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

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.

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(ID 9216)

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

(ID 9217)

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

.

(ID 9219)

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.]

(ID 9218)