Consider the Reynolds Transport Theorem in the following form: $$ \frac{d}{dt}\int_V{\rho}\boldsymbol{v} \ dV = \int_V{\frac{\partial(\rho\boldsymbol{v})}{\partial t}} \ dV + \int_S{(\rho\boldsymbol{v})(\boldsymbol{v}\boldsymbol{n})} \ dS $$ where $\boldsymbol{n}$ is the normal vector, $\rho$ density, and $\boldsymbol{v}$ velocity. $S$ and $V$ denote the surface and volume a fixed control volume.
For a steady flow of an incompressible fluid $ \frac{d}{dt}\int_V{\rho}\boldsymbol{v} \ dV $ is zero. Since $\frac{\partial(\rho\boldsymbol{v})}{\partial t} = \rho\frac{\partial\boldsymbol{v}}{\partial t}+\boldsymbol{v}\frac{\partial\rho}{\partial t} $, this terms should also be zero with the given conditions. This implies $\int_S{(\rho\boldsymbol{v})(\boldsymbol{v}\boldsymbol{n})} \ dS = 0 $. Which can not be generally true.
Where is flaw in this argumentation?
EDIT
The following example came up in the discussion: Consider a simple pipe with a 90 degree bend, constant $A$ and influx velocity $\vec{v}_{in}$ (something like this: http://thepipefitting.com/wp-content/uploads/2014/08/90-degree-elbow-drawing.jpg).
$$\int_S{(\rho\boldsymbol{v})(\boldsymbol{v}\boldsymbol{n})} \ dS = \int_{S_I} \begin{pmatrix}0\\v_{in}\end{pmatrix}\left[ \begin{pmatrix}0\\v_{in}\end{pmatrix} \begin{pmatrix}0\\-1\end{pmatrix}\right]dS_I + \int_{S_{II}}{ \begin{pmatrix}v_{out}\\0\end{pmatrix}\left[ \begin{pmatrix}v_{out}\\0\end{pmatrix} \begin{pmatrix}1\\0\end{pmatrix}\right]}dS_{II} = \rho A \begin{pmatrix}v_{out}^2\\-v_{in}^2\end{pmatrix} $$ which does not evaluate to $\begin{pmatrix}0\\0\end{pmatrix}$ since $v_{in}$ is a non-zero velocity.
