Why should the mass per second be constant in the equation of continuity? Why can't the mass accumulate inside a tube?
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3$\begingroup$ Which equation of continuity (there is one for every conserved quantity)? Also, what tube? $\endgroup$– ACuriousMind ♦Commented Mar 25, 2015 at 14:28
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1$\begingroup$ Krishna could you please edit all the necessary clarifying details into your question? $\endgroup$– David ZCommented Mar 25, 2015 at 14:33
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2$\begingroup$ Uh, why is there even a question as to what this post is about? The terms mass and continuity equation along with the tag fluid-dynamics should be a clear indicator as to what this question is asking. $\endgroup$– Kyle KanosCommented Mar 25, 2015 at 14:38
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1$\begingroup$ @KyleKanos: I should not have to look at the tags to figure out what a question is about, and I should not have to guess which equation is being talked about when it is perfectly possible to simply include the equation into the post. Note that I did not vote to close as unclear (because it is not really unclear what is being asked, agreed), but merely wanted to prod OP into making the question more accessible and easy to read and understand. $\endgroup$– ACuriousMind ♦Commented Mar 25, 2015 at 14:57
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1$\begingroup$ @ACuriousMind I disagree -- I think you should look at the tags to understand the context. That's what they are there for and that's why we don't want people to start putting "Fluid-dynamics: Why is mass constant" in the titles. Tags exist to explain the context (as well as allow searching etc). $\endgroup$– tpg2114Commented Mar 25, 2015 at 14:58
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4 Answers
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You are referring to the equation:
$$\frac{\partial \rho}{\partial t} + \frac{\partial \rho u_i}{\partial x_i} = 0$$
which is the conservative form of the continuity equation in Eulerian form (fixed domain, fluid moving through it). This can also be written as:
$$\frac{D \rho}{D t} = 0$$
which is the Lagrangian form (the density of a moving region of space remains constant).
The question you asked is "Why is mass per second constant" and the direct answer is, it isn't. It is equal to the negative of the convection of mass.
Now, to what I think your question is trying to ask -- why are there no source terms in the equation? What comes into the domain has to either stay in the domain or leave the domain. Both of those options are accounted for by the continuity equation when used in a time accurate sense (in the steady state case, you always have what comes in must leave unless you blocked off the domain or something). So if you inject more mass by either increasing density or increasing injection velocity, you will see mass accumulate inside the domain and eventually reach a new equilibrium. All of that is built into the equation.
Some things change when we consider a chemically reacting flow. There, instead of the traditional continuity equation, we have conservation of partial masses:
$$\frac{\partial \rho_i}{\partial t} + \frac{\partial \rho_i u_j}{\partial x_j} = \dot{\omega_i}$$
where now the total density is $\rho = \sum \rho_i$. Now we actually have a source term on the right hand side, $\dot{\omega}$ which says that the partial densities change based on chemical reactions and due to convecting them around. It is important to note though that since we cannot create nor destroy mass, even through these chemical reactions, $\sum \dot{\omega_i} = 0$ and so when you add up all of the chemically reacting partial density equations, you recover the continuity equation I first provided.
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If your tube has a hole in which mass can flow out, then we can lose mass at a rate $Q$. Similarly, if your tube has an inlet, then the mass can accumulate at a rate $P$. Thus, the time rate of change of mass would be $$ \frac{dm}{dt}=P-Q $$ However, with a tube, we usually consider it closed, such that $P=Q=0$, leading directly to $$ \frac{dm}{dt}=0 $$ from which the continuity equation can be derived.
answered Mar 25, 2015 at 15:12
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Do you mean this equation
$A_1v_1=A_2v_2$
In that case, we are supposing the liquid to be an ideal fluid i.e. it is incompressible fluid. That should mean that mass entering per second in the tube is equal to mass exiting per second.
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$\begingroup$ Then the equation of continuity doesn't hold. Then mass can be accumulated $\endgroup$– SaadCommented Mar 25, 2015 at 16:07
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As others have pointed out its an ideal fluid and it can't be compressed ..and all that ....here is an intuitive aid for you....Let us say I have a plastic pipe or a tube (a very odd one perhaps with lots of twists and turns..) Then let us say some volume of water goes inside it ..then don't you think that the wate volume coming out must be the same...also another example is when you were pouring water on your little garden plants you must have noticed..That if you closed the opening a little bit ...water begins to come out very fast..and that my dear friend is the equation of continuity ...how do you suposse the water or fluid is going to accumulate here...!! (After all its an ideal case...and so obviously you can't compress the liquid or make more assumptions and term it as IDEAL the conditions for fluid flow to be ideal has been clearly specified you can't alter it and still try to use the equation of continuity!!) ....also one thing that you yourself said is "laminar flow through a tube" ...but its not necessary it has to be a pipe..LAMINAR is the word here which shows why...which..is .because streamlines can never INTERSECT...so not necessary to take a pipe we can imagine a tube of flow also ...made of streamlines..
