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We know that for a system (a fixed amount of matter) the second law of dynamics is: $$\mathbf F_{sys}=\frac{d(\mathbf P_{sys})}{dt}$$ The general form of Reynolds theorem is: $$\frac{d( B_{sys})}{dt}=\frac{d}{dt} \iiint_{CV}\rho b\ d\tau \ + \ \iint_{CV} \rho b(\mathbf v_{rel} \cdot \mathbf n) \ d\Sigma$$ Where CV is a general control volume which can move around the space and warp itself, and at time $t=0$ coincides with the system we are considering.

Now writing: $$\mathbf B_{sys}=\mathbf P_{sys} \rightarrow \mathbf b=\mathbf v$$ Reynolds theorem gives us:

$$\mathbf F_{sys}=\frac{d(\mathbf P_{sys})}{dt}=\frac{d}{dt} \iiint_{CV}\rho \mathbf v\ d\tau \ + \ \iint_{CV} \rho \mathbf v(\mathbf v_{rel} \cdot \mathbf n) \ d\Sigma$$ Instead of this equation my book writes: $$\mathbf F_{CV}=\frac{d(\mathbf P_{sys})}{dt}=\frac{d}{dt} \iiint_{CV}\rho \mathbf v\ d\tau \ + \ \iint_{CV} \rho \mathbf v(\mathbf v_{rel} \cdot \mathbf n) \ d\Sigma$$ I don't understand why it puts $\mathbf F_{CV}=\mathbf F_{sys}$. I mean for time $t=0$ it is ok, since CV and the system are the same thing, but for $t>0$ the two follow different paths and they don't coincide anymore. Someone could explain this?

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  • $\begingroup$ I think conceptually it should be $B_{CV}$ in the general form of the Reynolds theorem as you are considering a control volume. Once integrating over all control volumes it becomes $B_{sys}$, i.e. $B_{sys}=\sum B_{CV}$ $\endgroup$ – nluigi Aug 31 '18 at 9:50
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The Reynold's Transport theorem is based on the application of a concept known as the '$material \ \ volume$'. The material volume is defined as a control volume that moves with the fluid's velocity field.

Consider an arbitrary control volume $V_m$ that exists in $\mathbb{R}^3$. Let $\bar{r}$ represent the position vector of $V_m$ at time $t$ such that $V_m = V_m(t)$.

The principle of conservation of linear momentum, utilizing Reynold's transport theorem, can be written in terms of the material volume as such: $$\frac{d}{dt}\int_{V_m(t)}p\bar{v}dV = \int_{V_m(t)}p\bar{b}dV + \int_{A_m(t)}\bar{t_{(n)}}dA$$

Where $\bar{b}, \ \ \bar{t_{(n)}}, \ \ A_m(t)$ represent the body force, point stress on surface, and the 'material surface' respectively. This relation is also known as Euler's first law.

To answer your question, the Control Volume and the System in your text should be interchangeable. The derivation of the Reynold's transport theorem is based on the utilization of the interchangeability between the Lagrangian and Eulerian descriptions of motion$^{[1]}$ to 'map' the Control Volume of interest back to $t = 0$. $$$$ Source [1]: Mei, C. C. Methods of Describing Fluid Motion (Lecture Notes). 2001. http://web.mit.edu/fluids-modules/www/basic_laws/1-1-LagEul.pdf

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