# Tag Info

## Hot answers tagged navier-stokes

15

Consider a standard volume of $1\textrm{ m}^3$ of air. This contains on the order of $10^{25}$ molecules of O2 and N2. If you needed to simulate or explain the physics occurring in that volume of air, would you want to model $10^{25}$ molecules and all the interactions between them or, say, 100x100x100 cells based on the Navier-Stokes equations? ...

10

Dear Ondřeji, a good question but a part of the answer is that your equation for the fluid is underdetermined. It treats $p,\rho$ as independent variables. But the physical system only knows how to behave if you also substitute some equation of state, i.e. a function $p=p(\rho)$ or $p=p(\rho,\vec v)$. Note that your Ansatz for the stress-energy tensor ...

9

Actually, there are two different viscosity coefficients. You can see this from the stress tensor $$\sigma_{ij} = -p_0 \delta_{ij} + \eta \left( \frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i} - \frac{2}{3} \delta_{ij} \frac{\partial v_k}{\partial x_k} \right) + \zeta \delta_{ij} \frac{\partial v_k}{\partial x_k}$$ which has the two ...

5

I belive you have it pretty much settled already. If I was to change anything, I would shrink instead of adding more items: Identify the relevant quantities of your system: Energy, Momentum, entropy, electric charge, mass ... Which may or may not be conserved. If you have boundary conditions, most probably you don't have energy and/or momentum ...

5

Frank White's Viscous Fluid Flow book contains a good list of these "exact" solutions. I am not sure if it is complete though. I've provided links to a few of the solutions. Steady flow between a fixed and moving plate Axially moving concentric cylinders Flow between rotating concentric cylinders Hagan-Poiseuille flow Combined Couette-Poiseuille flow ...

5

Strictly speaking, turbulence doesn't exist in two dimensions. The energy cascade required for turbulence to develop (transfer energy from large scales to small scales) is due to the (incompressible for illustration) vorticity equation: $\frac{D\vec{\omega}}{Dt} = \left(\vec{\omega}\cdot\nabla\right)\vec{v} + \nu\nabla^2\vec{\omega}$ specifically the ...

4

I don't think that such a computation of a theoretical limit of accuracy is possible. There are several sources of uncertainty in weather models: initial and boundary data, parameterizations, numerical instability, rounding and approximation errors of the numerical scheme employed to solve the Navier-Stokes equations for the atmosphere. The term ...

4

Solutions of the form $$cos(x_i)e^{-x_j}$$ are common specific solutions of the Navier-Stokes equations in simplified (not simple) problems. These are however problems where inertia is ignored, which you include. (Please note that I am using index notation, with $i,j\in\{1,2,3\}$). $x_j$ is then the wall normal direction. This is actually quite well ...

4

I don't know a good answer to your first question (I'd be interested in a good text for that myself), but I can answer the second. It's easier to explain if we temporarily imagine $\phi$ represents the concentration of some dye made up of little particles suspended in the fluid. The convective term (aka advective term) is transport of $\phi$ due to the ...

4

Turbulence is indeed an unsolved problem both in physics and mathematics. Whether it is the "greatest" might be argued but for lack of good metrics probably for a long time. Why it is an unsolved problem from a mathematical point of view read Terry Tao (Fields medal) here : ...

4

It's not that the random motion decreases when the flow rate increases. It is only that the random motion stays the same but the coherent motion dominates. If the diffusion velocity in a gas is 1 and the convective velocity of the flow is 1000 (units don't matter), then the diffusive action can be pretty safely ignored. The important thing to remember is ...

3

Turbulence is not one of the great unsolved problems in physics. Physics tells us exactly how turbulence emerges as a direct consequence of local mass and momentum conservation. We can create multiparticle computer models such as lattice gas automata that generate turbulence at large length and time scales. We can write down the equations that govern ...

3

My answer will focus just on the mathematical parts pertaining to partial differential equations. Scale invariance is the fact that some partial differential equations stay the same if you appropriately scale the variables. For example the heat equation (where $\boldsymbol{x}$ is the position vector in 1, 2 or 3D, doesn't matter) $$\partial_t ... 3 I am not sure how useful this "back of the envelope" calculation of reliability of Numerical Weather Prediction is going to be. Several of the assumptions in the question are not correct, and there are other factors to consider. Here are some correcting points: The Weather is 3 dimensional and resides on the surface of the planet up to a height of at ... 3 For Newtonian fluids (such as water and air), the viscous stress tensor, T_{ij}, is proportional to the rate of deformation tensor, D_{ij}:$$D_{ij} = \frac{1}{2}\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right)T_{ij} = \lambda\Delta\delta_{ij} + 2\mu D_{ij}$$where \Delta \equiv D_{11} + D_{22} + D_{33}. The ... 2 Let me here just derive the equation (6.11) that follows the sentence, you mention. The Navier-Stokes equation (6.6a) reads$$\partial_t v_i + v_j\partial_j v_i = -\partial_i p + f_i + \nu~\partial_j\partial_j v_i.$$The incompressibility condition (6.6b) reads$$\partial_jv_j=0. $$Hence we have in the unprimed and the primed point that$$\partial_t ...

2

There is something fishy about your result: all your quantities have a $e^{-\frac{1}{2}|t|}$ factor in them. So the movement those equations describe starts off at $t=-\infty$ completely stationary and with no pressure gradient, starts moving without an external force, and eventually dies out when $t=\infty$. That doesn't seem very consistent with what one ...

2

The Navier-Stokes equation to which you refer is more generally the first moment of velocity of the Boltzmann equation. In order to get a proper connection to heating, you need a second-velocity-moment Navier-Stokes equation. The Boltzmann equation keeps track of distributions of particles. This changes the question from "What is the density and flow of a ...

2

As I understand it from the statements in sections A.1 and A.2, this behaviour is due to the appearance of the constants in equation A.5. The equation is solved to linear order, i.e. for small perturbations of the solution. The terms in A.5 have to remain small in order for the solution to be valid. In those expressions, the Reynolds number enters in the ...

2

$(u \cdot \nabla)u$ is the so called advective acceleration term which arises when you consider the Navier-Stokes equations in an Eulerian frame of reference. It accounts for the effect that the we are following the particle as it moves around in the fluid, presumably to regions of the flow where the velocity is different. In contrast, if you consider the ...

2

First of all I am going to assume from your question that the viscosity is large enough, so the flow would be laminar (if it is not the case then the best answer would be numerical simulation or estimate based on the law of the wall velocity profile). Furthermore we will also neglect surface tension effects. Since we have a stationary problem the boundary ...

2

I'm far from expert and my advice is basically to read Terry Tao's piece linked in John Rennie's comment. (Read Terry Tao is always good advice.) But I would like to make a couple of points. Singularities are not impossible or unphysical Mainly, I wanted to clear up the idea that, if we find smooth solutions that became singular in finite time that would ...

2

Yes, it is incorrect. There should be a $\frac{1}{\rho}$ multiplying the $\nabla p$ term. This form is the incompressible form where it is assumed $\rho$ is a constant. This allows it to be factored out of the derivatives on the left hand side, then both sides are divided by $\rho$. This is why there is kinematic viscosity, $\nu$ on the right and not ...

2

Most importantly, the Navier-Stokes equations are based on a continuum assumption. This means that you should be able to view the fluid as having properties like density and velocity at infinitely small points. If you look at e.g. liquid flows in nanochannels or gas flows in microchannels you could be in a regime where this assumption breaks down. As far as ...

1

You don't want $1/R$ (although technically it means the same) but rather the full curvature term: $\Delta p=\sigma \kappa$. In fact you will get a source term in the Navier-Stokes equations that looks like this: $$\sigma \kappa \delta(n) \mathbf{n}$$ where $\delta(n)$ is the Dirac Delta function that only has a value at the interface and $\mathbf{n}$ is the ...

1

Discretize the equations for conservation of matter, momentum, energy, and any other relevant physical quantity, and solve them on a computer. Increase the resolution of the simulation until the results converge. (Then think really hard about the results, supplement the detailed calculations with analytic treatments of simplified analogous cases, etc. But ...

1

To allow for heat effects in a fluid, you need to couple the Navier-Stokes equations, (momentum conservation) which BTW contain the continuity equation for mass conservation too, to the energy (or temperature) equation (energy conservation). Momentum dissipation in the momentum equation  \frac{\partial v}{\partial t} + (\vec{v}\cdot\nabla)\vec{v} = ...

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