Consider that water, a perfect incompressible fluid, flow through the section $S_B$ of a cylindrical container with section $S$ and height $H$: figure (reference: "Jean Jacques Meister, exercices, fluides, Ecole Polytechnique Fédérale de Lausanne"). This container is opened at the top (not as in the figure). I would like to apply continuity equation. For me to do so, we have to chose a volume $\Omega$ of fluid that is contained in a fixed surface $\Sigma$. For example a surface around the sound the cylinder which bottom at the bottom of the cylinder and top at $H/2$. Here, I would choose $\Sigma$ to be the container. Then, I would have : $$\int_{\Sigma}\vec{J_m} \cdot \vec{d\sigma}=\int_{\text{section B}}\rho \vec{v}\cdot \vec{d\sigma}=\rho v_BS_B$$ since no fluid enters the surface, the only thing that is happening is fluid going out of the surface through the section $B$. $\vec{v}$ is the velocity of the fluid, $\vec{J_m}$ the mass density of the fluid and $\rho$ its density. But, the solution states that : $$v_AS=v_BS_B$$ and I don't understand this. Is it possible to apply the continuity equation on a surface $\Sigma$ which surround all the considered fluid, and which is moving with the fluid? enter image description here


1 Answer 1


You indeed have to consider a fixed spatial volume $V$ contained in a set of boundaries $\Sigma$. For steady-state flow and an incompressible fluid the continuity equation degenerates to the given term but there are two fluxes over the surface that need to be considered, one entering the control volume in $S_A$ and one exiting it in $S_B$. As for the integral conservation equations the normal vectors $\vec n_i$ are assumed to be pointing outwards of the considered control volume this results in a positive and a negative flux that in sum have to cancel out as there are no sources of mass in the control volume.

$$\int\limits_\Sigma \underbrace{\vec J_m}_{\rho \vec u} \cdot d \vec n = \int\limits_{S_A} \rho \vec u \cdot d \vec n_A + \int\limits_{S_B} \rho \vec u \cdot d \vec n_B = - \rho u_A S_A + \rho u_B S_B \overset{!}{=} 0$$

That there must indeed also be an incoming flux in $S_A$ can easily be seen when considering that for a liquid column of finite height after some time the surrounding gas would enter the control volume.

  • $\begingroup$ but here you suppose that $A$ does not move ? What is the control volume that you choose ? $\endgroup$
    – Dicordi
    Jan 3, 2020 at 13:57
  • $\begingroup$ @Dicordi I assumed your spatially fixed control volume, a cylinder with height $\frac{H}{2}$, and only looked at the fluxes through all the surfaces. The surrounding walls allow no mass fluxes though it but flow might exit through the surface $S_B$ and enter through $S_A$. $\endgroup$
    – 2b-t
    Jan 3, 2020 at 13:59
  • $\begingroup$ @Dicordi If you would drain the container completely you would notice that water is flowing out of $S_B$ with an average velocity $u_B$ and the water surface would move downwards with a velocity $u_A$ (in fact no matter which control volume you take). As the fluid can't be compressed $\rho = const$ everything exiting the volume has to enter it somewhere. $\endgroup$
    – 2b-t
    Jan 3, 2020 at 14:05
  • 1
    $\begingroup$ Ok that is now clear for me ! Because we can consider that the velocity of the fluid in the cylinder is the same on a current line, the velocity of the fluid that goes in the fixed volume is $v_A$. $\endgroup$
    – Dicordi
    Jan 3, 2020 at 14:08

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