# Steady state in fluid mechanics

1. Oct 3, 2013

### Nikitin

Hey! When a stream is steady-state, you can cancel the acceleration term in navier-stokes equation, right?

so:

$$\rho \vec{a} = 0 = - \nabla P + \rho \vec{g} + \mu \nabla ^2 \vec{V}$$

But there are many terms in the total acceleration which are not dependant on time! The acceleration term in navier stokes can be broken down into https://scontent-b-ams.xx.fbcdn.net/hphotos-prn2/1379985_10201506247874738_1799136895_n.jpg: Why do the acceleration terms with regards to distance (dx, dy, dz) disappear also? This makes no sense to me.

2. Oct 3, 2013

### dauto

Your premiss is incorrect. the acceleration won't necessarily be zero for a steady state flow.

3. Oct 3, 2013

### Nikitin

At 02:03 or so the professor says the flow is steady-state and removes the acceleration term.

Last edited by a moderator: Sep 25, 2014
4. Oct 3, 2013

Yes the $\frac{\partial \vec{V}}{\partial t}$ term will be zero on account of it is a time dependent quantity and is defined as zero in a steady-state flow. It is an acceleration term and it goes to zero, so in that sense you are right. But the larger "acceleration term" you are trying to zero out is the "convective acceleration" and is not necessarily zero.

5. Oct 3, 2013

### Staff: Mentor

Even for a steady state flow, an observer moving with the flow velocity along a streamline can be accelerating (Lagrangian frame of reference). His velocity vector is changing with time. The acceleration equation you wrote down (which is the so called Material Derivative of the velocity) determines his acceleration quantitatively.

Chet

6. Oct 5, 2013

### Nikitin

Okay, but why did the professor in the video in post #3 just say "steady state" and zero-out the
$$\frac{d\vec{V}}{dt}$$ term then? He said nothing about assuming the convective acceleration is zero..

EDIT: AND OOPS, I see I made a typing mistake in the picture in the OP. The velocity differentiated with regards to Z should be multiplied by w.

7. Oct 5, 2013

Because $\frac{d\vec{V}}{dt}$ is not the same as $\frac{D\vec{V}}{Dt}$. For a steady state flow, all $\frac{d}{dt}$ terms are zero, but not all $\frac{D}{Dt}$ terms are necessarily zero.

8. Oct 5, 2013

### Nikitin

oh crap. I just noticed that the professor wrote the PARTIAL derivative of u with respect to time, NOT the total derivative, as his acceleration term. How careless of me.

When I watched it I thought the professor wrote up the complete navier-stokes equation, but apparently he zeroed all the convective acceleration terms.

Okay, thanks everybody.