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

In summary, the centrifuge is having trouble calculating the pressure that will be developed at the end of the arm in the hydraulic lines. It appears that the acceleration due to gravity is not constant, and is not linear. The centrifuge user is unsure if their assumptions are correct or if they are wildly off base.
  • #1
Vr6Fidelity
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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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  • #2
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?
 
  • #3
Can anyone help me out on this? I really need a definitive answer.
 

1. What causes pressure distribution in a rotating pipe?

The pressure distribution in a rotating pipe is caused by the centrifugal force acting on the fluid inside the pipe. As the pipe rotates, the fluid is forced towards the outer edge of the pipe, creating a higher pressure on the outer wall and a lower pressure on the inner wall. This results in a pressure gradient across the cross-section of the pipe.

2. How does fluid viscosity affect pressure distribution in a rotating pipe?

The viscosity of the fluid affects the pressure distribution in a rotating pipe by influencing the frictional forces between the fluid and the pipe walls. Higher viscosity fluids will experience more resistance to flow, resulting in a more uniform pressure distribution across the cross-section of the pipe. Lower viscosity fluids, on the other hand, will experience less resistance and may have a more uneven pressure distribution.

3. What is the Coriolis effect and how does it impact pressure distribution in a rotating pipe?

The Coriolis effect is the apparent deflection of moving objects, such as fluids, due to the rotation of the Earth. In a rotating pipe, this effect causes the fluid to deflect towards the outer edge, resulting in a higher pressure on the outer wall and a lower pressure on the inner wall. This contributes to the pressure distribution in a rotating pipe.

4. What are the practical applications of studying pressure distribution in rotating pipes?

Understanding pressure distribution in rotating pipes is important in various engineering and industrial applications. This knowledge can be applied to design efficient and safe fluid transport systems, such as pipelines and pumps. It also plays a crucial role in the design and operation of rotating machinery, such as turbines and centrifuges.

5. How does the rotational speed of the pipe affect pressure distribution?

The rotational speed of the pipe has a direct impact on the pressure distribution. As the rotational speed increases, the centrifugal force acting on the fluid also increases, resulting in a higher pressure at the outer wall and a lower pressure at the inner wall. This means that the pressure distribution will be more pronounced at higher rotational speeds.

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