Describe surfaces of equal pressure in a rotating fluid

Click For Summary
SUMMARY

The discussion centers on determining the surfaces of equal pressure in a rotating fluid, specifically in a closed vessel filled with water rotating at a constant angular velocity, \(\Omega\). The conclusion drawn is that these surfaces form circular cylinders with a common axis positioned at a height of \(g/\Omega^{2}\) above the axis of rotation. The analysis utilizes the momentum conservation equation and simplifies it by considering inviscid flow and neglecting the Coriolis term, leading to the reduced pressure expression \(p - \frac{1}{2}\Omega^{2}r^{2}\).

PREREQUISITES
  • Understanding of fluid dynamics principles
  • Familiarity with the conservation of momentum in rotating systems
  • Knowledge of inviscid flow assumptions
  • Basic mathematical skills for manipulating differential equations
NEXT STEPS
  • Study the derivation of the momentum conservation equation in rotating fluids
  • Learn about inviscid flow and its implications in fluid dynamics
  • Explore the effects of angular velocity on pressure distribution in fluids
  • Investigate the Coriolis effect and its relevance in rotating reference frames
USEFUL FOR

Students and professionals in fluid dynamics, physicists studying rotational systems, and engineers working with rotating machinery or fluid systems.

amrasa81
Messages
2
Reaction score
0
Hi, I am trying to solve a basic question from a Fluid dynamics textbook. Could you help me with the answer? The question is as follows:

A closed vessel full of water is rotating with constant angular velocity \Omega about a horizontal axis. Show that the surfaces of equal pressure are circular cylinders whose common axis is at a height g/\Omega^{2} above the axis of rotation.

I don't know how to tackle this problem. Is there a technique in solving such theoretical questions?

Thanks,

P.S:- This is not a homework or coursework question. I am also new to Physics forum, and hence, my question may not be appropriate for this section. In that case please tell me in which section I should pose fluid dynamics questions.
 
Physics news on Phys.org
Interesting question... I only have a partial answer, working from Tritton's 'Physical Fluid Dynamics'. In it, he starts with:

\frac{Du}{Dt} =\frac{1}{\rho}\nabla p -\Omega \times \Omega \times r - 2\Omega \times u + \nu \nabla^{2} u +\rho g

So, assuming conservation of momentum, Du/Dt = 0. Also, the second term on the rhs can be written as
-\nabla (\frac{1}{2}\Omega^{2}r^{2})

and combined to give a reduced pressure

p - \frac{1}{2}\Omega^{2}r^{2}

Then, ignoring the Coriolus term and assuming inviscid flow, I can maybe see how you get the result you mention. Maybe...

hope this helps.
 
Thanks!
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 35 ·
2
Replies
35
Views
5K
  • · Replies 48 ·
2
Replies
48
Views
6K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K