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

[Vr6Fidelity]

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.

 

[Vr6Fidelity]

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?

 

[Vr6Fidelity]

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