Barometer, pressure and force

• fog37
In summary, the pressure under the column of mercury at the level of the mercury in the dish into which the column extends is equal to the pressure at the surface of the mercury exposed to the atmosphere.Thanks.f

fog37

Hello Forum,

I understand how the Torricelli barometer works. The atmospheric pressure p_0 is balance by the pressure generated by a column of mercury that is 76 cm tall. The shape and diameter of inverted tube containing the mercury does not matter. The size of the container does not matter either.

Pressure is force/area and gives an indication of how much force is acting over a certain area. If the free surface of the mercury was much larger than the cross-section of the inverted tube, it would seem, from a force point of view, that the total force produced by the atmosphere, which presses on every point of the mercury free surface, would be larger than the total force generated at the base of the inverted mercury tube dipped in the container.
As we know, the fluid is static, i.e. the fluid molecules are all in equilibrium...

Forces can be concentrated or distributed. If a concentrated force was applied to a portion of the free surface of a fluid, I am not sure if that force would transmit itself to all points in the fluids. I feel like only a distributed force that acts over the entire surface of the fluid would be transmitted to all internal fluid points...

Thanks,
fog37

Remember that the pressures at any horizontal points in an non compressible static fluid is the same. If it were not then the fluid would be moving. So the pressure under the column of mercury at the level of the mercury in the dish into which the column extends is equal to the pressure at the surface of the mercury exposed to the atmosphere.

Thanks.

In regards to a fluid in motion, I read that if a certain gauge pressure is applied at one end of a hose carrying water, when the water reaches the end of the hose it has close to zero pressure left as it exits... is that always true, regardless of the length and side of the hose?

When the water exits the faucet it has a certain exit speed. Usually we talk about the pressure exerted by the moving fluid on the walls of the pipe. Does the moving fluid exert a pressure also along its longitudinal direction of motion? Would the exiting fluid, moving at a certain speed, have pressure associated to it when it exits?

The pressure at the free surface of the water jet is atmospheric (zero gauge). Pressure is isotopic, meaning it is the same in all directions (Pascal's law). So water doesn't have a different value axially than it has radially. So, yes, the water pressure inside the jet is zero gauge.

In regards to a fluid in motion, I read that if a certain gauge pressure is applied at one end of a hose carrying water, when the water reaches the end of the hose it has close to zero pressure left as it exits... is that always true, regardless of the length and side of the hose?

When the water exits the faucet it has a certain exit speed. Usually we talk about the pressure exerted by the moving fluid on the walls of the pipe. Does the moving fluid exert a pressure also along its longitudinal direction of motion? Would the exiting fluid, moving at a certain speed, have pressure associated to it when it exits?
Some people (me) like to say that there is more than one kind of "pressure" (other people don't like that...). Generally when used alone, "pressure" refers to static pressure. It is the static pressure that must be zero (gauge) at the outlet of a hose. Yes, there is also pressure due to velocity (velocity pressure or dynamic pressure), though when water hits something that is the combination of velocity and static pressure, called "total pressure" (or stagnation pressure). Obviously, if velocity pressure is zero, then velocity pressure equals total or stagnation pressure. In a situation where there is no energy loss, the sum of these three is constant as the fluid flows. This is described by Bernoulli's equation:

Some people (me) like to say that there is more than one kind of "pressure" (other people don't like that...).
Hi Russ,

Please include me on the the list of people who like to say that, as far as they are concerned, there is not more than one kind of pressure (although I do understand what the terms static pressure, dynamic pressure, and total pressure refer to). These terms do have utility, but, in my humble opinion (for whatever that's worth), the concept that there is more than one fundamental kind of pressure only causes confusion among the uninitiated.

Chet

Please include me on the the list of people who like to say that, as far as they are concerned, there is not more than one kind of pressure (although I do understand what the terms static pressure, dynamic pressure, and total pressure refer to). These terms do have utility, but, in my humble opinion (for whatever that's worth), the concept that there is more than one fundamental kind of pressure only causes confusion among the uninitiated.
Fair enough. It's a trade-off: this user started out confused and hopefully this reduced it instead of adding to it...

Of course, we'll still have to discuss what to do about gauge and absolute pressure...

Chestermiller
Thank you. I would like to better understand these ideas and use only one type of pressure without having to distinguish between static and dynamic pressure.

1) Let's consider the public water supply and a closed horizontal pipe of uniform cross-sectional area A containing water. If the pipe was closed at both end the fluid will be static and the applied pressure p1 at the left pipe surface would be the same as the pressure at the right closing surface. What would happen if we opened the right pipe's end? The fluid will start flowing out. If the fluid was ideal (no viscosity) the gauge pressure at intermediate points along the pipe (which has no narrowing or down/up bends) will be the same in respect of Bernoulli's equation. Will that measured pressure be the same as the pressure p1 applied to the left surface to get the flow started? What happens in this ideal case? What would the pressure be at the very open end of the pipe where the fluid comes out? How would we even measure that? In the case of a water hose, if we put our thumb over the open end and let just a little bit of water come out, we surely reduce the flow rate (gmp) but increase the pressure? How would we measure the pressure at the open end? Water would be moving and have kinetic energy. If that moving water hit surface once it exited the hose we could talk about pressure on that surface...The measurement of pressure seems to require an impact with a surface...

2) If fluid was real and viscous, a pressure gauge located at intermediate points along the pipe would read smaller and smaller pressure values as we get more distant from the left end of the pipe. I know that pressure loss in piping without any size changes or fittings occurs due to friction between the fluid and the pipe walls.

3) Chestermiller mentioned that the pressure at the free surface of the water jet is atmospheric (zero gauge). Pressure is isotopic, meaning it is the same in all directions (Pascal's law). So water doesn't have a different value axially than it has radially. So, yes, the water pressure inside the jet is zero gauge.

So, neglecting the pressure increase due to depth, the pressure inside a fluid flowing in a river would be the same as atmospheric pressure the same as when the fluid is static? In the static case, the pressure at the free surface is p_0 (atmospheric pressure) and pressure increases linearly with depth according to p = p_0+rho*g*d. I agree that pressure is isotropic (same strength in all direction at a specific point inside the fluid). Spatial points at the same level has the same pressure...
And if the fluid is in motion? I would think that the if we placed a pressure measuring device inside the flow, we would measure a different pressure due to the kinetic energy of the fluid when it impacts with the device...

Thanks,
fog37

Thank you. I would like to better understand these ideas and use only one type of pressure without having to distinguish between static and dynamic pressure.

1) Let's consider the public water supply and a closed horizontal pipe of uniform cross-sectional area A containing water. If the pipe was closed at both end the fluid will be static and the applied pressure p1 at the left pipe surface would be the same as the pressure at the right closing surface. What would happen if we opened the right pipe's end? The fluid will start flowing out. If the fluid was ideal (no viscosity) the gauge pressure at intermediate points along the pipe (which has no narrowing or down/up bends) will be the same in respect of Bernoulli's equation.
You can't apply the conventional Bernoulli equation to this situation because it is an unsteady-state (transient flow).
Will that measured pressure be the same as the pressure p1 applied to the left surface to get the flow started? What happens in this ideal case? What would the pressure be at the very open end of the pipe where the fluid comes out?
Assuming the pipe is horizontal, fluid would start coming out the bottom portion of the pipe exit, and air would start flowing into the top half of the pipe. The behavior would be dominated by gravitiational effects. At the exit, the pressure would be atmospheric (at least at the free surfaces), and, if the fluid were considered incompressible, the pressure throughout the pipe would drop to nearly atmospheric (aside from the local hydrostatic head). As air entered, the greatest depth would be near the closed end, and the least depth would be at open end. Locally, the pressure variation would be nearly hydrostatic vertically, but, because of the depth variation, there would be horizontal pressure variations below the surface. This would provide the driving force for the liquid flow.
How would we even measure that? In the case of a water hose, if we put our thumb over the open end and let just a little bit of water come out, we surely reduce the flow rate (gmp) but increase the pressure? How would we measure the pressure at the open end? Water would be moving and have kinetic energy. If that moving water hit surface once it exited the hose we could talk about pressure on that surface...The measurement of pressure seems to require an impact with a surface...
At and after the exit, the pressure at the free surfaces would be atmospheric. Within the flow region after the exit, the fluid parcels would pretty much be in free fall, and the downward accelerations would cancel the hydrostatic varitation. So pressure would be nearly constant throughout the jet. The pressure could be measured by orienting the device normal to the velocity streamlines.

All this could not only be measured, but, for this complicated 2-3D flow, it could also be calculated using a fluid mechanics model.
2) If fluid was real and viscous, a pressure gauge located at intermediate points along the pipe would read smaller and smaller pressure values as we get more distant from the left end of the pipe. I know that pressure loss in piping without any size changes or fittings occurs due to friction between the fluid and the pipe walls.
Correct.
3) Chestermiller mentioned that the pressure at the free surface of the water jet is atmospheric (zero gauge). Pressure is isotopic, meaning it is the same in all directions (Pascal's law). So water doesn't have a different value axially than it has radially. So, yes, the water pressure inside the jet is zero gauge.

So, neglecting the pressure increase due to depth, the pressure inside a fluid flowing in a river would be the same as atmospheric pressure the same as when the fluid is static? In the static case, the pressure at the free surface is p_0 (atmospheric pressure) and pressure increases linearly with depth according to p = p_0+rho*g*d. I agree that pressure is isotropic (same strength in all direction at a specific point inside the fluid). Spatial points at the same level has the same pressure...
Saying that you are neglecting the pressure increase due to depth is the same as saying that you are neglecting hydrostatic pressure variations. I don't think you mean that.
And if the fluid is in motion?
If the fluid in a river is in motion, then the elevation is decreasing from upstream to downstream and the depth may also be decreasing. Vertically, the pressure variation will be nearly hydrostatic, but horizontally, there will be a pressure gradient at all depths below the surface that drive the flow downstream.

I would think that the if we placed a pressure measuring device inside the flow, we would measure a different pressure due to the kinetic energy of the fluid when it impacts with the device...
Again, you would vary the orientation of the pressure measuring device until it showed a minimum. It would then be oriented perpendicular to the streamlines, and there would be no stagnation effect.

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Hello and thanks again.

I found this picture in a plumbing book. The picture is supposed to show the pressure loss (drop) as the water flows in the pipe. What does it means that the pressure at the outlet is 42 psi? When the fixture is closed the pressure is 60 psi, the same as the inlet (head pressure)...

Hello and thanks again.

I found this picture in a plumbing book. The picture is supposed to show the pressure loss (drop) as the water flows in the pipe. What does it means that the pressure at the outlet is 42 psi?
It means that it is connected at the outlet to something else that is not shown.

ok, thanks.
the picture says "at fixture", let's say a faucet.

What if there was no pressure drop, i.e. pipe and fluid were ideal? Would we have 60 psi all the way up to the faucet and then a quick transition from 60 psi to atmospheric pressure?

ok, thanks.
the picture says "at fixture", let's say a faucet.

What if there was no pressure drop, i.e. pipe and fluid were ideal? Would we have 60 psi all the way up to the faucet and then a quick transition from 60 psi to atmospheric pressure?
Yes, provided you neglect the effect of the change in cross section at the spot labelled 57 psi.

What if there was no pressure drop, i.e. pipe and fluid were ideal? Would we have 60 psi all the way up to the faucet and then a quick transition from 60 psi to atmospheric pressure?
This is a self-contradictory set of constraints, leading to an answer that doesn't make sense. If you have no pressure drop you have no pressure drop and static pressure has to be atmospheric everywhere except when you have pipe sizing changes (then you apply Bernoulli's equation). Perhaps you meant no pressure drop in the piping and then a large obstruction in the faucet, in which case Chet's answer is correct. But then the question is basically asking "if you don't have pressure drop anywhere except where you have pressure drop, do you have pressure drop there?" Sure.

Either way, domestic piping systems are a prime example of where such assumptions lead to badly wrong answers about real life flow in them.

Hi russ_watters,

So, if the pipe and the fluid were both ideal, what would the pressures be at the different locations in the uniform cross-section pipe given that the head pressure is 60 psi?

Hi russ_watters,

So, if the pipe and the fluid were both ideal, what would the pressures be at the different locations in the uniform cross-section pipe given that the head pressure is 60 psi?
What does "ideal" mean? If you mean no friction losses, then all of the static pressure becomes velocity pressure at the outlet. Find the velocity/flow at the outlet with Bernoulli's equation, then work backwards for pressure and velocity in different sections. The flow rate will be nonsensically high though.

Is there a specific problem you are trying to solve of scenario you have in mind?

hi russ_watters,

By "ideal" I mean an inviscid, incompressible fluid flowing into a perfect pipe. Bernoulli principle applies under these conditions (and along the same streamline. Why does it need to apply only along the same streamline?)
If a force F is applied at one end of a pipe, the fluid will start accelerating and the flow rate Q=A v will be time dependent across a certain cross-section. In the figure below the pipe cross-section is not uniform. The pressure is P1 anywhere along the section of the tube with the same cross-section. Pressure P2<P1. At the exit end, the pressure is P2 or the atmospheric pressure P_0?

I have been trying to understand "dynamic pressure". As you mention, there is hydrostatic pressure (which I understand as being isotropic and linearly dependent on depth), dynamics pressure and total pressure (which is the sum of static pressure and dynamic pressure). Dynamic pressure seems to be measurable using Pitot tube which converts it into a static pressure...

What is your understanding of dynamic pressure?

thanks,
fog37

By "ideal" I mean an inviscid, incompressible fluid flowing into a perfect pipe. Bernoulli principle applies under these conditions (and along the same streamline.
Typically, that means lossless too, as Chet said. You can add terms to handle pressure (energy) loss, but the basic Bernoulli's equations assume conservation of energy.
Why does it need to apply only along the same streamline?)
A streamline follows the path of a row of individual particles as they travel. Bernoulli's principle is a conservation of energy statement describing to what happens to a particle as it travels, so it can only apply along the path of travel.
If a force F is applied at one end of a pipe, the fluid will start accelerating and the flow rate Q=A v will be time dependent across a certain cross-section. In the figure below the pipe cross-section is not uniform. The pressure is P1 anywhere along the section of the tube with the same cross-section. Pressure P2<P1. At the exit end, the pressure is P2 or the atmospheric pressure P_0?
At the end, the pressure must be atmospheric.

This example need not be time varying, as the force causes acceleration of the fluid through the funnel and a resulting constant/steady flow.
I have been trying to understand "dynamic pressure". As you mention, there is hydrostatic pressure (which I understand as being isotropic and linearly dependent on depth), dynamics pressure and total pressure (which is the sum of static pressure and dynamic pressure). Dynamic pressure seems to be measurable using Pitot tube which converts it into a static pressure...
Essentially yes. It's only really "felt" when something gets in the way of the flow. The actual interaction is pretty complicated (the air in a pitot tube isn't moving, but the pressure is generated by air flowing around it...), but that's the gist of it. That's why drag is dynamic pressure times cross sectional area times a coefficient (essentially an efficiency at interrupting the airflow).

Hi russ_watters,

I have been reflecting more about static and dynamic pressure and here are my thoughts:

1) Static pressure is the pressure that develops in a fluid at rest. Static pressure is isotropic at a point, is linearly dependent on depth, and is always perpendicular to a surface regardless of its orientation.

2) In the case of an incompressible, inviscid fluid flowing in a pipe, the "static" pressure decreases where the fluid velocity increase (Venturi effect). The pressure is called "static" in this case as well even if the fluid is actually moving. Why is it called static? Its value is surely different from the static pressure described in 1).
My explanation is that a manometer measuring this type of static pressure is formed by fluid contained in a tube that is perpendicular to the fluid flow direction. The fluid in the manometer is actually at rest and determines a pressure on an internal surface of the manometer. The value of this static pressure value depends on how fast the fluid is moving longitudinally along the pipe.

3) Total pressure for a moving fluid is equal to the static pressure + dynamic pressure. a Pitot tube can measure dynamic pressure by subtracting static pressure from the total pressure. The Pitot tube has two sections. One section is perpendicular to the fluid flow and works like a standard manometer measuring the "static" pressure as described in 2). This pressure is the p in Bernoulli equation. The other section of the Pitot tube is instead longitudinal to the flow. The moving fluid enters that longitudinal section which has a dead end. The fluid reaches the dead end and stops forming a stagnation point. The pressure at the stagnation point is related to the momentum of the fluid before it collides and stops. Why does the pressure at the stagnation point also have a static component? This static pressure is not the static type of pressure that would develop when the fluid is at rest...

Any correction?

thanks,
fog37