Hydrostatic pressure at a point inside a water tank that is accelerating

[Lnewqban]

The pressure would be statically indeterminate. But the pressure gradient would still form in the natural way corresponding to the apparent acceleration of gravity.
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Thank you @jbriggs444

 

[kuruman]

Its not curious. I probably have a misconception of some sort.

I think it's a preconception rather than a misconception. Most people are familiar with hanging objects, e.g. subway or bus straps, leaning "away" from the acceleration of the point of support when the vehicle is accelerating. This creates the false generalization that hanging objects will always lean away from the acceleration of the point of support. The preconception is reinforced when people see a child running while holding a string with a helium balloon tied to it.

 

[songoku]

View attachment 316268

$$P_{L-x} \neq P_x$$

If you've ever done the experiment when you are in the car with a helium balloon. You accelerate, and the balloon floats forward( which without considering the air in the car as a medium that feels the inertial forces should be unexpected). This is due to the pressure gradient created in the air of the car from the acceleration, and the fact that the balloon is buoyant.

If there is a lid, the pressures of the water at points along the lid are not all equal to atmospheric pressure.

We would do well to specify some details about the lid. I assume that:

1. The lid is rigidly attached to the tank. It stays horizontal when the tank is accelerated. It does not pivot in the middle, for instance to match the surface of the water. Nor does it slide up and down. Nor does it bend up and down significantly.

2. There is a vent on the right edge. So that the pressure of the water at the upper right corner of the rightward-accelerating tank is assured to match atmospheric pressure.

[We put the vent at the upper right corner so that we do not need to explain about negative gauge pressure, the absence of vacuum boiling and why water would not flow out of a small vent hole at the upper left. Or at the tank bottom or anywhere else. We use only one small vent hole because we want to make a static equilibrium possible. We use at least one vent hole because we want to avoid a scenario that is statically indeterminate].

In this circumstance, the pressure near the left side of the tank has a contribution from both the horizontal pressure gradient (from acceleration) and from the vertical pressure gradient (from gravity), no matter how you trace pressure differences along a path back to the reference point at the vent.

If I analyze from force perspective, is it correct to say the horizontal pressure comes from net contact force between the water and the left and right side of the tank where contact force from left side > contact force from the right side?

Thanks

 

[jbriggs444]

If I analyze from force perspective, is it correct to say the horizontal pressure comes from net contact force between the water and the left and right side of the tank where contact force from left side > contact force from the right side?

I think that I agree with what you are trying to say. But I have quibbles about the way that you have said it.

I think that you are saying that the horizontal pressure gradient results from the difference between the contact force (per unit area) on the left side and the contact force (per unit area) on the right side.

Yes. This is correct.

The force difference (per unit area) results in a particular acceleration rate. Obviously, that acceleration rate will vary depending on the fluid density and on the distance between the right and left walls.

Because we assume that the fluid in the tank is in equilibrium, the acceleration rate must be uniform throughout any horizontal column of water that is being accelerated rightward. It follows that the pressure gradient (the pressure change rate per unit horizontal distance) is also constant throughout the column. Each tiny cube of water must have the same pressure difference from left to right as any other tiny cube of same size. At least as long as the fluid is incompressible and equilibrium is to be maintained.

The actual pressure (not pressure gradient) at the center of the (hypothetical, rigid, closed-on-all-six-sides) tank might be estimated as the average pressure on the six sides, though I am sure that there are pathological shapes where that estimate would be wrong.

 

[songoku]

I think that I agree with what you are trying to say. But I have quibbles about the way that you have said it.

I think that you are saying that the horizontal pressure gradient results from the difference between the contact force (per unit area) on the left side and the contact force (per unit area) on the right side.

Yes. This is correct.

The force difference (per unit area) results in a particular acceleration rate. Obviously, that acceleration rate will vary depending on the fluid density and on the distance between the right and left walls.

Because we assume that the fluid in the tank is in equilibrium, the acceleration rate must be uniform throughout any horizontal column of water that is being accelerated rightward. It follows that the pressure gradient (the pressure change rate per unit horizontal distance) is also constant throughout the column. Each tiny cube of water must have the same pressure difference from left to right as any other tiny cube of same size. At least as long as the fluid is incompressible and equilibrium is to be maintained.

The actual pressure (not pressure gradient) at the center of the (hypothetical, rigid, closed-on-all-six-sides) tank might be estimated as the average pressure on the six sides, though I am sure that there are pathological shapes where that estimate would be wrong.

So if I interpret the question like post #38, it is not answerable because we don't know the exact position (as shown in post #41)?

Thanks

 

[jbriggs444]

So if I interpret the question like post #38, it is not answerable because we don't know the exact position (as shown in post #41)?

I am not following. What question are you talking about? And what position is not specified?

If one asks about the pressure 10 cm down and 10 cm right from the upper right corner of the sealed tank, I agree that the pressure is unknowable without knowing a reference pressure at some other known position.

 

[haruspex]

I am not following. What question are you talking about? And what position is not specified?

If one asks about the pressure 10 cm down and 10 cm right from the upper right corner of the sealed tank, I agree that the pressure is unknowable without knowing a reference pressure at some other known position.

Yes, post #55 was for the case of a full tank with lid, which I agree is almost certainly not the intent in post #1.

 

[songoku]

I am not following. What question are you talking about? And what position is not specified?

If one asks about the pressure 10 cm down and 10 cm right from the upper right corner of the sealed tank, I agree that the pressure is unknowable without knowing a reference pressure at some other known position.

Yes, post #55 was for the case of a full tank with lid, which I agree is almost certainly not the intent in post #1.

Maybe this is a little bit off topic but what should the question be if I want the answer to be:

Pressure at a certain point = $\rho g_{apparent} h$ where $\rho$ is density of the liquid, $g_{apparent}$ is apparent acceleration of gravity and $h$ is vertical distance from the surface of liquid to the point

Thanks

 

[Orodruin]

Maybe this is a little bit off topic but what should the question be if I want the answer to be:

Pressure at a certain point = $\rho g_{apparent} h$ where $\rho$ is density of the liquid, $g_{apparent}$ is apparent acceleration of gravity and $h$ is vertical distance from the surface of liquid to the point

Thanks

That will never be the case unless the apparent gravitation is vertical, i.e., acceleration is zero (or rather, parallel to the gravitational field).

 

[jbriggs444]

Pressure at a certain point = $\rho g_{apparent} h$ where $\rho$ is density of the liquid, $g_{apparent}$ is apparent acceleration of gravity and $h$ is vertical distance from the surface of liquid to the point

Make that "and $h$ is the distance to the surface of the liquid in the direction of the apparent vertical"

We may also need to include the caveat that the "surface" must be a free surface exposed to ambient pressure. A random point at the top of a sealed tank does not qualify.

It turns out that in the temperate latitudes, we are all exposed to a centripetal acceleration from the Earth's rotation. That acceleration has a horizontal component so that the direction of apparent gravity is tilted slightly from the direction of "true" gravity. The resultant apparent gravitation behaves in all respects like regular gravity and we routinely treat it as such.

But unlike the tank in this problem, we normally align our floors, tables and fish tanks (and lakes, rivers, oceans and landscape) to the apparent vertical and apparent horizontal so that no tilt is apparent. Spirit levels, plumb bobs and barometers work just fine in tilted apparent gravity.

 

[songoku]

Thank you very much for all the help and explanation erobz, kuruman, Orodruin, Lnewqban, haruspex, jbriggs444