Bernoulli from Navier-Stokes

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Bernoulli from Navier Stokes

Although it is quite easy to derive the Bernoulli-equation ( ) 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:

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

From Linear Algebra, it is known that ; using this equation (1) changes to

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

Furthermore, assume there is no friction inside the flow, so and, the body forces only consist of the gravity field, , and putting this in (2), we get

$$\nabla (\frac{1}{2}V^{2}+\frac{p}{\rho _{\infty }}+gz)=\vec{V}\times \vec{\xi }$$ (3)

Note that the part in brackets is a scalar field, and denoting it by , it's time for some more linear algebra - you can verify that . Multiplying both sides of (3) with the inner-product of , 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, , or is a constant, and , this equation (of course) holds.

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

1. Whenever is parallel to - i.e., comparing points on a streamline
2. Whenever is parallel to - i.e., comparing points on a vortex line
3. Whenever - i.e., there's no flow
4. Whenever - i.e., the vorticity in the flow is absent
5. Whenever is parallel to
6. Whenever the path of is in a plane which is perpendicular to (a Bernoulli surface)

Thus, whenever one of these six conditions apply, , 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.