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From the momentum equation of fluid mechanics: $${\frac{\partial}{\partial t} \int_{V_f(t)} \rho \vec{v} \text{d}\text{V}} = \int_{\sum_f(t)} \vec{\bar{\bar{\tau}}} \cdot \vec{n} \text{d}\sigma + \int_{V_f(t)}\rho \vec{f_m} \text{dV}$$

where $V_f$ and $\sum_f$ are the fluid volume and surface.

This equation means that the rate of momentum of the fluid volume equals the surface and volume forces on the system: the fluid volume. Now, if we take a control volume, use Reynolds transport theorem on the left and we express the stress tensor $\bar{\bar{\tau}}$ in a different way on the right it would lead to the following equation: $${\frac{\partial}{\partial t} \int_{V_c(t)} \rho \vec{v} \text{d}\text{V}} + \int_{\sum_c(t)}\rho\vec{v}\text{(}\vec{v}-\vec{v_c})\cdot\vec{n}d\sigma = -\int_{\sum_c(t)} p \vec{n} \text{d}\sigma + \int_{\sum_c(t)} \vec{\bar{\bar{\tau'}}}\cdot\vec{n}d\sigma + \int_{V_c(t)}\rho \vec{f_m} \text{dV}$$ where Vc and $\sum_c$ are the volume and the surface of the control volume. $\sum_c = \sum_f + \sum_1 + \sum_2$, where $\sum_1$ and $\sum_2$ are the surfaces needed to close the control volume.

I know that the surface and volume of the control volume are chosen to be equal to the system's volume at that instant. What I don't understand is the correspondence between the forces of the control volume and the fluid. You see, if you take a control volume you are studying the forces on the fluid surface plus the forces on the boundaries that are left. That would mean that the force $F_c = F_f + F_1 + F_2$, where $F_1$ and $F_2$ are the forces on the surfaces that close the control volume. But, according to momentum equation $F_c = F_f$, right? This is what doesn't make sense to me.

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  • $\begingroup$ What is F? Is it a force field? In that case, in your equation, isn't the control volume different from the fluid volume? How can the fields be equal if the volumes differ ..? $\endgroup$ – Emil Sep 18 '19 at 5:45
  • $\begingroup$ It is a force field and the control volume is supposed to be the same as the fluid volume. But I don't understand the correspondence between the surfaces and hence the surface forces $\endgroup$ – Patricia GC Sep 18 '19 at 6:58
  • $\begingroup$ Wait, why do you write needed to close the volume. The volume is already closed. Are you perchance trying to derive the transport equation? Then you should look in a reference. $\endgroup$ – Emil Sep 19 '19 at 6:23
  • $\begingroup$ Ok, I get it now. Since the fluid volume is already closed, the volumetric and surface forces of the fluid are the same as the control volume. I thought of the fluid volume as an open volume. $\endgroup$ – Patricia GC Sep 20 '19 at 10:03

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