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Bernoulli from Navier-Stokes

http://jsp.ds9a.nl/Jasper Spaans j@sp3r.net

Bernoulli from Navier Stokes

Although it is quite easy to derive the Bernoulli-equation ( $\frac{1}{2}\rho V^{2}+p+\rho gz=c$ ) using an infinitesimal small fluidum-element and the forces acting on it, deriving it ftrom the Navier-Stokes equation for an inviscid incompressible flow with body-forces is much more elegant and provides more insight into the usability of the Bernoulli-equation.

Starting with the Navier-Stokes equation for an incompressible, stationary flow, in vector form:

$\begin{displaymath} \vec{V}\cdot \nabla \vec{V}=\vec{f}-\frac{1}{\rho _{\infty }}\nabla p+\frac{\mu }{\rho _{\infty }}\nabla ^{2}\vec{V} \end{displaymath}$

(1)

From Linear Algebra, it is known that $\vec{V}\cdot \nabla \vec{V}=\frac{1}{2}\nabla (\vec{V}\cdot \vec{V})-\vec{V}\... ... (\nabla \times \vec{V})=\frac{1}{2}\vert V\vert^{2}-\vec{V}\times \vec{\xi }$ ; using this equation (1) changes to

$\begin{displaymath} \frac{1}{2}\nabla V^{2}+\frac{1}{\rho _{\infty }}\nabla p-\v... ...imes \vec{\xi }+\frac{\mu }{\rho _{\infty }}\nabla ^{2}\vec{V} \end{displaymath}$

(2)

Furthermore, assume there is no friction inside the flow, so $\mu =0$ and, the body forces only consist of the gravity field, $\vec{f}=\vec{g}=[0,0,g]^{T}=\nabla (-gz)$ , and putting this in (2), we get

$\begin{displaymath} \nabla (\frac{1}{2}V^{2}+\frac{p}{\rho _{\infty }}+gz)=\vec{V}\times \vec{\xi } \end{displaymath}$

(3)

Note that the part in brackets is a scalar field, and denoting it by $H(\vec{x})$ , it's time for some more linear algebra - you can verify that $dH=\nabla H\cdot d\vec{x}$ . Multiplying both sides of (3) with the inner-product of $d\vec{x}$ , we get

$\begin{eqnarray*} \nabla H\cdot d\vec{x} & = & \\ dH & = & (\vec{V}\times \vec{\xi })\cdot d\vec{x} \end{eqnarray*}$

Whenever both sides of this last equation are zero, $dH=0$ , or $H(\vec{x})$ is a constant, and $(\vec{V}\times \vec{\xi })\cdot \vec{x}=0$ , this equation (of course) holds.

Using vector algebra, we can identify 6 different situations in which the second part of this equation is zero:

Whenever $d\vec{x}$ is parallel to $\vec{V}$ - i.e., comparing points on a streamline
Whenever $d\vec{x}$ is parallel to $\vec{\xi }$ - i.e., comparing points on a vortex line
Whenever $\vec{V}=\vec{0}$ - i.e., there's no flow
Whenever $\vec{\xi }=\vec{0}$ - i.e., the vorticity in the flow is absent
Whenever $\vec{V}$ is parallel to $\vec{\xi }$
Whenever the path of $H(\vec{x})$ is in a plane which is perpendicular to $\vec{V}\times \vec{\xi }$ (a Bernoulli surface)

Thus, whenever one of these six conditions apply, $H(\vec{x})=c$ , or, written out,

$\begin{eqnarray*} \frac{1}{2}V^{2}+\frac{p}{\rho _{\infty }}+gz & = & c\Leftrightarrow \\ \frac{1}{2}\rho V^{2}+p+\rho gz & = & c \end{eqnarray*}$

holds. Note that just one of these six conditions is sufficient, so the domain of Bernoulli is much bigger than the classical 'only on a streamline' condition.

Bernoulli from Navier Stokes

Jasper Spaans 2001-05-06

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