Air Hockey - Fan rating? 1

[LittlePete]

Hi people, I’m designing my own Air Hockey table and I have been browsing these forums for some answers. I’m hoping someone could guide me on how to estimate the size of fan/s I’ll require. Originally I was just calculating the force each hole would need to provide given the area of the puck, and then calculate the flow rate for each hole and multiplying it by the total number of holes on the table. This way providing a value for the flow rate of the whole table, I was then just going to use this as the rate the fans needed to achieve. However reading through previous threads I find some people are talking about working with the pressure required in the area below the playing surface (and it's size). Am I missing some basic principles here? Any suggestion would be much appreciated.

 

[ChasBean1]

Pete, there was a couple extensive threads about this in the past but it looks like they're gone now.

Can you post your data?

Let's say by your method you get a value of 100 cfm. How do you specify the fan? How do you achieve design flow value through each hole without knowing the pressure on the underside?

Try the ASHRAE leakage equation - very simplified (1999 Applications 51.5):

Q = 2610 A dP^.5

Q in cfm
A in ft2
dP pressure across leakage boundary, in inches w.c.

Knowing desired flow (Q) and the total open area of all holes (A) might approximate a good plenum pressure (dP) for fan specification...

 

[LittlePete]

hi ChasBean1,
These aren't the finalised figures but I was using a hole diameter of 3mm spaced every 10mm to see what type of return I'd get. The playing surface is 1.72m^2 (almost a 2x1 meter square table), and the area below has a volume of 0.052m^3 (depth of 30mm).

The puck is a standard off the shelve puck.

Using the hole size above and a puck of 83mm diameter the puck would cover 61holes at any given time. The puck weighs 41g, so spread over 61holes the force required per hole would be 0.007N.

I used F = p A v^2 to calculate the velocity required, which equalls 28.72ms^-1.

Then using Q = A v , I calculated the flow rate per hole to be 2.03 x10^-4 m^3s^-1. Multiply this by the estimated 27,000 holes required in the table to calculate the table value of 5.481 m^3s^-1.
Although I am yet to convert this to cfm, I'm beginning to doubt if it'll return a sensible value!

As you say this does not take into consideration the pressure required, so perhaps I'll try the equation you have provided as see what values I produce.

 

[ChasBean1]

I like the science you've used, but somewhere we've gone astray. Haven't checked your equations or math - I will when I get a chance. This is 11,614 cfm, whereas I think we'd like to be in the 100-300 cfm range (sorry about the eng. units, this is how I think in airflow).

 

[ChasBean1]

Jury is back... overall thought: holes are way too big.

Next:

Find out the velocity pressure needed against the underside of the puck to counteract the weight force. I would pick a number such as three times the force of the puck's weight to keep it afloat on a reasonably thick cushion of air. Note that this is estimated. Puck force is .007 N, so use .02 N. The air velocity pressure is .02 N over the area (.00541 m2), so 3.7 Pa.

The air velocity required to create this pressure is:
v = ((2 * Pv)/rho)^.5, which equals 2.48 m/s (this is basically the pitot tube equation for using velocity pressure to find velocity). I would use this as the velocity out each hole. Each 3 mm hole (way too big by the way) has an area of 7.069 E^-6 m2, so air volume out each hole is 2.48 * 7.069 E^-6 = 1.753 E^-5 m3/s.

I don't know where you get 27,000 holes; I figure about 17,200, so total flow would be .302 m3/s, which is 639 cfm (yes, back to English units). Total hole area is 1.3 ft2 (again, too big. Picture a 1' by 1' hole in your table!).

Finally:
Q = 2610 A dP^.5

Q is in cfm, A is in ft2, and dP is in inches of water (ref: 1999 ASHRAE Applications, 51.5). With your known flow (639 cfm), rearrange and solve for fan outlet pressure, which is .035 in. w.c., or 8.7 Pa. This is the fan static pressure to specify. Note that this is very low because your holes are so big.

 

[LittlePete]

ChasBean1 you're a star!

I apoligise for my somewhat eratic calculations. My fault for not actually checking them over.

Thanks very much for your help, I'll work through the equations at the weekend and hopefully I should have enough information to finalise my table design.

I can't thank you enough for your input.

 

[ChasBean1]

Just re-check calcs - looking back, I'm not sure about the puck force based on your .007 N (that may be 1/61th of actual). If/when you finalize this table, could you please post back your results (e.g., what fan flow & pressure, other data, etc. and how it works). It would be nice to see how close calcs get us. Best of luck, -CB