Exploring the Physics of a Straw & Water: A Detailed Analysis

[MattGeo]

For quite a long time I had just gone by the traditional explanation that when you place a straw vertically into water without fully submerging it and then you withdraw the straw from the water, the water column stays inside because the pressure from the atmosphere acting on the bottom of the straw is larger than the pressure at the top of the straw, because at the top the air has been cut off from the rest of the atmosphere with the finger seal.

When I actually think about this in more detail though it doesn't make sense to me. The water column is in equilibrium after you remove the straw so there can be no net force acting on it. If the pressure at the top of the straw were less than at the bottom then the water would go further up into the straw. It doesn't though.

I know gravity is acting on the water to pull it down and because there is no way for air to be replaced the water is held in place. Sort of like a vacuum lock. At the same time, if the atmosphere is pushing up at the bottom of the straw it must push against the force of gravity but also it must be pushing on the column of air above the water column, and the air will not want to compress. So I feel like something here is connected, in the sense that the water can't fall and draw a vacuum because it is not heavy enough and the atmospheric pressure isn't large enough to compress the air above the water in the straw.

Let's also consider the case where one fully submerged the straw so that is is completely filled with water when you remove it. There is no air column. If the atmosphere is able to support a column of water approx 33 feet high, shouldn't the column of water in like 8 inch straw have an upward net force on it? Same case when there is air above the water column when the straw is partially filled. Shouldn't there be an upward net force on the water that is trying to compress the air in the head space?

So either there is a pressure imbalance which would mean water would move further up into the straw from higher pressure to lower pressure, which it doesn't.

Or, the pressure is the same above and below once you cover the top of the straw with your finger, but if that were the case then gravity should just drain the water from the straw, which it doesn't.

Can someone explain in a detailed way what is actually occurring from start to finish and keep account of the pressure and forces acting? I don't even need equations necessarily, just the conceptual explanation.

 

[anorlunda]

Sort of like a vacuum lock.

That's it. So the gauge pressure in the top goes negative -- vacuum.

It is no different than a mercury barometer.
https://en.wikipedia.org/wiki/Barometer#Equation

When atmospheric pressure is measured by a barometer, the pressure is also referred to as the "barometric pressure". Assume a barometer with a cross-sectional area A, a height h, filled with mercury from the bottom at Point B to the top at Point C. The pressure at the bottom of the barometer, Point B, is equal to the atmospheric pressure. The pressure at the very top, Point C, can be taken as zero because there is only mercury vapour above this point and its pressure is very low relative to the atmospheric pressure. Therefore, one can find the atmospheric pressure using the barometer and this equation:

Patm = ρgh

 

[MattGeo]

That's it. So the gauge pressure in the top goes negative -- vacuum.

It is no different than a mercury barometer.
https://en.wikipedia.org/wiki/Barometer#Equation
Patm = ρgh

The mercury is heavy enough to draw an actual vacuum though if you fill a tube, close it, and invert it. A column of mercury is raised approx 760 mm under atmospheric pressure near sea level. Water would be raised about 33 feet under the same conditions. So that is part of what confuses me. A 12 inch straw filled with water cannot draw a vacuum. There will be no space above the water in the straw. So shouldn't there be a net force pushing it up? If the water column were 33 feet tall the weight of the column would be balanced by atmospheric pressure. The mercury column is balanced by atmospheric pressure after it draws an actual vacuum and its level decreases to the point where its weight is equal to the force exerted by the atmosphere.

So if the water is less than 33 feet, the point where it achieves equilibrium with atmospheric pressure, then how can there not be a net force on it? As in the case with the straw?

 

[anorlunda]

A 12 inch straw filled with water cannot draw a vacuum.

A vacuum is not a binary yes/no thing. It is continuous. So yes the straw makes a vacuum. You can call it partial vacuum if you like.

But the physics of 33 feet of water or 0.33 feet of water are the same.

 

[MattGeo]

A vacuum is not a binary yes/no thing. It is continuous. So yes the straw makes a vacuum. You can call it partial vacuum if you like.

But the physics of 33 feet of water or 0.33 feet of water are the same.

Alright. but how can it be an equilibrium situation if the weight of water in my 12 inch straw is so much less than the weight that atmospheric pressure can support when it counters gravity? Take the mercury barometer as an example. The tube is filled with mercury. But THEN the mercury level decreases and you have empty space above the mercury. It draws a vacuum as the mercury descends, BUT, the mercury will stabilize at a certain level and stop descending once the atmospheric pressure equals the weight of the mercury column. How can that possibly be happening in my example with the small straw containing water? The force on the water from atmospheric pressure at the bottom of the straw is much larger than the weight of the water column.

In my partially filled straw there is air above the water. Does the water stay stationary because the atmosphere at the bottom can't force the water to compress the air above it? But if that were the case, the air pressure would go up and not down.

 

[jbriggs444]

Alright. but how can it be an equilibrium situation if the weight of water in my 12 inch straw is so much less than the weight that atmospheric pressure can support when it counters gravity?

Because it is not a perfect vacuum above. Only a partial vacuum -- reduced air pressure, not zero air pressure.

So you have this twelve inch straw. You are at sea level. You have pulled the straw one inch above the water. You seal the top with your thumb. At this point, before pulling the straw the rest of the way out of the water you have 1 inch of air in the tube at atmospheric pressure. You also have 11 inches of water in the tube.

With your thumb in place, you pull the straw free from the water and hold it there. What is the new situation?

A quick trip to Google says that atmospheric pressure at sea level is 407 inches of water. In order to support our 11 inch column of water, we need a pressure differential of $\frac{11}{407}$ of one atmosphere. By a stroke of good fortune, 11 divides 407 evenly. We need $\frac{1}{37}$ of an atmosphere.

But you complain: "The pressure in that one inch headspace in our straw was one atmosphere. Not $\frac{36}{37}$ of an atmosphere!"

Air can expand. Air does expand. If you look closely at the one inch headspace you should see that it is now 1 and $\frac{1}{36}$ ($\frac{37}{36}$ if you prefer improper fractions) of an inch instead of just 1 inch. After expansion by that much, its pressure is reduced to $\frac{36}{37}$ of an atmosphere.

[I am assuming an isothermal expansion of an ideal gas -- Boyle's law. The real world situation will match this acceptably well]

 

[russ_watters]

Might be useful to draw a diagram and label the forces.

 

[olivermsun]

"Let's also consider the case where one fully submerged the straw so that is is completely filled with water when you remove it. There is no air column. If the atmosphere is able to support a column of water approx 33 feet high, shouldn't the column of water in like 8 inch straw have an upward net force on it?"

I'm trying to think through why this example should have a clearly explainable answer.

As an aside, this assumption of 33 ft high maximum column is probably incorrect anyway: https://www.nature.com/articles/srep16790

Assume the straw is a narrow column, so we can think of it as a 1-d problem. For any water to escape from the bottom of the column, then even assuming there is no virtually air column on top of the water, then for the water to escape at the bottom it requires the water to expand. The net force is the hydrostatic pressure of an 8" column of water minus the ambient pressure (worth making the force diagram as russ_watters suggests above). If we don't consider phase changes or outgassing or anything else weird like that, then there is not much expanding the water is going to do. It follows that hardly any water is going to leak out.

 

[hmmm27]

I'm trying to think through why this example should have a clearly explainable answer.

Draw a free-body-diagram... and use an upside-down glass beaker for illustration - not a straw.

 

[olivermsun]

Draw a free-body-diagram... and use an upside-down glass beaker - not a straw.

Hmmm how does that change the problem?

 

[hmmm27]

Hmmm how does that change the problem?

Just an opinion : a lid at the top (eg: finger on straw) makes the problem look more complicated than it is ; also - unless you know the answer already - with a 1d diagram (simplifying the straw) you could miss something. The beaker needs a handle of some kind, of course.

Other than that, the answer to "But shouldn't there be net upwards pressure on the water inside the straw ?" is simply "Yes". [edit: ignore that ; brain's on vacation]

Think the FBD difference between a bucket being pulled out from underwater, right-side-up (closed bottom) vs. upside-down (closed "top").

 

[olivermsun]

Other than that, the answer to "But shouldn't there be net upwards pressure on the water inside the straw ?" is simply "Yes".

Think the FBD difference between a bucket being pulled out from underwater, right-side-up (closed bottom) vs. upside-down (closed "top").

The original post said, "The water column is in equilibrium after you remove the straw so there can be no net force acting on it. If the pressure at the top of the straw were less than at the bottom then the water would go further up into the straw. It doesn't though.
...
"If the atmosphere is able to support a column of water approx 33 feet high, shouldn't the column of water in like 8 inch straw have an upward net force on it?"

I think the asker understands that the water column doesn't have a net force on it, as shown by the fact that the water doesn't move further up the straw by itself. My reading of the question is, Why there isn't a net force on a short column given that the (same) atmospheric pressure is able to support even taller columns?

The upside-down bucket should be another example of the same physics.

 

[hmmm27]

Well, it's not moving so net force is zero ; a reasonable presumption (For some reason the phrases "net force" and "column of water" are tripping me up). I think the OP is trying to figure out why the straw doesn't go shooting off into the atmosphere.

Again, free-body-diagram : calculate the net force from the gross forces, not from the presumption.

But, I think the OP has left the building.