Results (
Thai) 1:
[Copy]Copied!
A. Sorkin et al. / Physica A 303 (2002) 13–26meansw|z=0;d = @2w@z2z=0;d= 0 (2.6)andthe rigidboundary condition which meansw|z=0;d = @w@zz=0;d= 0 : (2.7)Here, w = u3; z is the component of the velocity vector eldand d is the distancebetween the upper andlower planes.One may select appropriate combinations of these two conditions: (free–free, rigid–rigid, free–rigid). The following system is considered: a Guid is placed between thetwo planes, the lower plane in our case is the bottom of the beaker, andthe upperplane is a free boundary (open air), so we selected the free–rigid boundary condition.In our experiment, the S liquidrepresentedby sugar solute is placedabove andthe Tliquidrepresentedby salt solute is placedbelow.For the sake of simplicity, the T and S elds are treated as linear functions ofz coordinates (only the constant gradients of S and T are considered). A Cartesiansystem of coordinates is the most convenient choice for the description: the origin isplacedon the bottom plane andthe z-axis is directed perpendicular to the planes witha positive direction opposite to gravity acceleration vector.T(z) and S(z) can be representedaccordingly by the functionsT(z) = T(0)(1 + z) ; (2.8)S(z) = S(0)(1 + z) ; (2.9)where and are both constant andpositive values.Now let us investigate a small perturbation of our system that may lead to hydrody-namic instability. Only very small perturbations are considered, which means that allquadratic values (the second order perturbations and other higher order perturbations)are being neglected.ui = ˜ui + ui; i = 1; 2; 3 ; (2.10)T = T˜ + T ; (2.11)S = S˜ + S : (2.12)The corresponding equations are as follows: 1Pr@@t − ∇˜ 2-(∇˜ ˜u) = − RT∇2T + RS∇2S ; (2.13) @@t − ∇˜ 2-T˜ = − w ; (2.14) @@t − ∇˜ 2-S˜ = − w :
Being translated, please wait..
