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Water pressure in narrow containers

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mariusnz
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Feb27-14, 08:23 PM
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I know the general formula for water pressure is related to the height of the water column. But imagine the next situation: A vertical glass tube, 1.0 m high and 100 mm wide, full of water, can support the water inside even with a 0.5 or 1.0 mm thickness wall, but if I have a pool, 1.0 m high and 5 m diameter, I doubt the same wall will be enough to contain the water inside. Why?.The pressure will be the same, as for 1.0 m water column, so how is mass affecting that? I think the basic formula accounts for unlimited back-span. How can I calculate the required wall for a narrow container? What is the back-span from where the formula is OK?
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SteamKing
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Feb27-14, 09:39 PM
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The pressure of a fluid depends only on the depth of fluid and the density of the fluid.

However, when you are designing a container to hold X amount of water or other fluid, it is not sufficient to know that the pressures are the same in a small container versus a large container. The total load produced by the pressure on the container walls determines, in part, how strong the container must be made to prevent collapse.

You have a structural design problem to solve. The type of loading has a bearing on the solution, but it is not the only factor which must be considered.
Chestermiller
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Feb27-14, 09:47 PM
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Quote Quote by mariusnz View Post
I know the general formula for water pressure is related to the height of the water column. But imagine the next situation: A vertical glass tube, 1.0 m high and 100 mm wide, full of water, can support the water inside even with a 0.5 or 1.0 mm thickness wall, but if I have a pool, 1.0 m high and 5 m diameter, I doubt the same wall will be enough to contain the water inside. Why?.The pressure will be the same, as for 1.0 m water column, so how is mass affecting that? I think the basic formula accounts for unlimited back-span. How can I calculate the required wall for a narrow container? What is the back-span from where the formula is OK?
The force of the pressure is supported by tensile stress present within the walls. From a force balance, the tensile stress is equal to pR/t, where R is the radius of the container and t is the wall thickness. So to support a given pressure, the ratio of the radius to the thickness has to be the same for the pool as for the tube (if it is made of the same material). So a bigger radius requires a bigger wall thickness. The tensile stress must not exceed to ultimate stress of the material.

AlephZero
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Feb27-14, 10:04 PM
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Water pressure in narrow containers

The pressure only depends on the fluid depth.

If you have a cylindrical pipe or container, the stress in the radially-outwards direction varies from same as the fluid pressure, on the inside to zero on the outside. That component of the stress is very small and won't break anything.

But there is also a stress component acting around the circumference of the cylinder. That is much bigger and depends on the radius and thickness of the cylinder. Imagine you cut the pipe into two half-pipes with a vertical plane. The total "sideways" force of the fluid, on a thin vertical strip of thickness h, 2Prh where r is the pipe radius and P is the pressure. (Assume h is small enough so the pressure on the thin strip is constant).

That force has to be resisted by the tension in the pipe, which is 2Sth where S is the circumferential stress and t is the thickness. So S = Pr/t. (The above assumes the thickness t is small compared with the radius r. Otherwise, the math gets more complicated but the general idea is the same).

So, as a simple approximation, if you double the radius, you double the minimum thickness you need to resist the same water pressure, or the same water depth.

Chestermiller gave you the short version of this answer, while I was typing the long version!
SteamKing
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Feb27-14, 11:42 PM
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Quote Quote by AlephZero View Post
The pressure only depends on the fluid depth.
Don't forget the density of the fluid. 760 mm Hg makes a higher pressure than 760 mm H2O.
mariusnz
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Mar2-14, 01:18 PM
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OK, thank you very much.


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