# 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

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 ...

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

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 ...

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

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

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 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

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

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) $$\partial_t ... 4 It's not that the random motion decreases when the flow rate increases. It is only that the random motion stays the same but the coherent motion dominates. If the diffusion velocity in a gas is 1 and the convective velocity of the flow is 1000 (units don't matter), then the diffusive action can be pretty safely ignored. The important thing to remember is ... 3 Let me here just derive the equation (6.11) that follows the sentence, you mention. The Navier-Stokes equation (6.6a) reads$$\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 ...

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

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

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 ...

3

First, Navier-Stokes governs the fluid in your setup. So, anything apart from the fluid will be an external force in N-S equation. Body-force means an external force that applies in the bulk of the fluid, like gravity or a magnetic force. Interaction with a "body", as a wing, which is external to the fluid domain, is done through boundary conditions : the ...

2

Your use of $\Delta T$ is confusing you, as you have it, it is not the temperature difference between the top and the bottom of the film, but the difference between the temperature $T$ at a given point in the fluid, and a reference temperature $T_0$ at which the density of the fluid is $\rho_0$. Replace $\Delta T$ with $T-T_0$ first. You then need to ...

2

Here is the answer that I gathered from months of looking at these boundary conditions: (1) and (2) would mean that the slope is zero and the bending moment / curvature at the ends is zero. (1) and (3) mean that the slope is zero and the shear stress at the end is zero.

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

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

See for example Eq. 13.74 of here: http://www.pma.caltech.edu/Courses/ph136/yr2011/1113.2.K.pdf Basically the velocity shear tensor squared and multiplied by the viscosity.

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

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

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

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

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 ...

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