Mass conservation for water and oil flow

Question

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?

StephenG · Answer

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

Chet Miller · Answer

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:
$$frac{d}{dt}left[int{rho dV}right]=-int{rho (mathbf{v}centerdotmathbf{n})}dA$$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.

Mass conservation for water and oil flow

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