Talk:Reynolds transport theorem

Flagged as inaccurate
Revert. I have reverted back to the version of the article that preceded the revision of Administrator Arbitrarily0, an action motivated by the incorrect interpretations present in the version favored by the Administrator. Garethvaughan (talk) 00:37, 3 May 2012 (UTC)


 * Administrator Arbitrarily0 has restored an old version of this page, one that I think is incorrect.


 * I have invited Arbitrarily0 to justify his/her actions on this page.


 * In particular, Arbitrarily0 has not addressed the issue that the "general form" of the Reynolds transport theorem given here is actually the form specific to regions with material boundaries.

Garethvaughan (talk) 23:56, 9 March 2012 (UTC)
 * Hi Garethvaughan. On a quick review of the article history I found that the version from last July 2011 to be more comprehensive.  I do not have time at the moment to thoroughly check if this is indeed so, but for now I will take your word that it is not.  Feel free to undo any of my edits - I will try to take a more detailed look at the article next week.  Cheers mate, thanks for all your work.    A rbitrarily 0    ( talk ) 15:05, 10 March 2012 (UTC)


 * Thanks Arbitarily0, for your swift response, and your interest in this article.


 * I would prefer to attain a consensus before undoing your changes.


 * Note that the section titled "A special case" is a left-over from the article which I wrote.


 * After reviewing the differences in the 2 versions, and I am confident that the changes that I made were justified because we need to: 1) give the theorem correctly, 2) interpret the theorem correctly.


 * I advise reverting to the version of the article that I wrote, and adding worked examples to demonstrate its use.


 * 1)
 * For the theorem in the form that Reynolds published, see Reynolds (1903, vol 3, article 14: 'Moving Surface', equation 15) --- click the link in the article to read a pdf of this work --- The general form that I give in my version of the article is derived from this.


 * Reynolds notation is not modern and is difficult to digest I think: his proof even more so. It can all be understood by searching the preceding pages of vol 3 for the necessary definitions.


 * The general form, as it is given here is correct for the SPECIFIC case of material volumes. That it is incorrect generally is easily shown: consider the case where boundaries are stationary.


 * 2)
 * The intro and the figure are predicated on a confused (and incorrect) physical interpretation of the theorem: we don't need to constrain ourselves to integrating intensive physical properties and it is quite distinct from the fundamental principles of physics (i.e. conservation of mass/energy/momentum).


 * While it is (very) useful in deriving equations that are statements of these principles, it is more general than that: I have used it on a number of occasions for integrating over Lagrangian and semi-Lagrangian control volumes when discretising various governing equations--- not just the Navier-Stokes and Euler equations.


 * It is, more accurately, about a space balance: the time derivative of the the integral must equal integrals over the space that enters and leaves the domain of the integral, and the time derivative of the integrand.


 * The equations presented in the sections titled "mass/energy/momentum formulations" are more accurately described as "examples of integral forms of the mass, energy and momentum balance equations". See Bird, Stewart and Lightfoot "Transport phenomena" (1960) for a more comprehensive list: the integral forms change, depending on which mechanisms are significant in a system.  The presentation of the integral forms of the balance equations here is an example of an application of the theorem: I think these would be better placed in articles on those topics.

Garethvaughan (talk) 03:40, 11 March 2012 (UTC)

An Extensive Re-Write
I have performed this because: 1) the theorem was given incorrectly (see the following section) 2) after many years of minor edits, the article did not flow and had unnecessary material

I removed the material on "mass formulation" etcetera, because it does not appear in standard texts and it was unnecessary for the article. Garethvaughan (talk) 03:34, 25 August 2011 (UTC)

Reynold's Transport Theorem
There is much discussion on this page as to what Reynolds Transport Theorem actually is. The disagreement being what velocity should appear in the surface integral; whether $$\mathbf{v}$$, $$\mathbf{v}_r$$ or $$\mathbf{v}_b$$ or a linear combination of these. The confusion extends beyond this page: different variants appear in different books, so no one source is sufficient.

What Reynolds wrote
Perhaps Reynolds' writing provides the most authoritative reference. Interpreted into a more modern notation, he wrote (see page 13 of Volume 3),

\cfrac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega(t)} f~\text{dV} = \int_{\Omega(t)} \frac{\partial f}{\partial t}~\text{dV} + \int_{\Omega(t)} \nabla\cdot(f\mathbf{v}_b)~\text{dV} ~, $$ for $$\mathbf{v}_b$$ the velocity of the boundary. This becomes, after application of the divergence theorem

\cfrac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega(t)} f~\text{dV} = \int_{\Omega(t)} \frac{\partial f}{\partial t}~\text{dV} + \int_{\partial \Omega(t)} f\mathbf{v}_b\cdot~\mathbf{n}\text{dS} ~. $$

2 special cases:
If we take $$\Omega$$ to be constant with respect to time, then $$\mathbf{v}_b=0$$ and the identity reduces to

\cfrac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega} f~\text{dV} = \int_{\Omega} \frac{\partial f}{\partial t}~\text{dV} ~, $$ as expected. This simplification is not possible if an incorrect form of the Reynolds transport theorem is used.

If $$\partial \Omega(t)$$ comprises material surfaces, hence $$\Omega$$ is a material element, then $$\mathbf{v}_b\cdot~\mathbf{n}=\mathbf{v}\cdot~\mathbf{n}$$ and one obtains the variant given at the base of the article.

Interpretation and reduction to one dimension
It is the higher dimensional extension of Differentiation under the integral sign and should reduce to that expression in some cases. Suppose $$f$$ is independent of $$y-z$$, and that $$\Omega(t)$$ is a unit square in the $$y-z$$ plane and has $$x$$ limits $$a(t)$$ and $$b(t)$$. Then Reynolds transport theorem reduces to

\cfrac{\mathrm{d}}{\mathrm{d}t}\int_{a(t)}^{b(t)} f~\text{dx} = \int_{a(t)}^{b(t)} \frac{\partial f}{\partial t}~\text{dx} + \frac{\partial b(t)}{\partial t} f(b(t),t) -\frac{\partial a(t)}{\partial t} f(a(t),t) ~, $$ which is the expression given on Differentiation under the integral sign, except that there the variables x and t have been swapped.

Garethvaughan (talk) 10:55, 16 July 2011 (UTC)

Existing Discussion
I have no previous knowledge of the Reynold's Transport Theorem, but as a Physcist I make a big plea to whoever knows more on this topic to convert the intergral theorems into differential theorems. Integral theorems are much harder to apply and use than differential forms of the same equations. You can also "see" the physics more easily in differential form, than in integral form. At least, differential forms are taught these days, instead of integral forms.


 * The differential forms of these equations are known as the Navier-Stokes equations which you may have heard of. I have to disagree with your statements that the differential forms are more useful or more often used, though that may be true in your field.  I am a mechanical engineer, and the integral forms of these are both very powerful, and a lot clearer to me about what they describe.  Regardless, they have already been converted for you by some smart people a long time ago.  An important note: I added this page and it was written from the perspective of a mechanical engineer.  If there is something that could be done to make it more useful to other specialties, I welcome the additions. -EndingPop 17:27, 16 January 2006 (UTC)


 * On 2/8/2006 someone added "also magnetic" to the end of one of the sentences. That sentence ended in the words "electromagnetic fields".  Even though magnetic fields were already covered, if you think something is missing but don't want to add it in final form yourself make sure to put it on here on the Talk page. -EndingPop 02:03, 9 February 2006 (UTC)


 * I might add that the purpose of the Reynolds Transport Theorem, for the most part, was to allow differential forms of transport equations to be developed. One can't accomplish this feat without doing something with that nasty substantial derivative outside the integral (which, of course, has limits of integration that depend on time, and so can't just be "taken inside" the integral)  The Transport Theorem, by the way, is identical to the Leibniz integral rule for 1-dimensional domains.--71.98.78.28 04:39, 11 June 2007 (UTC)

If someone has the time, please consider simplifying the equations with $$\overrightarrow{V} = \overrightarrow{V_b} + \overrightarrow{V_r}$$. hoo0 05:06, 20 February 2006 (UTC)


 * This would eliminate generality and would be undesirable. Charlesreid1 (talk) 21:03, 24 October 2010 (UTC)


 * First, thanks for the additions to the page. I really like the introduction; it gives it a lot more depth.  Now, on the point of combining the equations.  I added them this way because it makes more sense from a teaching point of view.  The difference between Vb and Vr needs to be further explained, but I think that for a person looking for the equations would find this more useful than a condensed version.  The expanded versions simply showcases more about the physical phenomena.  If you still disagree, I won't revert it if you decide to make the change. - EndingPop 04:26, 21 February 2006 (UTC)

Does anyone know of a way to have a formula with a strike through it to symbolize the volume? You can see on this page it is just put there in normal text mode, and it looks terrible. - EndingPop 15:08, 21 August 2006 (UTC)
 * As far as I can tell you can't do strikethroughs in TeX without an extension package, which I don't know how to use on Wikipedia (if you even can). I'd suggest changing this to some other standard symbol for differential volume--$$dV$$, $$d\Omega$$, $$d^3x$$ or some such.  Starryharlequin 03:27, 5 October 2006 (UTC)
 * V is already used as velocity, and the strike through is what I have seen in academia as the standard. Is there another symbol that is used in thermofluids? -EndingPop 11:50, 5 October 2006 (UTC)
 * In my thermofluids class, as well as our book, script v is used for velocity and capital V for volume. This seems a pretty good compromise to me. Something alone the lines of:

$$\frac{DN_{sys}}{Dt} = \int_{c.v.}^{} \frac{\partial}{\partial t} (\rho \eta) dV + \int_{c.s.}^{} \rho \eta \vec{v_b}\cdot \widehat{n} dA+\int_{c.s.}^{} \rho \eta \vec{v_r}\cdot \widehat{n} dA$$
 * I used \vec with lowercase v, because I don't really know tex enough to figure out a script vector. !jim 18:45, 6 October 2006 (UTC)
 * I have always seen the script v as specific volume in thermofluids, but since that isn't used here I'll try to make that change. - EndingPop 12:57, 9 October 2006 (UTC)

Correctness?
Is the equation shown correct? Shouldn't it be:

$$\frac{DN_{sys}}{Dt} = \int_{c.v.}^{} \frac{\partial}{\partial t} (\rho \eta) dV -\int_{c.s.}^{} \rho \eta \vec{v_r}\cdot \widehat{n} dA$$

Can anyone check the reference? —The preceding unsigned comment was added by 193.1.100.105 (talk • contribs).
 * I'm the original author of the article, so maybe my assurance doesn't count, but I assure you the formula is correct. The term with "vb" in it deals with the more general case of deformable control volumes. - EndingPop 14:54, 3 January 2007 (UTC)


 * You might encounter your version of this equation in a basic thermo class. That equation (yours) assumes that the control volume is fixed in whatever reference frame you're using, so the vb is zero, and the third term goes away. The minus sign is just related to a different sign convention, I believe. For what it's worth, my Thermo I professor also did away with the third term in the more general equation. !jim 04:21, 4 March 2007 (UTC)


 * Why are the $$\int_{c.s.}^{} \rho \eta \vec{v_r}\cdot \widehat{n} dA$$ and the $$\int_{c.s.}^{} \rho \eta \vec{v_b}\cdot \widehat{n} dA $$ two integral instead of just one $$\int_{c.s.}^{} \rho \eta \vec{v}\cdot \widehat{n} dA $$ where $$\vec{v} = \vec{v_r} + \vec{v_b} $$ i.e. the fluid flow at the surface? —The preceding unsigned comment was added by 129.94.6.28 (talk) 03:03, 22 March 2007 (UTC).


 * This was actually discussed above. It was felt that leaving the two velocities seperate was more clear or provided more information to someone not necessarily familiar with this concept. See hoo0's comment above. !jim 06:03, 22 March 2007 (UTC)

If the equation is going to be kept the way it is now, shouldn't the final term on the right hand side have a negative coefficient? i.e.

$$\frac{DN_{sys}}{Dt} = \int_{c.v.}^{} \frac{\partial}{\partial t} (\rho \eta) dV + \int_{c.s.}^{} \rho \eta \vec{v_b}\cdot \widehat{n} dA -\int_{c.s.}^{} \rho \eta \vec{v_r}\cdot \widehat{n} dA$$,

since normal vector is pointing outward? I would've thought the outward net flux would be subtractive, whereas the expansion of the control volume would be additive to the system. —Preceding unsigned comment added by 129.67.151.47 (talk) 18:06, 16 February 2009 (UTC)


 * Use Vr = Vf - Vb. If the boundaries do not move (Vb = 0), then the flux terms reduce to just Vf. I apologize, I can't seem to come up with a good physical analogy for why this works. My suggestion is to try it on some examples to see how the math works out. Try a uniform 1D flow and play with different values of Vb at the inlet and outlet with the continuity equation. You'll see that it always balances out. - EndingPop (talk) 21:31, 16 February 2009 (UTC)


 * I've read the discussion of this and I think at the very least we need a definition for what is meant by Vr (it took me a while to work it out). I've added this in, on a new line so readers in other sections will find it easy to spot.219.89.247.21 (talk) 05:39, 7 July 2011 (UTC)

Mass Flow
Shouldn't the case where $$ \eta $$ = 1 be included? This will reduce to the mass flow conservation equations. I believe that they are important enough to warrant its own heading along with the momentum and energy conservation. 203.59.183.57 07:43, 18 April 2007 (UTC) Tim
 * Somehow I missed a deletion of that section several edits back. I am currently working on integrating past constructive edits into the last version that had the mass formulation in it. Thanks for the head's up! - EndingPop 14:22, 18 April 2007 (UTC)
 * There, I've updated the page with the mass formulation back in there. The initial removal was my fault, I reverted some vandalism and missed the edit before by the same user that removed the mass formulation section. All subsequent constructive edits should have been included, but I'd appreciate someone checking my work. - EndingPop 14:34, 18 April 2007 (UTC)

I believe the case where $$\frac{DN_{sys}}{Dt}=0 $$ for the mass flow equation only holds true in the steady flow conditions, and not in the general case. Consider a water tank with a hole, filling with water. The rate of change of the mass of the water tank is equal to the differences between the mass flow in and out. Also I think that the page should be edited so that it starts with the most general equation, then shows the equations with various simplifying forms. i.e. steady flow, fixed volume, incompressible flow, adiabatic flow, etc.
 * Mass of a system is always constant and therefore cannot change in time, meaning $$\frac{DM_{sys}}{Dt}=0 $$ is always true. - Hacktivist 15:58, 4 November 2007 (UTC)
 * In this formulation, "system" is defined as an identifiable and fixed quantity of matter. We aren't accounting for nuclear reactions, so the time-rate of change of mass in the system is necessarily zero. Don't confuse the system with the control volume (Lagrangian vs. Eulerian viewpoints) - EndingPop 15:42, 5 November 2007 (UTC)

Fluid Mechanics Text
In my fluid mechanics textbook, the Momentum Equation is slightly different, some of it is just placement, but I was wondering what assumption my text might be making that it neglects the 3rd integral? I'm not seeing how there equivalent.

$$ \begin{align} \sum_{} \vec{F}&=\int_{c.v.}^{}\frac{\partial}{\partial t}(\rho \vec\upsilon) dV + \int_{c.s.}^{} \rho \vec\upsilon(\vec\upsilon_b \cdot \widehat{n})dA + \int_{c.s.}^{} \rho \vec\upsilon(\vec\upsilon_r \cdot \widehat{n})dA \\ \sum_{} \vec{F}&=\frac{\partial}{\partial t}\int_{cv}^{} \vec\upsilon \rho d\vec{\mathbb{V}} + \int_{cs}^{} \vec\upsilon \rho \vec{V} \cdot d\vec{A} \end{align} $$

$$ \mathbb{V}$$ = Volume

$$\vec\upsilon$$ = Velocity relative to an inertial reference frame

$$\vec{V}$$ = Fluid velocity relative to the control surface

Zath42 (talk) 19:34, 2 March 2008 (UTC)


 * Your text assumes a nondeformable control volume. If you allow it to move, you get the extra integral. Use Vr = V - Vb to convert between the two. - EndingPop (talk) 13:55, 3 March 2008 (UTC)


 * Thanks, that clears things up. Zath42 (talk) 18:39, 3 March 2008 (UTC)
 * Could we add this form to the main article in its own section? Beginner fluid mechanics-ists only use this form of the equation.--عبد المؤمن (talk) 19:11, 16 January 2013 (UTC)


 * Thanks for your comments.


 * The choice of notation used in this article is arbitrary and not absolute, but it is rather common. MANY notations are used in the range of texts on the subject --- see the Reynolds reference for an archaic example.  Different lecturers/professors have different preferences as to which they prefer.


 * There are several differences between the forms:
 * 1) how vectors are denoted (bold face vs overline),
 * 2) the name of the region (cv & cs vs $$\Omega$$ & $$\partial \Omega$$,
 * 3) the use of $$\vec{v}_r$$ (the velocity relative to the surface).


 * My opinions:
 * 1) is minor,
 * 2) This equation is not limited to use with control volumes,
 * 3) There has been considerable discussion of whether or not $$\vec{v}_r$$ should be used. It is a matter of personal preference.
 * Garethvaughan (talk) 05:07, 20 January 2013 (UTC)

Formulation Error?
I believe there is still a problem though. In going from the general form to the specialized form for mass, you are taking the time derivative outside the integral over the c.v. - this is only true for a non-deformable c.v. For a deformable cv, from Liebnitz theorem,

del/delt(integral_over_cv(rho*eta*dV))=integral_over_cv(del/delt(rho*eta*dV))+integral_over_cs(rho*eta*v_b.normal*dS)

so, only the relative velocity integral should show up - this issue is covered in Potter and Foss' book that is cited as a reference —Preceding unsigned comment added by 128.138.64.39 (talk) 21:38, 23 May 2008 (UTC)


 * I'm not following you. Could you point out the alleged discrepancy/error? What, specifically, is wrong, and why? Also, a page number from Potter would be good too, since you said it was dealt with in there. - EndingPop (talk) 20:31, 27 May 2008 (UTC)


 * In the conversation below I noticed and understood what you were saying. The continuity equation has been updated. - EndingPop (talk) 00:21, 9 September 2008 (UTC)

Error in First Equation
I believe that there is an error in the surface integrals for the Reynolds Transport Theorem:
 * $$\int_{c.s.}^{} \rho \eta \vec\upsilon_b\cdot \widehat{n} dA+\int_{c.s.}^{} \rho \eta \vec\upsilon_r\cdot \widehat{n} dA$$

Definitions:
 * $$\vec{v}_b$$ = Velocity of the surface.
 * $$\vec{v}_f$$ = Velocity of the fluid.
 * $$\vec{v}_r = \vec{v}_f - \vec{v}_b$$ = Relative velocity at the surface.

I believe this is the intended definition in your equations. Now, if the absolute fluid velocity $$\vec{v}_f$$ is identical to the surface velocity $$\vec{v}_b$$ at all points on the surface, then the relative velocity $$\vec{v}_r$$ is identically zero. In this case, there should be no contribution from the surface integrals, since the surface flux is zero everywhere. However, your equation still shows a flux.

I think that the correct form should be:

$$\frac{DN_{sys}}{Dt} = \int_{c.v.}^{} \frac{\partial}{\partial t} (\rho \eta) dV + \int_{c.s.}^{} \rho \eta \vec\upsilon_r\cdot \widehat{n} dA$$.

Alternatively, if you want the surface velocity in there, then you can substitute $$\vec{v}_r = \vec{v}_f - \vec{v}_b$$ to obtain

$$\frac{DN_{sys}}{Dt} = \int_{c.v.}^{} \frac{\partial}{\partial t} (\rho \eta) dV + \int_{c.s.}^{} \rho \eta \vec\upsilon_f \cdot \widehat{n} dA -\int_{c.s.}^{} \rho \eta \vec\upsilon_b\cdot \widehat{n} dA$$

155.101.15.233 (talk) 19:48, 27 August 2008 (UTC)


 * Thanks for your thorough explanation. I think both formulations are correct, and it's really just a difference in notation. First, just look at the velocities. You have:


 * $$\vec{v}_r = \vec{v}_f - \vec{v}_b$$


 * I have:


 * $$\vec{v}_f = \vec{v}_r + \vec{v}_b$$


 * Substitute mine into yours and you get an equality.


 * Regarding your thought experiment, you're not summing over the whole control surface. On one side of the control surface the normal $$\widehat{n}$$ is positive, on the other it is negative (regardless of your coordinate system). The equation still comes out to zero. - EndingPop (talk) 22:57, 27 August 2008 (UTC)

I still think that there is an error. Continuing the thought experiment, if $$\vec{v}_f = \vec{v}_b$$ then $$\vec{v}_r=0$$. To be clear, there is no advective flux through the surface in this case, since the surface velocity identically matches the fluid velocity at all points on the surface. In this case your original equation becomes

$$\frac{DN_{sys}}{Dt} = \int_{c.v.}^{} \frac{\partial}{\partial t} (\rho \eta) dV + \int_{c.s.}^{} \rho \eta \vec\upsilon_b\cdot \widehat{n} dA$$,

and this is not consistent with the thought experiment, since in this case there should be no advective flux through the surface. The only way to maintain consistency in this situation is by using the equations I suggested above. Your form is not equivalent.

James Sutherland (talk) 04:01, 29 August 2008 (UTC)


 * I disagree. Sum the fluxes and you get zero. For instance, solve the problem where you have a rectangular control volume in a 1D flow from left to right (use continuity for simplicity). Use 3 cases: (1) The control volume is fixed, the flow moves at some velocity V. (2) The control volume is moving at V along with the flow. (3) One side (it doesn't matter which) moves with the flow at V and the other is fixed. In all three cases the equation holds correct.


 * Now, as I understand advection, the Vb term isn't an advection term. In case (3), above, there's zero flux at the moving boundary. It's there to balance the changing size of the control volume.


 * I did notice an error in the continuity equation, that the partial derivative wrt time should be inside the integral. I'll fix that shortly after this post. - EndingPop (talk) 14:43, 29 August 2008 (UTC)

Let me address each of the cases you propose:


 * Case 1: Fixed Volume. Here $$\vec{v}_b=0$$ by definition, which implies $$\vec{v}_r = \vec{v}_f$$ and your equation is correct.  However, it is only correct because $$\vec{v}_b=0$$, which is more clearly demonstrated by case 2.

$$\int_{c.v.}\frac{\partial \rho}{\partial t} + \int_{c.s.}\rho \vec{v}_f \cdot\hat{n} \, \mathrm{d}A = 0$$
 * Case 2: Surface moves with flow. Here $$\vec{v}_b=\vec{v}_f$$ and $$\vec{v}_r=0$$ at each point on the surface.  Your original equation then becomes (for continuity)
 * However, in this case (surface moves with flow), there is absolutely no movement across the surface. Therefore, continuity must reduce to a statement that

$$\int_{c.v.}\frac{\partial \rho}{\partial t} \mathrm{d}V = 0,$$
 * which is simply a statement that since the CV surface encloses the same mass (by virtue of the fact that $$\vec{v}_f=\vec{v}_b$$) then the mass in the volume must remain constant. Your equation does not respect this in general. For example, if the the velocity field is non-solenoidal $$\left( \nabla\cdot\vec{v} \ne 0 \right)$$ then your equation will show a net change in mass of the system in this case.
 * If we instead use the equation I propose:

$$\frac{DN_{sys}}{Dt} = \int_{c.v.}^{} \frac{\partial}{\partial t} (\rho \eta) dV + \int_{c.s.}^{} \rho \eta \vec{v}_f \cdot \widehat{n} dA -\int_{c.s.}^{} \rho \eta \vec{v}_b\cdot \widehat{n} dA$$
 * then for continuity in this case it reduces directly to the required form $$\int_{c.v.}\frac{\partial \rho}{\partial t} \mathrm{d}V = 0.$$
 * Note that in the simple case where the fluid velocity and properties (e.g. density) are constant in all space and time, then your equation still holds. But that does not make it correct.  It only holds in that case since the erroneous flux contributions cancel each other.  Again, if you consider a flow field with dilatation (say $$v_x=\sin(x), \; v_y=\mathrm{constant}, \; v_z = \mathrm{constant}$$) then you will be able to see more clearly that your equation breaks down since the fluxes on the two x-boundaries will not cancel each other in that case.


 * Case 3: One side fixed the other side moves with fluid. This case is not quite as clear as the previous two, but the same arguments remain.  On the fixed side, we have $$\vec{v}_b=0 \; \Rightarrow \vec{v}_r = \vec{v}_f$$ while on the moving side we have $$\vec{v}_b = \vec{v}_f \; \Rightarrow \; \vec{v_r}=0.$$

The key case to consider here is case 2. It is the one that most clearly demonstrates the error in the original equation.

James Sutherland (talk) 15:53, 29 August 2008 (UTC)


 * I'm going to ask that you go to the cited source (Potter) and check out the formulation of eqns 2.18, 2.23, and 2.24.
 * $$\frac{DN_{sys}}{Dt} = \frac{d}{dt}\int_{c.v.}^{} \rho \eta dV + \int_{c.s.}^{} \rho \eta \vec\upsilon_r\cdot \widehat{n} dA$$ (2.18)


 * It can be shown, by reformulating the first term on the RHS, that that term can be split into two more terms. The first accounts for compressiblity, etc., and the second accounts for a CV of changing size.
 * $$\frac{d}{dt} \int_{c.v.}^{} \rho \eta dV = \int_{c.v.}^{} \frac{\partial}{\partial t} (\rho \eta) dV + \int_{c.s.}^{} \rho \eta \vec\upsilon_b\cdot \widehat{n} dA  $$ (2.23)


 * Substitute this into 2.18 and you get the following...
 * $$\frac{DN_{sys}}{Dt} = \int_{c.v.}^{} \frac{\partial}{\partial t} (\rho \eta) dV + \int_{c.s.}^{} \rho \eta \vec\upsilon_b\cdot \widehat{n} dA+\int_{c.s.}^{} \rho \eta \vec\upsilon_r\cdot \widehat{n} dA$$ (2.24)


 * The final formulation (2.24) is the same as the original (2.18). Moving the d/dt term into the integral can only happen if you account for changes in CV volume some other way, in this case, the Vb term that's created. In fact, if you look back at your version (with d/dt inside the integral and no Vb term), I believe your equation does not work for my Case 3 (assuming steady, incompressible flow).


 * So, either you can have d/dt outside the integral and just the Vr flux term, or you put it in and get a Vb term as well.


 * As a side note, I'm not clear what you mean with the statement regarding $$\left( \nabla\cdot\vec{v} \ne 0 \right)$$. That's not true for an incompressible flow. Is that what you meant? Because I lost track of what was being assumed at that point. If the flow is compressible then first RHS term in 2.24 is nonzero. - EndingPop (talk) 00:17, 9 September 2008 (UTC)

Should be elobrated bit more
I read this article. I don't have in-depth knowledge about the topic but I want the article should elobrate so that one can build there thinking from "particle based computation to control volume based computation" and then corelate the importance of Reynolds-Transport equation. Too much math can take away the beauty of wikipedia article (My personal view..No offence :)). —Preceding unsigned comment added by 203.91.193.50 (talk) 12:10, 4 November 2008 (UTC)

Energy
¿Is there a missing term in the energy equation? first term of the RHS should be:

$$\int_{c.v.}^{} \frac{\partial}{\partial t} (\rho \eta) dV $$

If $$\eta = \frac{\upsilon^2}{2}+gz+\tilde{u}+\frac{p}{\rho}$$ then I believe there is a $$\frac{p}{\rho}$$ term missing in the first term on the RHS. —Preceding unsigned comment added by 200.27.56.84 (talk) 13:25, 25 February 2009 (UTC)

Material vs. Control Volume
Typically, a discussion of Reynolds Transport Theorem will include a discussion of material volume (changing volume and location, zero property flux) vs. control volume (fixed volume and location, arbitrary flux), also sometimes referred to as closed system vs. open system. I see that as a very important distinction (see e.g., Schobeiri's Fluid Mechanics for Engineers: A Graduate Textbook (2010) p. 42, Yamaguchi's Engineering Fluid Mechanics (2008) p. 35, Trif's Basics of Fluid Mechanics and Introduction to Computational Fluid Dynamics (2005) p. 13, Batchelor's classic text Introduction to Fluid Mechanics (1950) p. 131, etc etc. However, this is not mentioned in the article at all, primarily due to the fact that the article uses a single primary reference (Potter and Foss, Fluid Mechanics) and therefore does not incorporate other perspectives (this is the reason I added the "One source" template).  As all can see from the discussion on this page, the article's formulation isn't sufficiently general enough to satisfy all.  Incorporating additional references will certainly help with that - at the very least, it will provide more substantial justification when others disagree with the formulation.

I am planning on incorporating this distinction, and several of the references listed, into the article (particularly the classical text by Batchelor). --Charlesreid1 (talk) 08:48, 1 November 2010 (UTC)

Incorrect use of the material derivative
In the section "General Form" we had an equation with $$\frac{DN_\text{sys}}{Dt} $$, denoting the material derivative, but I think this should be

$$\frac{dN_\text{sys}}{dt} $$ denoting the total derivative, since $$N_\text{sys} $$ is a function solely of time.

This is because $$N_\text{sys} = \int_\text{c.v.}^{} \rho \eta dV $$ is not a function of x. Once we've defined the control volume we integrate over x: the variable no longer appears.

I have made a change to $$\frac{dN_\text{sys}}{dt} $$.219.89.247.21 (talk) 02:22, 7 July 2011 (UTC)
 * Apparently 'It is conventional in fluid mechanics to use D instead of d in time derivatives if it is the derivative of a Lagrangian quantity. ' --عبد المؤمن (talk) 20:21, 16 January 2013 (UTC)


 * The theorem applies to more than just integrals over Lagrangian volumes. The total derivative is appropriate.

Garethvaughan (talk) 05:46, 20 January 2013 (UTC)

Unclear
As an undergraduate studying fluid mechanics in my second year, this article is next to useless for me. It should at least include an 'unclear' template until someone improves it showing its significance in fluid mechanics in a way decipherable (and useful) to the passing engineer. The current intro (three alternative names, a reference to the rule in maths, the person it's named after, a statement that it is useful and a use of the word 'recast' which I have never encountered before) is not useful for those who use it. --عبد المؤمن (talk) 19:15, 16 January 2013 (UTC)

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Resurrection of the Relative/Boundary Velocity Issue
User 147.46.119.152 has made several changes that I have reverted: Thanks for taking an interest in this page and the time to edit it.

Your changes the fall into 2 areas: 1) alteration of the statement of the theorem 2) notation changes

I have reverted those changes in area 1 because I consider them to be incorrect. The central issue is what the velocity should be used. Some discussion of this issue is present in the article, and you can see how contentious this issue has been in the past by reading the talk page. Please note that the literature is inconsistent so no single source is sufficient (except, perhaps, the Reynolds citations that have been used in the article.)

I would request that you review the Reynolds citations before continuing with any further review of this article, and conduct a discussion on the talk page before further editing. Specifically, you described the velocity $v^{b}(x,t)$ incorrectly as being "relative". This velocity is the velocity of the boundary, and is quite independent of the fluid velocity. Please also read the section "A Special Case" of this article.

I have reverted the changes that fall into area 2 also, because of style and accuracy.

a) You incorrectly replaced a partial derivative with a total derivative b) You incorrectly moved that derivative operator outside the integral c) You used a notation not used elsewhere in this article without introduced it. d) You did not apply your change in notation consistently across the article.

Clarification of "Proof for a material element"
The time derivative of J is given by a formula involving the velocity function, which should be connected to the motion and deformation introduced in the subsection for clarity. Indeed, it's just given by the partial derivative of $$\frac{\partial x}{\partial t}$$ if I'm not mistaken, but one would have to check the reference to be sure. Further, in the same formula for the expression of $$\frac{\partial J}{\partial t}$$, it's not clear what the nabla stands for. It could be the derivative along $$x$$ or along $$X$$ coordinates, which is not obvious from the context. Again, if I'm not mistaken, I think it's the former. 88.215.102.65 (talk) 16:38, 29 November 2022 (UTC)


 * Hi, and thanks for your interest in this article.
 * It's been quite some time since I looked into that derivation. It was present in the article before I took an interest in it and it was taken, if I recall correctly, almost verbatim from another source (probably the Belytschko et al reference?)
 * I never quite got my head around that approach, instead preferring the Reynolds derivation in Eulerian coordinates (in-spite of the archaic notation used in those references).
 * I'd just like to encourage you to take the time to review the derivation and make any clarifications that you think necessary. Garethvaughan (talk) 21:45, 22 December 2022 (UTC)