Centrifugal force in high pressure vessel

AI Thread Summary
The discussion revolves around the effects of centrifugal force on a pre-pressurized cylindrical vessel spinning at 30,000 RPM. When the vessel is initially pressurized to 3,000 psi, the centrifugal force is theorized to create an additional pressure of 1,000 psi, leading to a total pressure of 4,000 psi. However, participants argue that the actual resultant pressure will not simply be additive due to the complexities of fluid dynamics and pressure distribution, as described by the Navier-Stokes equations. The pressure increases with the number of molecules, and calculations suggest that the final pressure could reach significantly higher values, potentially around 200,000 psi. The conversation emphasizes the need for precise calculations and understanding of gas behavior under rotation.
drewman13
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An enclosed cylindrical* (disc) vessel one inch thick and five inches in radius holding two pounds of fluid at ambient atmospheric pressure (14psi?)and being spun at 30,000 rpm's. Assume the centrifugal force acting on the fluid creates a pressure of 1,000 psi. The question is ...

"If the vessel is pre-pressurized, to say 3,000 psi, before it is spun up to 30,000 rpm's, will the centrifugal force add pressure to the already pre-pressurized (3,000 psi) vessel?"

Would the resultant force then be 4,000 psi? If not, what would the total psi be and why?
 
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Use Dalton's partial pressure law.
 
aekanshchumber said:
Use Dalton's partial pressure law.

He didn't say the vessel was pressurized with gas.
 
Tide said:
He didn't say the vessel was pressurized with gas.
If it is so, then the pressure of the gas will depend on the position from where it is being measured.
Due to rotations, the gas will be displaced toward the outer side. And the pressure of the gass will gradually increase when we move from ceter of rotation to the extreme of the daimeter.
 
Yes, I'm with aekanshchumber, and I'll give him an additional support for his opinion writing a few equations:

Inside the vessel, and choosing a reference frame spinning with it, the Navier-Stokes equations yield:

\frac{\partial P}{\partial r}=\rho\omega^2 r in Hydrostatic form.

Solving for P(r) you will obtain the pressure distribution. The constant resulting of the last equation has to depend on the initial value in the case of a closed vessel, where no boundary constraint is possible to satisfy.

Maybe, the unsteady process is harder [\B] of analyzing due to a progressive acceleration of the fluid is necessary. The kinematic field has to be solved firstly before doing any comment about the pressure at the very first instants. I don't think such a simple rule as drewman13 has stated would be correct for figuring the total pressure out.
 
The pressure will surely increase, but it will not be 4000psi.

Remember that the pressure is proportional on the no of molecules. So for a fixed volume (of the cylinder, V) at 1.4psi if the Number of moecules be N, then at 3000 psi it will be around 2000N.

Now if N molecules generated a 1000psi pressure when gyroscoped @ 30000rpm, then 2000N molecules will generate 2000 x 1000psi at same rpm. I imagine they are proportional however they may not be directly proportional. But logically and from Classical kinetic theory they are likely to be directly proportional.

So you can imagine the net pressure to be humongous 2,000,000 psi.
 
snbose said:
The pressure will surely increase, but it will not be 4000psi.

Remember that the pressure is proportional on the no of molecules. So for a fixed volume (of the cylinder, V) at 1.4psi if the Number of moecules be N, then at 3000 psi it will be around 2000N.

Now if N molecules generated a 1000psi pressure when gyroscoped @ 30000rpm, then 2000N molecules will generate 2000 x 1000psi at same rpm. I imagine they are proportional however they may not be directly proportional. But logically and from Classical kinetic theory they are likely to be directly proportional.

So you can imagine the net pressure to be humongous 2,000,000 psi.
The logic sounds good but the starting pressure was 14 psi, not 1.4 psi.

So, 3000 psi ---> 200N, not 2000N, and the final pressure will be 200,000 psi.
 
sal said:
The logic sounds good but the starting pressure was 14 psi, not 1.4 psi.

So, 3000 psi ---> 200N, not 2000N, and the final pressure will be 200,000 psi.

OPPS..thanks for correcting me.
 
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