Control volumes and Reynolds transport theorem

[etotheipi]

If we consider a system of fixed mass as well as a control volume which is free to move and deform, then Reynolds transport theorem says that for any extensive property $B_{S}$ of that system (e.g. momentum, angular momentum, energy, etc.) then$$\frac{dB_{S}}{dt} = \frac{d}{dt} \int_{CV} \beta \rho dV + \int_{CS} \beta \rho (\mathbf{V}_r \cdot \mathbf{n}) dA$$where $\beta := \frac{dB_{S}}{dm}$ is the quantity per unit mass and $\mathbf{V}_r$ is the relative velocity of the matter/fluid at the boundary w.r.t. the velocity of the control volume boundary. $CV$ and $CS$ denote control volume and control surface respectively.

My question is, what is the necessary relationship between the control volume and the system in order for that general relation to hold true? I have the feeling that the system and the control volume must coincide at time $t$ when the integrals are evaluated (but they will not necessarily coincide at $t+dt$), but I am not certain of this. My reasoning was because if we consider a case where the control volume and system are completely separated (no overlap), then that relation is definitely wrong.

Any clarification would be much appreciated, thanks!

 

[vanhees71]

Usually you choose a control volume enclosing a certain material amount of fluid, i.e., consisting of one and the same material fluid elements at any time. Then $\vec{V}_r$ is simply the fluid velocity (along the boundary of the control volume).

 

[etotheipi]

Usually you choose a control volume enclosing a certain material amount of fluid, i.e., consisting of one and the same material fluid elements at any time. Then $\vec{V}_r$ is simply the fluid velocity (along the boundary of the control volume).

That does make sense, e.g. at time $t$ we consider a system of fluid that is instantaneously inside our (in this case, fixed) control volume, and we find that for instance $\vec{F} = \frac{d\vec{P}}{dt} + \dot{\vec{P}}_{lost} - \dot{\vec{P}}_{gained}$, and like you say the $\vec{V}_r$ in $\dot{\vec{P}}_{lost} - \dot{\vec{P}}_{gained} = \int_{CS} \rho \vec{V}(\vec{V}_r \cdot \vec{n}) dA$ would just be the fluid velocity in the lab frame

But in the general case, must the boundaries of the control volume and system still coincide at time $t$?

 

[vanhees71]

Well, of course you can calculate volume integrals over any time-dependent volume you like, but then it's hard to interpret physically I'd say. Do you have a specific hydro problem, where you need this?

 

[etotheipi]

Well, of course you can calculate volume integrals over any time-dependent volume you like, but then it's hard to interpret physically I'd say. Do you have a specific hydro problem, where you need this?

I did learn this in the context of fluid dynamics but I was just thinking about how it could be applied to a rocket that is losing mass. I imagined that we could construct a control volume that is fixed w.r.t. the rocket, and that at some time $t$ we consider a system defined by all the particles in the control volume at time $t$. Then we would have from RTT$$\vec{F}_{ext} = \frac{d\vec{p}_S}{dt} = \frac{d\vec{p}_{CV}}{dt} + \frac{d\vec{p}_{lost}}{dt}$$If $\vec{v}_{rel}$ is the velocity of the fuel w.r.t. the rocket then$$\frac{d\vec{p}_{lost}}{dt} = -\dot{m}(\vec{v} + \vec{v}_{rel})$$so we end up with $$\vec{F}_{ext} + \vec{v}_{rel} \frac{dm}{dt} = m\frac{d\vec{v}}{dt}$$that is the expected formula. I believe we must also make the control volume arbitrarily large so that we can ignore interactions between ejected particles and the rocket. That would mean that we also need to assume the rocket has been traveling for long enough that there are exhaust particles crossing the far away boundary.

But I realized that for this argument to work, the control volume and system must coincide at time $t$. But luckily, I think that is a requirement for RTT

 

[vanhees71]

Great! That's really a very nice example for the application of Reynold's transport theorem!

 

[Chestermiller]

The way I understand this is as follows:

For a moving and deforming control volume, the general equation is:
$$\frac{d}{dt}\left(\int_{CV}\beta \rho dV\right)=\int_{CV}\frac{\partial (\beta \rho)}{\partial t}dV+\int{\beta \rho \mathbf{v}_{CV}}\centerdot \mathbf{n}dA$$For the special case in which every point on the control volume surface is moving at the same velocity as the fluid (i.e., with a fixed mass of fluid within the control volume), we define the material derivative as D/Dt:
$$\frac{D}{Dt}\left(\int_{CV}\beta \rho dV\right)=\int_{CV}\frac{\partial (\beta \rho)}{\partial t}dV+\int{\beta \rho \mathbf{v}_{fluid}}\centerdot \mathbf{n}dA$$Combining these two equations gives: $$\frac{d}{dt}\left(\int_{CV}\beta \rho dV\right)=\frac{D}{Dt}\left(\int_{CV}\beta \rho dV\right)+\int{\beta \rho \mathbf{v}_{r}}\centerdot \mathbf{n}dA$$where, at time t, the two control volumes coincide