- #1
Mr. Cosmos
- 9
- 1
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
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