Fluid Dynamics, free surface boundary condition derivation.
$begingroup$
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.
calculus fluid-dynamics
asked Jan 9 at 15:49
Oliver CohenOliver Cohen
185
$endgroup$
add a comment |
$begingroup$
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.
calculus fluid-dynamics
asked Jan 9 at 15:49
Oliver CohenOliver Cohen
185
$endgroup$
$begingroup$
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 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 at 18:04
add a comment |
$begingroup$
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.
calculus fluid-dynamics
asked Jan 9 at 15:49
Oliver CohenOliver Cohen
185
$endgroup$
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.
calculus fluid-dynamics
calculus fluid-dynamics
asked Jan 9 at 15:49
Oliver CohenOliver Cohen
185
asked Jan 9 at 15:49
Oliver CohenOliver Cohen
185
asked Jan 9 at 15:49
Oliver CohenOliver Cohen
185
asked Jan 9 at 15:49
Oliver CohenOliver Cohen
185
asked Jan 9 at 15:49
Oliver CohenOliver Cohen
185
185
$begingroup$
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 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 at 18:04
add a comment |
$begingroup$
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 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 at 18:04
$begingroup$
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 at 16:59
$begingroup$
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 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 at 18:04
$begingroup$
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 at 18:04
add a comment |
2 Answers
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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:
$$frac{D}{Dt} (z-h) = -h_t -uh_x +wz_z = -h_t-uh_z +w$$
answered Jan 9 at 18:32
RRLRRL
52k42573
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add a comment |
$begingroup$
$z=h(x,t)$ 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 $$w = h_t + u h_x$$
answered Jan 9 at 21:43
topspintopspin
737413
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
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$begingroup$
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:
$$frac{D}{Dt} (z-h) = -h_t -uh_x +wz_z = -h_t-uh_z +w$$
answered Jan 9 at 18:32
RRLRRL
52k42573
$endgroup$
add a comment |
$begingroup$
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:
$$frac{D}{Dt} (z-h) = -h_t -uh_x +wz_z = -h_t-uh_z +w$$
answered Jan 9 at 18:32
RRLRRL
52k42573
$endgroup$
add a comment |
$begingroup$
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:
$$frac{D}{Dt} (z-h) = -h_t -uh_x +wz_z = -h_t-uh_z +w$$
answered Jan 9 at 18:32
RRLRRL
52k42573
$endgroup$
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:
$$frac{D}{Dt} (z-h) = -h_t -uh_x +wz_z = -h_t-uh_z +w$$
answered Jan 9 at 18:32
RRLRRL
52k42573
answered Jan 9 at 18:32
RRLRRL
52k42573
answered Jan 9 at 18:32
RRLRRL
52k42573
answered Jan 9 at 18:32
RRLRRL
52k42573
52k42573
add a comment |
add a comment |
$begingroup$
$z=h(x,t)$ 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 $$w = h_t + u h_x$$
answered Jan 9 at 21:43
topspintopspin
737413
$endgroup$
add a comment |
$begingroup$
$z=h(x,t)$ 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 $$w = h_t + u h_x$$
answered Jan 9 at 21:43
topspintopspin
737413
$endgroup$
add a comment |
$begingroup$
$z=h(x,t)$ 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 $$w = h_t + u h_x$$
answered Jan 9 at 21:43
topspintopspin
737413
$endgroup$
$z=h(x,t)$ 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 $$w = h_t + u h_x$$
answered Jan 9 at 21:43
topspintopspin
737413
answered Jan 9 at 21:43
topspintopspin
737413
answered Jan 9 at 21:43
topspintopspin
737413
answered Jan 9 at 21:43
topspintopspin
737413
737413
add a comment |
add a comment |
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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 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 at 18:04