# Fan Pressure

1. Aug 3, 2009

### Darktimes

I'm currently building a device that will blow a ping pong ball vertically up a tube but don't know what fan I am going to need. Each fan gives a CFM flow rate but the problem is that I don't know how to calculate the required fan CFM i'll need in order to push the ball vertically. In other words I don't know how to calculate the pressure that a certain CFM fan will put on the ball in the pipe. Also the need to calculate against the force of gravity.

Ping Pong Ball Specs:
Weight= 2.7 grams
D= 40mm or 1.574 inches
SA/2 (because the air will only hit one side of the ball) = 15.558

Pipe Specs:
D= 1.75
L= between 4-10 inches

2. Aug 4, 2009

### Redbelly98

Staff Emeritus
Welcome to PF!

The fan wind speed should be greater than a ping pong ball's terminal velocity, which is about 9 m/s or 20 mph according to these links:

http://isaac.exploratorium.edu/snaktalk/hypermail/0091.html

You can figure out a fan's wind speed from the CFM and area of the pipe's cross section.

[wind speed] = [volume flow rate]/[pipe area] > 10 m/s​

or

[volume flow rate] > [pipe area] x 10 m/s​

There is some units conversion required here as well, post back if you have questions about how to do that.

You probably want the wind speed to be at least 3 times the 10 m/s number, to account for any inefficiencies and also to give the ping pong ball a real good boost.

If you have more questions, post again.

3. Aug 4, 2009

### minger

That wind speed would be correct if the ball was free falling. However, the required flow rate will not be velocity / area of the tube. It will be
Code (Text):

Q = \frac{V}{A_{\mbox{tube}} - A_{\mbox{ball}}}

As the area of the ball approaches the area of the tube, the pressure force on the ball can be thought of as the total pressure output of the fan. As this point, one can evaluate a simple pressure times area must be greater than weight of ball.

Having said that, there will be clearance, and associated uncertainty. The required fan won't be the one that can generate just enough total pressure, but it should be much less than one that flows area*10 m/s.

Last edited: Aug 4, 2009
4. Aug 4, 2009

### Redbelly98

Staff Emeritus
Guess that makes sense. Because of the tube constraining the airflow, the pressure on the underside of the ball is greater than in the freefall case.

What's with the [noparse]
Code (Text):
[/noparse] tags? :smile:  You're assuming the OP can decipher LaTex.
[INDENT]
$$Q = \frac{V}{A_{\mbox{tube}} - A_{\mbox{ball}}}$$[/INDENT]

[I]Q[/I] = wind speed
[I]V[/I] = volume flow rate
[I]A[/I] = area of tube, ball

And (correct me if I'm wrong) [I]A[sub]ball[/sub][/I] is [font="Times New Roman"][b]π[/b][/font] [I]r[sup] 2[/sup][/I] for our purposes here?

Last edited: Aug 4, 2009
5. Aug 4, 2009

### Darktimes

I appreciate the quick responses but I'm unsure on how to decipher what exactly your variables mean. You say the area of the tube, now are you referring to the "volume" of the whole tube or the overall area of the diameter. Also you label "Wind Speed" and "flow rate". Now flow rate is just simply the CFM but is the wind speed the PSI if converted (and if so how do you convert).

I guess I'm just trying to get the answer without fully using my brain but I don't exactly have room for trial and error because if the size is greater than 100 CFM (which I would assume it is) then I need to order it online which equals no returns. I tested a 85 CFM fan from Radio shack and no dice, but the ball did spin round and round.

6. Aug 4, 2009

### Bob S

Is there a possibility of a transverse instability here? If the airflow speed around the side of the ping pong ball is higher on the side that is closer to the pipe wall, then pressure will be less according to Bernoilli, and the ball might oscillate. See
http://hyperphysics.phy-astr.gsu.edu/Hbase/pber.html

7. Aug 4, 2009

### Darktimes

I don't think so. The ball is smooth and there is limited friction on certain areas of the ball.

I think my best bet would to go with a 150 CFM fan and hope for the best.

8. Aug 4, 2009

### Turv

Hi,

I am a fan engineer, what i need to know is how fast do you want to blow the ball and how high, the cfm is irrelevant until you know this.

9. Aug 4, 2009

### Darktimes

Speed isn't a concern. The length it would have to travel would hit about 6 inches. What I want the ball to do at the top of the pipe is to sit there and hover (which it should since all of the air will escape out of the end)

The full plan is to have the ball go down a pipe and feed into the "thrust pipe" I guess you could call it and then pop up to the top. My other question is will the fan create an air-wall and block the ball from going into the thrust pipe?

The reason I'm basing this off CFM is because I wanted to use a Muffin Fan (CPU sized fan) to do the job.

10. Aug 4, 2009

### Turv

If i think it is you want to do, it may be a bit awkward, if you have the feed pipe angled down then your get velocity pressure going back towards you, if you have the feed pipe sideways then your get static pressure blowing out, the easiest way would be to have a short feed pipe sideways then manually push the ball into the discharge pipe then push a cap over.

with regards your little fan, with a short pipe you will have minimal pressure loss so even a small fan will be capable of blowing a ping pong ball.

best of luck! if you get stuck message me, i'll try to assist.

11. Aug 4, 2009

### Redbelly98

Staff Emeritus
I'll work out the CFM rating using minger's equation from post #3. Important comment on units: since you want the answer in CFM or ft3/minute, we have to use feet and minutes for all the units. I.e., pipe diameter in ft, area in ft2, and speed in ft/min.

It's the cross-sectional area of the tube, which can be calculated from the tube inner diameter D using

Atube = (π/4) Dtube 2

And since Dtube=1.75" or (1.75/12)=0.146 ft, we get:

Atube = 0.0167 ft2

We want to subtract from that the area due to the ball, which is 0.0135 ft2. Doing that subtraction, we get

Atube - Aball = 0.003ft2

It's not a pressure or PSI, it's the speed of the wind. In this case, it would be the speed in feet/minute, since we are using volume flow rate in CFM or ft3/min.

We want the wind speed to be more than 10 m/s, which converts to 2000 ft/min in our feet-&-minutes units system.

So rearranging minger's formula, we get

V > Q × (Atube - Aball)
= (2000 ft/min) × 0.003 ft2
= 6 CFM

You can get a 40 mm fan with 15-20 CFM, so you can try one of these:

http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=P13591-ND

12. Aug 5, 2009

### Darktimes

I trust your math and all but if I used an 85 CFM fan and it didn't push it up, how will a 20 CFM fan work? That's back pedaling. What about gravity and the force it's putting on the ball as well.

13. Aug 5, 2009

### Redbelly98

Staff Emeritus
Oops, missed the part where you had tested an 85 CFM fan already. I guess at this point, go ahead and try the 150 CFM.

How big (diameter) are these 85 and 150 CFM fans? I ask because the 20 CFM fan I found was 40 mm, almost as big as your pipe.