Control volumes and Reynolds transport theorem

So, it seems like the relationship between the control volume and the system is that they must coincide at time t.
  • #1
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!
 
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  • #2
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).
 
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  • #3
vanhees71 said:
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##?
 
  • #4
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?
 
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  • #5
vanhees71 said:
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 :smile:
 
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  • #6
Great! That's really a very nice example for the application of Reynold's transport theorem!
 
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  • #7
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
 
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Related to Control volumes and Reynolds transport theorem

1. What is a control volume in fluid mechanics?

A control volume is an imaginary region in space used to analyze the flow of fluids. It can be any shape or size and is typically defined by a physical boundary, such as a pipe or a surface. It is used to study the flow of mass, energy, and momentum in and out of the volume.

2. What is the Reynolds transport theorem?

The Reynolds transport theorem is a fundamental concept in fluid mechanics that relates the change of a property within a control volume to the flux of that property across the control volume boundary. It is based on the conservation laws of mass, energy, and momentum and is used to analyze and solve problems involving fluid flow.

3. How is the Reynolds transport theorem derived?

The Reynolds transport theorem is derived from the continuity equation, which states that the rate of change of mass within a control volume is equal to the net mass flow into or out of the volume. By applying this equation to the conservation laws of energy and momentum, the Reynolds transport theorem is obtained.

4. What are the assumptions made in the Reynolds transport theorem?

The Reynolds transport theorem assumes that the fluid is incompressible, the flow is steady, and the control volume is fixed in space. It also assumes that there are no external forces acting on the control volume and that the properties of the fluid are continuous and differentiable.

5. How is the Reynolds transport theorem used in practical applications?

The Reynolds transport theorem is used in many practical applications, such as analyzing fluid flow in pipes, pumps, and turbines. It is also used in the design of aircraft and cars, as well as in environmental studies to model the transport of pollutants in air and water. It is a powerful tool for understanding and predicting the behavior of fluids in various systems.

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