Given a ballon and only one inlet, where some gas enters at a constant rate and the control volume of the balloon changes with time, why does conservation of mass hold?

To give some background, this is an example problem in the Raynold's Transport Theorem section.

What I don't understand and looking for clarification is that, since the mass is added to the system, how is the mass of the system constant. It changes with time, is it not? So how is the conservation of mass holds $\dfrac{dm_{sys}}{dt}=0$? (See "comments" section of the below image)

Schematic of control the volume of a balloon

Here is the original question:

Example problem 3.2

Where eq. 3.1 is $\dfrac{dm}{dt}=0$,

and the eq. 3.16 is the Raynold's Transport Theorem:

$$\frac{d}{dt}(B_{sys})=\frac{d}{dt}\biggl(\int_{CV} \beta \rho d\mathcal{V}\biggr) + \int_{CS} \beta \rho (\mathbf{V_r} \cdot \mathbf{n})dA$$

From this book:

Fluid Mechanics 7ed, Frank M. White

Thank you in advance!


The system is a collection of contents and, because of that, its mass does not change. Hence $Dm/Dt = 0$.

So, the problem you are facing has to do with: "since the mass is added to the system". The mass is not added to the system. The system is the contents of the balloon at the instant of observation (it is like you put a label on every molecule at that instant).

The control volume/surface corresponds here to the balloon and, as stated, it is deformable (it inflates).

After some time (i.e. after the instant of observation above), there will be more air in the balloon, but your system is still the same (i.e. the initial particles/molecules). This is, inside the balloon you will have molecules with label (the system) and molecules without label (the ones that entered afterwards).


The other answer is, unfortunately, missing the point. The Reynolds Transport Theorem relates a Lagrangian view (the system) with an Eulerian view (the control volume) and it does that for the instant they are coincident. This is what is done in the equation with the $B_{sys}$.

On the left side, you have the rate of change of property B in the system (for the instant the system and the cv coincide).

On the right side, you have two components:

(1) the time rate of change of that property in the control volume (for the instant the system and the cv coincide) + (2) net rate (in/out) of that property flow through the control surface (for the instant the system and the cv coincide)

And so, because it is in the instant the system and the cv coincide, the left side equals the right side of the equation.

Because in this case the property is mass, we know (by definition of the system - first part of my answer) that its rate of change (in the system) is zero.

  • $\begingroup$ Wait, I thought material derivative is the time derivative of a physical quantity? $\endgroup$ – Jek Denys Jul 2 '20 at 19:52
  • $\begingroup$ My mistake. White probably implies the material derivative in dm/dt while other authors prefer the Dm/Dt to the explicit about the Lagrangian approach. I will correct my answer. (White would never wrongly write dm/dt instead of the $\partial m / \partial t$, which was my first interpretation - sorry) $\endgroup$ – fdireito Jul 2 '20 at 20:25
  • $\begingroup$ Note what is said in the book answer. "Unsteady flow (the control volume mass increases)" So, not the system, because the collection of particles does not change. This unsteady part corresponds to a non-zero first component in the right-hand side of the equation (the one with the integral over cv). "By the conservation law of mass dm/dt = 0" $\endgroup$ – fdireito Jul 2 '20 at 20:40
  • $\begingroup$ Oh, that's what the Raynold's Transport Theorem does. Got it! Combining the Lagrangian and Eulerian views make a bit more sense. So to reiterate, in reality, the balloon's mass changes, but because we are considering a set quantity of gas in the ballon as a system, even though there is more gas coming in with time, after a while we are still considering the same set of molecules as the system, right? $\endgroup$ – Jek Denys Jul 2 '20 at 20:40
  • $\begingroup$ Yes, that's it! That's the reason why dm/dt = 0, as the book clearly states in the problem resolution. $\endgroup$ – fdireito Jul 2 '20 at 20:45

You're right, as long as gas particles are coming in through the pipe the total mass of the balloon will increase. Temperature, pressure, etc. don't even matter in this context, if even one molecule makes it into the balloon the mass has to, and will, increase. The only possible explanation is that your understanding of the defined system boundaries is lacking. If wherever the gas is coming from at the other end of the pipe is also part of the "system", then naturally the mass of the "system" won't change.

Edit with new information: As I thought, it was a misunderstanding of the defined system boundaries. The obvious solution to the question "how does the balloons mass change over time?" is "as much as stuff goes in, so rho1A1V1". The comment they added is indeed more of a sidenote and not as helpful as it should be. rhoAV is the actual solution. The equation they sat up is actually the balance over the whole system, including the pipe and naturally the change in mass of the entire "system" is how much heavier the balloon gets minus how much lighter the pipe gets... Interestingly enough they defined the system as "system mass within the balloon" so this balance is more or less pointless. The conservation of mass is only valid if no mass is transported over system borders, and in the case of their formula where they watched the entire "system" of pipe plus balloon, dm/dt is indeed 0 since the mass the balloon gains needs to come from somewhere. And if they set the equation value to 0 they can separate the variables and the constant terms to make it a bit cleaner. The whole comment is not relevant for the solution, though.

  • $\begingroup$ I've added the given assumptions of the system. Could you point out where I might have misunderstood it? I mean, I see that it says that the volume and the system expand together, but I have hard time visualizing this into the conservation of mass. If that's the misunderstanding. Could you paraphrase it for me may be? $\endgroup$ – Jek Denys Jul 2 '20 at 18:18
  • $\begingroup$ Doesn't it say in the assumptions that "the control volume mass increases"? In other words, the mass of the system (the balloon) is not constant. $\endgroup$ – passwortknacker Jul 2 '20 at 18:34
  • $\begingroup$ I've updated the question to include the bit where they talk about conservation of mass. So since in the assumption, it states that the mass of the system changes, I don't understand their comment about the conservation of mass. $\endgroup$ – Jek Denys Jul 2 '20 at 19:08
  • $\begingroup$ I updated my answer $\endgroup$ – passwortknacker Jul 2 '20 at 19:35

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