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Next: Normal modes Up: The stability problem Previous: Scaling of the equations

Boundary conditions

The fluid is confined between the planes $ z=0$ and $ z=1$. Regardless of the nature of these bounding surface we must require

$\displaystyle \theta=w=0$   at$\displaystyle \quad z=0,1,$ (21)

since these planes are maintained at constant temperature and normal component of velocity must vanish on these planes. There are other boundary conditions which depends on the type of surfaces distinguished by rigid surfaces on which no slip occurs and free surfaces on which tangential shear stresses vanish.

First consider a rigid surface where no slip condition holds. Hence $ u=v=0$ in addition to $ w=0$. This with the equation of continuity implies that

$\displaystyle \frac{\ensuremath{\partial}w}{\ensuremath{\partial}z} = 0$    on a rigid surface (22)

The condition on the free surface is that the tangential stress vanish i.e. $ \tau_{xz}=\tau_{yz}=0$ which imply

$\displaystyle \frac{\ensuremath{\partial}u}{\ensuremath{\partial}z}=\frac{\ensuremath{\partial}v}{\ensuremath{\partial}z} = 0
$

Using equation of continuity we get

$\displaystyle \frac{\ensuremath{\partial}^2 w}{\ensuremath{\partial}z^2} = 0$    on a free surface (23)

For the normal component of vorticity $ \omega_3$ we have

$\displaystyle \frac{\ensuremath{\partial}\omega_3}{\ensuremath{\partial}z} = 0$    on a free surface (24)

$\displaystyle \omega_3 = 0$    on a rigid surface (25)

In summary we have the following boundary conditions

$\displaystyle w=\frac{\ensuremath{\partial}w}{\ensuremath{\partial}z} = \theta = \omega_3 = 0$   (rigid boundary) (26)

$\displaystyle w=\frac{\ensuremath{\partial}^2 w}{\ensuremath{\partial}z^2} = \theta = \frac{\ensuremath{\partial}\omega_3}{\ensuremath{\partial}z}= 0$   (free boundary) (27)


next up previous
Next: Normal modes Up: The stability problem Previous: Scaling of the equations
A/C for Homepage of Dr. S Ghorai 2003-01-16