Force of air coming from a fan.

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Main Question or Discussion Point

if i had a circular fan with 8 blades and the diameter of 1 foot, how fast would it have to spin to push 125 Lbs of air?

if the angle of the blades would matter, it would be around 45°

if more specifications are needed, please let me know
 
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Answers and Replies

  • #2
Simon Bridge
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Angle does matter.
You can usually gain insight into these problems by reversing them:

If 125lbs of air is going through your fan (each second?) then how fast will that push the blades around?

This will be basic conservation of momentum if you have the final speed of that air.
Then it is a matter of spotting the important difference and adjusting for it.
 
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i think i understand what you're saying, and as for increments of time, it is to be for levitation off of the ground so im not sure how i would convert that into time increments, oh and thanks for the help.
 
  • #4
Simon Bridge
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You know you are moving 125lbs of air through the fan but you don't know how fast you want to move it? It is no good just to shift a particular mass of air - surely you understand that the time you take to move it is going to be important?

Loosely speaking:
You need to the change in momentum (downwards) of the air enough for the reaction to be able to offset the weight of the payload.

Do you want to build a hovercraft, a helecopter, or something more ducted-fan like a quad-copter? Best practice for you is to google for the machine closest to what you want to do and see how others have done it.
 
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  • #5
Simon Bridge
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May be able to do it back-of-envelope - guessing, due to 8 blades, that all the air inside the blades gets moved out (and new air sucked in) in 1/8th of a revolution ... or each rev moves 8x the volume of the fan. [this is very approximate.]

Let the fan be 1" ≈ 2.5cm effective thickness (adjust for what you have) then each revolution moves 8(0.025m)(0.15m)2π=0.014m3 (apologies for change in units - I think in metric). That is about 0.017kg of air (@1.2kg/m3) each revolution. To move 60kg (≈125lbs) each second you want the fan to rotate at 60/0.017 ≈ 3540 revolutions per second.

60kg per second through that 1' ≈ 30cm diameter fan suggests that the air moves through at speed v=60/(1.2)(0.152)π ≈ 700m/s wee! Unless I messed up, thats 42000N ≈ 16000lbs thrust!

If you just want to levitate 60kg against gravity, then try for 600N thrust (overkill but you want to be able to take off too). Maybe 8kg per second? That would be 470 revs.

Caution - this is all very approximate. Do not base a design on these calculations.
There are a lot of other factors like how directional the thrust is and how efficient the fan is.

You should back the predictions empirically.
 
  • #6
rcgldr
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Do a web search for static thrust calculators, these are just rough estimates, but it's a starting point.
 
  • #7
Simon Bridge
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I have a love/hate relationship with online calculators - but I'm inclined to agree. I was basically treating the fan as a rocket-thruster in the above.

Often, when someone comes up with this sort of question, they are trying to decide on whether or not to buy a more expensive motor (or something). This is actually an easier problem since you'll have the specs for the component and the question is whether the cheap version is capable of the desired performance.
 
  • #8
russ_watters
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Short answer: That's going to be tough for such a small fan.

Long answer:
Fan blades are curved, so they don't have a single angle, they have a pitch. Pitch is a measure of how far forward the fan would move if embedded in jello and rotated once. For the sake of example, lets assume a pitch of 1 ft.

There are two ways to figure this out: Bernoulli's Equation and momentum (as described). I'll give you the Bernoulli's version.

Using the simple form of Bernoulli's equation, you can calculate the stagnation pressure of air at a certain velocity....Or in this case go backwards and calculate the velocity given a desired pressure.

You asked for 125 lb of force, on a 1' diameter, or 159 psf.

The form of Bernoulli's equation for this is p=.5 rho v2

rho = .00238 slugs/ft3 (I love that unit!)

Solving/converting units gives me 365 ft/sec (248 mph) of air velocity (which is right at the limit for using an incompressible flow assumption).

365 ft/sec with a pitch of 1 ft per rotation equals 365 revs/sec (see what I did there? :wink: ) or 22,000 rpm.

This assumes perfection of course: a real-world propeller won't be perfectly effective. I'd guestimate you'd be lucky to achieve half of that performance. I once did a similar task: I built a little wind tunnel using a 1hp RC plane engine, rated at 16,000 rpm. I think I used a 10" dia, 11 inch pitch prop and measured 90something mph in the test section of a little wind tunnel using a homeade pito-static tube and the Bernoulli's calculation used above.
 

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