Solving the Rising Bubble Problem using Bernoulli's Equation

Click For Summary
SUMMARY

The discussion focuses on solving the rising bubble problem using Bernoulli's Equation, specifically for a spherical bubble of radius R rising in water. The derived rate of rise is U = (2/3)√(gR). Participants explored the application of Bernoulli's principle at the stagnation point and discussed integrating over the angle θ to account for gravitational forces. The final solution was achieved by assuming a small angle θ and expanding the equation, leading to the correct result.

PREREQUISITES
  • Understanding of Bernoulli's Equation
  • Familiarity with potential flow theory
  • Basic knowledge of fluid dynamics
  • Ability to perform calculus, particularly integration
NEXT STEPS
  • Study the derivation of Bernoulli's Equation in fluid mechanics
  • Learn about potential flow theory and its applications
  • Explore the concept of stagnation points in fluid flow
  • Practice integration techniques in the context of fluid dynamics problems
USEFUL FOR

Students and professionals in physics, particularly those studying fluid dynamics, as well as engineers working on buoyancy and fluid flow problems.

LoopQG
Messages
18
Reaction score
0

Homework Statement



Consider a spherical bubble of radius R,rising in water. Using Bernoulli's equation show that the rate of rise of the bubble is:

U=(2/3) \sqrt(gR)


Homework Equations



Bernoulli Equation
Potential Flow



The Attempt at a Solution



I have considered the problem from the bubble's frame so the rate of rise is just the velocity of the uniform flow around the sphere. I know that there is a stagnation point right at the top of the bubble. So Taking Bernoulli's at points on either side of the stagnation point is what I have been doing but I must be setting it up wrong because i don't see where to get that 2/3 from. Any help appreciated.

From Bernoulli's i get:

U= \sqrt(2gR(1-cos(\theta)))
 
Physics news on Phys.org
Try integrating over theta. :)
 
Do you mean take dU/d\theta

which equals

\sqrt(2gR)(1/2)(1-cos(\theta))^(-1/2) sin(\theta)

but then integrating that don't i just get the same thing back?

If not at what point should I consider integrating. I see where you are going with this because the flow velocity is zero at the boundary layers but I just am not sure how to apply it. Thanks for the help!
 
What does theta in your Bernoulli equation stand for?
 
The angle from the vertical axis of the sphere, grcos(\theta) is the gravitational force at that point
 
LoopQG said:
The angle from the vertical axis of the sphere, grcos(\theta) is the gravitational force at that point

So to get the total force you would have to integrate over the entire surface? :)
 
Thanks a lot for the help, I ended up assuming a small \theta then expnding and I get the right answer.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
805
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 7 ·
Replies
7
Views
4K