# Boundary Layer Equations: Neglecting the term for x-axis momentum

1. Apr 25, 2013

### gikiian

I'm having a little difficulty in the topic 'Boundary layer Equations' in Fluid Mechanics due to my weak math skills.

With reference to the figure in attachment, if we say that "we neglect $\frac{∂^{2}u}{∂x^{2}}$", does it mean that we will only consider the portion where $\delta(x)$ is almost constant?

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2. Apr 25, 2013

### Aero51

First, what answer would you suggest and why?

3. Apr 25, 2013

Well, sort of. $\delta$ is never constant. The statement in question is a statement about the assumptions made in deriving the equaitons. When non-dimensionalizing the equations, there is a $1/Re$ term multiplying the $\partial^2 u/\partial x^2$ term in the $x$-momentum equation. Since the boundary layer equations are derived under the assumption that $Re \to \infty$ and are therefore only valid in that region, that $\partial^2 u/\partial x^2$ term is small compared to the $\partial^2 u/\partial y^2$ term.

Of course, for the same reason, the boundary layer equations aren't valid near $x=0$ anyway since the Reynolds number is very small there and the assumptions in deriving the equations are no longer true. In this sense, your statement is nearly true. There is never (in general) a truly parallel flow region where $\delta$ is constant, but it is true that in areas of very high $d\delta/dx$ the boundary layer equations are not valid.

4. Apr 25, 2013

### gikiian

My answer is that we should consider the region before x=a. This is because if $\frac{∂^{2}u}{∂x^{2}}$ is equal to zero, then $\frac{∂u}{∂x}$ will generally be a constant, and hence $u$ will vary linearly with x. If that happens, then boundary layer thickness δ(x) will vary (because it is defined where $u$=0.99$u_{∞}$ which depends on $\frac{∂u}{∂y}$).

Last edited: Apr 25, 2013
5. Apr 25, 2013

### gikiian

My book[1] doesn't mention an assumption of $Re→∞$. However, what it says is something like: The boundary layer thicknesses are typically very small relative to the size of the object upon which they form, and the x-direction velocity must change from their surface to their free-stream values over this very small distance. Therefore, gradient normal to the surface is much larger than one along the surface. As a result, we can neglect terms that represent x-direction diffusion of momentum.

[1] Fundamentals of Heat and Mass Transfer, 6th Edition, by Frank P. Incropera, p365

Last edited: Apr 25, 2013
6. Apr 25, 2013

So, let's look at where these equations come from. You start with the two-dimensional Navier-Stokes equations,

$$\dfrac{\partial u^*}{\partial t^*} + u^*\dfrac{\partial u^*}{\partial x^*} + v^*\dfrac{\partial u^*}{\partial y^*} = -\dfrac{1}{\rho^*}\dfrac{\partial p^*}{\partial x^*} + \nu^* \left( \dfrac{\partial^2 u^*}{\partial x^{*2}} + \dfrac{\partial^2 u^*}{\partial y^{*2}} \right),$$
$$\dfrac{\partial v^*}{\partial t^*} + u^*\dfrac{\partial v^*}{\partial x^*} + v^*\dfrac{\partial v^*}{\partial y^*} = -\dfrac{1}{\rho^*}\dfrac{\partial p^*}{\partial y^*} + \nu^* \left( \dfrac{\partial^2 v^*}{\partial x^{*2}} + \dfrac{\partial^2 v^*}{\partial y^{*2}} \right),$$
$$\dfrac{\partial u^*}{\partial x^*} + \dfrac{\partial v^*}{\partial y^*} = 0.$$

Here the starred quantities denote dimensional quantities. We then want to nondimensionalize the equations. To do this, we use the following scaling laws:
$$x = \dfrac{x^*}{L^*},$$
$$y = \dfrac{x^*}{\delta^*},$$
$$u = \dfrac{u^*}{U^*},$$
$$x = \dfrac{v^*}{U^*}\dfrac{L^*}{\delta^*},$$
$$p = \dfrac{p^*}{\rho^* U^{*2}},$$
$$t = t^*\dfrac{U^*}{L^*}.$$

Here, $L^*$ is simply the dimensional horizontal length scale, $\delta^*$ is the dimensional boundary layer thickness (as opposed to displacement thickness) and $U^*$ is the dimensional free-stream velocity. So, if you then substitute these scales into the governing equations, we get the following:
$$\dfrac{\partial u}{\partial t} + u\dfrac{\partial u}{\partial x} + v\dfrac{\partial u}{\partial y} = - \dfrac{\partial p}{\partial x} + \dfrac{\nu}{UL}\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\nu}{UL}\dfrac{L^2}{\delta^2}\dfrac{\partial^2 u}{\partial y^2},$$
$$\dfrac{\partial v}{\partial t} + u\dfrac{\partial v}{\partial x} + v\dfrac{\partial v}{\partial y} = - \dfrac{L^2}{\delta^2}\dfrac{\partial p}{\partial y} + \dfrac{\nu}{UL}\dfrac{\partial^2 v}{\partial x^2} + \dfrac{\nu}{UL}\dfrac{L^2}{\delta^2}\dfrac{\partial^2 v}{\partial y^2},$$
$$\dfrac{\partial u}{\partial x} + \dfrac{\partial v}{\partial y}=0.$$

Then of course the definition of the Reynolds number is
$$Re = \dfrac{UL}{\nu}.$$

This leaves
$$\dfrac{\partial u}{\partial t} + u\dfrac{\partial u}{\partial x} + v\dfrac{\partial u}{\partial y} = - \dfrac{\partial p}{\partial x} + \dfrac{1}{Re}\dfrac{\partial^2 u}{\partial x^2} + \dfrac{1}{Re}\dfrac{L^2}{\delta^2}\dfrac{\partial^2 u}{\partial y^2},$$
$$\dfrac{\partial v}{\partial t} + u\dfrac{\partial v}{\partial x} + v\dfrac{\partial v}{\partial y} = - \dfrac{L^2}{\delta^2}\dfrac{\partial p}{\partial y} + \dfrac{1}{Re}\dfrac{\partial^2 v}{\partial x^2} + \dfrac{1}{Re}\dfrac{L^2}{\delta^2}\dfrac{\partial^2 v}{\partial y^2},$$
$$\dfrac{\partial u}{\partial x} + \dfrac{\partial v}{\partial y}=0.$$

Now, in talking about boundary layers, the assumption is always that the boundary layer, or the region where viscosity is important, is confined to a small region near the surface of the object. Alternatively, this means that the boundary layer is thin compared to the size of the object, or $\delta/L \ll 1$. So, looking at the first equation, we should realize that if we assume that $1/Re = O(1)$, then the $\partial^2 u/\partial x^2$ term would be extremely large and it would render the equation quite useless. By contrast, if we assume $1/Re \ll 1$ then we can say that
$$\dfrac{1}{Re}\dfrac{L^2}{\delta^2} = O(1).$$

In fact, the condition that $1/Re \ll 1$ (or alternatively $Re \to \infty$) is typically considered to be equivalent to saying that a boundary layer exists very close to the surface outside of which the flow is inviscid. Anyway, since we have now determined that this assumption is valid, we can say that
$$\delta = O(Re^{-1/2}L),$$
which gives you an idea of how the boundary layer thickness grows with respect to $x$.

It also lets us further simplify the equations by noting the O(1) terms. We also multiply the second equation by $1/Re$ in order to make sure we keep the pressure term:
$$\dfrac{\partial u}{\partial t} + u\dfrac{\partial u}{\partial x} + v\dfrac{\partial u}{\partial y} = - \dfrac{\partial p}{\partial x} + \dfrac{1}{Re}\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2},$$
$$\dfrac{1}{Re}\left(\dfrac{\partial v}{\partial t} + u\dfrac{\partial v}{\partial x} + v\dfrac{\partial v}{\partial y}\right) = - \dfrac{\partial p}{\partial y} + \dfrac{1}{Re^2}\dfrac{\partial^2 v}{\partial x^2} + \dfrac{1}{Re}\dfrac{\partial^2 v}{\partial y^2},$$
$$\dfrac{\partial u}{\partial x} + \dfrac{\partial v}{\partial y}=0.$$

Under the assumption that the Reynolds number is very large and therefore that there is a boundary layer in the traditional sense, that allows us to make the following simplifications to arrive at what are typically called simply the boundary layer equations:
$$\dfrac{\partial u}{\partial t} + u\dfrac{\partial u}{\partial x} + v\dfrac{\partial u}{\partial y} = - \dfrac{\partial p}{\partial x} + \dfrac{\partial^2 u}{\partial y^2},$$
$$- \dfrac{\partial p}{\partial y} = 0,$$
$$\dfrac{\partial u}{\partial x} + \dfrac{\partial v}{\partial y}=0.$$

Note that the second equation is the root of the reason why we typically say that the pressure is constant through the boundary layer normal to the surface. If you plug the scaling laws back in to re-dimensionalize the equations, you get the following:
$$\dfrac{\partial u^*}{\partial t^*} + u^*\dfrac{\partial u^*}{\partial x^*} + v^*\dfrac{\partial u^*}{\partial y^*} = - \dfrac{1}{\rho^*}\dfrac{\partial p^*}{\partial x^*} + \nu^*\dfrac{\partial^2 u^*}{\partial y^{*2}},$$
$$-\dfrac{1}{\rho^*}\dfrac{\partial p^*}{\partial y^*} = 0,$$
$$\dfrac{\partial u^*}{\partial x^*} + \dfrac{\partial v^*}{\partial y^*}=0.$$

So that is the typical way to do the order-of-magnitude analysis, and personally I find it more clear as to why you drop that term than the more vague physical reasoning. The end result is that, as I said before, the equations are not valid near the leading edge since the Reynolds number is not large in that region and it violates the order-of-magnitude assumptions made in deriving the equations. As a result, the equations are at their most accurate when $\delta$ is nearly constant, but the equations are valid at least to some degree in areas where it is not nearly constant. How far it is valid depends on how much error you wish to incur, otherwise you have to turn to the full Navier-Stokes equations ore another approximation based on less stringent assumptions.

7. Apr 26, 2013

### gikiian

But we do consider the variation of $\frac{∂u}{∂y}$ (by considering $\frac{∂^{2}u}{∂y^{2}}≠0$), which means the boundary layer thickness is changing throughout. But according to the equations, we do not consider the value of $\frac{∂^{2}u}{∂y^{2}}$ when $Re→∞$, which leads to a constant $\frac{∂u}{∂y}$.

So I'm one step closer to clarity!

8. Apr 26, 2013

No, it is $\partial^2 u/\partial x^2$ that we don't consider when $Re \to \infty$ because the term is multiplied by the $1/Re$ term, which goes to zero. $\partial^2 u/\partial y^2$ is important regardless since it is $O(1)$.

The boundary layer itself is most certainly growing, though that isn't neceesarily due to the fact that $\partial^2 u/\partial y^2 \neq 0$. While it would be unusual, you could certainly find a way to vary $\partial u/\partial y$ without changing $\delta$.

Interestingly enough, we earlier showed that in order to have valid equations, $\delta = O(Re^{-1/2}L)$. From this, by substituting the Reynolds number in place of $L$, you can show the qualitative growth of the boundary layer, which is
$$\delta = O(Re^{1/2}).$$
This shouldn't surprise you though because if you just look at the shape that the line showing $\delta$ makes, it looks an awful lot like a square root.

9. Apr 26, 2013

### gikiian

So the variation in $∂u/∂y$ does not mean that $δ$ also varies.

10. Apr 26, 2013

It is kind of funny to think about, but the condition that $\partial^2 u/\partial y^2\neq 0$ (which, as you correctly stated, implies that $\partial u/\partial y \neq \text{const.}$) does not necessarily mean the boundary layer is or isn't growing. It does mean that it is changing shape, but it is certainly possible for the boundary layer to change shape without $\delta$ changing. In practice, $\delta$ usually changes in this case, but not always.

11. Apr 26, 2013

### gikiian

I kind of get that, but I'm still quite skeptical due to my poor math skills.

12. Apr 26, 2013

Well, just imagine a typical boundary layer profile. Imagine it as a string or something. You could hold that point we call the boundary layer edge in place and wiggle the rest of it around in whatever way you want and get what is essentially a changing $\partial^2 u/\partial y^2$ without changing $\delta$. This isn't necessarily a common occurrence in physical reality, but there is nothing in the equations that says it can't happen.