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?


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.

  • $\begingroup$ This is a pretty common confusion. For a detailed explanation, see this answer $\endgroup$
    – rambi
    Commented Jun 25, 2023 at 12:28

1 Answer 1


The flaw is that, in steady flow, $\frac{d}{dt}\int_V{\rho}\boldsymbol{v} \ dV$ does not necessarily have to be zero. This quantity represents the time rate of change within the moving control volume. Each subsequent moving control volume might exhibit the exact same time rate of change. As reckoned by a stationary observer, this would mean that at any given spatial location, nothing would be changing with time.

  • $\begingroup$ The control volume is defined as fixed in the question. $\endgroup$
    – Ben L
    Commented Mar 9, 2017 at 12:47
  • 1
    $\begingroup$ Reynolds Transport theorem inherently implies a control volume moving with the velocity of the fluid. $\endgroup$ Commented Mar 9, 2017 at 14:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.