Bernoulli & Navier-Stokes: Relation Explained

[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?

 

[Studiot]

No the Bernoulli equation is an equation of energy conservation.

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

 

[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).

 

[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.

 

[Studiot]

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

 

[boneh3ad]

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

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.

 

[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

 

[boneh3ad]

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.

 

[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.