Crazy water pressure formula implication?

[mgb_phys]

if you want a really weird and counterintuitive result of water pressure - image a dry dock.
You have a 50,000ton battleship in a dry dock and let water in so that it floats.

Now imagine you had made the dry dock smaller - only about the same length and width as the boat, you would need less water but it would still float.

Now imagine you made a dry dock that was perfectly the size and shape of the boat - so that there was only a microscopic gap between the hull and the dock. Then you could put a teaspoon of water in the gap and the boat would float.

 

[stewartcs]

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.

If the tank is completely full and sealed with only a tube sticking out of the top, and water is added in the tube, then yes the pressure will increase and be equal to pgh, where h is the total height from the top of the fluid level in the tube to whatever point you are interested in. I'm 100% sure that is true, without a doubt.

Note that this is not the same case as dealing with an ocean...the ocean is open to atmosphere and will have room to rise whereas a sealed and completely full tank does not have room to rise. If the tank has room to rise, then the volume of water is spread out as stated before and the pressure increase is negligible.

If a submarine was in the tank as described, then adding a fluid column with a height of 10 km would certainly crush it.

There is no exception to this formula. It works for all cases.

CS

 

[Andy Resnick]

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

<snip>

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 haven't seen anyone mention this, but even though the pressure (and force applied to the container) will indeed grow very large, because of the conservation of energy the (virtual) displacement of pacific ocean against the container will be proportional to the area of the column. So as the column diameter shrinks, the container walls will only need to move a vanishingly small amount to compensate.

Put another way, as the container begins to bulge, the water in the column drops, reliving the applied pressure.

 

[stewartcs]

I haven't seen anyone mention this, but even though the pressure (and force applied to the container) will indeed grow very large, because of the conservation of energy the (virtual) displacement of pacific ocean against the container will be proportional to the area of the column. So as the column diameter shrinks, the container walls will only need to move a vanishingly small amount to compensate.

Put another way, as the container begins to bulge, the water in the column drops, reliving the applied pressure.

I think the OP assumed a perfectly rigid container in their thought experiment.

CS

 

[TheInversZero]

OK thank you for all the consrtuctive results of this thread.

New thread just not officially started yet.

Given the formula rho*g*h

A tube extending to space with a small diameter. What possible "not practical" applications could there be to launching a satellite into space using the massive pressure of water.

This is more theory then anything else.

Perhaps try including other potential applications of this.

Have fun let your minds wander.

 

[russ_watters]

Given the formula rho*g*h

A tube extending to space with a small diameter. What possible "not practical" applications could there be to launching a satellite into space using the massive pressure of water.

This is more theory then anything else.

Perhaps try including other potential applications of this.

Have fun let your minds wander.

Mind wandering...getting nowhere.

Huh?

 

[russ_watters]

Some wording issues make me wonder if there isn't just a misunderstanding (doubt it) about what some are saying. I'm not sure what "spreading out" is supposed to mean. So I'll refer you back to the challenge and example I gave waaaay back in post #22:

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.

No one has answered it, so I'll just give the answer (and a few more).

1. What is the pressure everywhere along the bottom of a 1 meter cubed, open water tank? Answer: 9800 n/m^2 (.097 atm)
2. If you put a top on the tank, install a 1m tall open pipe of 1 sq cm cross sectional area on top and fill it with water, what is the pressure at the bottom of the pipe? Along the entire top surface of the tank? Answer: 9800 n/m^2 (.097 atm). What is the pressure along the entire bottom surface of the tank? Answer 19600 n/m^2 (.194 atm).

If you have a 100m tall building with a hot water heating system in it, vented at the top, what is the pressure at the suction side of the pump in the basement with the pump off? Answer: 980,000 n/m^2 (9.7 atm)

Agree or disagree?

 

[TheInversZero]

You are very right Russ pressure is a value of, Newtons divided by Meters squared. N/Msquared

I am trying to figure out how many square meters are at the base of a nano tube. Perhaps you could enlighten us.

 

[Jobrag]

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.

But as soon as the submarine began to deform the level in the tube would fall and the pressure relieved, then you would have to refill the tube, little more deformation, by the time you had filled the tube often enough to destroy the submarine you would have used as much energy lifting the water to the top as you would have blowing the thing apart.

If you want an analogy think of a Van de Graaff generator lots and lots of volts but bugger all power.

 

[russ_watters]

I am trying to figure out how many square meters are at the base of a nano tube. Perhaps you could enlighten us.

Why would you care? It's irrelevant to the problem. p=mgh <-where do you see area in that equation?

 

[ericgrau]

Pressure at the bottom of the mile high tube will be incredibly high. Pressure in the ocean will remain unchanged. And then the water in the tube will fall to the level of the ocean, and the pressure in the tube will drop to that of the ocean. Pressure in the ocean wills still remain unchanged.