Pressure change when volume changes for incompressible fluid

[Chris F]

Homework Statement

Hi guys,

My first post here, I hope someone can help me out with a quick fluid mechanics problem.

I'm looking to calculate the pressure increase inside a closed tube full of water when the tube is crushed and therefore its volume reduced.

The tube is filled completely with water and pressurised to a nominal pressure of 10 bar. The starting volume is 2.1l and the volume after crushing is 90% of the initial volume.

2. The relevant equations

The tube remains sealed at all times so I think I can consider the fluid to be static and as its water, incompressible. However I am struggling to find the relevant equation to allow me to calculate the new pressure. All the static problems i have seen are concerned with open containers.

Can anyone help?

Thanks,

Chris

 

[Chestermiller]

Look up the BULK COMPRESSIBILITY of water. But a 10% decrease in volume seems very unlikely.

Chet

 

[Chris F]

Look up the BULK COMPRESSIBILITY of water. But a 10% decrease in volume seems very unlikely.

Chet

Ok thanks, so i can rearrange to solve for dP using the values i quoted above?

Sorry i should probably explained more thoroughly, I am crushing the tube with a press into a machined die as part of a hydroforming operation, the 10% is a rough approximation using the shape of the die as an indication of the form the tube will take.

Thanks,

Chris

 

[Chestermiller]

Ok thanks, so i can rearrange to solve for dP using the values i quoted above?Chris

yes.

 

[rezeew]

Hi Chris,

It sounds like you are dealing with a situation where the volume of an incompressible fluid, in this case water, is changing due to external pressure. In this scenario, the pressure inside the closed tube will indeed increase as the volume decreases. This can be explained by the relationship between pressure and volume for incompressible fluids, known as Boyle's Law.

Boyle's Law states that for an incompressible fluid, the pressure and volume are inversely proportional, meaning that as one increases, the other decreases and vice versa. In your case, as the volume is reduced to 90% of its initial value, the pressure inside the tube will increase by a factor of 10/9, or approximately 1.11 times the initial pressure.

To calculate the new pressure, you can use the equation P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume. Plugging in the values given in your problem, we get:

10 bar * 2.1 L = P2 * 0.9 * 2.1 L

Solving for P2, we get a final pressure of approximately 11.1 bar.

I hope this helps with your problem. Let me know if you have any further questions.

Best,