in meteorology. (Benard Problem). If a saturated porous layer is heated from below, similar instabilily and convection can occur. This is of basic interest to geothermal convection and suggests more studies of the more complex problem of convection induced by buried nuclear waste. Figure 6.6.1: A saturated porus medium in geothermal gradient Static state Hence Static equilibrium: hence u =, T S = T o + T t = ³ 1 z h dp h ³ S dz = ρ og T o β T 1 z i h ½ p S = p o ρ o g T o z + β T µ ¾ z h z (6.6.1) (6.6.)

2 Consider the perturbed state of small disturbances: u =+u, T = T S + T, P = p S + p (6.6.3) then u = (6.6.4) Non-dimensionalization = p µ k u + βgρ o T k (6.6.5) (ρc) m T t +(ρc) fu T S = K m T (6.6.6). Let K m, κ m = K m /(ρc) f be the conductivity and diffusivity of the mixture, and k the permeability. Define (x, y, z) h(x,y,z ),t σh t, κ m u κ m h u,t = T θ, p µκ m k p (6.6.7) Then, after omitting for brevity, we get u = (6.6.8) = p u + Ra θk (6.6.9) θ t w = θ (6.6.1) where Ra = ρ fgkβ Th Rayleigh number in a porous medium (6.6.11) µκ m is the Rayleigh number (ratio of buoyancy force to diffusive resistence) of the porous medium. In Benard s problem, Rayleigh number is defined as Ra = ρg Th4, Rayleigh number in a pure fluid (6.6.1) µκ Sometimes one calls D = k, Darcy number (6.6.13) h Darcy number so that Rayleigh number of a porous medium is the product of the traditional Rayleigh number and Darcy number. Taking the curl of (6.6.9), u = Ra(iθ y jθ x ) (6.6.14)

3 3 Note that the z compoent of the vorticity vector is zero. Taking the curl again and using u = ( u) u we get u = Ra [iθ xz + jθ yz k(θ xx + θ yy )] Taking the z component, we get w = Ra Hθ (6.6.15) where H = x + (6.6.16) y is the horizontal Laplacian. Equations (6.6.1) and (6.6.15) couple the two unknowns w and θ. The boundary conditions are w = θ =, z =, 1 (6.6.17) After they are solved the other velocity components and pressure can be found Solution for sinusoidal disturbances Let and (w, θ) =(W (z), Θ(z)) exp(ilx + imy iωt) (6.6.18) D = d dz then from (6.6.1), and from (6.6.15) where The boundary conditions are : iωθ W =(D a )Θ (6.6.19) (D a )W = a RaΘ (6.6.) a = ` + m (6.6.1) W = Θ =, z =, 1 (6.6.) Because the equations and the boundary conditons are homogeneous, the problem for W and θ is an eigenvalue problem. Note that l, m, a are related to dimensional wave numbers by µ 1 l = k x h = πh L x, m = k y h = πh L y,a= kh =πh L x + 1 L y 1/ with k = q k x + k y (6.6.3)

4 Principle of exchange of stabilities We shall first show that ω must be purely imaginary. Multiplying (6.6.) by W and integrating from z =toz = 1, we get, after partial integration and using the boundary conditions, ( DW + a W )dz = a Ra W Θdz (6.6.4) Similarly we multipy (6.6.19) by Θ and integrating from z =toz =1,andget Z ( DΘ + a Θ )dz = iω Θ dz W Θ dz (6.6.5) Taking the complex conjugate of the second equation Z ( DΘ + a Θ )dz = iω Θ dz W Θdz (6.6.6) Eqs (6.6.4) and (6.6.6) can be combined by eliminating the cross product terms,, Z ( DΘ + a Θ )dz = iω Θ dz 1 ( DW + a W )dz (6.6.7) a Ra Since all integrals above are real, iω = ω i iω r must also be real. We conclude that ω r =, hence iω = ω i (6.6.8) Marginal stability (the threshold of instability) occurs at ω r = ω i =. Ifω i >, the static state is unstable; ω i =,marginallystable;ifω i <, stable. A problem where the eigenfrequency is real so that marginal instabililty occurs when ω = is said to obey the principle of exchange of stabilities Solution to eigenvalue problem Consider the situation at marginal stability : ω =, (D a )W = a RaΘ (6.6.9) Eliminating Θ, we get subject to Expanding (6.6.31) W =(D a )Θ (6.6.3) (D a ) W = a RaW (6.6.31) W =, D W =, z =, 1 (6.6.3) D 4 W a D W + a 4 W = a RaW (6.6.33)

5 Clearly D 4 W =onz =, 1. Differentiating (6.6.33) twice we see that D 6 =onz =, 1. Repeating the process we find Therefore the eigensolution must be To satisfy (6.6.31) it is necessary that D (m) =, m =1,, 3,, on z =, 1 (6.6.34) W sin jπz (6.6.35) Ra = [j π + a ] a, for j =1,, 3..., (6.6.36) which is the eigenvalue condition. For any j, Ra becomes unbounded for both a and a and is curve concave upward in the plane of a (abscissa) vs, Ra (ordinate). The lowest threshold occurs at j =1,and Ra a = 5 i.e., or a = π (6.6.37) Ra c =4π =39.48 (6.6.38) Possible convection patterns This is similar to Benard s problem which has been exhaustively studied theoretically and experimentally. There are many possibilities. Let us consider the lowest mode only with j =1. -Dimensinal Rolls :(` = π,m=) Take w = cosπx sin πz (6.6.39) then from mass conservation, u = sin πx cos πz (6.6.4) The dimensionless wavelength is L x = L y =. Along lines x =, ±n, n =1,, 3, 4,..., u = but w 6=. Alongx =, ±m, w>, hence fluid rises verrically. Along x = ±m 1, w< hence fluids sinks vertically. Along z =and1,w =. On the bottom (z =),u> while on the top (z =1),if<x<1, <x<3, 4 <x<5,.... The streamlines are shown in Figure Experimetal images of rolls in a pure fluid (Benard problem) is shown in Figure

6 6 Figure 6.6.: Rolls in a period Rectangular cells: (` = m = π/ ). w =cos πx cos πy sin πz (6.6.41) From the z component of the vorticity equation (6.6.14), v x u y = (6.6.4) and from continuity By cross differentiation, we get u x + v y = w z (6.6.43) u x + u y = w z y = and v x + v y = w z x = These are easily solved to give π π sin πx cos πy cos πz (6.6.44) cos πx sin πy cos πz (6.6.45) u = 1 sin πx cos πy cos πz (6.6.46) and v = 1 cos πx sin πy cos πz (6.6.47) The streamlines in a horizontal plane is shown in Figure Hexagonal cells: See Chandrasekhar. See Figure

7 Figure 6.6.3: Rolls in a pure fluid layer heated from below. From van Dyke: An Album of Fluid Motion 7

8 8 Figure 6.6.4: Rectangular and square cells in a pure fluid heated from below. From Chandrasekhar

9 Figure 6.6.5: Hexagonal cells in a pure fluid headted from below. From van Dyke 9

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