Bernoulli & Navier-Stokes: Relation Explained

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Bernoulli's equation is derived from the Navier-Stokes equations under specific conditions: steady state, inviscid flow, and constant density. While Bernoulli's equation represents energy conservation, Navier-Stokes focuses on momentum and forces. The relationship between the two is established through integration along a streamline, leading to Bernoulli's principle. Additionally, the discussion highlights that both equations are forms of transport equations, with Bernoulli emerging from the energy-transport equation under certain assumptions. Understanding this relationship is crucial for fluid dynamics applications.
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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?
 
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No the Bernoulli equation is an equation of energy conservation.

Navier Stokes is an equation of motion - momentum or forces depending upon format.
 
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).
 
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.
 
And how do you guys get the navier stokes equations in the first place?
 
Studiot said:
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.
 
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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
 
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.
 
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.
 

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