Pressure distribution in rotating pipe. (dynamics and fluid mechanics)

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SUMMARY

The discussion focuses on calculating the pressure developed at the end of a centrifuge arm with a radius of 3 meters, subjected to a gravitational level of 100g. The user initially attempted to apply the static pressure head formula, P2 = pgh, but recognized that the acceleration due to gravity is not constant in this scenario. After correcting their approach, they derived the formula for pressure as P2 = p(w^2)(r^2), leading to a calculated pressure of 2589 kPa (375.5 PSI) using a hydraulic oil density of 880 kg/m³ and an angular velocity of 18.08 rad/sec. The user seeks confirmation on the accuracy of their calculations and assumptions.

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  • Understanding of fluid mechanics principles, particularly pressure calculations.
  • Familiarity with centrifugal force and its effects on fluid dynamics.
  • Knowledge of angular velocity and its relation to linear acceleration.
  • Basic proficiency in unit conversions, particularly between kPa and PSI.
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  • Research the derivation and application of the static pressure head formula in non-linear gravitational fields.
  • Learn about the effects of centrifugal force on fluid pressure in rotating systems.
  • Explore advanced fluid dynamics concepts, such as the Navier-Stokes equations.
  • Investigate methods for validating pressure calculations in hydraulic systems.
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Engineers, fluid mechanics students, and professionals working with hydraulic systems in rotating machinery will benefit from this discussion, particularly those involved in pressure calculations and system design in centrifuges.

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I have a centrifuge with hydraulic lines on the arm. I need to calculate the pressure that will be developed at the end of the centrifuge arm in the hydraulic lines.

The radius of the arm is 3 meters. At the end of the arm the G level will be 100g. Obvioulsy the g level at the center point is 0g in the radial direction. I tried to tie together the static pressure head formula, as well as some of my dynamics formulas but the resulats do not make sense. I have made an incorrect assumption somewhere along the line.

For regular old pressure, you have P2=pgh

where:
p="rho" the desity of the fluid
g= acceleration due to gravity, (9.81) A CONSTANT
h= depth of the fluid in meters.


Now my acceleration due to gravity is NOT constant, and is NOT linear. I tried simply substituting my g formula into the above aquation, noting that"

An=(w^2)(r)

a formula that would result in units of M/s/s. I then normalized this to "g's" by dividing by 9.81 to have:

g=((w^2)(r))/9.81

Substiuting both together, and nothing that in my opinion r=h you get:

P2= p [((w^2)(r))/9.81] (r)

which simplifies to: P2=p[((w^2)(r^2))/9.81]

now the density of hydraulic oil is 880 Kg/m^3

Omega is 18.08 Rad/sec

and R=3.

you get 263,908 somethings. I am not quite sure what the units are, but I am thinking kilopascals?

Is this correct? Are my assumptions correct or are they wildly off base? Either way I have thought about this for quite some time and I am very unsure.

please help.

Thank you.
 
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I thought about this a bit further, and the original pgh formula uses m/sec^2 as the units of g, so dividing this by 9.81 to go back to a unit of "g's" seems like a mistake now.

I think pressure = p(w^2)(r^2) should be correct but I would still appreciate input on this!

so it comes down to pressure = (880)(18.08^2)(3^2)= 2589KPA

=375.5 PSI.

Thoughts?
 
Can anyone help me out on this? I really need a definitive answer.
 

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