Difference in liquid column heights in a rotating U-tube

Click For Summary
SUMMARY

The discussion focuses on calculating the difference in liquid column heights in a rotating U-tube, specifically when one arm is aligned with the axis of rotation. The key equations involved are the acceleration due to rotation, expressed as a = w²r, and the hydrostatic pressure equation p = p(o) + density * g * h. The final derived formula for the height difference in the liquid columns is h = (L²ω²)/(2g), where ω is the angular speed and L is the length of the tube. The participants clarify the need to consider the varying acceleration across the tube's length when integrating pressure differences.

PREREQUISITES
  • Understanding of fluid mechanics principles, particularly hydrostatic pressure.
  • Familiarity with rotational dynamics and angular acceleration concepts.
  • Knowledge of calculus, specifically integration techniques.
  • Basic proficiency in physics equations related to pressure and density.
NEXT STEPS
  • Study the derivation of pressure differences in rotating fluids.
  • Learn about hydrostatic pressure in non-inertial reference frames.
  • Explore applications of fluid dynamics in engineering contexts.
  • Investigate the effects of angular velocity on fluid behavior in rotating systems.
USEFUL FOR

Students studying fluid mechanics, physics enthusiasts, and engineers interested in the dynamics of rotating systems will benefit from this discussion.

kidsmoker
Messages
85
Reaction score
0

Homework Statement



A U-shaped tube with a horizontal segment of length L contains a liquid. What is the difference in height between the liquid columns in the vertical arms if the tube is mounted on a horizontal turntable, and is rotating with angular speed w, with one of the vertical arms on the axis of rotation?

Homework Equations



a=w^2r towards the centre
p = p(o) + density*g*h

The Attempt at a Solution



A previous part to the question was finding the difference in height if the U-tube had acceleration a towards the right. I did this and found h=aL/g.

I was thinking I could just substitute a=w^2L into this equation but then I had second thoughts since the acceleration decreases as you get closer to the axis of rotation. Do you take the average acceleration intead?

Thanks.
 
Physics news on Phys.org
kidsmoker said:
I was thinking I could just substitute a=w^2L into this equation but then I had second thoughts since the acceleration decreases as you get closer to the axis of rotation. Do you take the average acceleration intead?

Hi kidsmoker! :smile:

In the first case (linear acceleration),

if the linear density (mass/length) is ρ, and the pressure gradient is Q, then a length dr in the middle will have mass ρ dr and acceleration a, and so mass x acceleration = aρ dr = pressure difference = Q dr,

so Q = aρ, and so total pressure difference = aρL.

Does that help you with the circular case? :smile:
 
Ah yeah I get it now thanks. :)
 
Although I took vector calculus, I didn't understand what you mean by "gradient pressure", tiny-tim. I think I should google for it.

For the first question I get that the answer is h=\frac{aL}{g} and the column of liquid that has a higher high is the right one.

I tried the second part of the question, namely the first question that kidsmoker asked, and reached h=\frac{L^2 \omega ^2}{2g} but I'm unsure of the integral I used.
The difference of pressure within a dr element is \omega ^2 r \rho dr. I integrated this expression with respect to r, from 0 to L.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
12
Views
7K
  • · Replies 2 ·
Replies
2
Views
6K
Replies
16
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 18 ·
Replies
18
Views
7K
Replies
4
Views
3K