Navier-stokes simplifications

In summary, the conversation discusses the Navier-Stokes equations in a boundary layer and the question of whether \frac{\partial^2 u}{\partial x^2} can be considered insignificant due to the smallness of \delta compared to L. It is mentioned that the assumption of \frac{\partial^2 u}{\partial x^2} being 0 is valid outside of the boundary layer, but there is uncertainty about its validity within the layer.
  • #1
member 428835
hey pf!

so i have a question concerning navier-stokes equations in a boundary layer, which, as a refresher, is [tex] \frac {D \vec{V}}{Dt} = - \nabla P + \nu \nabla^2 \vec{V}[/tex] where we know the x-component of [itex] \nabla^2 \vec{V}[/itex] may be re-wrote as [itex] \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}[/itex] (i dismiss the [itex]z[/itex] component due to two dimensions). define the vertical displacement as [itex] \delta[/itex] and the horizontal length scale as [itex] L [/itex]

okay, now for the question: we know [itex]\frac{\partial^2 u}{\partial x^2}[/itex] disappears, as it is not of much importance via order of magnitude. my question is, in boundary layer analysis, is this true from the fact that [itex]\delta < L \implies \delta^2 << L^2[/itex] and thus the [itex]\frac{\partial^2 u}{\partial x^2}[/itex] component can be thought of as insignificant (from the large denominator) compared to the [itex]\frac{\partial^2 u}{\partial y^2}[/itex] component?

if so, when we leave the boundary layer are we going to assume that the double partial over x is still insignificant, or are we allowed to assume this (assuming same flow and geometry, just outside the BL)

thanks!
 
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  • #2
Uhhh, it would be nice to have a figure of what your system is or some kind of information describing what's going on in your system. You can't always say that (d^2u/dx^2) is 0. If du/dx is constant or weakly dependent on x, then you may make that assumption.

Can I assume y is the direction to and from a plate, and x is the direction along the plate? So, that would seem the most logical to me. I'm not sure if the u velocity varies linearly inside the boundary layer or not, I honestly don't remember, it's been a while. However, outside of the boundary layer you can definitely assume the second derivative in the x direction is 0. That is, assuming flow on a plate as I described.
 

1. What are Navier-Stokes simplifications?

Navier-Stokes simplifications refer to the process of reducing the complexity of the Navier-Stokes equations, which are a set of mathematical equations that describe the motion of fluids. These simplifications are often necessary in order to make the equations more manageable and to better understand the behavior of fluids.

2. Why are Navier-Stokes simplifications important?

Navier-Stokes simplifications are important because the full Navier-Stokes equations are very complex and difficult to solve. By simplifying them, we can gain a better understanding of the underlying physical principles and make predictions about fluid behavior more easily.

3. What are some common Navier-Stokes simplifications?

Some common Navier-Stokes simplifications include assuming steady-state flow (no time dependence), incompressible flow (constant density), and inviscid flow (no viscosity). These simplifications can greatly reduce the complexity of the equations and make them easier to solve.

4. What are the limitations of Navier-Stokes simplifications?

Navier-Stokes simplifications have limitations in that they may not accurately capture certain fluid behaviors. For example, assuming inviscid flow would not be appropriate for highly viscous fluids such as honey. Additionally, simplifications may result in the loss of important details and nuances in fluid behavior.

5. Are there any alternative methods to Navier-Stokes simplifications?

Yes, there are alternative methods to Navier-Stokes simplifications, such as using computational fluid dynamics (CFD) techniques. CFD allows for more accurate and detailed simulations of fluid behavior by directly solving the full Navier-Stokes equations without the need for simplifications. However, CFD requires significant computational resources and may not be suitable for all situations.

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