Centripetal force on an industrial fan

[Turv]

I make industrial fans of which use impellers that can reach up to 80 m/s tip speed, i also learned that it shouldn't be tip speed but tip acceleration because the impeller is constantly changing direction it should be 80 m/s^2.

So to calculate the force should i use force = Ma or m x centripetal acceleration of which I've read is V^2/ radius?

Also can someone give me a good example how the force goes inwards because this is baffling me, i know you have probably heard this thousands of times but why does it go inwards?

 

[Doc Al]

I make industrial fans of which use impellers that can reach up to 80 m/s tip speed, i also learned that it shouldn't be tip speed but tip acceleration because the impeller is constantly changing direction it should be 80 m/s^2.

There's nothing wrong with saying the tip speed is constant, but you're right that the tip is accelerating since its direction is constantly changing as it is pulled into a circle.

So to calculate the force should i use force = Ma or m x centripetal acceleration of which I've read is V^2/ radius?

Sure. F = mv²/r would give you the net force acting on a piece of mass moving with speed v at a distance r from the center.

Also can someone give me a good example how the force goes inwards because this is baffling me, i know you have probably heard this thousands of times but why does it go inwards?

Imagine a piece of material at the tip. If no force acted on it, it would just keep going in a straight line. (This is Newton's first law.) But the material of the fan blade exerts an inward force on it that keeps it pulled into a circle.

Just like twirling a ball attached to a string in a horizontal circle. The string must pull the ball inward toward the center of the circle to keep it spinning. The faster the ball spins, the harder the string must pull. If the ball gets going too fast, the amount of force needed to keep it moving in a circle may end up greater than the string can produce--thus the string breaks.

 

[Turv]

Doc,

Thank you very much for your reply, much appreciated.

 

[Turv]

Doc,

Question,

So if a fan impeller is 50 kg and the tip speed is 80 m/s / tip acceleration is 80 m/s^2 would this mean that the centripetal force would be :

Mass = 50 kg

Velocity = 80 m/s

Radius = 0.535 m

so centripetal force = 50 x( 80^2/ 0.535) = 160000 N

Where does this force go? if the centripetal force goes inwards, am i right to assume the force of 160000 N will be on the hub of the impeller? In no way am i trying to argue with you but i just can't comprehend this, am i thinking wrong?

 

[Doc Al]

What's a fan impeller? If it's a small mass located at the tip of the rotating fan, then there's a centripetal force acting on it as you calculated (but check your arithmetic).

 

[Turv]

this is an impeller.

checked my equation lol yes it was wrong, it should be 598130 N.

does this force go to the hub of the impeller?

 

[rcgldr]

does this force go to the hub of the impeller?

It goes through the plates supporting the blades to the hub as tenstion, but assuming the fan is balanced, it doesn't create any radial force on the hub bearing.

 

[Turv]

Jeff,

My fan book suggests different, it states the radial force on the bearings will be weight in Newtons x (1 + ( center of impellor to first bearing/ center of the 2 bearings))

example:

an impellor weighs 50 kg = 490 N x (1 + ( 150mm / 600mm)) = 612.5 N

 

[bm0p700f]

Not all of the 50 kg of the impeller is moving at the tip speed.

acceleration is given by a = v^2/r where r is the distance from the centre. As you move inwards to the centre (r is decreasing) the acceleration decreases. As F = ma the force decreases as well. So the load on the bearings will not 16000 N!

If you define the time period of one rotation T then the angular velocity will be
w = 2*pi/T

The rotational speed of a paoint a distance r from the centre will be v = w*r

Thus the centripetal acceleration is a= w^2*r = 4*pi*r/T^2
Therefore force F = m*4*pi*r/T^2

If split your fan into points infinitesimally close together and do some calculus then you will get the total force provided by these infinitesimally close points. That is what your empirical formula is doing so use that.

 

[rcgldr]

radial force on the bearings

I meant due to rotation. If the fan is oriented horizontally, then gravity produces a side load on the beariing, but not a radial load.

 

[Turv]

Jeff,

A few years ago someone told me that the faster the impeller spins the force on the bearings decreases, is this what your saying or trying to imply or am i on a different wavelength?

 

[rcgldr]

A few years ago someone told me that the faster the impeller spins the force on the bearings decreases, is this what your saying or trying to imply or am i on a different wavelength?

I was only referring to the static load on the bearings. I was just trying to point out that the centripetal force of the blades doesn't create a load on the bearing unless the blades are imbalances.

The bearings and any mass of the fan is going to be affected by both gravity and centripetal force.