Fluid Dynamics, free surface boundary condition derivation.












1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














1












1








1


2



$begingroup$


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







share|cite|improve this question









$endgroup$




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






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










2 Answers
2






active

oldest

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2












$begingroup$

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:



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






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $z=h(x,t)$ 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 $$w = h_t + u h_x$$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

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      active

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      active

      oldest

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      2












      $begingroup$

      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:



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






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        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:



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






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          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:



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






          share|cite|improve this answer









          $endgroup$



          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:



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







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 18:32









          RRLRRL

          52k42573




          52k42573























              1












              $begingroup$

              $z=h(x,t)$ 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 $$w = h_t + u h_x$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $z=h(x,t)$ 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 $$w = h_t + u h_x$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $z=h(x,t)$ 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 $$w = h_t + u h_x$$






                  share|cite|improve this answer









                  $endgroup$



                  $z=h(x,t)$ 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 $$w = h_t + u h_x$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 9 at 21:43









                  topspintopspin

                  737413




                  737413






























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