# Bernoulli and N-S

1. Oct 31, 2012

### keng

hi all, i'm little confuse about the relation of these two equation.
is it right to say that Bernoulli's equation is just a case(incompressible,inviscid,steady) of navier stoke equation?

2. Oct 31, 2012

### Studiot

No the Bernoulli equation is an equation of energy conservation.

Navier Stokes is an equation of motion - momentum or forces depending upon format.

3. Nov 1, 2012

### cjl

The Bernoulli equation can be directly derived from the Navier Stokes relations, so the two are definitely closely related. A number of derivations can be found online, depending on your preferred form of the N-S equations (and whether or not you are comfortable with tensor notation).

4. Nov 3, 2012

### MrMatt2532

Hello,

If you start with the navier-stokes equation and assume steady state (drop ∂/∂t terms), inviscid flow (drop term with μ), and integrate over a streamline with density constant you will arrive at the bernoulli equation.

5. Nov 3, 2012

### Studiot

And how do you guys get the navier stokes equations in the first place?

6. Nov 3, 2012

Using Reynolds' transport theorem would be the easiest way to go about it or just a straight integral control volume analysis, which is intimately related to RTT.

At any rate, to the OP:

If you start with the Navier-Stokes equations and assume the flow to be inviscid, you get the Euler equation
$$\dfrac{\partial \rho}{\partial t} + \vec{V}\cdot\nabla\vec{V} = -\dfrac{1}{\rho}\nabla p$$

If you take the streamwise component of this equation, you get
$$u\dfrac{\partial u}{\partial s} = -\dfrac{1}{\rho}\dfrac{\partial p}{\partial s}$$

Integrating this along a streamline
$$\int\left(u\dfrac{\partial u}{\partial s} + \dfrac{1}{\rho}\dfrac{\partial p}{\partial s}\right) = 0$$

$$\dfrac{\partial}{\partial s}\int\left(u\;d u + \dfrac{d p}{\rho}\right) = 0$$

$$\dfrac{\partial}{\partial s}\left(\dfrac{u^2}{2} + \int\dfrac{d p}{\rho}\right) = 0$$

which is a statement of Bernoulli's principle where the integrand is constant over a streamline.

Last edited: Nov 3, 2012
7. Nov 29, 2012

### Melchizedek

I was solving N-S using SIMPLER algorithm and found that there is term u*dv/dy, where u=viscosity, v=vertical component of velocity, is u=0 for this term, since there won't be stress perpendicular to the surface. Need help

8. Nov 29, 2012

You really ought to start your own thread instead of hijacking one. At any rate, unless you have a good reason for doing so, you can't just call viscosity zero.

9. Nov 29, 2012

### Melchizedek

Thanks man! Below is something that I know about Bernoulli and N-S

Bernoulli and N-S are special cases of transport Equations. When the flow is steady and inviscid, the 'Energy(enthalpy)-Transport Equation' reduces to Bernoulli, i.e. Total energy is conserved. The Momentum-Transport Equation is called the N-S Equation, it is actually not a special, they are well known as Momentum Equation.