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Rayleigh-Bénard

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One of the widely studied instabilities is the so-called Rayleigh Bernard instability that describes the structure of cells that forms when a liquid enters convection in a temperature gradient.

>Model

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Rayleigh–Bénard instability

Equation

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When a liquid in a gravitational field with acceleration g is exposed to two horizontal plates with upper temperature T_t and lower temperature T_b, where the lower temperature is greater than the upper temperature, a spontaneous flow is generated if the Rayleigh number

R_c=\displaystyle\frac{g\alpha}{d\nu}(T_b-T_t)L^3

R_c=g*alpha*(T_b-T_t)*L^3/(d*nu)R_c=27*pi^4/4R=17610.39

exceeds a critical value. In this case, L represents the distance between the plates, d is the diffusivity, \alpha is the thermal expansion coefficient, and $

u$ is the kinematic viscosity.

You can find more information at this link: http://home.iitk.ac.in/~sghorai/NOTES/benard/benard.html

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Rayleigh-Bénard instability, limit free-free edges

Equation

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The eigenvalue problem becomes

(D^2-a^2)^3W=-a^2R_aW



subject to the boundary condition

$W=D^2W=D^4W=0 at z=0,1$



From this, we can show that

D^{2m}W=0

for

z=0,1

and

m = 1,2,\cdots



It follows that the required solution must be

W=A\sin(n\pi z)

with

n=1,2,3,\cdots



where A is a constant and n is an integer. Substituting W leads to the eigenvalue relationship

R_a=\frac{(n^2\pi^2+a^2)^3}{a^2}



For a given

a^2

, the lowest value of

R_c

occurs when

n=1

which is the lowest mode:

R_a=\frac{(\pi^2+a^2)^3}{a^2}



The critical Rayleigh number

R_c

is obtained by finding the minimum value of

R_a

as

a^2

varies.

\frac{dR_a}{da^2}=0\rightarrow a_c=\frac{\pi}{\sqrt{2}}



and the corresponding

R_c

is given by

R_c=\displaystyle\frac{27}{4}\pi^4

R_c=g*alpha*(T_b-T_t)*L^3/(d*nu)R_c=27*pi^4/4R=17610.39

ID:(9217, 0)



Rayleigh-Bénard instability, fixed-fixed edge limit

Equation

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If we set the origin at the midpoint of the chamber, the problem to be solved is given by

(D^2-a^2)^3W=-a^2R_aW



subject to the boundary condition

W=DW=(D^2-a^2)^2W=0

at

z=0,1

.

The problem is symmetric with respect to the two boundaries, so the eigenfunctions are divided into two distinct classes: (even mode), which has vertical velocity symmetry with respect to the midplane, and (odd mode), which has vertical velocity antisymmetry. The even mode has a single row of cells along the vertical, while the odd mode has two rows of cells. Let\'s assume the solution is of the form

W=e^{qz}



where the roots q are given by

(q^2-a^2)^3=-R_a^2



Let

R_a^2=\lambda^3a^6



then the roots are given by

q^2=-a^2(\lambda-1)



and

q^2=a^2\left[1+\frac{1}{2}\frac{\sqrt{3}}{2}\right]



Taking the square root again, the roots are \pm iq_0, \pm q, and \pm q^*, where q_0=a\sqrt{\lambda-1} and

re(q)=q_1=a\left[\frac{1}{2}\sqrt{1+\lambda+\lambda^2}+\frac{1}{2}\sqrt{1+\frac{\lambda}{2}}\right]^{1/2}



im(q)=q_2=a\left[\frac{1}{2}\sqrt{1+\lambda+\lambda^2}-\frac{1}{2}\sqrt{1+\frac{\lambda}{2}}\right]^{1/2}



Here, q^* denotes the complex conjugate of q. From these relationships, we have

(q_0^2+a^2)^2=a^4\lambda^2



(q^2-a^2)^2=\frac{1}{2}a^4\lambda^2(-1\pm i\sqrt{3})



Even mode solution

The even mode solution is expressed as

W=A\cos(q_0z)+B\cosh(qz)+C\cosh(q^*z)



Therefore, we have

DW=-Aq_0\sin(q_0z)+Bq\sinh(qz)+Cq^*\sinh(q^*z)



(D^2-a^2)^2W=A(q_0^2+a^2)^2\cos(q_0z)+B(q^2-a^2)+C(q^{*2}-a^2)^2\cosh(q^*z)



The boundary conditions provide

\left[\begin{array}{ccc}\cos(q_0/2) & \cosh(q/2) & \cosh(q^/2) \ \sin(q_0/2) & \sinh(q/2) & \sinh(q^/2) \ -q_0\tan(q_0/2) & q\tanh(q/2) & q^\tanh(q^/2)\end{array}\right]\left[\begin{array}{c}A \ B \ C\end{array}\right]=0



For a nontrivial solution (after some manipulations), we must have

\displaystyle\left\vert\begin{array}{ccc} 1 & 1 & 1 \ -q_0\tan(q_0/2) & q\tanh(q/2) & q^\tanh(q^/2) \ \frac{1}{2}(i\sqrt{3}+1) & \frac{1}{2}(i\sqrt{3}-1) & -1\end{array}\right\vert=0



which, in simplified terms,

im\left[(\sqrt{3}+i)q\tanh(q/2)\right]+q_0\tan(q_0/2)=0



which can be written as (with further simplification)

-q_0\tan(q_0/2)=\frac{(q_1+q_2\sqrt{3})\sinh(q_1)+(q_1\sqrt{3}-q_2)\sin(q_2)}{\cosh(q_1)+\cos(q_2)}



This equation must be solved using the trial and error method: for a given value of a, we need to find the value of \lambda and then find the value of R_a. The critical values of a and R_a (Reid & Harris, Phys of Fluids, Vol-1) are given by

$a_c=3.117 and R_c=1707.762$



Taking A_0=1 and C=B^*, we can find W and \Theta.

Impar Solution

The odd solution is represented as

W=A\sin(q_0z)+B\sinh(qz)+C\sinh(q^*z)



Proceeding similarly, we obtain

q_0\cot(q_0/2)=\displaystyle\frac{(q_1+q_2\sqrt{3})\sinh(q_1)-(q_1\sqrt{3}-q_2)\sin(q_2)}{\cosh(q_1)-\cosh(q_2)}



In this case, the minimum Rayleigh number occurs at a=5.365, and the corresponding value of the Rayleigh number is

R=17610.39

R_c=g*alpha*(T_b-T_t)*L^3/(d*nu)R_c=27*pi^4/4R=17610.39

.

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Rayleigh-Bénard instability, fixed-free edge limit

Equation

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The solution for the case where the upper surface is free and the lower surface is rigid can be deduced from the odd solution of the rigid-rigid box.

The problem is defined by:

\frac{(D^2-a^2)^3}{W}=-a^2R_aW



subject to the boundary condition:

W = DW = (D^2-a^2)^2W = 0

at

z = 0



W = D^2W = D^4W = 0

at

z = 1



The boundary conditions at the mid-height for the odd solution hold. Therefore, an odd solution for the rigid-rigid limit at depth

d

provides a solution for the rigid-free limit at depth

d/2

. Thus, using the stability results from the rigid-rigid case, we have:

a_c = 5.365 / 2 \approx 2.682

and

R_c = 17610.39 / 2^4 \approx 1100.65



[Note: From the dimensional form of

\Delta_1^2f + k^2f

, we have

a = kL

(where

L

is the scale length) as the dimensionless wavenumber.]

$$

R_c=g*alpha*(T_b-T_t)*L^3/(d*nu)R_c=27*pi^4/4R=17610.39

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