Hydrostatic pressure on top of a cone filled with water

[JABAS]

TL;DR

Water cone hydrostatic pressure vacuum

Does the cone shape and extra water below the colum of water of 1 cm increase the pressure/vacuum effect at "B" or "A"?

Is it simular to an inverted cone dependent on angle of sides of cones, when looking for pressure at the bottom of a cone or cylinder?
In simple terms, how?
Basic maths only please.

 

[russ_watters]

Shape does not matter, only height: https://en.m.wikipedia.org/wiki/Vertical_pressure_variation

 

[JABAS]

So the pressire or gravity or draw of the weight of water dropping does not change from the 1 cm coluum of water? No matter the shape or quanity of water below A? So just the colume of 1cm pulls at "b" or "a"? Thank you so much for you time. Signed simplton!

 

[kuruman]

First imagine the water in the cylindrical column that you drew outlined by the dotted line as a real cylindrical vessel made from steel. The pressure at a given depth is the same on a horizontal plane, i.e. a plane perpendicular to the acceleration of gravity. The pressure variation $p$ with depth $d$ is given by $$p=p_0+\rho_{\text{water}}~g~d$$where $~p_0=$ pressure at the free surface of the water.

The water does not spill out of the steel cylinder because, at any depth, the cylinder exerts an equal and opposite force on the water. Of course that is the case whether the cylinder is made of steel, wood or plastic. Now imagine the dotted cylindrical container made from water with a wall thickness that extends farther from the dotted line. The pressure variation inside its walls will be exactly the same as in the water it contains. Balancing the pressure at the dotted line means balancing the forces on a mass element of fluid which means no sideways motion.

 

[JABAS]

That would make the next 1 cm colume next to the center one totally supported bt the angled roof above. How can there be no side ways force to some percent on the next over colume?

I've read the papers seen the math. Makes no sense

 

[kuruman]

That would make the next 1 cm colume next to the center one totally supported bt the angled roof above.

I cannot follow your reasoning here. The roof does not support the liquid but the bottom of the vessel does. Remove the bottom of the vessel keeping the roof in place and see what happens.

How can there be no side ways force to some percent on the next over colume?

I never said there is no sideways force. I said that the forces are balanced which means there is no sideways motion.

I've read the papers seen the math. Makes no sense

What papers and what math? Please provide references and we will try to explain whatever it is that makes no sense to you.

 

[JABAS]

So, a 10 meter tall tube of water of any thickness is able to over come atmospheric pressures and draw/pull/or create a vacuum to create steam or kind of boil water at room temperature.

My question is... can a vessel with a small top and larger bottom (""holding the same or greater volume or weight of water"") boil/create a vacuum at a lower height?

By papers I meant things I have read in either Physics Forum or the internet. Most pertaining to pressures at the bottom of a container.

Also the container is closed on top but not closed on the bottom but submerged in a container of water. Hence the atmopheric pressures.

 

[jbriggs444]

So, a 10 meter tall tube of water of any thickness is able to over come atmospheric pressures and draw/pull/or create a vacuum to create steam or kind of boil water at room temperature.

Yes. Or a 760 millimeter tube of mercury. The assembly you describe is often used as a barometer.

My question is... can a vessel with a small top and larger bottom (""holding the same or greater volume or weight of water"") boil/create a vacuum at a lower height?

The answer is no.

By papers I meant things I have read in either Physics Forum or the internet. Most pertaining to pressures at the bottom of a container.

We need a specific reference so that we know what specifically is not making sense to you.

Also, not just the reference. We need to know what specifically within that reference is making sense and what is not.

 

[russ_watters]

So, a 10 meter tall tube of water of any thickness is able to over come atmospheric pressures and draw/pull/or create a vacuum to create steam or kind of boil water at room temperature.

Yes. Or a 760 millimeter tube of mercury. The assembly you describe is often used as a barometer.

This is also a major problem when pumping water. You can't draw suction to pump water from deeper than 10m. Sort of an unintentional water-based barometer.

 

[russ_watters]

@JABAS is this purely a theoretical question about hydrostatics or do you have an application in mind? You mentioned boiling water; vacuum distillation is a thing. You can't do it with a static column of water though, because you need a way to get the water vapor out. You need a vacuum pump.

 

[JABAS]

Thank you all for your replies and your time! It is very appreciated! Please see attachment.

 

[russ_watters]

Thank you all for your replies and your time! It is very appreciated! Please see attachment.

You're welcome...Diagram isn't attached.

 

[sophiecentaur]

The roof does not support the liquid

The roof prevents the liquid from flowing upwards - that's enough. The wall prevents the liquid from getting out. The floor supports the liquid. AND the water that's in the column going from floor to surface keeps the water from leaving the sides. Take any small cube of water, anywhere and the forces acting on opposite sides will balance out.

Remember we are discussing Hydrostatics which involves no motion and so the pressure is the same in all directions at any point. Draw the microscopic box in any orientation and that still applies.

 

[JABAS]

Here is the attachment. It didnt load before


 

[JABAS]

Also, sorry didnt mention before, not static but opened at the bottom and a little submerge in a liquid.

 

[sophiecentaur]

So atmospheric pressure is acting on the surface and the same pressure is acting upwards on an imaginary plane coplanar with the water / air interface. That pressure is enough to support the column of liquid up to the top. It will only support up to 10m (approx).
If not, water in the tube will drop until it is. Pressure works everywhere and you can't chose a static case where it doesn't.

 

[jbriggs444]

Also, sorry didnt mention before, not static but opened at the bottom and a little submerge in a liquid.

The liquid is not moving or accelerating. So it is indeed static.

 

[russ_watters]

Here is the attachment. It didnt load before:

You ask if the pressure is different at point "A" in from one diagram to the other. The answer is no.

 

[JABAS]

So, no. Ok thank you.
Sopiecentaur, you said

"It will only support up to 10m (approx).
If not, water in the tube will drop until it is."

So if ,in the last drawing, if they were 15 meter tall. Filled static and closed. And then opened would water in the first would drop until its weight equals that of atmophereic pressure. 10meter (approx) And the water in the second would drop to the same level? Or further due to more mass?

Thank you so much everyone. I am finally asking the right questions to get my answer.

 

[A.T.]

So if ,in the last drawing, if they were 15 meter tall. Filled static and closed. And then opened would water in the first would drop until its weight equals that of atmophereic pressure. 10meter (approx) And the water in the second would drop to the same level? Or further due to more mass?

They drop to the same level.

Note that all this assumes the liquid cannot have negative absolute pressure. However, if you can prevent evaporation somehow, liquids can form higher columns than atmospheric pressure can support. This is how trees get water higher than 10m:

 

[sophiecentaur]

would water in the first would drop until its weight equals that of atmopheric pressure

It's not the weight that counts; it's the pressure that affects whether water will flow or not.

This is how trees get water higher than 10m:

This is a great video but must be viewed with care by the questioner. For water to be 'sucked up' by a tall tree, you need very special conditions which just don't exist in the sort of experiments that we've been discussing here. There is no cognitive dissonance.

 

[jbriggs444]

So if ,in the last drawing, if they were 15 meter tall. Filled static and closed. And then opened would water in the first would drop until its weight equals that of atmophereic pressure. 10meter (approx) And the water in the second would drop to the same level? Or further due to more mass?

You seem to still not understand what "static" means here.

Yes, if we have these two 15 meter inverted containers that are both sealed on the bottom and let them sit until the water is still, that would be a static situation. It is static because the water is motionless and not accelerating. It is in an equilibrium state.

If we open one of the containers at the bottom, the water will flow away from the top and out the bottom, leaving a near vacuum at the top.

[For room temperature water, the "air space" at the top will be occupied by water vapor at about 3% of one atmosphere of pressure so the surface of the water will be low by about 1/3 of a meter from where one might expect. In addition to problems with the required building height, the lower saturated vapor pressure of Hg at room temperature is a reason why we traditionally use mercury in barometers instead of water]

While the water is flowing out the bottom, the situation is not static. It is dynamic. The inertia of the water as it flows out may cause the water surface to bounce up and down a bit before settling into its position 10 meters up.

Once the water has settled down so that the flow has ceased and the surface holds a steady position, we have a static situation again. No water is moving. No water is accelerating. It is just sitting there.

Pascal's principle requires that the fluid be at rest in a static equilibrium.

In this situation, the variation of pressure from one position in the fluid to the next is given by $\rho g h$ where $\rho$ is the fluid density (assumed to be constant), $h$ is the vertical distance between the two positions and $g$ is the local acceleration of gravity. The size, shape and capacity of the container do not enter in.

The net force supporting a greater volume of water in a wider container will be greater than the net force supporting a lower volume of water in a narrower container.

The pressure (force per unit of surface area) difference across 10 meters of depth will be identical.

 

[sophiecentaur]

[For room temperature water, the "air space" at the top will be occupied by water vapor at about 3% of one atmosphere of pressure so the surface of the water will be low by about 1/3 of a meter from where one might expect.

Fire crews know that they can't get away with anything like 10m for getting water from sunken sources. If they need it from great depth then they would need two pumps in series with an intermediate reservoir (or a force pump at the bottom).

 

[jbriggs444]

Fire crews know that they can't get away with anything like 10m for getting water from sunken sources. If they need it from great depth then they would need two pumps in series with an intermediate reservoir (or a force pump at the bottom).

In part, I would assume that this is the result of the low flow rate that would result. You can pump a fair volume of water through a five inch hard suction hose with one atmosphere (100 kPa or 15 PSI) of available head.

Inertial effects (Bernouilli) mean that the flow velocity (and therefore the flow rate) will scale as the square root of the available pressure head. If the effective pressure head is only 0.03 atmospheres then the flow rate will be reduced by a factor of about 6.

Viscous effects (Poiseuille) mean that the flow rate will scale linearly with the available pressure head. If the pressure head is only 0.03 atmospheres then the flow rate will be reduced by a factor of about 33.

So the reduction in hose performance if one goes all the way to 9.6 meters of suction (10 meters minus 0.3 meters for vapor pressure minus another 0.3 meters to provide some actual head) will be somewhere between a factor of 6 and a factor of 33.

This is before we start worrying about the effects of cavitation at or near the pump.

However, the above is armchair theoretical stuff. Here is some practical information:

A fire engine in fairly good condition can lift water two-thirds of the theoretical lift, 2/3 × 33.9 = 22.5 feet. This height is called the maximum attainable lift. With an increase in elevation above sea level, atmospheric pressure decreases, thus reducing the vertical distance from the water source where drafting can be done effectively.

 

[sophiecentaur]

Here is some practical information:

It's common to need river water and 22ft is not a lot on a tidal stretch. Plus, just below the tide line, the water can shelve annoyingly slowly.