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

[jbriggs444]

can't be the traditional ρgh (pressure gradient in the vertical direction).

Note that $\rho g h$ is not the pressure gradient. It is the pressure. More correctly, it is the pressure difference between the test point and a reference pressure at a point on the surface directly above.

A "gradient" is a rate of change. It is a vector. Its direction is the direction in which pressure increases most rapidly with displacement. Its magnitude is the rate at which pressure increases with displacement in that direction. In the drawing that you provide (one gee of vertical gravity and one gee of horizontal acceleration), the direction of the gradient is 45 degrees down and to the left. Its magnitude is $\sqrt{2}\rho g$. As you clearly agree.

But we do not much care how rapidly pressure is increasing in a direction 45 degrees down and to the left. Our reference point is directly above our test point. We care about how rapidly pressure is increasing in the purely vertical direction. We want the vertical component of the pressure gradient.

If you wanted, you could calculate that vertical component as $\sqrt{2}\rho g \cos 45 \text{ degrees}$ That simplifies to $\rho g$.

Or you could just use the obvious fact that the vertical component is $\rho g$. Just like the horizontal component is $a g$ (with $a = g$ in the case you portray).

That point is supporting the force (each mass x common diagonal acceleration) of all the molecules that are aligned with the resultant acceleration vector and located between that point and the surface.

View attachment 315891

Right. So if you have a column of water $\frac{h}{\sqrt{2}} \text{meters}$ in diagonal extent so that our test point is $\frac{h}{\sqrt{2}}\text{ meters}$ below a reference point at the surface in a direction 45 degrees up and to the right. It will be at a pressure delta of $\sqrt{2} \rho g h\frac{1}{\sqrt{2}}$ above surface pressure.

Equivalently, it is $h \text{ meters}$ below a reference point on the surface directly above and is at a pressure delta of $\rho g h$ above surface pressure.

Both calculations yield the same result. But one is simpler than the other.

 

[kuruman]

Here is an explanation in the accelerated frame. First refer to the picture on the left below. The apparent acceleration of gravity is the vector sum of the horizontal $(\mathbf{-a})$ and the vertical acceleration of gravity $\mathbf{g}$. The fluid surface (blue line) will be perpendicular to the apparent acceleration of gravity. In the non-inertial frame the hydrostatic pressure will be $p=\rho~g_{\text{app}}~d$ where $d$ is the perpendicular distance to the surface from the point of interest. That was my approach in post #3 where I suggested that one needs to find $d$ and $g_{\text{app}}$. However, that is unnecessary extra work.

Refer to the picture on the right where the perpendicular distance to the surface is shown as $d$. Note (what I missed at first until @Orodruin pointed it out) that from the picture on the left $$g = g_{\text{app}} \cos \theta.$$
Then the hydrostatic pressure at perpendicular distance $d$ from the surface is $$p=\rho~g_{\text{app}}~d=\rho~\frac{g}{\cos\theta}~d=\rho~g~h.$$On edit: The same argument stated differently. Planes parallel to the surface are planes of constant hydrostatic pressure. So when you go from point A to point B, the pressure change will be the same as when go from A directly to C because there is no change in pressure when you go along the surface from B to C.

On second edit: If the picture on the right were rotated counterclockwise by $\theta$, so that the apparent gravity is aligned from the top of the screen to the bottom, the proposition that the pressure difference between A and B is the same as between A and C would be perfectly acceptable to "inside the box" thinkers.

 

[Orodruin]

Incompatible with being compressed, right?

Indeed. I blame auto complete.

 

[Lnewqban]

Note that $\rho g h$ is not the pressure gradient. It is the pressure. More correctly, it is the pressure difference between the test point and a reference pressure at a point on the surface directly above.
...
Both calculations yield the same result. But one is simpler than the other.

Thank you much for the detailed explanation, excellent as usual.

Yes, I referred to the natural rate of pressure change in the vertical direction as pressure gradient, being the measuring point at one singular point of it.

It is just a simply trigonometric thing: what we gain in acceleration in that direction, we gain in number of molecules in a direction 45 degrees down and to the left.
Sorry about wasting yours, @erobz, @kuruman time, I should have noted that dependence.

Respectfully, I still prefer to see what actually happens in the direction it happens.

 

[Lnewqban]

@songoku , this two problems are similar:
https://www.physicsforums.com/threads/acceleration-of-u-pipe-containing-water.999370/

I also think they are similar but I can not connect the dots

@songoku
I believe that this link could help you connect those missing dots:
https://www.me.psu.edu/cimbala/Learning/Fluid/Rigid_body/rigid_body.htm

 

[Orodruin]

Respectfully, I still prefer to see what actually happens in the direction it happens.

The more intuitive approach is the full vector approach given in post #4. You get to use the full pressure gradient and there is no need to pick a reference point other than what is given by the problem.

 

[songoku]

I have read the replies and the link given many times but I still have things I don't understand. Let me try one by one.

It removes any doubt re the meaning of depth here. It would also mean the top of the water above the point is not necessarily at atmospheric pressure.

It seems that the case without lid and with lid will produce different answers but I don't know why. Without lid, the top of water will always be at atmospheric pressure and with lid the pressure can be at different value than atmospheric pressure. But why this affect the answer?

And how can the information about lid removes any doubt about meaning of "depth" here?

Thanks

 

[haruspex]

I have read the replies and the link given many times but I still have things I don't understand. Let me try one by one.It seems that the case without lid and with lid will produce different answers but I don't know why. Without lid, the top of water will always be at atmospheric pressure and with lid the pressure can be at different value than atmospheric pressure. But why this affect the answer?

And how can the information about lid removes any doubt about meaning of "depth" here?

Thanks

Full with a lid means the depth of the water is constant. The surface does not slope.

 

[erobz]

Full with a lid means the depth of the water is constant. The surface does not slope.

@songoku
The ambiguity its doesn't remove is which wall to measure horizontally from.

 

[songoku]

Full with a lid means the depth of the water is constant. The surface does not slope.

Ah ok I see now.

But why the answer would be different with lid?

Thanks

 

[erobz]

Ah ok I see now.

But why the answer would be different with lid?

Thanks

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

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.

 

[jbriggs444]

But why the answer would be different with lid?

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.

 

[kuruman]

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).

It is curious that you mention this. If, at the same time, you hang a plumb bob from the ceiling of the car, the plumb bob would lean backwards. Knowing that helium balloons always stretch strings opposite to plumb bobs, the observation should not be unexpected.

 

[erobz]

It is curious that you mention this. If, at the same time, you hang a plumb bob from the ceiling of the car, the plumb bob would lean backwards. Knowing that helium balloons always stretch strings opposite to plumb bobs, the observation should not be unexpected.

Its not curious. I probably have a misconception of some sort. Its not the presure gradient in the cabin air that pushes the balloon forward? Surely there is a pressure gradient across the balloon that pushes it opposite EDIT: [in] the direction of acceleration.

What I'm trying to convey, ( perhaps inadequately for an actual physics analysis) is that to people like me (without physics degrees) things in the car are generally pushed in the direction opposite the acceleration. The balloon is special, and surprising in that case. If it were just a regular mass on an inverted pivot, (not a balloon) we expect it to go backward like everything else...and it would?

 

[jbriggs444]

What I'm trying to convey, ( perhaps inadequately) is that things in the car are generally pushed in the direction opposite the acceleration.

When I am in a car, I am pushed in the direction the car is accelerating as are the content of my pockets. This is fortunate because I dislike being ejected from automobiles and scuffing across the pavement. Nor do I like stopping to go back and fetch my wallet.

 

[erobz]

When I am in a car, I am pushed in the direction the car is accelerating as are the content of my pockets. This is fortunate because I dislike being ejected from automobiles and scuffing across the pavement. Nor do I like stopping to go back and fetch my wallet.

Yeah, I get that. But in the frame of the car we feel as though we are being pushed backwards.

 

[jbriggs444]

Yeah, I get that. But in the frame of the car we feel as though we are being pushed backwards.

Surely there is a pressure gradient across the balloon that pushes it opposite the direction of acceleration?

Ahhh, so by the "direction of acceleration" you mean the direction of the pseudo-gravitational "downward" acceleration, not the direction of the car's acceleration.

Yeah. Helium balloons move oppositely to most other bodies such as plumb bobs and this is due to the pressure gradient. Aka buoyancy.

 

[erobz]

Ahhh, so by the "direction of acceleration" you mean the direction of the pseudo-gravitational "downward" acceleration, not the direction of the car's acceleration.

Yeah. Helium balloons move oppositely to most other bodies such as plumb bobs and this is due to the pressure gradient. Aka buoyancy.

I'm getting all screwed up. I meant to say:

Surely there is a pressure gradient across the balloon that pushes it in the direction of acceleration?

 

[Lnewqban]

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.
...
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.

In the hypotetical case that we could remove all the interior air, the proposed vent, and make the walls and lid of the closed reservoir absolutely rigid, would a pressure gradient form under the influence of the total acceleration?

Let’s imagine the reservoir and liquid being located inside an industrial elevator and being vertically accelerated upwards at a substantial rate of acceleration.

Then, let’s imagine the thing in free fall.

I am not sure about the correct answer, but it may help @songoku understand this without the “noise” of the OP’s lateral acceleration.

Thank you.

 

[jbriggs444]

In the hypotetical case that we could remove all the interior air, the proposed vent, and make the walls and lid of the closed reservoir absolutely rigid, would a pressure gradient form under the influence of the total acceleration?

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

To be pedantic, it is statically indeterminate if one treats water as an incompressible fluid. Real water is compressible, so once you compare the quantity of water (in kilograms or moles), the temperature and the exact volume of the container, the pressure will pop out.

 

[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.
...

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.