# Crazy water pressure formula implication?

kotreny
Hey everyone, I just thought of a mind-boggling "thought experiment":

As a reminder, the formula for water pressure acting on an object due to weight is:

(Water density)*(Acceleration of gravity)*(Depth of object underwater)

Note that the shape of the container doesn't matter.

Imagine a water-filled tank the size of the Pacific Ocean (exact measurements won't be necessary). The tank is completely filled with water and is sealed shut, so any pressure other than water pressure is absent. Now imagine that at one corner of the tank there is an incredibly narrow "chimney" about ten nanometers in diameter and extending about ten kilometers above the top of the tank. This tube is connected directly to the big tank and is sealed shut, though the water level can be changed at will. I didn't do the math, but I'm pretty sure a few drops of water would raise the water level by at least hundreds of meters.

The weird part: According to the formula, water pressure throughout a container is dependent only upon the height of the container, assuming gravity and density remain constant. Technically, our imaginary tank is over ten kilometers tall, since the chimney is part of the tank. Does this mean that adding just a few drops of water can create enormous pressure everywhere in the ocean-sized tank? Seems to violate the law of conservation of energy, doesn't it?

Yes, I know that pressure is force/area, so the weight of the drops is concentrated like a needle. But can anyone tell me how--preferably on a molecular level--that force is distributed to every part of a body of water as big as the PACIFIC OCEAN?

Thanks

Nice question but the increased pressure at the bottom of the narrow tube will cause water to be forced out until equilibrium is reached .In fact for a tube so narrow surface tension effects will play a major part and only some of the water will flow out.

The weird part: According to the formula, water pressure throughout a container is dependent only upon the height of the container, assuming gravity and density remain constant.

The pressure at any point in the container is equal to $\rho g h$. However, the h refers to the height of the water column directly above the point. That is, the water pressure is only the same at points that have the same fluid column heights above them, and not the tank's total height.

CS

kotreny
Nice question but the increased pressure at the bottom of the narrow tube will cause water to be forced out until equilibrium is reached .In fact for a tube so narrow surface tension effects will play a major part and only some of the water will flow out.

Surface tension is irrelevant to my question, but thanks for the reality check anyway.

So basically, the weight of the water in the tube pushes the water below it aside? Then the water that was just pushed aside pushes other molecules aside, then those molecules push others aside, etc., spreading the pressure like a chain of dominoes? Maybe!

At first, I disregarded the domino analogy because I thought that the molecules would eventually run out of steam (pun not intended), while dominoes would keep going because of gravity. However, thinking twice, I was reminded of two things:

1. The molecules are pushing through a vacuum, so friction is absent (hydrogen bonding and van der waal's forces might complicate things though).

2. More importantly, the molecules DO have gravity on their side, since they're being propelled by the weight of the tube's water!

These might seem like simple insights, but they're good enough for me.

One more question. If what I said is true, then I wonder how fast the pressure spreads through our ocean-sized tank? Instantaneously, or slower?

The pressure at any point in the container is equal to $\rho g h$. However, the h refers to the height of the water column directly above the point. That is, the water pressure is only the same at points that have the same fluid column heights above them, and not the tank's total height.

CS

If my understanding is correct, pressure is scalar and applies equally to all parts of a container, regardless of shape or size.

If my understanding is correct, pressure is scalar and applies equally to all parts of a container, regardless of shape or size.

Pressure is transmitted equally and undiminished in all directions, but the actual value of the pressure due to the hydrostatic head pressure of the fluid at that point varies with the height of the fluid column.

For example if the pressure at h1 = 1 psi and h2 = 2 psi, then an increase in pressure at h1 of 3 psi (h1 now is 4 psi), h2 will be increased equally to 5 psi.

Does that help?

CS

One more question. If what I said is true, then I wonder how fast the pressure spreads through our ocean-sized tank? Instantaneously, or slower?

Pressure waves in fluid travel at different speeds depending on the fluid properties. The maximum they will travel is the speed of sound for that particular medium.

$$c = \sqrt{\frac{E}{\rho}}$$

where,

E is the bulk modulus of elasticity of the fluid
rho is the fluid density

CS

Mentor
The pressure at any point in the container is equal to $\rho g h$. However, the h refers to the height of the water column directly above the point. That is, the water pressure is only the same at points that have the same fluid column heights above them, and not the tank's total height.

CS
I think you may have said that in a way you didn't intend - the "...directly above them..." part doesn't apply. The fluid height is measured from the level of the chosen point to the top of the tank, regardless of where horizontally those two points are wrt each other.

kotreny
Pressure is transmitted equally and undiminished in all directions, but the actual value of the pressure due to the hydrostatic head pressure of the fluid at that point varies with the height of the fluid column.

For example if the pressure at h1 = 1 psi and h2 = 2 psi, then an increase in pressure at h1 of 3 psi (h1 now is 4 psi), h2 will be increased equally to 5 psi.

Does that help?

CS

Oh! Yes, I knew that. I must have misinterpreted what you said earlier. For example, the pressure at the top of the tube is zero, and at the bottom it'd be $${\rho}{g}$$(10000m) if surface tension didn't play a role. Thanks anyway.

Pressure waves in fluid travel at different speeds depending on the fluid properties. The maximum they will travel is the speed of sound for that particular medium.

$$c = \sqrt{\frac{E}{\rho}}$$

where,

E is the bulk modulus of elasticity of the fluid
rho is the fluid density

CS

Hmm! So I take it my domino model was pretty much right?

I think you may have said that in a way you didn't intend - the "...directly above them..." part doesn't apply. The fluid height is measured from the level of the chosen point to the top of the tank, regardless of where horizontally those two points are wrt each other.

Yeah that probably was a very poor way for me to word that! Long day at work! The total column height above that point (like in the attachement) is what I meant. Thanks for catching that Russ. I hate adding confusion to threads!

Photo courtesy of: http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Pressure/HydroStatic.html [Broken]

CS

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Idoubt
I'm no expert, but wouldn't a chimney that small cause capillary action? Due to adhesive forces I think.

You are right Idoubt.This point has already been mentioned with reference to surface tension but capillary action is probably a better way of describing it.

TheInversZero
Given most other factors as assumed irrelevant for this application.

It would be prudent to take the the column of water for that chimney as a single column of many columns with in the whole tank.

You would then take the the area of that chimney column and find how many others of that same area would fit in the tank than take an average of all to find the net increase in tank pressure.

Mentor
Sorry, but that's completely wrong. Pressure works as we said: it is just p=rho*g*h.

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BenchTop
Hey everyone, I just thought of a mind-boggling "thought experiment":

As a reminder, the formula for water pressure acting on an object due to weight is:

(Water density)*(Acceleration of gravity)*(Depth of object underwater)

Note that the shape of the container doesn't matter.

Imagine a water-filled tank the size of the Pacific Ocean (exact measurements won't be necessary). The tank is completely filled with water and is sealed shut, so any pressure other than water pressure is absent. Now imagine that at one corner of the tank there is an incredibly narrow "chimney" about ten nanometers in diameter and extending about ten kilometers above the top of the tank. This tube is connected directly to the big tank and is sealed shut, though the water level can be changed at will. I didn't do the math, but I'm pretty sure a few drops of water would raise the water level by at least hundreds of meters.

The weird part: According to the formula, water pressure throughout a container is dependent only upon the height of the container, assuming gravity and density remain constant. Technically, our imaginary tank is over ten kilometers tall, since the chimney is part of the tank. Does this mean that adding just a few drops of water can create enormous pressure everywhere in the ocean-sized tank? Seems to violate the law of conservation of energy, doesn't it?

Yes, I know that pressure is force/area, so the weight of the drops is concentrated like a needle. But can anyone tell me how--preferably on a molecular level--that force is distributed to every part of a body of water as big as the PACIFIC OCEAN?

Thanks

I calculate that the mass of water in the column is @ 0.078539816339744830961566084581988 grams
at the base of the column (the surface of the enclosed sea), then, the pressure will be increased by that amount of mass divided by the total square inches of the surface covering the sea.
(please forgive any misplaced decimals if they don't really affect the practical outcome)

Wouldn't that put the highest pressure somewhere around half way up the column?

kotreny
I calculate that the mass of water in the column is @ 0.078539816339744830961566084581988 grams
at the base of the column (the surface of the enclosed sea), then, the pressure will be increased by that amount of mass divided by the total square inches of the surface covering the sea.
(please forgive any misplaced decimals if they don't really affect the practical outcome)

Wouldn't that put the highest pressure somewhere around half way up the column?

I don't think so. If I'm not mistaken, you're using the area of the ocean to calculate pressure when you should be using the area of the chimney (pi*25nm^2).

The purpose of this thread was to illustrate through hyperbole the intuitive paradox that pressure at a given point depends exclusively on the weight density of the fluid and the depth below the surface. I find it hard to imagine that the ocean's pressure change created by the water in this unbelievably minuscule chimney is exactly the same as that created by water in a gargantuan chimney of equal height. (Discounting forces like capillary action.)

This is the crux of my experiment:
Does this mean that adding just a few drops of water can create enormous pressure everywhere in the ocean-sized tank? Seems to violate the law of conservation of energy, doesn't it?

I can just picture a scuba diver swimming peacefully along in the Pacific Ocean, when suddenly, someone adds a single, tiny drop of water to the tube, crushing the diver to bits. Pardon my sadism...

BenchTop
I did the area and volume and 1 g/cc.

You can't get PSI without putting Ps on top of SIs.
When you add a tiny amount of mass, the sealed, enclosed horizontal surface where the little column empties represents the SI you added the minuscule fraction of a P to. At that level, every SI of the surface shares the load of the mass above, which is practically nothing in the first place.
I bet Archimedes would eureke on that one.

I understand that much but I don't understand where the point of highest pressure must be in that system.

Or if I'm off, I hope somebody smarter will clear it up.

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Mentor
You are completely wrong. The weight of the water column does not get spread out over the area of the surface of the tank.

Again, p=rho*g*h. That's it. There is no more to it than that. The pressure is therefore highest at the bottom of the tank and the pressure is equal to the density of water times g times the vertical distance from the bottom of the tank to the top of the colum. The pressure at any point is equal to rho*g* the vertical distance between that point and the top of the colum.

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BenchTop
I think I get it now.
It really does take that kind of pressure to lift that column.
Thanks.

TheInversZero
Ya russ i am aware of that formula perhaps you misunderstood the statement i made.

I use that formula for each column then apply an average to find the new pressure from the old supposedly know pressure.

The single small tube of great height would not pruduce a massive amount of overall pressure increase it would apply to only that single column.

And since water pushes out in all directions it would be spread throughout the whole.

If water didnt push to the side it wouldn't run off a table.

Does this mean that adding just a few drops of water can create enormous pressure everywhere in the ocean-sized tank? Seems to violate the law of conservation of energy, doesn't it?

How do you figure this seems to violate conservation of energy?

If water is added to a wide tube any momentary pressure increase at the base of the tube causes the water to flow out until equilibrium is reached the pressures at all points at the same level becoming equal.If the tube is narrow enough for surface tension effects to be appreciable then there will be a partial outflow until equilibrium is reached.In this latter case the water in the tube will be at a higher level than the water outside the tube but at equilibrium the pressure at all points at the same level in the liquid will again be the same..The reason for this is that when you move up the tube the pressure decreases as in accordance with... Atmospheric pressure minus h*density*g.The pressure is still equal to atmospheric pressure above the surface but this pressure is higher than the pressure just below the surface because of the angle of contact the water makes with the glass,it can be shown that pressure on the concave side of a curved liquid surface exceeds that on the convex side by( 2*surface tension)/radius of surface.

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Mentor
Ya russ i am aware of that formula perhaps you misunderstood the statement i made.

I use that formula for each column then apply an average to find the new pressure from the old supposedly know pressure.

The single small tube of great height would not pruduce a massive amount of overall pressure increase it would apply to only that single column.

And since water pushes out in all directions it would be spread throughout the whole.

If water didnt push to the side it wouldn't run off a table.
Repeating it doesn't make it right. Everything you said there is completely wrong.

As a hypothetical, consider a 1 meter cubed tank with a 1 square centimeter column of 1m height sticking out the top. Calculate for me the pressure at the top of the tank and provide a reference for your method of doing the calculation.

Consider also a real example I work on every day: the heating water piping system in a building. How would you go about calculating the static head pressure at the bottom of the system? I use p=rho*g*h and it works.

Here is a link that explains the concept more: http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Pressure/HydroStatic.html [Broken]

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BenchTop
I just stuck a piston in the column and I had most of a hydraulic jack.
That cleared it up for me.

TheInversZero
OK i do understand why you are using the rho*g*h

However what method would you use to determine the wall thickness of a tank of water or the wall thickness of a water pipe.

Or would their be no pressure pushing the water out of the sides.

Could you use a pipe with a wall thickness of 1 micrometer of wax paper to contain the water in a 20 meter radius pipe of 100 meter height.

Or could you have a fish tank made of glass the same thickness as a small houshold thermometer if it was a tank big enough to hold a whale.

The pressure in a system of water is not only in one direction.

If you have a large tank and you add more water do the sides of the tank not require a material of a thickness to be able to resist a pressure.

So how would you calculate that pressure.

I am not trying to be argumentative i really would like to know.

TheInversZero
You are completely wrong. The weight of the water column does not get spread out over the area of the surface of the tank.

Again, p=rho*g*h. That's it. There is no more to it than that. The pressure is therefore highest at the bottom of the tank and the pressure is equal to the density of water times g times the vertical distance from the bottom of the tank to the top of the colum. The pressure at any point is equal to rho*g* the vertical distance between that point and the top of the colum.

The pressure P is the same on the bottom of each vessel.

Why the pressure does not depend upon the shape of the vessel or the amount of fluid in the vessel rests upon three things:
a. Pressure is force per unit area and this is not same as the total weight of the liquid in a vessel.
b. A fluid can not support itself without a container. Thus the walls of the container exert a pressure on the fluid equal to the pressure of the fluid at that depth.
c. The pressure at given level is transmitted equally throughout the fluid to be the same value at that level.

This exerpt is from the link you sent me russ.

The pressure formula is for pressure at the bottom of the tank only

Point A is indicating the total weight of the liquid is a factor for side pressure.

Point B supports that.

Point C says a pressure at a given level is equally spread throughout the fluid.

If you care to clarify what you meant in the quote when you said
The weight of the water column does not get spread out over the area of the surface of the tank.

The pressure P is the same on the bottom of each vessel.

Why the pressure does not depend upon the shape of the vessel or the amount of fluid in the vessel rests upon three things:
a. Pressure is force per unit area and this is not same as the total weight of the liquid in a vessel.
b. A fluid can not support itself without a container. Thus the walls of the container exert a pressure on the fluid equal to the pressure of the fluid at that depth.
c. The pressure at given level is transmitted equally throughout the fluid to be the same value at that level.

This exerpt is from the link you sent me russ.

The pressure formula is for pressure at the bottom of the tank only

Point A is indicating the total weight of the liquid is a factor for side pressure.

Point B supports that.

Point C says a pressure at a given level is equally spread throughout the fluid.

If you care to clarify what you meant in the quote when you said
The weight of the water column does not get spread out over the area of the surface of the tank.

The formula for hydrostatic pressure is good for any point, not just the bottom of the tank.

As long as the height is the same at any two points, the pressure at those points will be the same. Again, note that it is the total fluid column height as shown in the link provided.

CS

TheInversZero
The formula for hydrostatic pressure is good for any point, not just the bottom of the tank.

As long as the height is the same at any two points, the pressure at those points will be the same. Again, note that it is the total fluid column height as shown in the link provided.

CS

You are correct the formula will give the pressure at the BOTTOM of a column of water even if that BOTTOM is not at the BOTTOM of the tank itself.

There still seems to be no one that has stated a formula to say what the presssure is at the side of a tank or pipe and in what way the nano tube presented in the origin of this thread was referring to as the increase in pressure due to the extreme hieght of said tube

" and to refer again that it is not needed to restate that water would be drawn up the tube as this is a thought xcersise and is dealing only with the formula at hand "

cmon people. present solutions to a problem before extending the scope of the problem.
Then modify the parts to which you take issue and present a solution to that or pose another question for someone else to solve.

You are correct the formula will give the pressure at the BOTTOM of a column of water even if that BOTTOM is not at the BOTTOM of the tank itself.

It will give the pressure at ANY point not just the bottom.

There still seems to be no one that has stated a formula to say what the presssure is at the side of a tank or pipe and in what way the nano tube presented in the origin of this thread was referring to as the increase in pressure due to the extreme hieght of said tube

Yes we have. It's the same as we've been saying all along. It's the $\rho gh$, nothing more. As pointed out previously, pressure is transmitted, undiminished, in ALL directions at the point of interest. So if your point of interest is right beside the wall, the pressure on the wall will be the same as you point of interest, which is $\rho gh$.

cmon people. present solutions to a problem before extending the scope of the problem.
Then modify the parts to which you take issue and present a solution to that or pose another question for someone else to solve.

The solution has been presented. You are the one extending the scope by making assertions about taking some average of the weight to determine the pressure (or whatever it was that you wrote).

CS

If you care to clarify what you meant in the quote when you said
The weight of the water column does not get spread out over the area of the surface of the tank.

It's pretty clear what he meant...the weight is not a controlling factor in determining the pressure. Only $\rho gh$. Forget this non-sense about the weight of the fluid in determining pressure. It only works out numerically sometimes, whereas $\rho gh$ always works.

CS

I thought the question has already been answered or am I missing something here?You can and do get pressure fluctuations at a particular point in a liquid due to,as another example,surface variations such as waves.These fluctuations tend to cause a flow within the liquid and the pressure becomes equal at each point at the same horizontal level within the liquid but only when and if an equilibrium is reached,the surface becoming level.

I thought the question has already been answered or am I missing something here?You can and do get pressure fluctuations at a particular point in a liquid due to,as another example,surface variations such as waves.These fluctuations tend to cause a flow within the liquid and the pressure becomes equal at each point at the same horizontal level within the liquid but only when and if an equilibrium is reached,the surface becoming level.

In the OP, if I'm not mistaken, he is referring to fluid in a tank, hence this a fluid statics problem and any dynamic effects are to be neglected.

CS

TheInversZero
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Pressure/HydroStatic.html [Broken]

Go to this part it is a near match for the image described in the original question

Vessel C: Again we could divide the water into three sections. The middle section is similar to that of vessel A or B. Since the height of the fluid in section 1 or 3 is not high enough to produce the same pressure as the height of the fluid in 2, how does the pressure on the bottoms of section 1 and 3 get to be the same as that of 2 ?
The answer is that top of the container's walls in sections 1 and 3 produce a downward pressure that is equal to the fluid pressure in the middle section at the same level. If you poked a hole in the top of the container in sections 1 or 3, water would fountain upwards from the hole under pressure. From Pascal's principle, this pressure has to be that of the fluid in the middle section at the same level.

Hey everyone, I just thought of a mind-boggling "thought experiment":

As a reminder, the formula for water pressure acting on an object due to weight is:

(Water density)*(Acceleration of gravity)*(Depth of object underwater)

Note that the shape of the container doesn't matter.

Imagine a water-filled tank the size of the Pacific Ocean (exact measurements won't be necessary). The tank is completely filled with water and is sealed shut, so any pressure other than water pressure is absent. Now imagine that at one corner of the tank there is an incredibly narrow "chimney" about ten nanometers in diameter and extending about ten kilometers above the top of the tank. This tube is connected directly to the big tank and is sealed shut, though the water level can be changed at will. I didn't do the math, but I'm pretty sure a few drops of water would raise the water level by at least hundreds of meters.

The weird part: According to the formula, water pressure throughout a container is dependent only upon the height of the container, assuming gravity and density remain constant. Technically, our imaginary tank is over ten kilometers tall, since the chimney is part of the tank. Does this mean that adding just a few drops of water can create enormous pressure everywhere in the ocean-sized tank? Seems to violate the law of conservation of energy, doesn't it?

Yes, I know that pressure is force/area, so the weight of the drops is concentrated like a needle. But can anyone tell me how--preferably on a molecular level--that force is distributed to every part of a body of water as big as the PACIFIC OCEAN?

Thanks

OK now he presents the problem that using the formula it gives a MASSIVE amount of pressure.

I said right off that the pressure would be spread out evenly if you dissagree with that point say so.

And if you do dissagree go look at the web link above.

If you use only that one formula it does not apply to the process of the pressure distribution it ONLY applies to the pressure at the bottom of the column.

We are dealing with columns of 2 different heights not just 1

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http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Pressure/HydroStatic.html [Broken]

Go to this part it is a near match for the image described in the original question

Vessel C: Again we could divide the water into three sections. The middle section is similar to that of vessel A or B. Since the height of the fluid in section 1 or 3 is not high enough to produce the same pressure as the height of the fluid in 2, how does the pressure on the bottoms of section 1 and 3 get to be the same as that of 2 ?
The answer is that top of the container's walls in sections 1 and 3 produce a downward pressure that is equal to the fluid pressure in the middle section at the same level. If you poked a hole in the top of the container in sections 1 or 3, water would fountain upwards from the hole under pressure. From Pascal's principle, this pressure has to be that of the fluid in the middle section at the same level.

OK now he presents the problem that using the formula it gives a MASSIVE amount of pressure.

I said right off that the pressure would be spread out evenly if you dissagree with that point say so.

And if you do dissagree go look at the web link above.

If you use only that one formula it does not apply to the process of the pressure distribution it ONLY applies to the pressure at the bottom of the column.

We are dealing with columns of 2 different heights not just 1

No, that one formula works everywhere, not just at the bottom of the column of fluid.

However, the height (at any point) is measured from the fluid surface (i.e. the total height to that point whether at the bottom or not). Take a look at the very last picture on that website...it shows why it is only pgh and nothing more...if you still don't get it...look at it again and again until you do...or take a fluid mechanics class.

In the OP, it was asked whether or not a few drops in a tube would cause the tank level to raise. Well, technically it would, but it would be essentially infinitesimal due to the vast volume of the tank (assumed to be the size of the ocean). He then stated that according to the theory, pgh, it would raise the ocean by hundreds of meters. He stated that because he didn't understand at the time the relationship between hydrostatic pressure and depth.

The tank would not be raised hundreds of meters, the volume of the drop of fluid would be spread through out the tank, there by resulting in a very very small pressure increase (i.e. the total tank height would not increase but a very very small amount which results in the very very small pressure increase AT ANY POINT IN THE TANK).

Again, the pressure is only a function of pgh in a static fluid such as the one described.

CS

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TheInversZero
actually he is saying that it is a full tank and a few drops are added to fill the column otherwise why make the column so high.

So stated again could only a few drops of water ACTUALLY increase the pressure by such a massive amount or is the formula perhaps flawed in applications of this extreme type of scenario.

Not every formula works for every situation, perhaps this could be an exception.

If there were a submarine floating in that tank the formula rho*g*h would cause that sub to be crushed by a few drops of water. Can you be sure that it would happen is there no doubt whatsoever.