Convective derivative N-S

Question

This is probably an easy answer, but I've not been able to find it yet -

Why in some formulations of the N-S equations (for example here https://www.grc.nasa.gov/www/k-12/airplane/nseqs.html), is the $u$ term left IN the brackets (for the convective terms) and others its able to be taken out front. Its clear why the density can be factored out (ie. for incompressible flows where it doesnt change with position).

I learnt fluid mechanics using the 'engineering' approach (i.e. with a physical explanation of the gradient changing with position, hence $u$ out front), but got confused when i found the N-S equations in summation notation, and expanded them out, and i got to the form on the GRC page above. Since $u$ is changing with position, its not clear how to get the $u$ component out front or alternatively why the $u$ is left within the brackets.

If its because one of the $u$'s is constant and the other is a vector, how can the above page combine them into a $u^2$ term?

Apologies in advance for the simple question.

Answer 1

there are many ways to write down the NS equations. I think the form you are referring to is (1) $$ \rho (\frac{\partial \vec u}{\partial t} +(\vec u \cdot \vec \nabla)\vec u) = \vec f $$ (in $\vec f$ you have the usual pressure, viscous and external forces) compared to (2) $$ \frac{\partial \rho\vec u}{\partial t} +\vec \nabla \cdot (\rho \vec u \otimes\vec u) = \vec f $$ Actually, you don't need the incompressible assumption to go from one to the other, it is more general and only relies on the conservation of mass: $$ \frac{\partial \rho}{\partial t} +\vec \nabla \cdot \rho\vec u = 0 $$ which you can check directly by expanding the time derivative in (2) and injecting the conservation of mass.

In fact (2) is the direct translation of momentum conservation, which is the vector analogue of the continuity equation, which is why it is very general.

Hope this helps, and tell me if you find some mistakes.