# Pressure in a U-shaped Tube

1. Mar 2, 2007

### e(ho0n3

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

$$P_0 + \rho h g$$

where $P_0$ is the atmospheric pressure and $\rho$ is the density of the fluid. The pressure on the other side is

$$P_0 + \rho (x - h) g$$

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?

2. Mar 2, 2007

### apelling

You are wrong when you assert that the upward pressure is due to:

$$P_0 + \rho (x - h) g$$

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. Mar 2, 2007

### e(ho0n3

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. Mar 2, 2007

### Dick

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?

Last edited: Mar 2, 2007
5. Mar 2, 2007

### apelling

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)

Last edited: Mar 2, 2007
6. Mar 2, 2007

### e(ho0n3

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.

Last edited: Mar 2, 2007
7. Mar 2, 2007

### apelling

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. Mar 3, 2007

### e(ho0n3

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

$$P_0 A + \rho g (x - x_2) - (P_0 A + \rho x_2 g) = \rho g (x - 2x_2)$$

Since the net force is 0, then $x_2 = x / 2$. 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 $x / 2$. Why is this?

9. Mar 3, 2007

### Mentz114

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. Mar 4, 2007

### e(ho0n3

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. Mar 4, 2007

### Mentz114

I missed that.

I don't understand your question. If the fluid is in equilibrium all forces cancel out.

12. Mar 5, 2007

### e(ho0n3

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. Mar 5, 2007

### Mentz114

I don't think I need to. If nothing is moving - there are no forces.

14. Mar 5, 2007

### e(ho0n3

That is certaintly not true. Consider gravity. What force is "canceling out" gravity?

15. Mar 6, 2007

### Mentz114

The weight of the liquid on the left cancels out the weight on the right - so no movement.

16. Mar 6, 2007

### e(ho0n3

How do you know the weight of the liquid on the left is the same as on the right?

17. Mar 6, 2007

### Mentz114

Because the levels are the same left and right - and - there is no movement.

18. Mar 6, 2007

### e(ho0n3

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. Mar 6, 2007

### e(ho0n3

The above has an error. It should be:

$$P_0 A + \rho g (x - x_2)A - (P_0 A + \rho g x_2 A) = \rho g (x - 2x_2)A$$

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

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

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. Mar 6, 2007

### Mentz114

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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