Determine particle position knowing the velocity field

In summary, the conversation discusses the process of solving the Navier-Stokes equations in an Eulerian description and using particles to characterize fluid flow. The transformation between Euler and Lagrange coordinates is also mentioned, along with the need to distinguish between partial time derivatives and material time derivatives. The conversation also touches on the concept of solving an initial-value problem to determine the position of a particle at different time steps. An example of using an analytical solution for the velocity field is given to illustrate the process.
  • #1
Zoli
20
0
Dear all,

I solved the Navier-Stokes equations in Eulerian description. I would like to illustrate it as follows:
I thought to place particles in the domain which will characterize the fluid flow. However I must know the particle position in the Lagrangian specification. As I place the particles at specified positions at t=0, I know the initial state. But how to step further?

Thanks,
Zoli
 
Physics news on Phys.org
  • #2
The idea is to integrate the velocity of the particle u(x0) for a time step. Then the particle has reached a new position (x1) where it has a new velocity u(x1), and so you integrate that velocity, and so on…

Many fancier schemes abound for doing this estimation more accurately, but that's the basic idea.
 
  • #3
This is given by the transformation between Euler and Lagrange coordinates:
[tex]\vec{r}=\vec{r}(t,\vec{R}).[/tex]
It describes the motion of a fluid element that was at the initial time, [itex]t=0[/itex] at position [itex]\vec{R}[/itex]. This mapping between [itex]\vec{r}[/itex] and [itex]\vec{R}[/itex] is invertible as long as there are no singularities like shock waves etc. Then you can express every quantity in both the Lagrange coordinates [itex]\vec{R}[/itex] and Euler coordinates [itex]\vec{r}[/itex]. E.g., the velocity of a fluid element is given by
[tex]\vec{v}(t,\vec{R})=\partial_t \vec{r}(t,\vec{R}).[/tex]
You can as well take the velocity field as function of the Euler coordinates. Then the relation is
[tex]\vec{v}(t,\vec{R})=\vec{v}[t,\vec{r}(t,\vec{R})] \equiv \vec{v}(t,\vec{r}).[/tex]
Then you have to distinguish between partial time derivatives (at fixed Euler coordinates [itex]\vec{r}[/itex]) and "material time derivatives (time derivatives at fixed Lagrange coordinates [itex]\vec{R}[/itex]. E.g., for the equations of motion (Euler perfect fluid or viscous Navier Stokes etc) you need the acceleration of a given fluid element, i.e., the material time derivative
[tex]\mathrm{D}_t \vec{v}=\frac{\partial}{\partial t} \vec{v}(t,\vec{r})+\frac{\partial \vec{r}(t,\vec{R})}{\partial t} \cdot \vec{\nabla}_{\vec{r}} \vec{v}(t,\vec{r})=\frac{\partial}{\partial t} \vec{v}(t,\vec{r})+(\vec{v} \cdot \vec{\nabla}_r) \vec{v}(t,\vec{r}).[/tex]

If you have the velocity field in Euler coordinates, [itex]\vec{v}(t,\vec{r})[/itex], you can thus simply calculate the trajectories of the fluid elements by solving the set of coupled differential equations as an initial-value problem:
[tex]\frac{\mathrm{d} \vec{r}}{\mathrm{d} t}=\vec{v}(t,\vec{r}), \quad \vec{r}(t=0)=\vec{R}[/tex]
This gives you the Euler coordinates in terms of the Lagrange coordinates, [itex]\vec{r}=\vec{r}(t,\vec{R}).[/itex]
 
  • #4
So if I am not mistaken I should do the following for each particle:
1. Solve the initial-value problem to gain [itex]\vec{r}(t,\vec{R})[/itex] at different time steps. For this I must solve an IVP consisting of 2 equations because of the 2D-problem. But I have [itex]\vec{v}[/itex] at discrete points at each time.
2. How to continue?

Could you please specify algorithmically how to solve it? An example would help me a lot. E.g. let us regard I have the analytical solution [itex]\vec{v} = e^{-t}\cos(x)\sin(y)[/itex] on [itex][-1,1]\times[-1,1][/itex] in Euler coordinates given at [itex](x_1,y_1),\ldots, (x_N,y_N)[/itex] at [itex]t_0,t_1,\ldots,t_M[/itex]. How could I determine the position of the particle (that is [itex]\vec{R}(t)[/itex]) starting at [itex]\vec{R}(t=0)=(0,0)[/itex]?
 
  • #5
vanhees71 said:
This is given by the transformation between Euler and Lagrange coordinates:
[tex]\vec{r}=\vec{r}(t,\vec{R}).[/tex]
It describes the motion of a fluid element that was at the initial time, [itex]t=0[/itex] at position [itex]\vec{R}[/itex]. This mapping between [itex]\vec{r}[/itex] and [itex]\vec{R}[/itex] is invertible as long as there are no singularities like shock waves etc. Then you can express every quantity in both the Lagrange coordinates [itex]\vec{R}[/itex] and Euler coordinates [itex]\vec{r}[/itex]. E.g., the velocity of a fluid element is given by
[tex]\vec{v}(t,\vec{R})=\partial_t \vec{r}(t,\vec{R}).[/tex]
You can as well take the velocity field as function of the Euler coordinates. Then the relation is
[tex]\vec{v}(t,\vec{R})=\vec{v}[t,\vec{r}(t,\vec{R})] \equiv \vec{v}(t,\vec{r}).[/tex]
Then you have to distinguish between partial time derivatives (at fixed Euler coordinates [itex]\vec{r}[/itex]) and "material time derivatives (time derivatives at fixed Lagrange coordinates [itex]\vec{R}[/itex]. E.g., for the equations of motion (Euler perfect fluid or viscous Navier Stokes etc) you need the acceleration of a given fluid element, i.e., the material time derivative
[tex]\mathrm{D}_t \vec{v}=\frac{\partial}{\partial t} \vec{v}(t,\vec{r})+\frac{\partial \vec{r}(t,\vec{R})}{\partial t} \cdot \vec{\nabla}_{\vec{r}} \vec{v}(t,\vec{r})=\frac{\partial}{\partial t} \vec{v}(t,\vec{r})+(\vec{v} \cdot \vec{\nabla}_r) \vec{v}(t,\vec{r}).[/tex]

If you have the velocity field in Euler coordinates, [itex]\vec{v}(t,\vec{r})[/itex], you can thus simply calculate the trajectories of the fluid elements by solving the set of coupled differential equations as an initial-value problem:
[tex]\frac{\mathrm{d} \vec{r}}{\mathrm{d} t}=\vec{v}(t,\vec{r}), \quad \vec{r}(t=0)=\vec{R}[/tex]
This gives you the Euler coordinates in terms of the Lagrange coordinates, [itex]\vec{r}=\vec{r}(t,\vec{R}).[/itex]

Using this idea, I managed to perform what I wanted. Thank you!
 

What is the significance of determining particle position based on the velocity field?

Determining particle position based on the velocity field allows scientists to track the movement of individual particles in a fluid or gas. This information can be used to understand the flow patterns and dynamics of the system, as well as to make predictions about future behavior.

How is the velocity field measured?

The velocity field is typically measured using a variety of techniques, such as particle image velocimetry (PIV), laser Doppler velocimetry (LDV), or hot-wire anemometry. These methods involve tracking the movement of small particles or using sensors to measure the velocity of the fluid at different points.

What factors can affect the accuracy of determining particle position from the velocity field?

There are several factors that can affect the accuracy of determining particle position from the velocity field. These include the resolution and precision of the measurement techniques, as well as any external factors that may disrupt the flow patterns, such as turbulence or obstacles in the system.

What are some applications of determining particle position from the velocity field?

Determining particle position from the velocity field has a wide range of applications in various fields, including fluid dynamics, aerodynamics, and biomedical engineering. It can be used to study the behavior of fluids in pipes and channels, the flow of air over aircraft wings, and the movement of blood cells in the body, among others.

Are there any limitations to determining particle position from the velocity field?

While determining particle position from the velocity field is a valuable tool, it does have some limitations. For example, it may not be suitable for highly complex or turbulent systems, and it may be challenging to accurately measure the position of very small particles. Additionally, the accuracy of the results may be affected by the size and shape of the particles being tracked.

Similar threads

Replies
10
Views
565
Replies
1
Views
933
Replies
2
Views
621
Replies
14
Views
2K
  • Mechanics
Replies
11
Views
2K
Replies
4
Views
859
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Mechanics
Replies
6
Views
2K
Replies
6
Views
2K
Back
Top