Force Produced from Fluid Motion

[Mr. Cosmos]

Dear all,
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,

$$\frac{\partial}{\partial t} \int_{V\llap{-}} \rho \vec{V} \ d V\llap{-} + \int_{\mathcal{S}} \rho \vec{V} \left(\vec{V} \cdot \hat{n} \right) \ d\mathcal{S}= \sum F_{external}$$
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 not momentum flux to or from the control volume and the net momentum flux is exactly zero. Thus, our equation reduces to,

$$\frac{\partial}{\partial t} \int_{V\llap{-}} \rho \vec{V} \ d V\llap{-} = \sum F_{external}$$
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,

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

 

[Cutter Ketch]

Why would that violate any laws? The fluid in the box has momentum, hits the wall, and the box accelerates. That's legal. If you've ever sloshed liquid back and forth in a container you can get the amplitude high enough that when you let go the whole box will actually slide back and forth on the table a few times. No laws were broken. If the sloshing box was in space the momentum of the whole system, box and water, would move at a constant rate, but the box, the defining volume, would appear to move jerkily accelerating and decelerating as the center of mass of the water sloshed back and forth relative to the center of mass of the box. The water would be experiencing net forces and accelerations, the box would be experiencing net forces and accelerations, and the whole system would still be obeying Newton's laws.

 

[Delta2]

I am not very good at fluid dynamics but anyway here is my opinion:

Your reasoning up to the final equation looks correct to me, but then when you look at that equation and try to understand what it tells you I believe you misinterpret the whole thing. It seems to me that when you look at that equation you forget the integral symbol and you interpret it like it was

## \frac{\partial \vec{V}}{\partial t} = \frac{1}{\rho} \sum F_{external}##.

The integral makes a huge difference, it tell you that when we don't have external forces, whatever unsteady motion exists inside the volume, it will be canceled by an equal and opposite unsteady motion so that the integral(sum) of the two motion is zero as zero is the external force.

(I hope this is helpful but I doubt it, I didn't sleep so well so I might as well don't know what I am talking about).

 

[A.T.]

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

No, it doesn't imply that. Non-zero vectors can still sum up to zero.

 

[Mr. Cosmos]

No, it doesn't imply that. Non-zero vectors can still sum up to zero.

But it is certainly possible to have a flow situation in a closed volume where the velocity vectors do not sum to zero.

 

[Nidum]

There can certainly be flow circulation in a closed box with no clearance volume .

There can also be pressure wave oscillation .

Whether there is an active flow circulation or pressure wave oscillation depends on initial conditions .

In the absence of viscous losses the circulation or pressure wave oscillation could theoretically go on forever .

 

[A.T.]

But it is certainly possible to have a flow situation in a closed volume where the velocity vectors do not sum to zero.

The formula is not integrating velocity vectors, but accelerations. If the acceleration integral is non-zero, then there must be a non-zero net external force. If the integral is zero, then the net external force is also zero.