I came across the following lines that appear after the derivation of equation of continuity for the steady flow of an ideal liquid in Resnick, Halliday, Kranes's Fundamentals of Physics:

The equation of continuity states that if within any volume element of space (not volume of fluid) there are no sources(where additional matter is introduced into the flow) or sinks(where matter is removed from the flow),then the total mass within the volume element must remain constant. In more general cases, if sources or sinks are present, the equation of continuity gives the mathematical representation of the very reasonable assertion that the rate of outflow OR inflow of matter is equal to the rate at which the mass contained in the volume element is changing.

Now, the statement in bold is what is troubling me. Is it not true that the mass contained in a volume element is simply ρdV where ρ is the density of the liquid at the location of some volume element of size dV. Now, since steady flow is assumed, ρ does not change with time. Hence, regardless of whether there are sources or sinks in the element, the mass of the fluid within the element also should not change with time ( as the mass is a function of only ρ and dV which is also constant). So where comes the question of "mass contained in the volume element changing"(refer blockquote)?

Also note the usage of OR in the passage. Does it mean the the rate of outflow is equal to rate of inflow? Are these two rates equal even when there are sources or sinks in the element?

Can someone please elaborate?

  • $\begingroup$ I think the first sentence of the quoted passage is important here. It emphasises that they are talking about volume elements of space, not volume elements of the fluid. $\endgroup$ – jm22b Nov 25 '17 at 10:54
  • $\begingroup$ I am sorry, I am unable to figure out where I have failed to take that into consideration. $\endgroup$ – AVU Nov 25 '17 at 11:12
  • $\begingroup$ It is like the mass "is labelled". Flow is assumed. $\endgroup$ – Alchimista Nov 25 '17 at 12:06

Well, if there are sources or sinks, the density can change in time since mass is being added or removed directly inside the volume, that is, not by just entering it from outside or leaving it through the surface of the volume.

If there are no sources or sinks, however, if the mass inside the volume is to change, it needs to get inside or outside through the surface.

Concerning the "OR": It can happen that there are parts of the surface through which fluid is entering the volume element and at the same time there are other parts of the surface through which fluid is leaving the volume. However, if you sum over all contributions along the whole surface, there can be either a net flow to the inside OR to the outside. The net flow cannot be to the inside AND to the outside at the same time.

  • $\begingroup$ The OR issue is cleared. But what if the fluid is ideal,i.e incompressible. How can the density ever change? $\endgroup$ – AVU Nov 25 '17 at 11:03
  • $\begingroup$ Well, in this case both terms in the continuity equation are zero separately. See also en.wikipedia.org/wiki/… $\endgroup$ – Photon Nov 25 '17 at 11:12
  • $\begingroup$ May I request you to please put it in simple terms. I am unable to reach a conclusion :). $\endgroup$ – AVU Nov 25 '17 at 11:15
  • $\begingroup$ In an ideal fluid the density is the same everywhere and at every time, so we have $\partial_t\rho=0$ and $\nabla\rho=\vec 0$ and therefore also $\nabla\cdot \vec v=0$. This does not contradict to the continuity equation which then states that $0+0+0=0$. :) $\endgroup$ – Photon Nov 25 '17 at 11:20
  • $\begingroup$ Just one more thing to be clarified: If there are sources or sinks in a volume element, can the flow ever be steady? I don't think this is possible as you clearly said : "if there are sources or sinks, the density can change in time since mass is being added or removed directly inside the volume". Does this also mean that the equation of continuity can never be applied to unsteady flow ( as the derivation of the same was done assuming steady flow)? $\endgroup$ – AVU Nov 27 '17 at 6:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.