Imagine we have a U-tube with constant diameter which is filled with water and oil in its equilibrium state (where water and oil are fully separated).

Then, we start inserting oil with constant speed at the end of the U-tube where the oil is.

My intuition says that if I take a control volume around the separation boundary between oil and water, the boundary will start moving towards the water and eventually my control volume will be full with oil. It feels just like the oil is a piston pressing the water out of the U-tube.

This implies that if I add $\frac{dV_{oil}}{dt}$ oil volume flow , I would get the same volume of water at other end of the U-tube, i.e., $\frac{dV_{oil}}{dt} = \frac{dV_{water}}{dt}$.

But this seems to contradict the mass conservation in my control volume because:

$$ \rho_{oil} \cdot \frac{dV_{oil}}{dt} = \rho_{water} \cdot \frac{dV_{oil}}{dt} \implies \rho_{oil} = \rho_{water} (WRONG) $$

Now imagine that the U-tube is full with water but the left half has very high temperature and the right half has very low temperature so that we have again the same densities as the previous example, i.e.: $\rho_{water left} = \rho_{oil}$ and $\rho_{water right} = \rho_{water previous}$. Now we insert water from the left side with volume rate $\frac{dV_{waterleft}}{dt}$.

In this case, I feel that the conservation of mass holds because the same material "get transformed" into water with different density. The mass conservation holds but with different speeds: $$ \rho_{waterleft} \cdot \frac{dV_{waterleft}}{dt} = \rho_{waterright} \cdot \frac{dV_{waterright}}{dt} \implies $$ $$ \rho_{waterleft} \cdot u_1 \cdot A = \rho_{waterright} \cdot u_2 \cdot A \implies $$ $$ u_2 = \frac{\rho_{waterleft}}{\rho_{waterright}} \cdot u_1 $$

Could anyone explain to me what I am doing wrong? Or, if I am not doing something wrong, how can I reconcile the first example with the mass conservation?


2 Answers 2


You appear to violate conservation of mass in your oil/water example because you have not included the accumulation/depletion of mass term in the mass balance for the U tube. The total mass in the control volume is $$M=\rho_WV_W+\rho_OV_0$$ The mass balance including the accumulation term is $$\frac{dM}{dt}=(\rho_O-\rho_W)\dot{V}$$where $\dot{V}$ is the volume rate of flow of oil in (and the volume rate of flow of water out). That is, $$\frac{dV_O}{dt}=-\frac{dV_W}{dt}=\dot{V}$$

The general equation we are dealing with here, applicable to both examples, is the "macroscopic mass balance equation" on a control volume: $$\frac{d}{dt}\left[\int{\rho dV}\right]=-\int{\rho (\mathbf{v}\centerdot\mathbf{n})}dA$$where V is the control volume, A is the surface surrounding the control volume, $\mathbf{v}$ is the fluid velocity vector, and $\mathbf{n}$ is an outwardly directed unit normal to the surface surrounding the control volume. The left hand side of this equation is the rate of change of mass within the control volume, and the right hand side is the net rate at which mass is entering the control volume.

  • $\begingroup$ Hello Chester. Thank you for your answer. You are right that there is accumulation of mass due to the fact that a more dense material fills in the tube. But what makes this case fundamentally different from the second case where the liquid is "transformed" to a denser liquid? I am sorry if this is some trivial question that I probably don't know. My background is on electrical engineering and I am approaching this using the electromagnetics viewpoint where we have the similar conservation of charge. $\endgroup$
    – CuriousNik
    Nov 6, 2017 at 18:34
  • $\begingroup$ There if I would see such an accumulation of charge, I would say immediately that there is a source somewhere. Additionally, If I were to insert a denser "current", then the outgoing current would adjust so that again the charge conservation holds. What makes this different in fluid dynamics? Sorry, again, I am studying fluid dynamics on my own and I cannot find a good mental model to reconcile it with EM experience. $\endgroup$
    – CuriousNik
    Nov 6, 2017 at 18:35
  • $\begingroup$ Your mass balance in the 2nd example is incorrect also. The volume in has to match the volume out, so the velocity in also has to match the velocity out. But, here again, the mass in does not have to match the mass out, because of accumulation/depletion of mass within the control volume. So your 2nd example is exactly the same as your first example, except for the presence of an observable interface. If you wanted to observe the change in mass with time, you could always weigh the U tube and see the scale reading go up (or down). $\endgroup$ Nov 6, 2017 at 19:11
  • $\begingroup$ Sorry for insisting too much. But is it true that $v_{in} = v_{out} = v$ for the 2nd case? This implies that we are accumulating/losing mass with time, although there hasn't been any change in the consistency of the fluid in the tube while the time passes. I mean: at $t = 0$ we have $m_{left}(0)= \rho_{1} * V_{1}$, at any other arbitrary time $t$, we have again $m_{left}(t)= \rho_{1} * V_{1}$. Same for the right side with corresponding density and volume. The total mass is preserved. $\endgroup$
    – CuriousNik
    Nov 6, 2017 at 20:01
  • $\begingroup$ One more point of clarification from my part here: left side let's say has very high temperature $T_1$ and right side has very low temperature $T_2$. The densities and volumes depend on the temperature on each side, so for example $\rho_1 = \rho_1(T_1)$. There is still an interface but this time due to temperatures difference. $\endgroup$
    – CuriousNik
    Nov 6, 2017 at 20:21

It's a U-Tube. What defines the fluid balance is pressure.

If we assume the pressure on both ends is the same, then the only thing that defines the balance is how much oil and water is on both sides.

Now you have defined a "control volume" but this does not factor in the pressure balancing requirements of the U-tube, so you don't really have a full picture of the system.

In effect you have taken an isolated part of the system and tried treating it as a whole.

But this seems to contradict the mass conservation in my control volume because

There is no mass conservation in the (arbitrary) control volume. The control volume gives you no connection to the mass in it. It's just a volume you have picked.

if I take a control volume around the separation boundary between oil and water, the boundary will start moving towards the water and eventually my control volume will be full with oil. It feels just like the oil is a piston pressing the water out of the U-tube.

The oil is not pressing the water out of the U-tube, it's only pushing the water out of the control volume.

If you'd chosen another control volume you'd see no change in the level of oil or no change in the level of water

  • $\begingroup$ Hello Stephen. Thank you for the answer and your time. I have a few questions that I don't understand in your answer: a) Why there is no mass conservation in arbitrary control volume? b) I could take as control volume the whole U-tube, still the question applies as I am inserting an oil flow from one end. c) Why does it apply in the second case that I presented? What is fundamentally different? $\endgroup$
    – CuriousNik
    Nov 4, 2017 at 11:50
  • $\begingroup$ You are assuming the change of mass in your control volume equals the change of mass for the entire system. That's an invalid assumption. Why this is wrong is because, as I explained, you are ignoring the effect of forces (pressure) within the tube to distribute the change in mass across the system (and that it is the system that this balance applies to, not the necessarily control volume). So change in mass of the control volume can be quite different from the entire system. $\endgroup$ Nov 6, 2017 at 3:57
  • $\begingroup$ Hey Stephen, what if I take the whole tube as my control volume? Then, my control volume is my entire system and there is influx of mass from one side and outflow from the other. Still the boundary between oil and water will move conserving volume but not mass. Chester's answer below explains this. $\endgroup$
    – CuriousNik
    Nov 6, 2017 at 18:39

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