Fluid Heights in an Open U-Tube

[annamal]

Homework Statement

A U-tube with both ends open is filled with a liquid of density to a height h on both sides (See attachment). A liquid of density is poured into one side and Liquid 2 settles on top of Liquid 1. The heights on the two sides are different. The height to the top of Liquid 2 from the interface is and the height to the top of Liquid 1 from the level of the interface is . Derive a formula for the height difference.

Relevant Equations

$p = p_0 + \rho gh$

Pressure on side with liquid 1 = $p_0 + \rho_1gh_1$
Pressure on side with liquid 2 = $p_0 + \rho_2gh_2$
The solution sets the two pressures equal. I am confused how the two sides have the same amount of pressure, one would have more than the other.

 

[Orodruin]

If one side had more pressure then you would not have hydrostatic equilibrium in the bend of the tube.

 

[Lnewqban]

No attachment.
A balance can't be in horizontal position if we put different weights on each side of it.

 

[sysprog]

What does "a liquid of density" mean without saying density 'something'? non-zero-density? $-$ I searched on some words in your question, and found this:

and more definitively, from: https://opentextbc.ca/universityphysicsv1openstax/chapter/14-2-measuring-pressure/

Fluid Heights in an Open U-Tube​

A U-tube with both ends open is filled with a liquid of density

to a height h on both sides (https://opentextbc.ca/universityphysicsv1openstax/chapter/14-2-measuring-pressure/#CNX_UPhysics_Figure_14_02_UtubeTwoDe). A liquid of density

is poured into one side and Liquid 2 settles on top of Liquid 1. The heights on the two sides are different. The height to the top of Liquid 2 from the interface is

and the height to the top of Liquid 1 from the level of the interface is

. Derive a formula for the height difference.

Figure 14.14 Two liquids of different densities are shown in a U-tube.

Strategy​

The pressure at points at the same height on the two sides of a U-tube must be the same as long as the two points are in the same liquid. Therefore, we consider two points at the same level in the two arms of the tube: One point is the interface on the side of the Liquid 2 and the other is a point in the arm with Liquid 1 that is at the same level as the interface in the other arm. The pressure at each point is due to atmospheric pressure plus the weight of the liquid above it.

[End of post edited by a Mentor]

 

[Helios047]

The reason why we can set both sides of the equation to p0 is because of the boundary layers that forms when a liquid comes into a contact with air. The deeper down we go in elevation, higher the pressure. Since we're assuming that tube isn't that tall, we can approximate the pressure in one opening of the tube to be equal to the other opening.

 

[annamal]

What does "a liquid of density" mean without saying density 'something'? non-zero-density? $-$ I searched on some words in your question, and found this:

View attachment 299598

and more definitively, from: https://opentextbc.ca/universityphysicsv1openstax/chapter/14-2-measuring-pressure/

Fluid Heights in an Open U-Tube​

A U-tube with both ends open is filled with a liquid of density

to a height h on both sides (https://opentextbc.ca/universityphysicsv1openstax/chapter/14-2-measuring-pressure/#CNX_UPhysics_Figure_14_02_UtubeTwoDe). A liquid of density

is poured into one side and Liquid 2 settles on top of Liquid 1. The heights on the two sides are different. The height to the top of Liquid 2 from the interface is

and the height to the top of Liquid 1 from the level of the interface is

. Derive a formula for the height difference.
View attachment 299599Figure 14.14 Two liquids of different densities are shown in a U-tube.

Strategy​

The pressure at points at the same height on the two sides of a U-tube must be the same as long as the two points are in the same liquid. Therefore, we consider two points at the same level in the two arms of the tube: One point is the interface on the side of the Liquid 2 and the other is a point in the arm with Liquid 1 that is at the same level as the interface in the other arm. The pressure at each point is due to atmospheric pressure plus the weight of the liquid above it.

[End of post edited by a Mentor]

Yes, that is where I got this question from. The solution is unclear/don't understand why we can set the two pressure sides equal.

 

[sysprog]

Yes, that is where I got this question from. The solution is unclear/don't understand why we can set the two pressure sides equal.

There are 3 fluids involved: air, Liquid 1, and Liquid 2. The air pressure in the 2 sides of the tube is negligibly different, because the height of each of the two sides of the tube is very short compared to that of the atmosphere. But the density of Liquid 2 is less than that of Liquid 1, so Liquid 2 floats on Liquid1.

We're not 'setting' anything equal; we're recognizing the fact that the pressure at the height of the interface between the two liquids is equal to the pressure at the same height in the other side of the tube.

The amount of Liquid 1 above that height in the Liquid 1 side has less volume than the amount of Liquid 2 in the other side has, but it has equal weight $-$ that's because Liquid 1 has greater density than Liquid 2 has.