# Continuity in fluid flow

1. Sep 22, 2008

### rupam_iit

Please consider a U-tube filled with an incompressible fluid as in the attached figure. Piston P divides the fluid in two segments. When P moves, the fluid particles on immediate vicinity of either face (points marked 1 and 2) will have same velocity.

Does this mean, they may considered to be the same point on a streamline ? Or, can the piston be neglected and the entire fluid mass be considered continuous from fluid mechanics point of view ?

#### Attached Files:

• ###### piston.pdf
File size:
18.1 KB
Views:
66
2. Sep 22, 2008

### Andy Resnick

It's tough to say exactly- the surface of the piston introduces a boundary condition (no slip) that is not present otherwise, so I don't think it can be neglected.

Also, I'm not sure it's true that points '1' and '2' will have the same velocity, even up to a sign difference. On one side of the piston there is compressive stress, on the other tension. Since (the way you have drawn the figure) no fluid can move around the piston, the dynamics of each arm will be different, especially if gravity is present.

3. Sep 22, 2008

### rcgldr

A streamline (in the Bernoulli sense) implies that no work is being done, but the piston would be peforming work if there's any resistance to the movement of the fluid, such as gravity as pointed out by the previous post.

4. Sep 22, 2008

### rupam_iit

Thanks Jeff and Andy,

I will now put the actual problem which prompted me to clear up the above confusion. Please refer to fig. "piston2.bmp". In both the cases, equal columns of water are on either side of the piston with unequal working areas. In Case 1, the connection is rigid while in Case 2, it is a spring coupling.

Now, for equilibrium, will the forces (h*A) have to be equal on both sides (here, h is the height of the water column), or just pressure (h) must be equal, irrespective of the piston working area.

#### Attached Files:

• ###### piston2.bmp
File size:
120.2 KB
Views:
82
5. Sep 23, 2008

### Andy Resnick

This second problem is also a little unclear, but it looks like you are studying how a U-tube manometer works:

http://www.practicalphysics.org/go/Experiment_878.html;jsessionid=alZLdQlAHb1 [Broken]

At equilibrium, the sum of the forces and torques are zero. In terms of the fluids, it means there is no pressure gradient present, and no flow- each piston opposes the same pressure (force/area). This is not a contradiction with balancing the total force on each face of the two pistons, which you can verify for yourself. Replacing the rigid connection with a spring will introduce an enormous amount of complexity to the dynamical problem, although it could be a good homework problem....

Last edited by a moderator: May 3, 2017
6. Sep 23, 2008

### stewartcs

If the fluid columns are the same height (and the same density of fluid is used in each), the pressure at the center of each piston face is equal. Since F = PA, and P is the same, the force (F) will be greater for the piston with the larger surface area. Thus, they will not be in equilibrium if the fluid columns are of the same height.

For example, if the pistons are locked in place, and fluid is placed in each tube to the same height, once the pistons are unlocked, the one with the larger area will push the smaller one causing the fluid columns to adjust their heights until equilibrium is achieved (i.e. the smaller piston's column will rise and the larger piston's column will fall).

Whether it is a rigid bar or spring connecting them makes no difference.

BTW, the pressure is equal pgh (p is fluid density, g is gravitational acceleration, h is height). Also, I'm neglecting frictional effects of the piston seals.

Hope this helps.

CS

7. Sep 23, 2008

### rupam_iit

Thanks a lot, the problem is cleared