hydraulic head of a bubbly mixture 2

[Leclerc]

This is not homework.

A. would argue that the hydraulic head at a point in a bubbly mixture is equivalent to the average density x g x height of liquid surface above the point. Period.

B.(Leclerc)would respond that he is not so sure. He would argue the hydraulic head at a point in a bubbly mixture, where the liquid exists as a continuous path between the point and the surface is density of liquid x g x height of liquid surface above the point. He would explain his argument by saying that in a lightly bubbled liquid where there is a liquid continuum, buoyant bubbles are not "tied" to the liquid, but are free to rise and slip past the liquid with an accelerating velocity. This slippage is the point of difference: cases of hindered bubble rising and stable foams would be as A.

A. might argue that it is all a matter of degree: all liquids are, to some extent, viscous, and therefore there is always friction between bubbles and liquid. He might go on to say that in a vertical cylinder, for example, the mass of the bubbly mixture is the mass of the liquid fraction x g gives the force downwards on the cross sectional area of the cylinder.

B. would still say he wasn't sure; what about conical flasks ...

Gentlemen, who's right?

 

[hacksaw]

Part of the problem is that you need to re-examine how you define density.

Personally I like the Guiness approach to generating the fine distrubution of bubbles, some of which are heavier than the liquid...

 

[DickRussell]

Back to Leclerc's last post. We agree on the buoyancy issue. That is just Archimedes' principle. The upward force equals the weight of the fluid displaced, and for a submerged object the net upward pressure is the difference between pressures at the top and bottom of the object, which arises from the weight of the fluid between those two elevations.

As to the pressure inside the bubble being that at half way up the bubble, where did you get that? I argue that the pressure would be that of the fluid at the bottom. Suppose you had a gas-filled cylinder open at the bottom but bolted to the bottom of the tank, with a thin membrane separating the gas from the liquid. Wouldn't the gas pressure be just that of the liquid at the elevation of the membrane?

As to the height of the liquid in the level glass outside the tank, start with no gas bubbles at all. The two levels are the same. Then introduce a significant volume of small massless bubbles into the liquid, well-dispersed so that the liquid phase is indeed continuous. The level of the liquid at the top of the tank will become higher, according to the collective volume of the bubbles, yet the total mass of fluid above the bottom of the tank remains the same. The pressure at the bottom must be the same (total weight/total area - straight sides in the vessel). Therefore the level in the level glass outside (no bubbles in that fluid) will remain the same, and below the level of the bubbly fluid inside the tank. Over time, as the bubbles rise to the top and disperse, the volume occupied by the remaining bubbles in the liquid diminishes, and the top level of the liquid gradually drops to that in the level glass outside - which never changes.

If you are going to argue to the contrary, please address the issue of total mass/area. If you are correct and I am wrong, show me convincingly and I'll concede.

 

[Compositepro]

The question here is one I pondered many years ago and can seem very complicated but is actually not. You must remember that Archemedes Principle and other equations are based on the concept of equilibrium. In classical mechanics the analysis of systems in equilibrium is called "Statics". That does not mean that nothing is moving - it means that all forces are in balance and that the net force is zero so the system is not accelerating.
Much of the confusion is based on changing the definition of the "system" or equilibrium and not realizing it.

All forces must balance out to zero or else something is accelerating and "Statics" no longer applies. In the case of a rising bubble in a large tank it first accelerates and then reaches a constant velocity. There is a period where Dynamics is used and not Statics. When you have a steady stream of bubbles rising in single file then you can apply Statics. And what is the answer? The height of the liquid in the the tank will be slightly higher at the point where the bubbles are rising and there will be an upward flow of bubbles and liquid in the center of the tank and a downward flow of liquid at the walls. On the liquid surface there will be a flow from the center to the walls which results in a height gradient. The viscous drag forces at the wall balance-out the viscous drag of the rising bubbles. So, in equilibrium, when all of the flow patterns in the tank are steady, the pressure at the bottom-center of the tank is very slightly lower than at the bottom-wall where the level gage is.

So, you just have to be careful that the "system" you describe is, in-fact, in equilibrium. On one scale it may be but on a smaller scale it may not be. Just remember to balance the forces. F=ma and in equilibrium a=0.

 

[Leclerc]

Dick Russell,
I believe that it is personal preference for me to measure hydrostatic pressure in a liquid as height (or depth) x liquid density x g. I can see how, within your vessel with vertical walls of A cross section area, the force, ie mass x acceleration = height x A x density x g. When this force is applied downwards on the area, A, the force/ unit area is height.A.density.g/A, which is the same as the expression I use. However, liquid hydrostatic pressure does not merely act in a downwards direction; it acts sideways and upwards and can be transmitted through the continuous phase, if there is a continuous phase. If the vessel were not a right circular vessel but had a much more complicated shape and such that, say, no part of the bottom could be seen from the liquid surface when viewed vertically down, then I am able to visualise and calculate hydrostatic pressure directly from a knowledge of the vertical distance below the surface and the density of the continuous phase.


I think that Compositepro's thread is close to the heart of the problem and is helping me to better visualize what is happening and where I may be going wrong. Reading my opening thread again, I seem to argue that there is hindered bubbling and then there is bubbles within a liquid continuous phase. Compositepro's argument is that apart from an initial acceleration period, there is always hindered bubbling, since viscous forces restrict bubble velocity to terminal values and these forces will affect hydrostatic pressure.

The only outstanding piece of not-understanding I now have is this: given that Compositepro's explanation is approaching the truth of the matter, how come that people know to calculate pressure simply as depth x average density x g? What is the connection between the insight into what is happening and the terms in the equation?