Ian K said...

Ian. You can't just drop a complicated equation like that without explaining it (it drives me around the twist because I will spend hours trying to understand it). I know the mathematics, but can't relate it to it's application (i'm electronics. I'm not a mechanic).

The left equation. What does 'p' (rho) mean ?. Is this mass ?

'v' is velocity ?.. and 't' time ?

The partial differentiation of velocity I understand (acceleration), but what does the dot product of velocity and the gradient of velocity mean ?

People need to be hung for doing this

Thanks J

jn1 said...

Ian K said...

The left equation. What does 'p' (rho) mean ?. Is this mass ?

'v' is velocity ?.. and 't' time ?

The partial differentiation of velocity I understand (acceleration), but what does the dot product of velocity and the gradient of velocity mean ?

People need to be hung for doing this

Thanks J

Sorry J, yes p is the density and when the equation is based on volume units that is equal to the mass.

The left side of the equation allows the maths to follow a unit of fluid and gives its acceleration, so yes the left side is ma. The right side is the forces.

The left side is called the substantive derivative. The dot product is there because you want to multiply the x component of velocity by the derivative of the x component of velocity in the x direction, and the same in the y and z direction. But to understand it just think in one dimension for a bit. The component of velocity in the x direction is traditionally called u, but I'll use v.

You want to know the acceleration, dv/dt while following a particle of fluid that is going in the x direction, but you want to describe this given the value of the velocity specified at points in space. ie co-ordinates fixed in space rather than co-ordinates that move with the fluid particle. That's called a Lagrangian co-ordinate system rather than an Eulerian system which is generally not as easy to do calculations with.

v*(dv/dx) converts a Lagrangian specified velocity field to the acceleration of the particle. Because 1 second later the particle has moved a distance v and it's new velocity is given by multiplying that distance by the gradient wrt distance.


If the whole velocity field is changing with time you need to add the partial derivative dv/dt. Note, that being a partial derivative, it is not following the flow but at the one spot with x,y,z held constant. In steady flow the partial derivative dv/dt is zero so you leave it out.

I see. I see.

Macroscien said...

Ian K said...
The equation as written holds even in turbulent flow, but then is even more impossible to solve

Could you provide us with calculations for drag force created by our windsurfing fin say 42cm at 50 km /h ?
I only guess that level of turbulence depends of fin quality so to simplify we could assume laminar flow (unless we could work out from this equitation that
at such speed flow must be somehow turbulent).
If the topic ( fin drag) was already discussed on SB (for sure was ) it could be nice to see how N-S works in our everyday situation anyway ( and who is to blame here (N_S ))

No. Solving the Navier Stokes equation is a job for specialists.

They probably do it numerically, dividing the region around the fin into a finite-sized grid and letting a computer do a lot of number crunching. And as you say, you'll need a bit of empirical input/higher theory to get the Reynolds stress in the right ballpark. You'd need an infinitely fine grid to model turbulence explicitly and even the biggest computers aren't up to that yet as far as I know?

Really good fins might have laminar flow up until the wide point but after that the pressure gradient reverses and you'll get turbulence for sure. I recall that the Reynolds stress associated with turbulence will be much greater than the simple viscosity of laminar flow so a laminar model won't be very useful.

Here's the cue for some of the fin modelling experts out there to jump in and let us know, straight from the horse's mouth, how it's done??? And what is the lift to drag ratio of the typical 42cm fin at 50kph? I'd guess 17:1 does anyone out there know for sure?

Hi I

So, just to summarise what you said (correct me if I'm wrong).

The equation expresses a bunch of particles, but lets just look at one particle. Here is an expanded left side of the equation (sorry about my crappy pics).



The left side of the equation expands to two terms.



The first term is force (mass times acceleration) that most of us are familiar with. The 2nd term converts the force (the 1st term) from a relative reference system to an absolute reference system.

If a boat is going down a river. If I'm on the boat, my view is a relative view point (relative reference system). If I am standing on the bank of the river, then my view point is an absolute view point (absolute reference system).

I must admit Ian. My knowledge of vector geometry is basic at best. I know how they work in pictures, but I certainly don't use them day-to-day. So I'm not going to bother agonising over the 2nd term.

Okay, that makes things a bit clear with the left side, but what about the right side of the equation ?. We have a 'p', a 'mu' (ie: a funny 'u' character), a 'v' (velocity ?), and a 'f'. We have 3 terms, and more vector geometry !. Could you talk us through the right side of the equation Ian ? (one little particle, not an entire field) :)

J

You're getting there Jn1, I remember spending a lot of time trying to get the n-s equation on board. It's worth the battle, it's only f=ma for fluids.

Getting the system of reference sorted is the first step. It helps to look ahead to how this equation will be used in a computer because that is why it is expressed like it is.

If you were modeling the flow around a fin for instance you'd set up a grid of a whole lot of little boxes around the fin. As the computer program runs each little box will have two lots of data.

the pressure of the fluid in that box
the 3d velocity vector of fluid passing through that box.

For boxes that happen to be at the surface of the fin you'd constrain the velocity to zero. You'd let the computer fiddle with pressure in each of these boxes, because of course the pressure distribution over the surface of the fin is what we want.

Now back to the first term on the left hand side. The first term is the partial time derivative of the 3d velocity vector at a fixed location. ie in each box, how much does the velocity vary in the next time step of this computer program? For steady state flow the velocity will remain unchanged, and for modeling steady state flows, such as the pressure distribution on the fin at a steady 40 knots, this term will be zero. (It may be non-zero before the program settled down to give the answer, although computer programmers may have tricks to get the flow up to speed without worrying about this. Not sure)

So for steady state flow all the acceleration as the fluid particle moves is given by that 2nd term in the brackets on the LHS. The partial derivatives of velocity with respect to distance are of course easy for the computer to calculate just by looking at the neighbouring boxes.





The first term on the right is the gradient of pressure. Looking at our computer program that's making sure all the values in the boxes add up according to the n=s equation we can see if the pressure on one side of a grid box is greater than that on the other side of the the box there is a force on the fluid particle that is passing through the box at this particular instant of computer time.

The 2nd term on the right is the viscous force. The mu is the viscosity coefficient, honey vs water. If the fluid above is sliding forward more than the fluid below is dragging there will be a viscous force acting on the fluid in the box. It's the 2nd derivative because the first derivative. ie sliding forward above = drag back below it cancels out. The computer can work out the second derivative of the velocity components quite easily just by looking at the velocity data in neighboring boxes. This is where the Newtonian assumption comes in, shear stress is linearly proportional to velocity gradient.

The last force is the body force gravity, or something electrostatic, call it zero.

This view is all based on a computer doing what's called a finite difference method. You let the computer step through time in very small fractions of a second. At each time step it looks at each box in the grid and lets it interact with its neighbors according to the n-s equation. The computer should settle down and give the stable smooth flow. Mostly it crashes. These days there are more sophisticated, less intuitive ways of solving this equation in a computer

Although the Equation is for incompressible flow, incompressible flow is not ensured by the equation. There's an extra equation to look after this.

We're probably the only two still tuned into this thread jn1 At leat we're not getting red thumbs.

jn1 said...

Correct me if I'm wrong Ian. Here goes:

The left side of the equation describes a particle (I'll talk about one particle for simplicity). The ∂v/∂t term describes the acceleration of the particle in absolute reference, and right term converts it to relative reference (so it's now from the point of view of the particle's last position in time).

The ∂v/∂t term describes the time dependant variation of the velocity field at the point of interest x,y,z. For partial derivatives we hold all the other variables constant. For steady state flow this term is zero. The 2nd term on the left is all there is to the acceleration term in steady state flow. It is an addition rather than a conversion of the most often zero first term.

(For a sort of a electronic analogy check out Ampere's Law for a steady current and how it was modified by Maxwell for a time dependent current/electrical field.)

jn1 said...


Looking at the right side of the equation, the first term is the pressure (force) pushing against the particle (that is described in the left hand side of the equation) as it moves through a viscous-"less" fluid.

Note: The negative sign in the 1st term tells us it's an "against" force.

The pressure is pushing on the particle regardless of the viscosity. It is negative because the force is in the direction from higher to lower pressure.

jn1 said...

The 2nd term (viscous force) is a transformer. It converts the force caused by the viscous-less fluid described in the 1st term, to a force of a viscous fluid (as set by the viscosity coefficient - μ).

It's an additional force dependant on the rate of deformation of the fluid and the μ



You're in the ball park. You need to look at that 2nd term on the left with your partial derivatives sorted out. Let each of the 4 components vary one at a time, with the other 3 held constant.