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I am super confused about two different formulations of RTT (Reynolds transport theorem) that yield two different results when used in the same class of problem. The first is found on wiki and continuum mechanics books:

$$\frac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega(t)}\mathbf{f}\,\mathrm{d}V=\int_{\Omega(t)}\frac{\partial\mathbf{f}}{\partial t}\,\mathrm{d}V+\int_{\partial\Omega(t)}\left(\mathbf{v}_b\cdot\mathbf{n}\right)\mathbf{f}\,\mathrm{d}A\tag{1}$$ The second I see often on fluid mechanics books: $$\left(\frac{\mathrm{d}N}{\mathrm{d}t}\right)_\text{sys}=\frac{\mathrm{d}}{\mathrm{d}t}\underset{\text{control volume}}{\iiint}\eta\rho\,\mathrm{d}v+\underset{\text{control surface}}{\iint}\eta p\left(\vec{v}_r\hat{n}\right)\,\mathrm{d}A\tag{2}$$

The problem arises when I try to apply the first formulation to a constant control volume like a tube, where fluid enters and leaves the boundaries, according to the wiki, it yields, $$\frac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega}\mathbf{f}\,\mathrm{d}V=\int_{\Omega}\frac{\partial\mathbf{f}}{\partial t}\,\mathrm{d}V.$$

But this result implies that the system is closed with no inflow or outflow, which doesn't make sense, because the system is clearly open. So, what am I missing here? What is the difference between the two formulations?

Also professors often warn that in formulation 2) you can only move the derivative inside the control volume (CV) if the CV has constant volume. But in 1), the derivative is already inside the integral. So, clearly there is a difference between the two and how to use them, which I am struggling to spot.

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I've got a partial answer to my doubt after researching on many books. So, in the first formulation, which I'll rewrite as:

$$\frac{d}{dt} \int_{V(t)}{bdV} = \int_{V(t)}{\frac{\partial b}{\partial t}dV} + \int_{\partial V(t)}{b(v\cdot n)dA} $$

So, according to the books of Gurtin, Ingers, Leal, Shivamoggi and others, this specific formulation of RTT applies to the volume $V(t)$ when this volume is a material volume identical to the fluid

So if you're tracking a single droplet, the volume is identical to the volume of the droplet and will respond to any deformations. If you're tracking the flow of water out of a tank then your Volume is the entire volume of water, both the one inside the tank and the outflow, moving with this "fluid body".

In a sense, formulation 1 is a "global" rate of change of a certain extensive property. If it's mass, for instance, then it becomes

$$\frac{d}{dt} \int_{V(t)}{\rho dV} = \int_{V(t)}{\frac{\partial \rho}{\partial t}dV} + \int_{\partial V(t)}{\rho(v\cdot n)dA} = 0 $$

In the case of an incompressible flow, the entire fluid body has constant volume. Every change of volume increasing in some direction has to be compensated by a decrease of volume by the same amount in some other direction. Like the emptying tank, the fluid body is deforming it's volume to occupy the outside space by a certain amount and such amount has to reduce by the same magnitude in the inside of the tank. So the equation becomes:

$$ \int_{V(t)}{\frac{\partial \rho}{\partial t}dV} = 0 $$

Constant density is the consequence of both volume and mass being constant.

Now, going to the second formulation, it has a distinct application. Often we can't track the entire mass of fluid moving around and we're interested normally in only a certain volume where the fluid flows, this control volume is not identical to the entire body of fluid and can move, grow and shrink independently. Compared to the global case, we're tracking a local part of the flow and dealing with a different volume. Putting all of this together we get formulation 2 :

$$\frac{d}{dt} \int_{V_{cv}(t)}{bdV} = \frac{\partial}{\partial t}\int_{V_{cv}(t)}{bdV} + \int_{\partial V_{cv}(t)}{b(v\cdot n)dA} $$

Which, in the case of the emptying tank, the density is constant as conditioned by the incompressible flow, but the volume inside the tank isn't, so we use formulation two to obtain:

$$0 = \rho \frac{\partial V}{\partial t} + \int_{\partial V_{cv}(t)}{\rho(v\cdot n)dA} $$

$$-v_{out}A =\frac{\partial V}{\partial t} $$

Which agrees with the intuitive accumulation = inflow - outflow.

What remains in question to me is if it's possible to transform formulation 1 into formulation two mathematically, i've tried to no avail, it seems that the term $\int_{V(t)}{\frac{\partial b}{\partial t}dV}$ already implies that the volume moves with the flow, so the derivation for a control volume with possible independent movement has to begin from scratch.

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  • $\begingroup$ I think you should name the second one a "continuity equation", because it is really what it is about. See here. What's funny is that on the english page, there is no mention of the Reynold theorem, but in the french one it is in the first paragraph ! $\endgroup$
    – rambi
    Commented Jun 22, 2023 at 16:01
  • $\begingroup$ en.wikipedia.org/wiki/Leibniz_integral_rule (Higher dimensions), We apply the Ostrogradski formula to the 2nd equation, we obtain the continuity equation. $\endgroup$
    – The Tiler
    Commented Jun 22, 2023 at 20:46
  • $\begingroup$ @HVAC I don't see why: in the continuity equation the volume doesn't change right ? I didn't notice the first time but I don't understand why the border changes since it is a "control volume" $\endgroup$
    – rambi
    Commented Jun 23, 2023 at 12:18
  • $\begingroup$ If we take a balloon and make a circular cut $dS$, the amount of air that will come out is $\;\;dQ=\rho v_{n}dtdS$, the total amount that comes out during a unit time is: $Q =\int\int_{S}\rho v_{n}dS$, the quantity of air occupying the volume (v) limited by (S) is expressed by $\,\;\int\int\int_{v}\rho dv \;\;$ and during the time interval of this quantity varies by $dt \int\int\int_{v}\frac{\partial \rho }{\partial t}dv$ for a unit of time $\;\;Q=\int\int\int_{v}\frac{\partial \rho }{\partial t}dv\;\;$, for a closed surface this quantity is negative, i.e. $\endgroup$
    – The Tiler
    Commented Jun 23, 2023 at 14:21
  • $\begingroup$ we have $Q=\int\int_{S}\rho v_{n}dS=\int\int\int_{v}\vec{\nabla}({\rho\vec{v})} dv=-\int\int\int_{v}\frac{\partial \rho }{\partial t}dv\;\;,$from where: $\frac{\partial \rho }{\partial t}+\vec{\nabla}({\rho\vec{v})}=0$ $\endgroup$
    – The Tiler
    Commented Jun 23, 2023 at 14:21
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Short answer

The two formulations are indeed equivalent, but not in a trivial way: deriving one from the other requires a change of perspective that is not obvious if you are not used to it.

To apply Reynolds Theorem, you always need to consider a material volume (meaning, a moving volume defined by some portion of the fluid), but if you have a control volume $V$ (a fixed volume through which the fluid flows), you can think about the material volume that coincide with $V$ at time $t$, and apply Reynold's theorem to it

Detailed answer

To explain the topic in full detail, I will have to talk about three things:

  1. Leibniz integral rule, the mathematical fondation of Reynolds Theorem
  2. The original formulation of Reynolds Theorem
  3. The conservation equation, a consequence of Reynolds Theorem for a control volume

Leibniz integral rule

https://en.wikipedia.org/wiki/Leibniz_integral_rule

Let $V(t)$ be a time-dependant volume of border $\partial V(t)$
We consider the field $g(M,t)$, that also depends on time

Let $G(t)=\iiint_{V(t)}g(M,t)dV$ That means $G(t)$ is the integral of $g$ over all the volume $V(t)$, volume that depends on time.

Leibniz integral rule says: $$ \frac{ \partial G }{ \partial t } = \iiint_{V(t)}\frac{ \partial g }{ \partial t }(M,t)dV + \iint_{\partial V(t)} g(M,t) \vec{v_{b}}(M,t)\cdot \vec{dS} $$

Here, $\vec{v_{b}}(M,t)$ has nothing to do with a fluid: it is just the speed of the border at position $M$ and at time $t$

Let's get give name to the 3 terms:

  • the left-hand side of the equation is the total variation
  • the first right term is the local variation
  • the last term is the convective term

This equation looks pretty similar to Reynolds Transport Theorem, clearly. But before writing it down, I have to clarify the physical interpretation of each one of these terms.

Reynolds transport theorem

https://en.wikipedia.org/wiki/Reynolds_transport_theorem

Now, let's study a fluid. This fluid has a velocity at each point, $\vec{v}(M,t)$
Let's choose an extensive property of the fluid denoted by the letter $g$ (it can be $\rho$ for mass, $u$ for energy, or even the constant $1$ for volume ...)

To apply Leibniz integral rule, we have to chose a 3D region that changes with time. We could really take whatever region we want, but to give a physical meaning to the total variation, the clever idea is to chose a region that moves with the fluid: a material volume. That way, we know the integral of $g$ over the region is the quantity of $g$ in a specific system, namely the portion of fluid that you chose.

In that setup, we can write:

$$ \frac{ D G }{ D t } = \iiint_{V(t)}\frac{ \partial g }{ \partial t }(M,t)dV + \iint_{\partial V(t)} g(M,t) \vec{v}\cdot \vec{dS} $$

Here, the term $\frac{ D G }{ D t }$ represents the variation of $g$ in the system, and I wrote it with the capital letter $D$ to show it is a variation for a moving (or material) system. It is a quite common convention (see here )

So to make sense of Leibniz integration rule, we had to chose a material volume. But to study a volume that does not move, how could we do ?

The continuity equation

https://en.wikipedia.org/wiki/Continuity_equation

I will use the name "conservation equation" since it is really what it is about

Let's study the same property $g$ of a fluid, but now what we want is $$ \frac{d}{dt} \iiint_{V} g(M,t) $$ with $V$ a fixed volume.

Hopefully, we can use the previous theorem with a smart trick. Chose a time $t_{0}$

I will define $\tilde V(t)$ as:

  • $\tilde V(t_{0})=V$
  • $\tilde V(t)$ is a material volume that follows the particles of fluid.

In other words $\tilde V$ is the material volume that coincides with $V$ at time $t_{0}$

If we define $\tilde G(t)=\iiint_{\tilde V(t)}g(M,t)dV$, we can then use Reynolds transport theorem to write:

$$ \frac{ D \tilde G }{ D t }(t_{0}) = \iiint_{V}\frac{ \partial g }{ \partial t }(M,t)dV + \iint_{\partial V} g(M,t) \vec{v}\cdot \vec{dS} $$

Now, if I note $P$ the variation of $g$ in the system of the material volume, I can rearrange the terms to get

$$ \frac{d}{dt}\iiint_{V}g(M,t)dV = P - \iint_{\partial V}g(M,t) \vec{v} \cdot \vec{dS} $$

Which gives me the variation of $g$ in my control volume !

And the last equation is exactly the general form of a conservation law: For example if $g=\rho$ and there is no source (meaning the mass in the material volume is constant), we get: $$ \vec{\nabla} \cdot (\rho \vec{v}) = - \frac{ \partial \rho }{ \partial t } $$

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Short Answer

I can't really follow your notation in the original question, but Reynold's Transport Theorem is the Leibniz Integral Theorem applied to a region whose surface has the same velocity as the flow velocity of a fluid/field.

Long Answer

Imagine some region of space $R$ filled with a field (e.g. a fluid). For an arbitrary region $\Omega(t)$ in $R$ consider a scalar quantity X defined by:

$$ X = \int_{\Omega(t)} \rho \,dV $$

where $\rho$ is the X-density field. The notation $\Omega(t)$ is meant to emphasize that the shape, volume, and position of the region may change over time as it moves in $R$. The time derivative of X is:

$$\frac{dX}{dt} = \frac{d}{dt}\left(\int_{\Omega(t)} \rho \,dV \right) $$

We'd like to be able to swap the derivative and integral, but can't because the domain of integration is time dependent. To get around this we need to use the three dimensional version of "Leibniz integral theorem" (L.T.). If we then use the Divergence Theorem we can write your equation (1):

$$ \begin{align} \frac{dX}{dt} &= \frac{d}{dt}\left(\int_{\Omega(t)} \rho \,dV \right) \\ &\stackrel{L.T.}{=} \int_{\Omega(t)} \left( \frac{\partial\rho}{\partial t} + \nabla \cdot (\rho \mathbf{v}) \right) \,dV \\ &\stackrel{D.T.}{=} \int_{\Omega(t)} \frac{\partial\rho}{\partial t} \,dV + \int_{\partial\Omega(t)} \left( \rho \mathbf{v} \cdot \mathbf{n} \right) \, dA \end{align} $$

Here $\mathbf{v}$ is the velocity (field) of the surface of $\Omega(t)$, denoted $\partial \Omega(t)$.

If $\Omega(t)$ always "contains the same material" then $\mathbf{v}$ is called the "flow velocity" and the Leibniz integral theorem is called the Reynold's Transport Theorem. To be clearer about what that means it is helpful to derive a "Balance Equation" from this result. We'll assume we can express the rate of change of X in terms of sources or sinks in the X-density field:

$$\frac{dX}{dt} = \int_{\Omega(t)} \gamma\,dV $$

If $X$ is a conserved quantity (e.g. mass, energy, etc.) then $dX/dt = 0$, or equivalently $\gamma = 0$. When we substitute this into the volume integral above we get:

$$ \int_{\Omega(t)} \gamma\,dV = \int_{\Omega(t)} \left( \frac{\partial\rho}{\partial t} + \nabla \cdot (\rho \mathbf{v}) \right) \,dV \implies 0 = \int_{\Omega(t)} \left( \frac{\partial\rho}{\partial t} + \nabla \cdot (\rho \mathbf{v}) - \gamma \right) \,dV $$

The only way the integral on the RHS can be zero for an arbitrary region is if the integrand is itself zero: $$ 0 = \frac{\partial\rho}{\partial t} + \nabla \cdot (\rho \mathbf{v}) - \gamma $$ This is a local field equation, in the sense that it applies at every point within the region $R$, including all those in $\Omega(t)$.

Now consider some other deformable region $K(t)$ in $R$ called the control volume. The velocity (field) of $\partial K(t)$ (the control surface) is $\mathbf{w}$. If we integrate the field equation over $K(t)$ and apply the Divergence Theorem we get:

$$ \begin{align} 0 &= \int_{K(t)} \left( \frac{\partial\rho}{\partial t} + \nabla \cdot (\rho \mathbf{v}) - \gamma \right) \,dV \\ &\stackrel{D.T.}{=} \int_{K(t)} \frac{\partial\rho}{\partial t} \,dV + \int_{\partial K(t)} \left( \rho \mathbf{v} \cdot \mathbf{n} \right) \, dA - \int_{K(t)} \gamma \,dV \end{align} $$

Independently, if we apply the Leibniz theorem and Divergence theorem to $K(t)$ we get:

$$ \begin{align} \frac{dX_{CV}}{dt} &= \frac{d}{dt}\left(\int_{K(t)} \rho \,dV \right) \\ &\stackrel{L.T.}{=} \int_{K(t)} \left( \frac{\partial\rho}{\partial t} + \nabla \cdot (\rho \mathbf{w}) \right) \,dV \\ &\stackrel{D.T.}{=} \int_{K(t)} \frac{\partial\rho}{\partial t} \,dV + \int_{\partial K(t)} \left( \rho \mathbf{w} \cdot \mathbf{n} \right) \, dA \end{align} $$

We can combine these results to obtain:

$$ \frac{dX_{CV}}{dt} = - \int_{\partial K(t)} \left( \rho \left(\mathbf{v} - \mathbf{w}\right) \cdot \mathbf{n} \right) \, dA + \int_{K(t)} \gamma \,dV $$

I think this is what you are getting at in your Equation 2. This is a macroscopic balance equation that describes the rate of change of X in a control volume due to flow across the control surface and sources (or sinks) of X inside the control volume.

The above equation holds regardless of if the control volume change its shape, volume, or position. Note that $\mathbf{v} - \mathbf{w}$ is the relative velocity of the flow with respect to the surface of the control volume. Generalizations of this result exist for vector and tensor fields--see the references below for details.

Consider the following limiting cases.

  1. $\mathbf{w} = 0$. The control volume is stationary with respect to our frame of reference:

$$ \frac{dX_{CV}}{dt} = - \int_{\partial K(t)} \left( \rho \mathbf{v} \cdot \mathbf{n} \right) \, dA + \int_{K(t)} \gamma \,dV $$

  1. $\mathbf{w} = \mathbf{v}$. The control volume and the flow have the same velocity with respect to our frame of reference (i.e. the control volume is stationary with respect to the flow):

$$ \frac{dX_{CV}}{dt} = \int_{K(t)} \gamma \,dV $$

In case 2 nothing flows through the boundary of the control volume, so it is often called a material volume.


More Resources:

  1. Transport Phenomena by Bird, Stewart, and Lightfoot.
  2. Advanced Transport Phenomena by Slattery.
  3. Most continuum mechanics textbooks.
  4. Leibniz Integral Theorem (This is sometimes called Feynman's trick. The 3D Version is described in Ref. 1)
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