Question How does pressure increase with depth in water?

[Tracy]

I think the pressure will settle between the initial readings of x and y.

I'm sending the drawing my with son to his high school physics class tomorrow.

Eventually I'm going to have to do this experimentally. Unless @Tracy wants to do it first (like How does moisture enter tanks? ).

I could mock it all up, but I can save you the hassle and just tell you the depth gauges will read the same in that drawing.

 

[huwporter]

Which argument

Let me restate my initial argument using a visual aid.

View attachment 808882

In the right bottom corner of each body of water is a depth gauge. One is labeled x, the other y.

My argument is that x will show a shallower depth than y.

Edit: I should note that the "roof" of the x chamber is magically self-supporting in this scenario and so adds no pressure to the water below it.

It bakes my noodle too, but your diagram convinced me that the pressure at X and Y will indeed be the same.

A pressure gauge at the very bottom of the slender pipe just above where it opens up will read the same as a pressure gauge at the same depth in the wide pit. The diameter of the pipe clearly doesn't matter at that point.

So, for X and Y to be different, at what point do you think the pressure would then diverge as you move the two gauges deeper? Would the pressure suddenly drop as soon as you moved the gauge down into the wide part of the left chamber? Or what?

 

[lowwall]

A pressure gauge at the very bottom of the slender pipe just above where it opens up will read the same as a pressure gauge at the same depth in the wide pit. The diameter of the pipe clearly doesn't matter at that point. So, for X and Y to be different, at what point would the pressure then diverge as you move the two gauges deeper? Would the pressure suddenly drop as soon as you moved the gauge down into the wide part of the left chamber? Or what?

That is a very interesting question. My hypothesis is that the pressure would drop to just a bit more than 1 atmosphere as you moved the gauge just under the "roof".

Think of it this way. Fix your pressure gauge in position just under the roof somewhere away from the pipe. Take a reading. Now double the cross section of the pipe, so it's only half as high. Did your reading change?

Now excavate just enough of the roof for the entire column of water in the pipe to collapse into the pit. The gauge is now under a couple of cm/inches of water. Did the reading change?

I would expect the pressure reading to stay the same in all three scenarios. How could it be different? The all have the same weight of water above the gauge. Why would the shape of the column matter?

 

[stiebs]

I would expect the pressure reading to stay the same in all three scenarios. How could it be different? The all have the same weight of water above the gauge. Why would the shape of the column matter?

This is indeed a mind bending scenario. But take a look at the link below. The formula for pressure includes the fluid density, effect of gravity and the depth of the fluid. It does *not* include the mass or surface area of the fluid.

Pressure

 

[Brett Hatch]

That is a very interesting question. My hypothesis is that the pressure would drop to just a bit more than 1 atmosphere as you moved the gauge just under the "roof".

Think of it this way. Fix your pressure gauge in position just under the roof somewhere away from the pipe. Take a reading. Now double the cross section of the pipe, so it's only half as high. Did your reading change?

When you say "double the cross section, so it's only half as high," you have in mind that the increased cross section causes the same amount of water to spread out a bit horizontally, right? In that case, yes, the pressure reading decreases.

Now excavate just enough of the roof for the entire column of water in the pipe to collapse into the pit. The gauge is now under a couple of cm/inches of water. Did the reading change?

If you dig out a little piece of the ceiling, the water in the pipe will go down a bit. So yes, again, the gauge will read a smaller quantity of pressure. If you continue to dig out the ceiling, bit by bit, the gauge will continue to decrease. Eventually, the volume of the ceiling will exceed the original volume of water in the pipe. At that point, further excavation will have no effect. So there is a diminishing return here, that asymptotically approaches zero

I would expect the pressure reading to stay the same in all three scenarios. How could it be different? The all have the same weight of water above the gauge.

It can be different if the weight of water above the gauge is not the relevant quantity.

Why would the shape of the column matter?

It doesn't. That's the OP's question. What matters is the height delta between the gauge and the water surface.

Maybe an example of a flexible container will help illustrate -- we're all divers here, we've thought about the underwater balloon. Consider a flexible, sealed balloon with a fixed amount of air in it, say the size of a volleyball at the surface. That's about 8 inches in diameter. I think we would be in agreement that if you submerged that balloon, the pressure would increase, and it would cause the air inside the balloon to compress, so the balloon would shrink. Say you bring it down 10m / 33ft, the balloon should have shrunk to about half the size, about 4 inches in diameter, like a grapefruit.

If that point at 10m/33ft in your picture is where the thin pipe meets the large basin below, what happens if we move the balloon around:
1) If we move it to the right 10m/33ft , does it change size?
2) If we move it straight down another 10m/33ft, does it change size?
3) If we first move it to the right, then move it down, does it change size?

It sounds like in your mind, the balloon would have to decrease in size as you move it to the right. Have I understood you correctly?

In my mind, 1) is no, 2) is yes, and 3) is no for the horizontal movement, and yes for the vertical movement

 

[Tanks A Lot]

You are certainly not alone in finding this not straight forward and I too struggled when I tried to wrap my head around it a long time ago. There is even a proper name for it, the hydrostatic paradox.
Allegedly, Blaise Pascal did not trust his intuition either and did and experiment on this - Pascal's barrel.

I'm not entirely sure if my reasoning is correct and people with a better understanding of physics will correct me, but here is how I convinced myself that my intuition, that lowwals scenario is correct, is indeed wrong:

Pressure is defined as follow: Pressure (P) = Force (F) / Area (A)

P = F / A

Force is supplied by the mass water column (m) and the gravitational constant (g): Force (F) = Mass Water (m) * Gravitational Constant (g)

F = m * g

We can substitute the mass of the water (m) with the volume (V) and density (d): Force (F) = Volume (V) * Density (d) * Gravitational Constant (g).

F = V * d * g

The Volume depends on the height (h) and area (A), so we can substitute those as well: Force (F) = Height (H) * Area (A) * Density (d) * Gravitational Constant (g).

F = H * A * d * g

We are finally at the stage where we can substitute this derived formula in our first one.

Pressure (P) = (Height (H) * Area (A) * Density (d) * Gravitational Constant (g)) / Area (A)
P = (H * A * d * g) / A

As can be easily observed now, the area will cancel each other out, it becomes irrelevant. The pressure will only be effected by the effective height of the water column, and not the area.
The above of course only applies to static scenarios. If there is any flow of liquids, that flow will have of course effects on pressure according to Bernoulli's principle.

The above is in fact exactly how hydraulic pumps achieve their incredible power for the little work they put up.

Some reading which may shed further light on this:
Communicating vessels
Pressure in liquids

 

[stiebs]

Consider a flexible, sealed balloon with a fixed amount of air in it, say the size of a volleyball at the surface. That's about 8 inches in diameter. I think we would be in agreement that if you submerged that balloon, the pressure would increase, and it would cause the air inside the balloon to compress, so the balloon would shrink. Say you bring it down 10m / 33ft, the balloon should have shrunk to about half the size, about 4 inches in diameter, like a grapefruit.

Not quite right. If the diameter is 8" at the surface, at 10m/33ft the volume will halve, not the diameter. So when the pressure doubles, the diameter will actually be 6.35" (Using for the formula (4/3) × π × r³)

 

[stiebs]

Here's a YouTube video will explains reasonably well.


The mind-bending concept here comes about when trying to apply the physics of a solid to a fluid. If you have an oddly shaped solid object, the entire mass of the object is concentrated on the surface on which it is resting.

With fluids, there is an additional dimension of pressure - the pressure keeping the fluid in the container, which is exerted on the sides of the container. And the in the narrow pipe scenario, the ceiling of the large chamber is what exerts the pressure against the water, which is needed to keep it in the container.

 

[stiebs]

One thing that certainly is obvious, is that this is not a concept that is easily grasped. The interwebs are full of forum posts and physics educational discussions trying to explain and understand it!

 

[lowwall]

Pascal's law supplies to enclosed systems like hydraulic rams. I do not believe it applies to my example above.

I don't have any trouble with Pascal's law. As I stated above, it's a lever except you replace a metal bar with a contained incompressible fluid.

Edit: going to bed. I'll watch the YouTube video tomorrow.