Why is the Pressure in a U-shaped Tube Constant?

In summary, the U-shaped tube is filled with fluid and is still because the atmospheric pressure pushing on both openings of the tube are the same. The pressure on any cross-section of the fluid in the tube is 0. When considering a cross-section of fluid at some distance below the surface, the pressure on one side is equal to the pressure on the other side. This is because the pressure at a point in a fluid is a scalar and is the same in all directions. The formula p = density * g * depth can be used to calculate pressure at a certain depth, and only the fluid above the cross-section counts towards this pressure. The weight of the fluid above the column and the force due to atmospheric pressure are the main forces acting on
  • #1
e(ho0n3
1,357
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Let's say there is a U-shaped tube, both ends of the tube open, filled with some fluid. The fluid is still because the atmospheric pressure pushing on both openings of the tube are the same.

Suppose the column of fluid, when streched into a straight-tube, measures x. Given any cross-section of the fluid in the tube, the pressure on it measures 0. Let's look at this in detail:

Consider a cross-section of fluid at some distance h below the surface of the fluid. The pressure on one side of the fluid is

[tex]P_0 + \rho h g[/tex]

where [itex]P_0[/itex] is the atmospheric pressure and [itex]\rho[/itex] is the density of the fluid. The pressure on the other side is

[tex]P_0 + \rho (x - h) g[/tex]

Since the pressure on both sides are equal, equating the two equations above yields h = x - h. This, of course, is only possible if the cross-section is in the middle of the tube (h = x/2). Hmm...did I miss something here?
 
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  • #2
You are wrong when you assert that the upward pressure is due to:

[tex]P_0 + \rho (x - h) g[/tex]

This implies that the remainder of the fluid is all above the cross section in question. This is true at the mid point only so your calculations work there. If the cross section is not at the mid point then some of the fluid is below the cross section. Fluid below the cross section does not contribute to the pressure. The only fluid above the cross section is the bit in the other arm of the u tube which is the same depth as the fluid in the first arm.

Hence the fluid is in equilibrium at all points.
 
  • #3
apelling said:
Fluid below the cross section does not contribute to the pressure. The only fluid above the cross section is the bit in the other arm of the u tube which is the same depth as the fluid in the first arm.

There has to be some pressure on the fluid below the cross-section, or otherwise, the top pressure would make the liquid move. Let A denote half of the tube, and B the other. You're saying that because the pressure atop the cross-section in A is the same as that on the parallel cross-section at the same depth in B, the column of water between these two cross-sections doesn't move. In other words, you're saying that the pressure below the cross-section in A is equal to the pressure atop the parallel cross-section in B. I guess that makes sense intuitively, but how can you derive this from first-principles? Specifically, why do I have to consider the parallel cross-section at the other half of the tube?
 
  • #4
The pressure at a point in a fluid is a scalar. It doesn't point in any direction. It's the same in all directions. Is that enough of a first principle?
 
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  • #5
The formula p= density *g*depth works out pressure at depth. Only fluid above the cross section counts. The pressure increase is due to the weight of the fluid above the area over which pressure is created. Consider a column of fluid of crossectional area A density rho and height H.
Its mass is volume*density= A*rho*H
Its weight is mass*gravity = A*rho*g*H
pressure = weight/area = rho*g*H

Fluid below the cross section creates a reduction in pressure if its in the same side as the cross section. This balances out with pressure from the fluid below the cross setion in the other arm. Therefore no net contribution from fluid below the cross section.

(Pressure at a point is a scalar but the force exerted by pressure on a surface is a vector normal to that surface)
 
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  • #6
It seems that the concept of pressure is the cause of my confusion, so for now let's forget about pressure. Consider a column of fluid in the tube. What are the forces acting on this column of fluid? The sum of the forces pushing against the top of the column are the weight of the fluid above the column and the force due to atmospheric pressure right? What about the forces pushing on the bottom side of the column? What are they?

Oh, and let's not forget about the weight of the column itself.
 
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  • #7
e(ho0n3 said:
Consider a column of fluid in the tube. What are the forces acting on this column of fluid? The sum of the forces pushing against the top of the column are the weight of the fluid above the column and the force due to atmospheric pressure right?

The weight of fluid above column (if its just open to the air) is what causes atmospheric pressure. The atmospheric pressure on the average person is caused by the tonnes of air each of us supports above us.

If the column rests on something ie its in a sealed container, then a reaction force from the base holds it up.

If we have an open bottom to the tube then the fluid can be held up if the top of the tube is sealed. The sealed region above the column could contain a vacuum so there is no downwards force on the liquid other than its own weight. This can be held up by the force created by atmospheric pressure pushing up on the bottom of the column. Atmospheric pressure can support a 10m (30 foot) column of water. This is how most trees get water to their top branches.

The height of the supported column varies from day to day because air pressure does too. Some measurements of pressure are linked to the height of fluid supported by air pressure eg mmHg (millimeters of Mercury).

If there were some air in the region above the column of fluid then this would exert a downward force. If this force is larger enough it could push the column down. But as the column moves down the air expands and its pressure drops until an equilibrium is reached (assuming we have a long tube).
At this point the weight of the trapped air plus the weight of the fluid column is equal to the force created by atmospheric pressure upwards on the base of the fluid column.
 
  • #8
e(ho0n3 said:
Consider a column of fluid in the tube. What are the forces acting on this column of fluid? The sum of the forces pushing against the top of the column are the weight of the fluid above the column and the force due to atmospheric pressure right? What about the forces pushing on the bottom side of the column? What are they?

Oh, and let's not forget about the weight of the column itself.

To simplify my analysis, permit me to stretch the tube out so that it becomes straight. Let [tex]x[/tex] be the height of the fluid in the straightened tube. Let [itex]x_2 - x_1[/itex] be the height of the column of fluid. The net force on this colum of fluid is:

[tex]P_0 A + \rho g (x - x_2) - (P_0 A + \rho x_2 g) = \rho g (x - 2x_2)[/tex]

Since the net force is 0, then [itex]x_2 = x / 2[/itex]. This is exactly the same kind of result I derived in my first post. Hmm...I guess it is pointless to analyse the forces on a column of fluid that does not have one of its ends at exactly [itex]x / 2[/itex]. Why is this?
 
  • #9
I guess it is pointless to analyse the forces on a column of fluid that does not have one of its ends at exactly at x/2. Why is this?

You can ignore any part of the tube above the liquid. In the pic attached, the situation is identical. So the bottom of the U is always at the center of the effective length.
 

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  • #10
Mentz114 said:
You can ignore any part of the tube above the liquid. In the pic attached, the situation is identical. So the bottom of the U is always at the center of the effective length.

Yes, I agree. In my analysis, I don't mention what the length of the tube is. I only work with the column of fluid. x/2 is the center of the column of fluid.

I'm still at a loss though as to why it is not possible to calculate the forces on any small column of fluid within the whole column of fluid.
 
  • #11
Yes, I agree. In my analysis, I don't mention what the length of the tube is. I only work with the column of fluid. x/2 is the center of the column of fluid.

I missed that.

I don't understand your question. If the fluid is in equilibrium all forces cancel out.
 
  • #12
Mentz114 said:
I don't understand your question. If the fluid is in equilibrium all forces cancel out.

Would you prove your last statement, i.e. would you show how the forces on a small column of fluid in the U-tube cancel out. Thanks.
 
  • #13
I don't think I need to. If nothing is moving - there are no forces.
 
  • #14
Mentz114 said:
I don't think I need to. If nothing is moving - there are no forces.
That is certaintly not true. Consider gravity. What force is "canceling out" gravity?
 
  • #15
The weight of the liquid on the left cancels out the weight on the right - so no movement.
 
  • #16
Mentz114 said:
The weight of the liquid on the left cancels out the weight on the right - so no movement.

How do you know the weight of the liquid on the left is the same as on the right?
 
  • #17
Because the levels are the same left and right - and - there is no movement.
 
  • #18
Mentz114 said:
Because the levels are the same left and right - and - there is no movement.

Please show me how the weight of the fluid above and below the column of fluid being analyzed is related to the levels of the fluid. I fail to see this relationship.

Also, what if the levels weren't the same? How would that affect the analysis of the forces on the column of fluid?
 
  • #19
e(ho0n3 said:
To simplify my analysis, permit me to stretch the tube out so that it becomes straight. Let [tex]x[/tex] be the height of the fluid in the straightened tube. Let [itex]x_2 - x_1[/itex] be the height of the column of fluid. The net force on this colum of fluid is:

[tex]P_0 A + \rho g (x - x_2) - (P_0 A + \rho x_2 g) = \rho g (x - 2x_2)[/tex]

The above has an error. It should be:

[tex]P_0 A + \rho g (x - x_2)A - (P_0 A + \rho g x_2 A) = \rho g (x - 2x_2)A[/tex]

But either way, that is wrong. I just had an "Aha!" moment. The fluid below the column of fluid from [itex]x_2[/itex] to [itex]x[/itex] is pushing against the column of fluid from [itex]x/2[/itex] to [itex]x[/itex] and the air beyond. Thus, the force pushing the column of fluid from below is:

[tex]P_0 A + \rho g Ax/2 - \rho gA(x/2 - x_2) = P_0A + \rhoA x_2 g[/tex]

This is exactly the same amount of force pushing the column of fluid from the top and hence, the sum of the forces is 0.
 
  • #20
You got it

That's it. I was coming to post this pic and point out that at the cut the downward force is
r.a.x

and the upward is

r.a.h - (r.a.h-r.a.x) = r.a.x
 

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  • #21
Mentz114 said:
That's it. I was coming to post this pic and point out that at the cut the downward force is
r.a.x

and the upward is

r.a.h - (r.a.h-r.a.x) = r.a.x

I don't know what you mean by r.a.h and r.a.x but thansks anyways.
 
  • #22
r.a.h = rho*area*height
r.a.x = rho*area*x

OK ?
 

What is U-shaped tube?

A U-shaped tube is a laboratory instrument commonly used to measure pressure or pressure differences in fluids. It consists of a long, narrow tube bent into the shape of a "U" with one end closed and the other end open.

What is the principle behind pressure in a U-shaped tube?

The principle behind pressure in a U-shaped tube is based on the fact that fluids will always seek to reach a state of equilibrium. This means that if one side of the tube has a higher pressure than the other, the fluid will flow from the area of high pressure to the area of low pressure until the pressures are equal.

How is pressure measured in a U-shaped tube?

Pressure in a U-shaped tube is measured using a manometer, which is a device that consists of a U-shaped tube filled with a liquid, usually mercury, and a scale for measuring the height of the liquid in the tube. The difference in height between the two sides of the tube represents the difference in pressure.

What factors affect pressure in a U-shaped tube?

The factors that affect pressure in a U-shaped tube include the type of fluid used, the density of the fluid, the height of the fluid column, and the shape and diameter of the tube. Changes in any of these factors can result in a change in pressure measured by the manometer.

What are some practical applications of pressure in a U-shaped tube?

Pressure in a U-shaped tube has many practical applications, including measuring the pressure in a closed system, such as a tire or a pipe, and determining the difference in pressure between two points, such as the top and bottom of a water tank. It is also commonly used in chemistry experiments to measure gas pressures and in medical devices such as blood pressure cuffs.

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