Navier-Stokes equations: conservation of momentum

The first Navier-Stokes equation (conservation of mass) says: $\vec \nabla \cdot \vec v=0$

For a stationary flow, the l.h.s of the second equation is (conservation of momentum): $\rho \frac{D\vec v}{Dt}=\rho (\underbrace{\frac{\partial \vec v}{\partial t}}_{=0} + (\vec v\cdot \vec \nabla) \vec v)=\rho ( \underbrace{(\vec \nabla \cdot \vec v)}_{=0??} \vec v)\stackrel{??}{=}0$

I find that the l.h.s of the conservation of momentum equation is always equal to zero for a stationary field. I know this isn't true but where am I wrong in this reasoning ?

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Note that $\vec\nabla\cdot\vec v=0$ only for incompressible flow. The real first Navier-Stokes equation is the conservation of mass, $$\partial_t\rho + \vec\nabla(\rho\vec v)=0.$$ And what the stationary momentum equation gives you (since $\rho\neq0$) is $$\underbrace{(\vec v\cdot\vec\nabla)}_{=\frac\partial{\partial\vec v}}\vec v=0,$$ i.e. $\vec v$ is constant along its own direction –  Tobias Kienzler Jun 11 '13 at 13:08

Your mistake here is to assume that the multiplication $\vec v\cdot \vec \nabla$ is commutative. It is not; the dot product here is just a convenient mathematical notation. This part of the Wikipedia article on Navier-Stokes equations explains how to interpret this term.