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This might be an incredibly simple question, but I haven't been able to figure it out on my own. I'm an engineer by trade, so please forgive my unfamiliarity with vector calculus. I'm interested in the inviscid, incompressible Navier-Stokes equation and how it relates to the Bernoulli equation. The form of the NS equation I'm interested in is as follows:

$$ \frac{\partial {\mathbf u} }{\partial t} + ({\mathbf u} \cdot \nabla ){\mathbf u} = -\frac{1}{\rho} \nabla p + {\mathbf g} $$

which can relatively easily be manipulated into the Bernoulli equation if the flow is assumed to be steady-state.

$$ \frac{p}{\rho} + \frac{1}{2}\| {\mathbf u} \| ^2 +gh = const. $$

The convective term of NS $({\mathbf u} \cdot \nabla ){\mathbf u}$ turns into the velocity term in Bernoulli, $\frac{1}{2}\| {\mathbf u} \| ^2$. This is obviously a very important part of the Bernoulli equation. However, if the fluid is incompressible, we have the continuity equation which states that $\nabla \cdot {\mathbf u}=0$. Since the dot product is commutative, doesn't this imply that $({\mathbf u} \cdot \nabla ){\mathbf u} = {\mathbf 0}$? Obviously this is not the case, I am just wondering where I am going wrong, since vector calc is not my strength.

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Since the dot product is commutative, doesn't this imply that $(\mathbf u\cdot\nabla)\mathbf u=0$?

No because $\nabla$ is an operator that acts on the object to its right. The term in the parenthesis is equal to, $$ \mathbf u\cdot\nabla=u_x\partial_x+u_y\partial _y+u_z\partial_z $$ and this is applied to each term in the vector $\mathbf u$, $$\left(\mathbf u\cdot\nabla\right)\mathbf u=\left(u_x\partial_x+u_y\partial_y+u_z\partial_x\right)\mathbf u=\left(\begin{array}{c}u_x\partial_xu_x+u_y\partial_yu_x+u_z\partial_xu_x \\ u_x\partial_xu_y+u_y\partial_yu_y+u_z\partial_xu_y \\ u_x\partial_xu_z+u_y\partial_yu_z+u_z\partial_xu_z\end{array}\right)$$

See also,

And probably others on this site.

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