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If a fluid flow, such as water, is incompressible then the convective derivative term in the material derivative is equal to zero. How then, in a Bernoulli flow where there is an increase in velocity at the narrowing of a pipe, is the convective derivative zero? The local derivative shouldn't matter for a flow that speeds up with respect to position, because the local derivative is time-dependent.

I know that del*v (the divergence of the velocity field) is zero for incompressible fluids. This makes sense because the divergence of the velocity field tells us the rate of change of volume per unit volume of a finite control volume. But how does this relate to things like Bernoulli's equation when fluids are speeding up or slowing down in a manner that is not time-dependent? I have really confused myself over this. Can some explain to me what these derivatives mean in such a scenario?

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    $\begingroup$ an incompressible fluid means the $\nabla \cdot \vec{u} = 0$. it does not mean that the convective term ($\vec{u} \cdot \nabla \vec{u}$) is zero. one is the divergence of $\vec{u}$ and the other is $\vec{u}$ dotted with the gradient of $\vec{u}$ $\endgroup$
    – Ragnar
    Feb 12, 2019 at 1:49
  • $\begingroup$ the dot product is commutative though. (delu)u is the same as (udel(u) ). in which case we have 0 times u. this is how I became confused originally. $\endgroup$
    – MattGeo
    Feb 12, 2019 at 1:54
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    $\begingroup$ What does it mean to take the gradient of a vector? Are you aware that this is a 2nd order tensor, and not a vector? $\endgroup$
    – Chet Miller
    Feb 12, 2019 at 13:25
  • $\begingroup$ Yes, but the dot product is commutative and associative. So you could look at it as the divergence of velocity multiplied by a vector, just the same as you could look at it as the velocity multiplied by the gradient of the velocity. They're equivalent I thought, yet in the former case, we get zero since div(V) = 0 $\endgroup$
    – MattGeo
    Feb 12, 2019 at 13:49
  • $\begingroup$ the dot product is commutative but not associative (see the wikipedia page). Therefore the two terms are not equivalent. $\endgroup$
    – Ragnar
    Feb 12, 2019 at 19:17

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