Navier-stokes equation (fluid mechanics)

AI Thread Summary
The discussion centers on confusion regarding the Navier-Stokes equation for viscous fluid flow, particularly the term v(del squared)u, where v represents kinematic viscosity and u is the velocity field. Clarification is provided that (del squared)u can indeed apply to a vector field, as the differential operator can be used in this context. The conversation also touches on the convection acceleration term V.delV, which is a significant source of non-linearity in the equation. There is a distinction made that the original inquiry was about the viscous dissipation term rather than the convective term. Overall, the discussion emphasizes the complexities of fluid mechanics and the importance of understanding these terms in the Navier-Stokes equation.
alsey42147
Messages
22
Reaction score
0
i'm revising for my exams, and i didn't go to many of my fluids lectures, now I'm well confused. in the navier-stokes equation for viscous fluid flow, there is a term:

v(del squared)u

where v is the kinematic viscosity and u is the velocity field of the fluid. at this point in my notes, the lecturer seems to start doing crazy things which don't make sense.

first of all, its (del squared)u, not (del squared)(dot)u. i thought (del squared)u only had any meaning if u is a scalar field, but its not, its a vector field. what does this mean?
 
Physics news on Phys.org
\nabla^{2} is a differential operator that perfectly well can be applied to a vector.
 
V.delV is convection accelaration term in NSE it is the major source for non-linearity of the equation

You can work it out by

(V.del)V or V.(del V) both methods are same
 
altruistic said:
V.delV is convection accelaration term in NSE it is the major source for non-linearity of the equation

You can work it out by

(V.del)V or V.(del V) both methods are same

I believe the OP was asking about the viscous dissipation term not the convective term.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...

Similar threads

Back
Top