What are the internal forces in fluid
Viscosity  how to calculate the internal force in a fluid
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Hi,
I'm looking for a little bit of help in this matter, I'm trying to put up a general formula of the internal force/volume in a fluid in a spot, as a function of how the velocity varies locally around that spot.
What I started with was a formula I found in a book called "Physics Handbook  for Science and Engineering":
Anyway, from what was in the book, the acceleration that comes from the viscosity (hence exluding all external forces), is
[tex]\frac{\partial v_y}{\partial t}=\frac{\eta}{\rho}\ \frac{\partial^2 v_y}{\partial x^2}[/tex]
Multiplying with the density gives us
[tex]\frac{dm}{dV}a_y=\eta\frac{\partial^2 v_y}{\partial x^2}[/tex]
where m is the mass, V is the volume, and a_{y} is the acceleration in ydirection. Since m*a=f, we have:
[tex]\frac{df_y}{dV}=\eta\frac{\partial^2 v_y}{\partial x^2}[/tex]
Now, if you have [tex]v_y=v_y(x,z,t)[/tex] instead, I guess the equation would look like
[tex]\frac{df_y}{dV}=\eta\left(\frac{\partial^2 v_y}{\partial x^2}+\frac{\partial^2 v_y}{\partial z^2}\right)[/tex]
And to what I suppose, the force created (only by the inner friction of the liquid, i.e. the viscosity) would be the same, even if [tex]v_y[/tex] is a function of both x, y, z and t.
What I realized was this:
[tex]\frac{df_x}{dV}=\eta\left(\frac{\partial^2 v_x}{\partial y^2}+\frac{\partial^2 v_x}{\partial z^2}\right)[/tex]
[tex]\frac{df_y}{dV}=\eta\left(\frac{\partial^2 v_y}{\partial x^2}+\frac{\partial^2 v_y}{\partial z^2}\right)[/tex]
[tex]\frac{df_z}{dV}=\eta\left(\frac{\partial^2 v_z}{\partial x^2}+\frac{\partial^2 v_z}{\partial y^2}\right)[/tex]
or
[tex]\frac{d\vec{f}}{dV}=\eta\left(\begin{array}{ccccc}
0&+&\frac{\partial^2 v_x}{\partial y^2}&+&\frac{\partial^2 v_x}{\partial z^2}\\ \\
\frac{\partial^2 v_y}{\partial x^2}&+&0&+&\frac{\partial^2 v_y}{\partial z^2}\\ \\
\frac{\partial^2 v_z}{\partial x^2}&+&\frac{\partial^2 v_z}{\partial y^2}&+&0
\end{array}\right)[/tex]
Now, is there any way to simplify this expression? Anyway, I worked a bit on it on my own and I found that it equals to
[tex]\eta\left(\begin{array}{ccccc}
\frac{\partial^2 v_x}{\partial x^2}&+&\frac{\partial^2 v_x}{\partial y^2}&+&\frac{\partial^2 v_x}{\partial z^2}\\ \\
\frac{\partial^2 v_y}{\partial x^2}&+&\frac{\partial^2 v_y}{\partial y^2}&+&\frac{\partial^2 v_y}{\partial z^2}\\ \\
\frac{\partial^2 v_z}{\partial x^2}&+&\frac{\partial^2 v_z}{\partial y^2}&+&\frac{\partial^2 v_z}{\partial z^2}
\end{array}\right)

\eta\left(\begin{array}{c}
\frac{\partial^2 v_x}{\partial x^2}\\ \\
\frac{\partial^2 v_y}{\partial y^2}\\ \\
\frac{\partial^2 v_z}{\partial z^2}
\end{array}\right)[/tex]
I also found that
[tex]\left(\begin{array}{ccccc}
\frac{\partial^2 v_x}{\partial x^2}&+&\frac{\partial^2 v_x}{\partial y^2}&+&\frac{\partial^2 v_x}{\partial z^2}\\ \\
\frac{\partial^2 v_y}{\partial x^2}&+&\frac{\partial^2 v_y}{\partial y^2}&+&\frac{\partial^2 v_y}{\partial z^2}\\ \\
\frac{\partial^2 v_z}{\partial x^2}&+&\frac{\partial^2 v_z}{\partial y^2}&+&\frac{\partial^2 v_z}{\partial z^2}
\end{array}\right)
=
(\vec{\nabla}\cdot\vec{\nabla})\vec{v}[/tex]
[tex]\text{(And also }=\vec{\nabla}(\vec{\nabla}\cdot\vec{v})\ \ \vec{\nabla}\times(\vec{\nabla}\times\vec{v})\ )[/tex]
so that basically, one can write
[tex]\frac{d\vec{f}}{dV}
=
\eta(\vec{\nabla}\cdot\vec{\nabla})\vec{v}

\eta\left(\begin{array}{c}
\frac{\partial^2 v_x}{\partial x^2}\\ \\
\frac{\partial^2 v_y}{\partial y^2}\\ \\
\frac{\partial^2 v_z}{\partial z^2}
\end{array}\right)[/tex]
Now I have showed you my work, how long I have come so far. I would really appreciate if someone could help me with the last part. The thing I have done here is that I have removed almost all x, y and z and written it as a vector field instead of as three different scalar fields. Hence trying to make it undependent of which coordinate system that is used. I just can't find a way to write that last vector without using x, y and z, and without making it depend on which coordinate system you use. I want to minimize it into one row as I have done with the bigger vector.
If someone knows a way to do this, I would be really thankful if that one could tell me. Or, if someone knows how the thing I wanted to know from the begining (how to calculate [tex]\frac{\partial \vec{v}}{\partial t}[/tex] if [tex]\vec{v}[/tex] is a function of x, y, z and t) usually is written, that would be fine too.
I'm looking for a little bit of help in this matter, I'm trying to put up a general formula of the internal force/volume in a fluid in a spot, as a function of how the velocity varies locally around that spot.
What I started with was a formula I found in a book called "Physics Handbook  for Science and Engineering":
Now, do anyone know how to calculate [tex]\frac{\partial \vec{v}}{\partial t}[/tex] if [tex]\vec{v}[/tex] is a function of x, y, z and t?Differential equation for a Newtonian liquid flowing in the ydirection with speed v_{y} = v_{y}(x,t)
[tex]\frac{\partial v_y}{\partial t}=\frac{\eta}{\rho}\ \frac{\partial^2 v_y}{\partial x^2}+\frac{f}{\rho}[/tex]
[tex]f[/tex]=external force in ydirection per volume
[tex]\eta[/tex]=coefficient of viscosity
Anyway, from what was in the book, the acceleration that comes from the viscosity (hence exluding all external forces), is
[tex]\frac{\partial v_y}{\partial t}=\frac{\eta}{\rho}\ \frac{\partial^2 v_y}{\partial x^2}[/tex]
Multiplying with the density gives us
[tex]\frac{dm}{dV}a_y=\eta\frac{\partial^2 v_y}{\partial x^2}[/tex]
where m is the mass, V is the volume, and a_{y} is the acceleration in ydirection. Since m*a=f, we have:
[tex]\frac{df_y}{dV}=\eta\frac{\partial^2 v_y}{\partial x^2}[/tex]
Now, if you have [tex]v_y=v_y(x,z,t)[/tex] instead, I guess the equation would look like
[tex]\frac{df_y}{dV}=\eta\left(\frac{\partial^2 v_y}{\partial x^2}+\frac{\partial^2 v_y}{\partial z^2}\right)[/tex]
And to what I suppose, the force created (only by the inner friction of the liquid, i.e. the viscosity) would be the same, even if [tex]v_y[/tex] is a function of both x, y, z and t.
What I realized was this:
[tex]\frac{df_x}{dV}=\eta\left(\frac{\partial^2 v_x}{\partial y^2}+\frac{\partial^2 v_x}{\partial z^2}\right)[/tex]
[tex]\frac{df_y}{dV}=\eta\left(\frac{\partial^2 v_y}{\partial x^2}+\frac{\partial^2 v_y}{\partial z^2}\right)[/tex]
[tex]\frac{df_z}{dV}=\eta\left(\frac{\partial^2 v_z}{\partial x^2}+\frac{\partial^2 v_z}{\partial y^2}\right)[/tex]
or
[tex]\frac{d\vec{f}}{dV}=\eta\left(\begin{array}{ccccc}
0&+&\frac{\partial^2 v_x}{\partial y^2}&+&\frac{\partial^2 v_x}{\partial z^2}\\ \\
\frac{\partial^2 v_y}{\partial x^2}&+&0&+&\frac{\partial^2 v_y}{\partial z^2}\\ \\
\frac{\partial^2 v_z}{\partial x^2}&+&\frac{\partial^2 v_z}{\partial y^2}&+&0
\end{array}\right)[/tex]
Now, is there any way to simplify this expression? Anyway, I worked a bit on it on my own and I found that it equals to
[tex]\eta\left(\begin{array}{ccccc}
\frac{\partial^2 v_x}{\partial x^2}&+&\frac{\partial^2 v_x}{\partial y^2}&+&\frac{\partial^2 v_x}{\partial z^2}\\ \\
\frac{\partial^2 v_y}{\partial x^2}&+&\frac{\partial^2 v_y}{\partial y^2}&+&\frac{\partial^2 v_y}{\partial z^2}\\ \\
\frac{\partial^2 v_z}{\partial x^2}&+&\frac{\partial^2 v_z}{\partial y^2}&+&\frac{\partial^2 v_z}{\partial z^2}
\end{array}\right)

\eta\left(\begin{array}{c}
\frac{\partial^2 v_x}{\partial x^2}\\ \\
\frac{\partial^2 v_y}{\partial y^2}\\ \\
\frac{\partial^2 v_z}{\partial z^2}
\end{array}\right)[/tex]
I also found that
[tex]\left(\begin{array}{ccccc}
\frac{\partial^2 v_x}{\partial x^2}&+&\frac{\partial^2 v_x}{\partial y^2}&+&\frac{\partial^2 v_x}{\partial z^2}\\ \\
\frac{\partial^2 v_y}{\partial x^2}&+&\frac{\partial^2 v_y}{\partial y^2}&+&\frac{\partial^2 v_y}{\partial z^2}\\ \\
\frac{\partial^2 v_z}{\partial x^2}&+&\frac{\partial^2 v_z}{\partial y^2}&+&\frac{\partial^2 v_z}{\partial z^2}
\end{array}\right)
=
(\vec{\nabla}\cdot\vec{\nabla})\vec{v}[/tex]
[tex]\text{(And also }=\vec{\nabla}(\vec{\nabla}\cdot\vec{v})\ \ \vec{\nabla}\times(\vec{\nabla}\times\vec{v})\ )[/tex]
so that basically, one can write
[tex]\frac{d\vec{f}}{dV}
=
\eta(\vec{\nabla}\cdot\vec{\nabla})\vec{v}

\eta\left(\begin{array}{c}
\frac{\partial^2 v_x}{\partial x^2}\\ \\
\frac{\partial^2 v_y}{\partial y^2}\\ \\
\frac{\partial^2 v_z}{\partial z^2}
\end{array}\right)[/tex]
Now I have showed you my work, how long I have come so far. I would really appreciate if someone could help me with the last part. The thing I have done here is that I have removed almost all x, y and z and written it as a vector field instead of as three different scalar fields. Hence trying to make it undependent of which coordinate system that is used. I just can't find a way to write that last vector without using x, y and z, and without making it depend on which coordinate system you use. I want to minimize it into one row as I have done with the bigger vector.
If someone knows a way to do this, I would be really thankful if that one could tell me. Or, if someone knows how the thing I wanted to know from the begining (how to calculate [tex]\frac{\partial \vec{v}}{\partial t}[/tex] if [tex]\vec{v}[/tex] is a function of x, y, z and t) usually is written, that would be fine too.
Answers and Replies
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There's a lot to this post, I'll tackle it piecemeal, and see how far I get before having to take a break.Hi,
<snip>
What I started with was a formula I found in a book called "Physics Handbook  for Science and Engineering":
Now, do anyone know how to calculate [tex]\frac{\partial \vec{v}}{\partial t}[/tex] if [tex]\vec{v}[/tex] is a function of x, y, z and t?
Ok, first that equation is a simplified version of a=F/m, written for continua. Normally, the dv/dt term is written as Dv/Dt, because v = v(x,y,z,t), and so v can change either by you standing somewhere and looking at it (dv/dt) or by you moving somewhere else (v *grad(v)), where '*' is a dot product. So, this text has not only assumed that the flow is linear (so the v*grad(v) term is gone), but also v = v(x,t), and only in the 'y' direction recall v is a vector field.
There's nothing that bizarre about the equation shown, the usual way to deal with it is by separation of variables.
I didn't see anything obviously wrong, but I think the origin of your confusion is that you are not treating ma=F as a vector equation right from the beginning; there are three (one for each component). So you end up with this mishmash of expressions. That, plus starting from a 1D simplified expression and trying to generalize to 3D.Anyway, from what was in the book, the acceleration that comes from the viscosity (hence exluding all external forces), is
[tex]\frac{\partial v_y}{\partial t}=\frac{\eta}{\rho}\ \frac{\partial^2 v_y}{\partial x^2}[/tex]
<snip>
Now, is there any way to simplify this expression? Anyway, I worked a bit on it on my own and I found that it equals to
<snip>
Now I have showed you my work, how long I have come so far. I would really appreciate if someone could help me with the last part. The thing I have done here is that I have removed almost all x, y and z and written it as a vector field instead of as three different scalar fields. Hence trying to make it undependent of which coordinate system that is used. I just can't find a way to write that last vector without using x, y and z, and without making it depend on which coordinate system you use. I want to minimize it into one row as I have done with the bigger vector.
If someone knows a way to do this, I would be really thankful if that one could tell me. Or, if someone knows how the thing I wanted to know from the begining (how to calculate [tex]\frac{\partial \vec{v}}{\partial t}[/tex] if [tex]\vec{v}[/tex] is a function of x, y, z and t) usually is written, that would be fine too.
Look up the NavierStokes system of equations and how (say) Pouiseuille (sp?) flow is derived from the general expression. That should help.
Right or wrong, such a pretty post.
Ok found one error. Or two, depending on how you count. First, [itex]\rho\neq\frac{dm}{dV}[/itex]. That would be d[itex]\rho[/itex], IIRC. And then also, a[itex]\neq\frac{\partial v}{\partial t}[/itex], and df[itex]\neqHi,
<snip>
[tex]\frac{\partial v_y}{\partial t}=\frac{\eta}{\rho}\ \frac{\partial^2 v_y}{\partial x^2}[/tex]
Multiplying with the density gives us
[tex]\frac{dm}{dV}a_y=\eta\frac{\partial^2 v_y}{\partial x^2}[/tex]
where m is the mass, V is the volume, and a_{y} is the acceleration in ydirection. Since m*a=f, we have:
[tex]\frac{df_y}{dV}=\eta\frac{\partial^2 v_y}{\partial x^2}[/tex]
<snip>
{\partial m}a[/itex]
[for understand your problem you begin with the cinematic of fluid.
for describe any fluid it exixt two types of description lagrangien and euleurien descriptions. where any propriety is write as f=f(x,y,z,t) like velocity v=(x,y,z,t).
for calculate the acceleration you use the particular derivative = the locale acceleration +convective term.
for the viscosity you can see :dynamic of ''real fluid''
good luck
for describe any fluid it exixt two types of description lagrangien and euleurien descriptions. where any propriety is write as f=f(x,y,z,t) like velocity v=(x,y,z,t).
for calculate the acceleration you use the particular derivative = the locale acceleration +convective term.
for the viscosity you can see :dynamic of ''real fluid''
good luck
Okay, your post is confusing me. I really don't see why it shouldn't be the case. Isn't [itex]\rho=\frac{dm}{dV}[/itex] like the definition of density? If you mean that [itex]\frac{dm}{dV}[/itex] should equal [itex]d\rho[/itex] then you really have to explain why  if you count the d:s and [itex]\partial[/tex]:s in your equation it doesn't add up. The same thing with the acceleration. Why shouldn't that be the case (you also got to consider that this is only the acceleration which is caused by the inner friction of the liquid  not depending on any external forces; then you also have to subtract v*grad(v) as you stated before)? And df[itex]\neq{\partial m}a[/itex], where have I written that they are equal?Ok found one error. Or two, depending on how you count. First, [itex]\rho\neq\frac{dm}{dV}[/itex]. That would be d[itex]\rho[/itex], IIRC. And then also, a[itex]\neq\frac{\partial v}{\partial t}[/itex], and df[itex]\neq
{\partial m}a[/itex]
Concerning your last post, you say that what is causing my confusion is that I treat F not as a field but rather as three different components  don't you think that I deal with that later? If you have read the entire post you will know that I derived those equations from one single equation that was in a book called Physics Handbook, which had simplified it all so that it was valid to a 1D field only, hence it wasn't really possible for me to go ahead in any other way. If you have any suggestions on how I could have done it instead, you can gladly tell me.
But thanks for your reference to NavierStokes equations, I think I have found something useful there.
Kristofer
The density [itex]\rho[/itex] is m/V (i.e. kg/m^3). I haven't seen density used in a differential form, even though we sometimes write dm = [itex] \rho dV[/itex]. But note, that's different than what you wrote, because there's an integral floating around somewhere to get back to 'm' and 'V'.
The quantities 'dm' and 'dV' are different than the mathematical sense of 'dm/dV'. It makes sense to talk about infinitesimal portions of mass and volume (related by the density), but writing dm/dV and treating that as a welldefined mathematical operation (d/dV (ma)) is questionable.
The quantities 'dm' and 'dV' are different than the mathematical sense of 'dm/dV'. It makes sense to talk about infinitesimal portions of mass and volume (related by the density), but writing dm/dV and treating that as a welldefined mathematical operation (d/dV (ma)) is questionable.
Still, if you have a suggestion on how I can do it instead, I would be happy to hear it. So far you have only come up with problems that didn't even exist from the beginning (imo).
dear sir, you are trying to solve the well known NavierStokes equation for the flow of newtonian liquids. You can only solve this equation analytically in the cas of the laminar steady flow in a duct of simple cross section shape (circular, rectangular...).
For other cases : non steady, turbulent... you need to use numerical methods like kepsilon or direct numerical simulation (DNS). You can find easy tu use numerical codes like polyflow...
Best regards
For other cases : non steady, turbulent... you need to use numerical methods like kepsilon or direct numerical simulation (DNS). You can find easy tu use numerical codes like polyflow...
Best regards
Well, I am really not trying to solve any equation, I am just looking for another way to write (what I concider to be) [tex]\displaystyle{\frac{d\vec{F}}{dV}}[/tex], that is undependent of the coordinate system. If you look at my first post, you can see that I have only managed to get the expression to
[tex]\frac{d\vec{f}}{dV}=\eta(\vec{\nabla}\cdot\vec{\nabla})\vec{v}\eta\left(\begin{array}{c}\frac{\partial^2 v_x}{\partial x^2}\\ \\ \frac{\partial^2 v_y}{\partial y^2}\\ \\ \frac{\partial^2 v_z}{\partial z^2}\end{array}\right)[/tex]
Where the last term is still problematic. I want to express it in another way, just like I have done with the [itex]\eta(\vec{\nabla}\cdot\vec{\nabla})\vec{v}[/itex] term, to get completely rid of the parameters (in this case x, y and z).
Kristofer
[tex]\frac{d\vec{f}}{dV}=\eta(\vec{\nabla}\cdot\vec{\nabla})\vec{v}\eta\left(\begin{array}{c}\frac{\partial^2 v_x}{\partial x^2}\\ \\ \frac{\partial^2 v_y}{\partial y^2}\\ \\ \frac{\partial^2 v_z}{\partial z^2}\end{array}\right)[/tex]
Where the last term is still problematic. I want to express it in another way, just like I have done with the [itex]\eta(\vec{\nabla}\cdot\vec{\nabla})\vec{v}[/itex] term, to get completely rid of the parameters (in this case x, y and z).
Kristofer
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