Hydrostatic pressure and consequences of pascal's law

[blurrscreen]

I've had a doubt about the following (somewhat contradictory) statements.
1) pressure applied to an incompressible fluid is equally transmitted at all points.
2) pressure at points at different heights in a fluid placed in a container in a gravitational field are different.

Gravity applies a pressure on the fluid. If the fluid transmits the pressure equally at all points, then how is pascal's law valid?

 

[UltrafastPED]

For many purposes the hydrostatic pressure is equal everywhere.

But as the scale of the containment increases it must be realized that the weight of the water column increases the pressure ... and more so the further down you go.

This is very significant in the oceans; insignificant in a barrel of water.

You have found an example of physical laws with limitations to their applicability; this happens all the time.

 

[Chestermiller]

I've had a doubt about the following (somewhat contradictory) statements.
1) pressure applied to an incompressible fluid is equally transmitted at all points.
2) pressure at points at different heights in a fluid placed in a container in a gravitational field are different.

Gravity applies a pressure on the fluid. If the fluid transmits the pressure equally at all points, then how is pascal's law valid?

Number (1) is stated incorrectly. It should read "pressure within an incompressible fluid acts equally in all directions at a given point in the fluid." This is Pascal's law observation.

Chet

 

[blurrscreen]

If it acts equally in all directions at a given point, then why isn't the net pressure at all points in a fluid zero?

 

[Buckleymanor]

It's at a given point not at all points.

 

[paisiello2]

I've had a doubt about the following (somewhat contradictory) statements.
1) pressure applied to an incompressible fluid is equally transmitted at all points.
2) pressure at points at different heights in a fluid placed in a container in a gravitational field are different.

Gravity applies a pressure on the fluid. If the fluid transmits the pressure equally at all points, then how is pascal's law valid?

I think 1) is a reasonable restatement of Pascal's Law.

For 2) you are essentially summing up all the pressures from the weight of the fluid. So as you work your way down the effect is cumulative.

I don't see a contradiction between the two, rather 2) follows as a consequence of 1).

 

[blurrscreen]

Alright, consider one point at a certain depth in a fluid. If the pressure along all directions is equal, then the net pressure there is zero. So why do fluids have pressure at all?

 

[blurrscreen]

Paisiello, thanks for the response. But imagine gravity pushing down on the fluid, thus applying a pressure to it which, according to pascal's law is equally transmitted at all points. Then why do points at different heights in the fluid have different pressures.

 

[Chestermiller]

If it acts equally in all directions at a given point, then why isn't the net pressure at all points in a fluid zero?

If you take a tiny cube of fluid at a given location in the fluid as a free body and the fluid is static, then the pressure on each of the 6 sides of the cube is p. The cube is in static equilibrium, and the net force on the cube is zero, but that doesn't mean that the force on each of its sides is equal to zero.

Chet

 

[blurrscreen]

True, but the NET pressure at every point is zero. Then why does fluid pressure exist?

 

[Chestermiller]

True, but the NET pressure at every point is zero. Then why does fluid pressure exist?

If you have a solid cube (e.g., not submerged in liquid), and you apply equal normal forces on all six of its sides so that the NET force on the cube is zero, are you asking why force exists?

Chet

 

[russ_watters]

Alright, consider one point at a certain depth in a fluid. If the pressure along all directions is equal, then the net pressure there is zero. So why do fluids have pressure at all?

Pressure is a scalar so it doesn't sum to zero, but even if it did, summing to zero doesn't mean the force is zero, it just means there is no motion. If you put your hand in a vise and crank it down, the forces sum to zero too - but it still hurts!

 

[russ_watters]

Paisiello, thanks for the response. But imagine gravity pushing down on the fluid, thus applying a pressure to it which, according to pascal's law is equally transmitted at all points. Then why do points at different heights in the fluid have different pressures.

In all directions, not at all points.

Do you feel more pressure sitting on top of a pile of bricks or under it.

 

[russ_watters]

True, but the NET pressure at every point is zero. Then why does fluid pressure exist?

There is no such thing as "net pressure". Pressure is a scalar, not a vector.

 

[UltrafastPED]

There is no such thing as "net pressure". Pressure is a scalar, not a vector.

I always thought that pressure is defined as force per unit area; thus pressure can have a direction.

See http://hyperphysics.phy-astr.gsu.edu/hbase/press.html

This is usually most important at the boundaries - for example, inside the bubble the pressure is pointing outwards at the surface, and the air pressure is pointing inwards.

For ball under water the external pressure is all pointing inwards - until the ball (or submarine) is crushed.

 

[Chestermiller]

Here is another example to illustrate item (1). Suppose you have a dam. At any given depth (location) right next to the dam, the water pressure is not only acting up and down. It is also pressing sideways (horizontally) on the face of the dam. This is all that item (1) is saying.

Another example is a submarine hull. The water pressure doesn't only press down on the top of the hull and up on the bottom of the hull. It also presses sideways on the sides of the hull.

The same happens to you when you go down to the bottom of a swimming pool at the deep end. If you are standing on the bottom at the deep end (say, held down by weights), the water presses horizontally on your sides as well as on the top of your head, even though your sides are nearly vertical.

Chet

 

[Chestermiller]

I always thought that pressure is defined as force per unit area; thus pressure can have a direction.

See http://hyperphysics.phy-astr.gsu.edu/hbase/press.html

This is usually most important at the boundaries - for example, inside the bubble the pressure is pointing outwards at the surface, and the air pressure is pointing inwards.

For ball under water the external pressure is all pointing inwards - until the ball (or submarine) is crushed.

Actually, if we are getting technical, pressure is the isotropic part of the (second order) stress tensor. Under hydrostatic conditions, the stress tensor is equal to p times the identity tensor. To get the force per unit area acting on a surface, one simply dots (contracts) the stress tensor with a unit normal vector to the surface. This yields the pressure times the unit normal. So the pressure always acts normal to surfaces, as UltrafastPED has indicated.

Chet

 

[Buckleymanor]

Paisiello, thanks for the response. But imagine gravity pushing down on the fluid, thus applying a pressure to it which, according to pascal's law is equally transmitted at all points. Then why do points at different heights in the fluid have different pressures.

Gravity does not apply pressure equally it's stronger the nearer you are to the body that's attracting. Pascal's law allows pressure to be to be equally transmitted when gravity is ignored or pressure is applied in it's absence.

 

[Chestermiller]

Gravity does not apply pressure equally it's stronger the nearer you are to the body that's attracting. Pascal's law allows pressure to be to be equally transmitted when gravity is ignored or pressure is applied in it's absence.

Variation of gravitiational attraction with distance between the attracting bodies is typically not a significant contributor to hydrostatic pressure variations in practice (on the scale of a swimming pool or a glass of water, or even the depth of the ocean). The main contributor is the weight of the overlying fluid. Irrespective of this, I stand by what I said. The pressure at any given location is pushing equally in all directions. This applies whether the pressure is the result of a hydrostatic column of liquid, or whether the liquid is in a cylinder being squeezed by a piston. In fact, it applies not only to incompressible liquids but also to compressible gases.

Chet

 

[Buckleymanor]

Variation of gravitiational attraction with distance between the attracting bodies is typically not a significant contributor to hydrostatic pressure variations in practice (on the scale of a swimming pool or a glass of water, or even the depth of the ocean). The main contributor is the weight of the overlying fluid. Irrespective of this, I stand by what I said. The pressure at any given location is pushing equally in all directions. This applies whether the pressure is the result of a hydrostatic column of liquid, or whether the liquid is in a cylinder being squeezed by a piston. In fact, it applies not only to incompressible liquids but also to compressible gases.

Chet

What causes the variation in the pressure of the overlying fluid.
Why is there more pressure at the bottom of a column of water than the top.I agree that at any given location or point the pressure is equall but if you measured the pressure at a piont at the bottom of a column of water and then at another point at the top won't the two locations be pushing with different amounts of force when compared.

 

[Chestermiller]

What causes the variation in the pressure of the overlying fluid.
Why is there more pressure at the bottom of a column of water than the top.

Suppose you have a cylindrical tank full of water. The pressure at the top surface of the tank is just atmospheric pressure. The pressure at the bottom of the tank is atmospheric pressure plus the weight of the water in the tank divided by the area of the base A. The weight of the water in the tank is equal to the mass of water in the tank times g. The mass of water in the tank is equal to the density of water ρ times the volume of water in the tank. The volume of water in the tank is equal to the area A of the base times the depth of the water h. So,

P_bottom=P_atmospheric+ (ρAh)g/A =P_atmospheric+ ρgh

I agree that at any given location or point the pressure is equall but if you measured the pressure at a piont at the bottom of a column of water and then at another point at the top won't the two locations be pushing with different amounts of force when compared.

Sure. But that's not what item (1) is saying. Item (1) is saying that, at any point near the top of the tank, the pressure is the same in all directions and, at any point near the bottom of the tank, the pressure is the same in all directions.

Chet

 

[Ken G]

But imagine gravity pushing down on the fluid, thus applying a pressure to it which, according to pascal's law is equally transmitted at all points. Then why do points at different heights in the fluid have different pressures.

The answer to this is that gravity does not apply pressure, it is a force with a very different nature than pressure. Gravity is what is known as a "body force", which means, if you think in terms of time rate of change of momentum, gravity acts to remove momentum from every gram of material at a fixed rate. Since it in some sense "acts on the body of the system", since it removes momentum per gram, it is a body force. What this means is, you should think of it as source or sink of momentum that does not need to move any momentum through any boundaries, it just plain deposits or removes it (depending on the sign you are imagining) at every point in the body, like magic if you will.

Pressure is quite different, because it is a surface force. The key difference is that when you think about how pressure adds or removes momentum from something, it always has to pass that momentum through the surface of that something, it cannot just make momentum appear like magic the way gravity does.

In the problem you are thinking about, like water in a pool, all the surfaces we need to think about are horizontal surfaces, and the only direction we need to worry about is vertical. This simplifies the situation, we don't need to worry about the fact that pressure is isotropic, we can just look at what it is doing with momentum in the up and down directions, and that suffices to see what is going on in a pool.

So imagine any horizontal surface inside the water in a pool, and ask, what kind of momentum is passing through that surface? The answer is, pressure is causing upward momentum to pass upward through that surface, and an equal amount of downward momentum to pass downward (that's the "isotropic" aspect). Let's call this equal rate of momentum crossing the surface in either direction P₁A/2 (the notation is motivated by the fact that P is pressure and A is area, but it doesn't matter, it's just a number I'm talking about). It doesn't matter how it is doing that, this is just what pressure does, but if you want to know how, realize it is the action of all the little water molecules that are moving up and down across that surface all the time-- an upward molecule is obviously going to transport upward momentum upward across the surface, and a downward molecule is going to transport downward momentum downward across the surface. Now your question is, if an equal amount of upward momentum is moving upward as downward momentum is moving downward, why does anyone care?

Good question, and it has a good answer. Next imagine a second surface that is a bit deeper down below that first one. Again the same amount of upward momentum is crossing upward as downward momentum that is crossing downward, but the numerical value of this rate of crossing of momentum is different from before, now call it P₂A/2, and you will not be surprised to hear that P₁ < P₂, since I mentioned that P is really pressure here.

Now for the reason you care about these P: think about an imaginary box that is bounded above by the first surface, and below by the second surface, and think about the rate that momentum is entering this box. There are 4 terms you have to add up, two from the top surface and two from the bottom. The top surface has P₁A/2 rate of downward momentum coming down into it, so that's a net accumulation of downward momentum. It has the same rate of upward momentum leaving it upward, but here's the reason they don't cancel-- upward momentum has the opposite sign from downward momentum! So having upward momentum leave has the same effect as having downward momentum enter. Thus the two terms add up to a total of P₁A, not to zero. In short, there is downward momentum entering the imaginary box through its upper surface, which is a downward force on the box, due to the P at its upper boundary.

Play the same game at the lower boundary, and you will find that P₂A is the rate that upward momentum is entering from below, but that's the same as -P₂A rate of downward momentum, since momentum is a vector. So the top and bottom momentum fluxes are the things that might cancel out, not the momentum passing through a given surface into our imaginary box. But if there's gravity, that body force that is adding downward momentum to the mass in the interior of our box, then we can only get force balance if pressure is removing downward momentum (or adding upward momentum, it's the same thing), and that's what the surface momentum fluxes that pressure controls is doing. That's why P₁A < P₂A, and that's why pressure matters, even though it is locally isotropic. It matters because it is different in different places, and Pascal's law says that once you have set up those differences, if you exert additional pressure (like adding weight to the top, for example), it can't affect the differences in pressure, because those were already what they needed to be to move a net flux of momentum around as needed to balance gravity.

 

[blurrscreen]

Thanks for all the help! Hydrostatics was a little tough for me to grasp.

 

[blurrscreen]

New doubt!

Consider a point at a certain depth in an ideal fluid. We say that pressure at that point is due to the weight of the fluid above it. But since the fluid is static, the fluid layer at that depth must exert some sort of reaction force to the weight of the fluid. Why don't we consider this when we calculate pressure at a given point in a fluid.

 

[russ_watters]

Consider a point at a certain depth in an ideal fluid. We say that pressure at that point is due to the weight of the fluid above it. But since the fluid is static, the fluid layer at that depth must exert some sort of reaction force to the weight of the fluid. Why don't we consider this when we calculate pressure at a given point in a fluid.

All forces come in pairs, so you'd be counting the same force twice. For example, if you squeeze your thumb and finger together with 10N force, that requires one finger pushing one way at 10N and the other pushing in the opposite direction with the same force.

 

[Ken G]

But since the fluid is static, the fluid layer at that depth must exert some sort of reaction force to the weight of the fluid. Why don't we consider this when we calculate pressure at a given point in a fluid.

We do, that's what pressure is. Many people think pressure is a force per unit area, but not always-- it only becomes a force when you strip away half the stuff, leaving the force from the other way unbalanced. But pressure usually isn't like that, it's only like that at a boundary and only if you look at only have the story. So you can think of pressure two different ways-- either say pressure is a force per unit area, and only exists at a boundary, so when you talk about it in the interior, you imply you have stripped away one side and created a boundary. That's what we mean when we say "a piston exerts a pressure", so it's best used for external pressure. Or, we can say that pressure is a momentum flux per unit area, coming from both sides, so exists in the interior of the fluid or solid, and is best thought of as an interior pressure. Your confusion is between these two rather different, yet quantitatively equal, meanings of pressure.

 

[blurrscreen]

Exactly, but anything kept in the middle(in this case the point in the fluid) must experience both forces. The pressure hence must be twice what we experience. Why isn't it so?

 

[Ken G]

Exactly, but anything kept in the middle(in this case the point in the fluid) must experience both forces. The pressure hence must be twice what we experience. Why isn't it so?

Excellent question. Consider a crowd of people hitting tennis balls against a wall, and bouncing back. Now consider a similar crowd on the other side of the wall, doing the same thing. There is no net force on the wall, by symmetry, but there is "pressure", and you can get that pressure by looking at the rate the tennis balls are "depositing momentum" into the wall from either side (not both, there is no net deposition from both as there is no net force). Now remove the wall-- what changes? Nothing! The balls look exactly the same on average. There is still no net force in the middle plane, just as there was not before, and there is still pressure there, just as there was before! You don't even need the balls to collide with each other, pressure has no direct requirement for collisions.

Now your question is, but wait a minute, when you have the wall, we are only counting the pressure from half the balls, and when we remove the wall, we are counting all the balls. Do we not make a factor of 2 error? No, we do not, because when there is a wall, the balls from one side are bouncing, so they deposit their momentum twice. When there is no wall, the walls do not bounce, but there are twice as many that participate in the counting. This is that "two different ways" to think about pressure, that give the same answer.

 

[blurrscreen]

Alright. So when we measure pressure without an object at that level, we must consider both forces, right?

 

[russ_watters]

Exactly, but anything kept in the middle(in this case the point in the fluid) must experience both forces. The pressure hence must be twice what we experience. Why isn't it so?

It does experience both forces, but the situation hasn't changed any: now you have four forces instead of two, all of which are balanced and equal to 10N.

Instead of 10=10, you have 10=10=10=10.

In order for forces to be additive, they have to be applied at the same point and pointed in the same direction. Here, the forces are opposite each other.

 

[russ_watters]

Alright. So when we measure pressure without an object at that level, we must consider both forces, right?

There is no "both forces" with pressure in a fluid: pressure is just pressure. It is a scalar. At a macroscopic level, it is continuous: acting in all directions, not just two.

 

[Ken G]

I think what is troubling you is that you have been told that pressure is a force per unit area, but a force is a vector, so has a direction. If you put an infinitely thin plate inside a fluid, where is the force per unit area? Well, it is on both the top and bottom of that plate, but it cancels, so you wonder, what's the point?

The point is, now imagine the plate is not infinitely thin, it has an interior that is empty, it's thin flat metal box. Now you have a force per unit area on the top, pointing down, and a force per unit area on the bottom, pushing up. That's what pressure does-- it tries to crush empty boxes you put in there! Those two forces per unit area are both P, but you'd never add them to get 2P, because for the net force on the box, they add up to zero (no net force on the box), but on each surface of the box, there is a big force, and it better be a strong box or it will crush.

Also, if the pressure changes with height, due to gravity, then the pressure at the top of a box with a finite width is a little less then the pressure at the bottom. That results in a force that does not cancel out, but only because the box has a nonzero width-- the force is proportional to the width of the box. That is called the buoyancy force, and it is not a pressure, it is a force-- but it's there because the pressure changes, not only because there is pressure, and that force goes to zero as the width of the box goes to zero. Pascal's theorem is the recognition that there is a big difference between the pressure, and the change in pressure. If you put more water above the box we are talking about, the pressure will increase, and the box better be even stronger to avoid crushing, but the buoyancy force is not changed, because that deals not in the pressure but in the change in pressure, and Pascal's theorem says the changes in pressure won't be different if you just crank up the pressure by some external action you do.

 

[Chestermiller]

Consider a point at a certain depth in an ideal fluid. We say that pressure at that point is due to the weight of the fluid above it. But since the fluid is static, the fluid layer at that depth must exert some sort of reaction force to the weight of the fluid. Why don't we consider this when we calculate pressure at a given point in a fluid.

This is very much the same thing as the tension in a string. At any point in the string the tension is T, but if you imagine a planar cut at any point within the string, the tension on the left surface of the cut is T acting to the right, while the tension on the right surface of the cut T is acting to the left. Why does pressure act in a way so much similar to tension? Because they are both asoects if the second order stress tensor. The pressure is a compressive stress, and the tension in the string is a tensile stress. The stress tensor encompasses both of these.

Chet

 

[blurrscreen]

OK, got it. Just one last question. If I have a fluid in a container in outer space with an object in the fluid and I seal the container tight, is there any pressure acting on the object?

 

[Chestermiller]

Yes. If there is liquid water and water vapor in the container, then the pressure on the object is equal to the equilibrium vapor pressure of water at the temperature in the container. If there is only liquid water present in the container, then the pressure on the object is higher than the equilibrium vapor pressure of water. The magnitude of the pressure then depends on how much the container is squeezing on the liquid, which, in turn, depends on how the container was filled.

Chet