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Re: [Phys-L] Another Fluid/Density 'Problem'





-----Original Message-----
From: Phys-l [mailto:phys-l-bounces@phys-l.org] On Behalf Of Chuck Britton
Sent: Wednesday, January 29, 2014 4:52 PM
To: Phys-L@Phys-L.org
Subject: [Phys-L] Another Fluid/Density 'Problem'

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An Old Fashioned Glass Milk Bottle with a narrow neck is delivered to your
doorstep in the morning and because it is no homogenized, the Cream has
risen up into the narrow(er) region.

The fluid pressure exerted on the bottom of the bottle is (rho) g h.
Where (rho) is the average density of the column of milk/cream that extends
from the top surface to the bottom.

I'm going to try this without looking at what others are saying.

All quantities (volume/VOLUME, mass/MASS, rho/RHO) refer to the milk and cream, NOT to the bottle. I define average density as mass of liquids/volume of liquids. For the separated liquids, this is a squirrely idea, but is IMHO the best definition. (rho_cream+rho_milk)/2 is the worst.

rho = total mass (of milk and cream)/volume of milk & cream, but we have to distinct regimes: milk and cream. If they have different densities, then the pressure at the bottom depends on the height of each regime, no matter what order they are in. when everything is "truly" mixed then only the total height of the mixture matters. The profile of the bottle means that a given volume of cream will produce a lower (partial) pressure in the wider part than in the narrow part (because the column will be higher in the narrow part). But, upon mixing, we are raising some of the more dense milk to a higher position which could increase pressure.

Now we will thoroughly MIX the bottle of milk and we note that the average
density of this same column of milk is GREATER than (rho). Let's call this new
density (RHO).
RHO=MASS/VOLUME
mass (before) = MASS (AFTER)
If the density increase is a statement of fact then RHO = MASS/ VOLUME (by definition) means that the VOLUME is less than original volume. Okay, alcohol and water do this. I wasn't aware that milk and cream do. Interesting. Because the shape of the bottle doesn't change, but the VOLUME is less, the HEIGHT of the mixture is less.


How can the pressure have increased while the area of the bottom has
stayed constant?
Pressure at the bottom has nothing to do with the area. It has to do with density and height. We must compare (rho)g(height) with (RHO)g(HEIGHT). Drop the g's because they don't make any difference between the 2 situations.

rho = mass/volume and RHO= mass/VOLUME, so now we compare mass*height/volume with mass *HEIGHT/VOLUME. Drop the mass because they are the same.

If the height/volume ratio is less than HEIGHT/VOLUME, then the pressure has increased. If the bottle was a cylinder, this wouldn't be possible because the ratio would be 1/(pi*r^2) for both, but the bottle has a neck. I suspect (haven't proved it yet) that this change in profile causes the ratio to increase as the height decreases. I've tried the calculation for a truncated cone (r = a-by where b*height and b*HEIGHT are both less than a) and I found that the pressure would decrease (!), if I did my analysis correctly. Maybe a different profile would behave differently.

OTOH... The changing radius portion of the bottle does produce a downward force on the liquid which results in a force greater than mg from the bottom. Then the milk/cream is mixed, this changes the density at the "neck in" portion of the bottle which in turn COULD produce more down force, depending on how much the density increases, versus how fast the height drops.

How can we accept a greater amount of downward fluid force caused by
'just' mixing the milk?

See previous statement.

(This is reprised from back in the last century.) Probably presented better
back then.
;-)
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