# External Forces from Fluid Motion

I have a question that has eluded explanation in fluid mechanics textbooks and even some of my colleagues. Suppose we have a general control volume. The application of linear momentum conservation will yield an equation of the form,

$$\frac{\partial}{\partial t} \int_{V\llap{-}} \rho \vec{V} \ d V\llap{-} + \int_{\mathcal{S}} \rho \vec{V} \left(\vec{V} \cdot \hat{n} \right) \ d\mathcal{S}= \sum \vec{F}_{external}$$

which simply states the summation of external forces on the control volume is equal to the time rate of change of momentum within the control volume, plus the net momentum flux across the control surfaces. Suppose our system of interest is a closed system. Meaning there is no momentum flux to or from the control volume and the net momentum flux is exactly zero. Thus, our equation reduces to,

$$\frac{\partial}{\partial t} \int_{V\llap{-}} \rho \vec{V} \ d V\llap{-} = \vec{F}_{external}$$

Now, at this point the mass in the system is fixed. Additionally, If we further assume that the fluid in question is incompressible and the volume of the system is fixed, then we obtain,

$$\int_{V\llap{-}} \frac{\partial \vec{V}}{\partial t} \ d V\llap{-} = \frac{1}{\rho} \sum \vec{F}_{external}$$

This simple expression would imply that if there were unsteady fluid motion inside the control volume, such that $\partial \vec{V}/\partial t \neq 0$, when $\vec{V}=\vec{V}(x,y,z,t)$, then there must be an observed non-zero reaction force on the the control volume. However, does this not violate Newton's 3rd law of motion? For instance, the equation would imply that some internal unsteady motion in the system will cause an external force and hence acceleration of the system. How is this possible though? So an example would be applying this equation to water sloshing around in a closed container. This equation would suggest that the unsteady momentum of the water inside the container will yield an observed external force on the container. I would love to here some feedback regarding this particular question.

Thanks

• I suppose yes, in general. For instance, consider a cylindrical container. Inside of it - and attached to it -, an electric motor (sub-system 1), which rotates some (frictionless) paddles (2), which, in turn, transfer momentum to the fluid inside (3); all axes aligned. Starting at rest, angular momentum is null, which means $\sum I_i \vec\omega_i = \vec 0$, where $I_i$ and $\omega_i$ are moment of inertia and angular velocity. So, $\omega_1$ might be nonzero (ie, observable acceleration). Make sense? Commented Nov 20, 2016 at 4:56
Whatever force is exerted by contents inside the control volume is internal, and does not count among external forces, $\textbf{F}_{external}$. The equation you have written says that if $\textbf{F}_{external}=0$, then L.H.S. of your equation must be zero (mind the if-then structure of the sentence). In other words, in the absence of momentum flux, $\textbf{F}_{external}$ is the sole cause and momentum change inside the control volume is the consequence. This means that if $\textbf{F}_{external}=0$, then in the absence of momentum flux, $\int_V\frac{\partial \textbf{u}}{\partial t}=0$. Even though it is possible that $\frac{\partial \textbf{u}}{\partial t}\neq 0$ at every point inside the control volume, its vector sum (in this case integral) over all points in the control volume $V$ can turn out to be zero,and in fact momentum conservation assures you that it will turn out to be zero (in this particular case).
@mmagos has stated an excellent example from real life. Consider egg's surface as your control-surface, so that everything interior to it becomes your control-volume. When the egg is initially spun, it has a certain momentum, and the rate at which it is being lost equals $\textbf{F}_{friction}$. When the egg is stopped by hand, its total momentum does not become zero immediately, because its internal contents are still in motion. But the internal contents are certainly losing momentum, and the rate at which this is happening equals $\textbf{F}_{hand}$. Friction force between egg shell and the surface on which it rests, $\textbf{F}_{friction}$, is zero right now. If you remove the hand before egg has lost all its momentum, then depending on how much momentum it has, and the magnitude of friction force, it may begin spinning again, losing its momentum at the rate $\textbf{F}_{friction}$ again.