# Homework Help: Water Pressure problem

1. Dec 23, 2005

### bruceflea

A siphon is used to empty a large vessel of water (see picture). Calculate/state the intial (absolute) pressures at points 0, 1, 2, 3, 4 and 5.

http://img499.imageshack.us/img499/8104/siphon1oq.th.jpg [Broken]

P0 = 1.013 x 10^5 Pa - the pressure exerted on the water by the atmosphere

P1 = Po + density x gravity x height of water

P2 = Po + density x gravity x height of water

The answer for pressure at point 3 is 1.013x10^5 Pa but I'm not sure why that is.

P4 = P3 - density x gravity x height of water

P5 = 1.013 x 10^5 Pa - the minimum pressure required to overcome atmospheric pressure.

Could someone please explain the pressure at point 3 for me?

Thanks

Last edited by a moderator: May 2, 2017
2. Dec 23, 2005

### Astronuc

Staff Emeritus
Assuming that P3 is at the same elevation as P1, then the pressure P3 = P1, which would seem to contradict the pressure due to the height (2 m) of water above between P4 and P3.

I'll have to get back to this later.

3. Dec 23, 2005

### Clausius2

One knows that Bernoulli problems are tricky if we don't pay attention to the physics underlying. There is no contradiction because pressure at point 1 is different than pressure at point 3 (assuming 1 is far away from the hole). Pressure at point 1 is the Hydrostatic pressure, whereas pressure at point 3 is a Static pressure influenciated by the dynamic fluid motion (the fluid is being accelerated!!!!!). Moreover, as Astronuc pointed out, pressure at point 3 must be the same than pressure at point 5 if there is no transversal section variation and 5 and 3 are at the same altitude (apply Bernoulli). Pressure at point 5 must satisfy the mechanical constraint P=Patm, therefore the pressure at 3 must be also Patm. The Static pressure lost from 1 to 3 is due to the transformation of Hydrostatic pressure in Dynamic pressure.

4. Dec 23, 2005

### Astronuc

Staff Emeritus
I was pondering whether the fluid was moving or is the question referring to the moment just prior to movement of the fluid.

There must be a gradient of pressure for the fluid to move.

5. Dec 23, 2005

### Clausius2

If there is no motion, then P1=P3=Hydrostatic pressure. Let's think of how the water is put into motion. In order to get the siphon working one must suck at the siphon exit to generate the motion (similarly when trying to take the gas out of the car gas deposit). I mean, the siphon does not start to work spontaneously.

6. Dec 23, 2005

### bruceflea

Thanks, makes perfect sense.

Just to make sure I understand this here's another siphon situation.

http://img427.imageshack.us/img427/1121/siphon22ho.th.jpg [Broken]

The entrance of the siphon is at the bottom of the tank and the exit is below the tank.

For this situation would the pressure at the exit be Patm and so the pressure at the entrance would be Patm - dgh?

Assuming that the exit pressure is Patm, with respect to bernouilli's problems, do fluids always leave (or its reasonable to assume so) pipes/hoses/siphons at atmospheric pressure, whether the exit is up, down or horizontal?

Last edited by a moderator: May 2, 2017
7. Dec 23, 2005

### Astronuc

Staff Emeritus
The surrounding air is at atmospheric pressure, so the water at the siphons exit has to be atmospheric pressure, and actually is might be slightly higher due to interatomic forces which are responsible for surface tension.

8. Dec 24, 2005

### Clausius2

What is "h"?. If h is the difference of altitude between the exit and entrance then you're right.
About the boundary condition P=Patm for an outflowing jet
The boundary condition P=Patm is usually employed in bernoulli problems as a mechanical constraint at outflowing jets into the atmosphere. This boundary constraint is only valid at high Reynolds number, when $$\partial P/\partial r$$ is negligible. The assumption underlying is the boundary layer flow. The exiting jet flow would be described by Prandtl boundary layer equations, developing an annular boundary layer due to air viscosity. At high Re, this boundary layer is thin enough to assume the internal pressure in the jet core is the same than the external surrounding pressure (pressure remains constant radially across the layer). The existance of this layer is unavoidable. Because of that one always assume a flow discharge as an irreversible process, because this layer is an entropy generator, and some kinetic energy is ultimately dissipated.
The condition P=Patm would not be longer valid at small Re (Stokes flow), or at very high Mach (in supersonic flow) where the hydrostatic boundary condition is not felt by some fluid particles due to the hyperbolicity of the flow.