# Viscosity - how to calculate the internal force in a fluid

1. Aug 10, 2009

### TriKri

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":

Now, do anyone know how to calculate $$\frac{\partial \vec{v}}{\partial t}$$ if $$\vec{v}$$ is a function of x, y, z and t?

Anyway, from what was in the book, the acceleration that comes from the viscosity (hence exluding all external forces), is

$$\frac{\partial v_y}{\partial t}=\frac{\eta}{\rho}\ \frac{\partial^2 v_y}{\partial x^2}$$

Multiplying with the density gives us

$$\frac{dm}{dV}a_y=\eta\frac{\partial^2 v_y}{\partial x^2}$$

where m is the mass, V is the volume, and ay is the acceleration in y-direction. Since m*a=f, we have:

$$\frac{df_y}{dV}=\eta\frac{\partial^2 v_y}{\partial x^2}$$

Now, if you have $$v_y=v_y(x,z,t)$$ instead, I guess the equation would look like

$$\frac{df_y}{dV}=\eta\left(\frac{\partial^2 v_y}{\partial x^2}+\frac{\partial^2 v_y}{\partial z^2}\right)$$

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 $$v_y$$ is a function of both x, y, z and t.

$$\frac{df_x}{dV}=\eta\left(\frac{\partial^2 v_x}{\partial y^2}+\frac{\partial^2 v_x}{\partial z^2}\right)$$
$$\frac{df_y}{dV}=\eta\left(\frac{\partial^2 v_y}{\partial x^2}+\frac{\partial^2 v_y}{\partial z^2}\right)$$
$$\frac{df_z}{dV}=\eta\left(\frac{\partial^2 v_z}{\partial x^2}+\frac{\partial^2 v_z}{\partial y^2}\right)$$

or

$$\frac{d\vec{f}}{dV}=\eta\left(\begin{array}{ccccc} 0&+&\frac{\partial^2 v_x}{\partial y^2}&+&\frac{\partial^2 v_x}{\partial z^2}\\ \\ \frac{\partial^2 v_y}{\partial x^2}&+&0&+&\frac{\partial^2 v_y}{\partial z^2}\\ \\ \frac{\partial^2 v_z}{\partial x^2}&+&\frac{\partial^2 v_z}{\partial y^2}&+&0 \end{array}\right)$$

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

$$\eta\left(\begin{array}{ccccc} \frac{\partial^2 v_x}{\partial x^2}&+&\frac{\partial^2 v_x}{\partial y^2}&+&\frac{\partial^2 v_x}{\partial z^2}\\ \\ \frac{\partial^2 v_y}{\partial x^2}&+&\frac{\partial^2 v_y}{\partial y^2}&+&\frac{\partial^2 v_y}{\partial z^2}\\ \\ \frac{\partial^2 v_z}{\partial x^2}&+&\frac{\partial^2 v_z}{\partial y^2}&+&\frac{\partial^2 v_z}{\partial z^2} \end{array}\right) - \eta\left(\begin{array}{c} \frac{\partial^2 v_x}{\partial x^2}\\ \\ \frac{\partial^2 v_y}{\partial y^2}\\ \\ \frac{\partial^2 v_z}{\partial z^2} \end{array}\right)$$

I also found that

$$\left(\begin{array}{ccccc} \frac{\partial^2 v_x}{\partial x^2}&+&\frac{\partial^2 v_x}{\partial y^2}&+&\frac{\partial^2 v_x}{\partial z^2}\\ \\ \frac{\partial^2 v_y}{\partial x^2}&+&\frac{\partial^2 v_y}{\partial y^2}&+&\frac{\partial^2 v_y}{\partial z^2}\\ \\ \frac{\partial^2 v_z}{\partial x^2}&+&\frac{\partial^2 v_z}{\partial y^2}&+&\frac{\partial^2 v_z}{\partial z^2} \end{array}\right) = (\vec{\nabla}\cdot\vec{\nabla})\vec{v}$$

$$\text{(And also }=\vec{\nabla}(\vec{\nabla}\cdot\vec{v})\ -\ \vec{\nabla}\times(\vec{\nabla}\times\vec{v})\ )$$

so that basically, one can write

$$\frac{d\vec{f}}{dV} = \eta(\vec{\nabla}\cdot\vec{\nabla})\vec{v} - \eta\left(\begin{array}{c} \frac{\partial^2 v_x}{\partial x^2}\\ \\ \frac{\partial^2 v_y}{\partial y^2}\\ \\ \frac{\partial^2 v_z}{\partial z^2} \end{array}\right)$$

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 $$\frac{\partial \vec{v}}{\partial t}$$ if $$\vec{v}$$ is a function of x, y, z and t) usually is written, that would be fine too.

2. Aug 17, 2009

### Andy Resnick

There's a lot to this post, I'll tackle it piecemeal, and see how far I get before having to take a break.

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 mish-mash of expressions. That, plus starting from a 1-D simplified expression and trying to generalize to 3-D.

Look up the Navier-Stokes system of equations and how (say) Pouiseuille (sp?) flow is derived from the general expression. That should help.

3. Aug 17, 2009

### Cyrus

Right or wrong, such a pretty post.

4. Aug 18, 2009

### Andy Resnick

Ok- found one error. Or two, depending on how you count. First, $\rho\neq\frac{dm}{dV}$. That would be d$\rho$, IIRC. And then also, a$\neq\frac{\partial v}{\partial t}$, and df$\neq {\partial m}a$

5. Aug 18, 2009

### sarah1

[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

6. Aug 19, 2009

### TriKri

Okay, your post is confusing me. I really don't see why it shouldn't be the case. Isn't $\rho=\frac{dm}{dV}$ like the definition of density? If you mean that $\frac{dm}{dV}$ should equal $d\rho$ then you really have to explain why - if you count the d:s and $\partial[/tex]:s in your equation it doesn't add up. The same thing with the acceleration. Why shouldn't that be the case (you also got to consider that this is only the acceleration which is caused by the inner friction of the liquid - not depending on any external forces; then you also have to subtract v*grad(v) as you stated before)? And df[itex]\neq{\partial m}a$, where have I written that they are equal?

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 1-D 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 Navier-Stokes equations, I think I have found something useful there.

-Kristofer

7. Aug 19, 2009

### Andy Resnick

The density $\rho$ is m/V (i.e. kg/m^3). I haven't seen density used in a differential form, even though we sometimes write dm = $\rho dV$. 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 well-defined mathematical operation (d/dV (ma)) is questionable.

8. Aug 20, 2009

### TriKri

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).

9. Aug 23, 2009

### delplace

dear sir, you are trying to solve the well known Navier-Stokes 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 k-epsilon or direct numerical simulation (DNS). You can find easy tu use numerical codes like polyflow...
Best regards

10. Aug 24, 2009

### TriKri

Well, I am really not trying to solve any equation, I am just looking for another way to write (what I concider to be) $$\displaystyle{\frac{d\vec{F}}{dV}}$$, 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

$$\frac{d\vec{f}}{dV}=\eta(\vec{\nabla}\cdot\vec{\nabla})\vec{v}-\eta\left(\begin{array}{c}\frac{\partial^2 v_x}{\partial x^2}\\ \\ \frac{\partial^2 v_y}{\partial y^2}\\ \\ \frac{\partial^2 v_z}{\partial z^2}\end{array}\right)$$

Where the last term is still problematic. I want to express it in another way, just like I have done with the $\eta(\vec{\nabla}\cdot\vec{\nabla})\vec{v}$ term, to get completely rid of the parameters (in this case x, y and z).

-Kristofer