Ball Floating on Air Stream

[BenB8691]

Hi all,

I am working on a prototype and I have a question that I am not sure how to answer, thus the reason I am here.

I have a piece of 4" pipe that I will vertically mount and I am reducing down to 1". Inside the tubing I have a racquetball. I am going to be sending air through the pipe and I want to know how to calculate at what flow rates the ball will float on the air and how high the ball will go.

I know the inside diameter of the pipe is 4" and the ball is 2.25". I am guessing I will need the weight of the ball, but I don't know of an equation that will tell me the following: 1.) How much air is required to lift the ball inside of the pipe? and 2.) How high the ball with varying amounts of air passing through the tube?

Can anyone help me with this? Any help is greatly appreciated.

Thank you,
BenB8691

 

[jbriggs444]

I know the inside diameter of the pipe is 4" and the ball is 2.25". I am guessing I will need the weight of the ball, but I don't know of an equation that will tell me the following: 1.) How much air is required to lift the ball inside of the pipe? and 2.) How high the ball with varying amounts of air passing through the tube?

If the vertical tube is uniform then if the racket ball lifts at all it will go all the way to the top and stick in the outlet hole.

How fast the air needs to be moving to make this happen seems to me to be a matter for experiment. Turn it on and see.

 

[AlephZero]

You can get a rough estimate of the air velocity be saying that the drag force on the sphere (see http://en.wikipedia.org/wiki/Drag_coefficient) equals its weight.

But since the ball is blocking about 30% of the area of the pipe, that will change the air flow pattern around the ball and the assumptions in the drag calculation won't very accurate.

As jbriggs said, if the pipe has a uniform diameter you won't be able to get the ball in a stable position part way along the pipe, unless you do something like make small holes along the length of the pipe to leak some of the air out, so the air velocity at the top is smaller than at the bottom.

 

[adimantium]

air pressure is measured in psi, (pounds per square inches) Measure the surface area of the ball, divide it by 2. Now measure the weight of the ball in pounds. Pounds divided by the surface area equals the force needed to keep the ball in place. Let's say half the surface area is 10 square inches, and it weighs 2 pounds. You would need 0.2psi

 

[jbriggs444]

air pressure is measured in psi, (pounds per square inches) Measure the surface area of the ball, divide it by 2. Now measure the weight of the ball in pounds. Pounds divided by the surface area equals the force needed to keep the ball in place. Let's say half the surface area is 10 square inches, and it weighs 2 pounds. You would need 0.2psi

That is a reasonable estimate for how much pressure loss you need between the bottom of the ball and the top of the ball in the rising air stream. How much velocity you need to attain this pressure loss across the ball is less clear. AlephZero's suggestion applies. The need to maintain that velocity through the vertical pipe and get an appropriate flow rate out to the ambient atmosphere through the exhaust nozzle on top will dictate the inlet pressure and flow rate required at the bottom.