# Equation of continuity

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?

• 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. – jm22b Nov 25 '17 at 10:54
• I am sorry, I am unable to figure out where I have failed to take that into consideration. – AVU Nov 25 '17 at 11:12
• It is like the mass "is labelled". Flow is assumed. – Alchimista Nov 25 '17 at 12:06

• 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 $0+0+0=0$. :) – Photon Nov 25 '17 at 11:20