- #1

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\frac{\partial u}{\partial t}+(u\cdot\nabla)u=f-\nabla p+\nu\Delta u

\nabla\cdot u=0

I couldn't figure out how to actually get a LaTeX output. I have to give a presentation on them, and I can't find anything off the web.

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- Thread starter adamabel
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- #1

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\frac{\partial u}{\partial t}+(u\cdot\nabla)u=f-\nabla p+\nu\Delta u

\nabla\cdot u=0

I couldn't figure out how to actually get a LaTeX output. I have to give a presentation on them, and I can't find anything off the web.

- #2

cristo

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I got your tex working (click on the equations to see the code; you just missed out the tex tags)

[tex]\frac{\partial u}{\partial t}+(u\cdot\nabla)u=f-\nabla p+\nu\Delta u \hspace{1cm} \nabla\cdot u=0 [/tex]

I find it quite hard to believe that you found**nothing** on the Navier Stokes equations on the internet! What precisely are you looking for; I did a quick google search and came up with lots of websites!

[tex]\frac{\partial u}{\partial t}+(u\cdot\nabla)u=f-\nabla p+\nu\Delta u \hspace{1cm} \nabla\cdot u=0 [/tex]

I find it quite hard to believe that you found

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- #3

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It seems to me, that from Classical Mechanics, (thinking just in one dimension), that that would translate to [tex]v\frac{dv}{dx}[/tex]. But then that equals [tex]\frac{dv}{dt}[/tex]. So it sort of seems like double counting, on the left side. On the right side, I think that [tex]\nu\Delta u[/tex] is some sort of friction, and [tex]\nabla p[/tex] is the force due to pressure--I don't see why, though.

But yes, there are loads of Google hits; but most of them seem to have a different form of the equations, and none that I found gave a derivation.

- #4

cristo

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There is one important difference between the viewpoint of classical mechanics, and the viewpoint taken here in fluid mechanics; it has to do with the Eulerian and Lagrangian descriptions of the motion. I don't know how much you know about this, but let's consider some scalar field f.

The Eulerian viewpoint says that we take the time derivative [itex]\partial / \partial t[/itex] holding x fixed; so that [itex]\partial f/\partial t[/itex] is the time rate of change of f(**x**,t) **at a fixed point in space, x**. This is the usual description of motion taken in CM problems.

However, in fluid mechanics, this viewpoint isn't of much use to us. It is much more useful to consider the Lagrangian description of the motion. This is to describe the motion of the fluid as it is**viewed by a given particle**. We "label" each particle with its starting postion **X**, so we can parametrise our scalar field as f(**X**,t). Now, we want to take the time derivative of the scalar field holding **X** fixed; i.e. the time derivative following the flow, say, D/Dt. To work with this description is a little awkward (conservation laws, etc. do not necessarily hold in a Lagrangion description), but it turns out it is quite straight forward to transform from a Lagrangian frame of reference to an Eulerian one, where we can then use all the familiar laws of physics. Using the chain rule, we obtain:

[tex]\frac{D f}{Dt}=\frac{\partial f}{\partial t}+\frac{d x_1}{dt}\frac{\partial f}{\partial x_1}+\frac{d x_2}{dt}\frac{\partial f}{\partial x_2}+\frac{d x_3}{dt}\frac{\partial f}{\partial x_3}= \frac{\partial f}{\partial t}+u_1\frac{\partial f}{\partial x_1}+u_2\frac{\partial f}{\partial x_2}+u_3\frac{\partial f}{\partial x_3}=\frac{\partial f}{\partial t}+(\bold{u}\cdot\nabla)f[/tex] where**x**=(x_{1},x_{2},x_{3}), and **u** is the velocity in each direction.

If we let the scalar function be temperature, then intuitively we can see why this derivative is needed, since the temperature of the fluid can change either by the temperature of the part of the room the fluid is in changing, or the particle moving to another part of the room where the temperature is different.

An analagous "derivation" hold for vector fields, and so the acceleration is [tex]\bold{a}=\frac{D\bold{u}}{Dt}=\frac{\partial \bold{u}}{\partial t}+ (\bold{u}\cdot\nabla)\bold{u}[/tex]

The Eulerian viewpoint says that we take the time derivative [itex]\partial / \partial t[/itex] holding x fixed; so that [itex]\partial f/\partial t[/itex] is the time rate of change of f(

However, in fluid mechanics, this viewpoint isn't of much use to us. It is much more useful to consider the Lagrangian description of the motion. This is to describe the motion of the fluid as it is

[tex]\frac{D f}{Dt}=\frac{\partial f}{\partial t}+\frac{d x_1}{dt}\frac{\partial f}{\partial x_1}+\frac{d x_2}{dt}\frac{\partial f}{\partial x_2}+\frac{d x_3}{dt}\frac{\partial f}{\partial x_3}= \frac{\partial f}{\partial t}+u_1\frac{\partial f}{\partial x_1}+u_2\frac{\partial f}{\partial x_2}+u_3\frac{\partial f}{\partial x_3}=\frac{\partial f}{\partial t}+(\bold{u}\cdot\nabla)f[/tex] where

If we let the scalar function be temperature, then intuitively we can see why this derivative is needed, since the temperature of the fluid can change either by the temperature of the part of the room the fluid is in changing, or the particle moving to another part of the room where the temperature is different.

An analagous "derivation" hold for vector fields, and so the acceleration is [tex]\bold{a}=\frac{D\bold{u}}{Dt}=\frac{\partial \bold{u}}{\partial t}+ (\bold{u}\cdot\nabla)\bold{u}[/tex]

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- #5

cristo

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Now, a precise derivation of the N-S equations is quite long, and would require a lot of tex! But the basic idea is thus. We write an expression for the momentum of a fluid contained in a certain volume. We note that there are two types of forces that act on the fluid; external forces such as gravity, and internal forces, which are forces exerted on the fluid by itself. We then use Newton's second law which states that the time rate of change of momentum is equal to the sum of the applied forces.

Now, from here depends on whether you want to consider a viscous fluid or not; i.e. do you want to take into account friction? The derivation of the Euler equations are easier, since there are no off diagonal terms in the stress tensor. I don't really know what level you are at, so I'll leave it here for now. Feel free to ask anything you're not unsure of, and let me know if you want to include friction or not.

Also, what sort of presentation do you need to do; i.e. is it mainly interested in the Physics or the Maths of the equations? Anyway, if you have any thoughts when reading what I've put above, feel free to ask.

Now, from here depends on whether you want to consider a viscous fluid or not; i.e. do you want to take into account friction? The derivation of the Euler equations are easier, since there are no off diagonal terms in the stress tensor. I don't really know what level you are at, so I'll leave it here for now. Feel free to ask anything you're not unsure of, and let me know if you want to include friction or not.

Also, what sort of presentation do you need to do; i.e. is it mainly interested in the Physics or the Maths of the equations? Anyway, if you have any thoughts when reading what I've put above, feel free to ask.

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- #6

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I don't know what a tensor is, but I do have an unread book on them. Maybe I should read it. A derivation of either the Euler or the Navier-Stokes equations would be fine (preferrably N-S). I do want to take into account friction, if this is all possible without tensors.

I need to do a presentation in a math class-none of the other students have had too much physics (except maybe in highschool), so I don't want to go into that in too much detail. Just a general derivation. I mean, I can introduce terms like viscosity and such, but it probably wouldn't be a good idea to go into tensors, even if I figured out what they were by the time I present. If nothing else, (if something like this works), I could just say: "the left side is the acceleration, and on the right side, [tex]f[/tex] is external forces, [tex]\nabla p[/tex] is the force due to pressure, and [tex]\nu\Delta u[/tex] is a force due to a kind of friction. The second equation just says the fluid is incompressible."

- #7

cristo

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To me, this sounds like you can't be bothered to read the book, and expect me to explain it to you! However, this may not be the case, and if no-one in your class is familiar with tensors then there is no need to use themThanks! At least now the left side makes sense.

I don't know what a tensor is, but I do have an unread book on them. Maybe I should read it.

You can derive the NS equations without tensors since, in this case, the tensor can be written as a matrix.A derivation of either the Euler or the Navier-Stokes equations would be fine (preferrably N-S). I do want to take into account friction, if this is all possible without tensors.

I need to do a presentation in a math class-none of the other students have had too much physics (except maybe in highschool), so I don't want to go into that in too much detail. Just a general derivation.

Well, in my opinion there is no such thing as a "general derivation". Especially if this is a Maths class, then a derivation is where equations are derived from first principles. I think you are after a brief(ish) explanation of why the equations come about.

I mean, I can introduce terms like viscosity and such, but it probably wouldn't be a good idea to go into tensors, even if I figured out what they were by the time I present. If nothing else, (if something like this works), I could just say: "the left side is the acceleration, and on the right side, [tex]f[/tex] is external forces, [tex]\nabla p[/tex] is the force due to pressure, and [tex]\nu\Delta u[/tex] is a force due to a kind of friction. The second equation just says the fluid is incompressible."

What you've put in quotes here is not a derivation, or explanation, it is just a statement of the equation, and I don't think that this will do.

I would start with Newton's second law, in the form: time rate of change of momentum = sum of external forces.

From here it again depends how much maths you know. For example, have you come across triple integrals, divergence theorem?

Have you found any derivations on the internet? If you find one, and link to it, and say where your having difficulties, this will be a lot easier than me writing it out here!

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- #8

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http://www.allstar.fiu.edu/aero/Flow2.htm

For now, I have only read the bit on the continuity equation, and I don't see why, if the fluid is incompressible, the density is constant in space and time.

As for math: (though it might not matter if I did find a website), I've gone through multi. variable calculus, differential equations, and abstract algebra

Thanks for all the responses.

- #9

cristo

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I did just recently find a derivation:

http://www.allstar.fiu.edu/aero/Flow2.htm

For now, I have only read the bit on the continuity equation, and I don't see why, if the fluid is incompressible, the density is constant in space and time.

By definition, an incompressible fluid is one whose density is constant.

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