Exploring Smoke Rings: Investigating Project Questions

[AttilaTheNun]

Hi all,

I'm trying to answer the following questions related to a project on smoke rings.

We are in an introductory class, so this project is more for fun and just applying the little we have learned about point, rigid body, and fluid motion to something less understood like vortex rings.

Essentially we have contsructed a smoke ring device that, guessed it, blows smoke rings. I have already experimentally shown that by blowing two different sized rings, you can find that the larger ring moves at a slower pace than the smaller (radius) ring.

However, I am having trouble answering the following questions:


8. Explain mathematically why the speed of propagation of the
ring should vary roughly as the reciprocal of its radius.

9. Show (from calculations) that the air in the center of the
ring is sucked through at a speed of 4 times the propagation
of the ring itself (it's like some weird flying, sucking thing!)

Any help is greatly appreciated, even just a quick explanation off the top of your head.

thanks

 

[eczeno]

how do you generate the different sized rings?

 

[AttilaTheNun]

Similar to the first one with the trash can in this video:




we made two separate cannons, one with a 5 inch hole and one with a 10 inch hole at the end. We made sure to pull back the bungie cord at the opposite end the same distance (8 inches) each time to approximate the same force for both cases. This resulted in two different sizes of vortex rings.

 

[eczeno]

ok, then the speed difference is easy to explain. you are pushing the same volume of air out of the hole in each case, and this is happening in about the same amount of time. so, the air that has a smaller hole to escape from must come out faster. you can test this by pulling the bungie different lengths for the same size hole. the farther you pull it back, the faster the ring will come out.

 

[AttilaTheNun]

Thanks


What equation would I use to verify this though?

I mean, how does this principle relate back to the ideas learned in fluid mechanics? or even all the way back Newton's laws of motion?

 

[eczeno]

vol/time = area(length/time) = area * velocity

 

[AttilaTheNun]

Does this look right for number 8?

v = (1/r)*Pressure

I'm guessing I'm way off.

 

[eczeno]

well, the units on each side do not match, so something is amiss.

i don't think you are going to find a nice expression that depends on the pressure (which will not be constant). but if you want to compare the speeds for two different sized holes, i think considering the volume flow rate (vol/time) is the way to go; at least as a very simple first approach to help get a feel for the problem.

i will consider the cannon as a uniform tube, of cross-section A with a hole, of area a, at one end and a movable piston at the other. if we pull the piston back a distance d, then we have increased the volume of air in the cannon by V = Ad (note this is the increase in volume, not the total volume of the cannon), which we can just call V from here on. now, let's say that when we release the piston, it moves back to its original position in time t. that means that air with a total volume V moves out of the cannon in a time t. thus, the volume flow rate is V/t. now, if we assume that the air comes out in a cylinder of area a and length l, then some dimensional analysis shows us that:

V/t = a (l/t), and since that cylinder of air moved a distance l in a time t, l/t = v, so

V/t = constant = a * v, and so, by my (overly simple, granted) calculation, i get that

v = k/r^2 instead of k/r, as you report.

now, that is not to say your data is wrong. my calculation ignored the fact that these vortex rings are being generated, which would require some pretty sophisticated fluid mechanics to deal with. but, it does show that, in general, the speed should decrease as the size of the hole.

this is a very interesting problem.

i am going to think about #9 as well, but right now I haven't a clue. :)

cheers

 

[AttilaTheNun]

Thanks for the long write up eczeno

I wish I could post the video I have of my professor talking about this portion of the project but it is too big of a file. You would like him, really cool guy .

anyhow, he used the angular momentum formula:

L = mvr

where the conservation of momentum and mass (assumption) holds. v and r are always inversely proportional leading to v = 1/r. Then he went further into the mechanics of the smoke ring.

The outside air particles with a larger pressure hold the smoke ring intact. Should the pressure inside the ring be the same as the atomspheric pressure though?

Essentially the outer most particles in the ring have velocity v, then the further in towards the center, the larger in magnitude velocity gets. apparently it only should get to four times the velocity of the outer most particles though.

 

[eczeno]

hmm... i will have to play with that angular momentum idea.

thanks for giving me such a wonderful thing to think about. :)

 

[AttilaTheNun]

I should aslo mention I found this statement for question #9 on yahoo answers. (a different classmate must asked I guess)

If the ring moves forward by a distance v t in time t, then the air has to move back that distance, front twice that distance and back again once that distance. So 4 times the distance in the same time. Hence a fourfold speed

- Dr. Zoro


not entirely sure I understand though

 

[AttilaTheNun]

No one able to explain #9?


I was dissecting the angular momentum formula a bit more finding L = m * (w*r) * r.
so I guess the angular speed (w) is inversely proportional to r^2. Does that allow me to arrive at 4 times the speed? It doesn't look like w depends on the radius though, so shouldn't it be the same thru out?

Also, if w*r is the velocity of the ring itself, the angular speed is only twice the actual velocity in my example.