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I'm stuck on part of the derivation for a boundary condition for the free surface of a fluid. The fluid concerns a rectangular trough of water whose surface is at a height $$ z=h(x,t) $$ (Problem in 2D)

Now another condition stated earlier (which I understand), states the surface moves with the fluid. The surface given by $ S =z-h(x,t) $.

This means $$\frac{D}{Dt}(z-h)=0$$

Evaluated on the surface $z=h$ , where I'm referring to the material derivative.

Fully expanding this out i get (with subscript derivative notation):

$$z_t-h_t+uz_x+vz_y+wz_z-uh_x-vh_y-wh_z=0$$

Here u,v and w are the x, y and z components of the flow respectively.

The final condition is said to be.

$$-h_t-uh_x+w=0$$

I can't work out how so many of the derivatives are zero. The derivatives of z and what exactly z refer to seems to be my main area of confusion. Is it the coordinate system itself or the height of the fluid or something else.

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  • $\begingroup$ Some notes on free surface and its derivation for a rotating liquid in a cylindrical vessel. The free surface is also the z dimension of a Cartesian coordinate or cylindrical coordinate system. Take a look at this link and it will give some clarity. But I still don't understand your idea of derivation. Please post the full problem ( I do not understand what you mean by "as I stated above" without any statement.en.wikipedia.org/wiki/Free_surface $\endgroup$ – Satish Ramanathan Jan 9 '19 at 16:59
  • $\begingroup$ Apologies for the vagueness although this isn't centred on a specific problem per say. By derivation, I simply mean how you start out at the condition that the free surface moves with the fluid and end with the last equation I wrote which can be then used when your trying to find for the fluid potential/flow field. Will look at the link you sent, thanks. $\endgroup$ – Oliver Cohen Jan 9 '19 at 18:04
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The variables $x$,$y$ and $z$ are Cartesian coordinates and are independent variables -- both spatially and temporally

Thus,

$$z_t = \frac{\partial z}{\partial t} = 0, \,\,\,z_x = \frac{\partial z}{\partial x} = 0, \,\,\, z_y = \frac{\partial z}{\partial y} = 0, \,\,\, z_z = \frac{\partial z}{\partial z} = 1$$

Since $h$ is a function of only $t$ and $x$, the only nonzero partial derivatives are $h_t$ and $h_x$.

Substituting into your general expression we see that only three nonzero terms remain in the free-surface (kinematic) condition:

$$\frac{D}{Dt} (z-h) = -h_t -uh_x +wz_z = -h_t-uh_z +w$$

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$z=h(x,t)$ is the equation of the free surface . each fluid particle on the free surface has two coordinate $x , z$ and since the fluid particle moves with time . clearly $z$ is a function of $x , t$

The first condition on the free surface is called impermeability condition or no slip condition

$$\frac{dz}{dt}=\frac{dh(x,t)}{dt}=\frac{\partial h}{\partial t} + \frac{d h}{dx}\frac{dx}{dt}$$

which implies that $$w = \ h_t \ + u h_x$$

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