what's the speed of blades of ceiling fan in miles/hour,where the RPM of the fan is..

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and now assume if i want to calculate how much air speed the blade experiences as it runs at 300 RPM in a room

the blades creates the airspeed against as it rotates..
Yes.
and now assume if i want to calculate how much air speed the blade experiences as it runs at 300 RPM in a room
I'm not sure if you actually want to calculate the air speed relative to the fan blade, but I'm absolutely sure that you aren't able to calculate it.

It's a complex three-dimensional stream, which can be best approximated in a finite element solver. At best you get some qualitative insights about the flow profile.

so if we take the linear velocity of the blade tip,which is 41 miles/hour,the blade tip really doesn't experience 41miles/hour airspeed ,does it?
You can assume that the relative speed of blade to surrounding air will be lower than the blade speed, because the blade is accelerating the air in the direction of it's movement. But in exact terms it depends on blade profile, angle of attack, pressure difference across the fan, existence of guide plates etc.

You'll find details in technical fluid mechanics text books.
 


i'm just trying to learn and clear my confusion and ready to learn from learned one.. may be i'm not that good explaining my question...

you said "then the blade tip is moving at 41MPH relative to the air" if the blade tip moves 41Mph relative to the stationary air,then the air speed which is experienced by the same tip is 41Mph? like every action has equivalent and opposite reaction,does that rule apply to this?
 

the blades creates the airspeed against as it rotates.
Airspeed is the speed of an aircraft relative to the air; I don't think it's a term applied to fans.
as the linear velocity decreases as the point on blade is closer to the center,then the airspeed decreases with it..
You are correct. Airspeed is the speed of an object relative to the air. It does not matter if the object is moving or the air is moving. The relative motion is what counts. It is certainly applied to an aircraft and also for the blades of an helicopter which is the same case as the fan. The tip of the blades experiment a higher airspeed than the roots.
 



This is a snapshot of the fan hanging on the ceiling of my room. It shows one blade. Most fans are made like this. The whole blade is at an angle of 5 deg. Further the greyish area is bend at around 15 deg. You can see it is more towards center than towards ends. As air is not blown in a pipe like jet stream some air is pushed upwards near ends. As you can see blade is not perpendicular but at an angle of only 15 deg. The effective downward thrust is much less. Most central part has slow speed so more area of blade is used for downward thrust. This is to equalize air flow in full span of blade. As some air is pushed upwards near the ends of blade. 5 deg angle of the whole blade helps to stream air downwards. When there is some flow of air upwards from sides. it would have caused to push air upwards by blade when air come upwards from sides if blade would be at 0 deg horizontal. This is implying practically according to situation, the difference of theory and practical. Practical cause to make and revise theory after seeing a point overlooked.
 
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so in a heli rotor,there would be more lift at the tip and the lift decreases as you go towards the center...

so geometric twist are applied on the blades,where the AOA is decreased from the center as you go towards the tip,as to maintain equal lift on all points on the blade..
 

sorry I repeated the post...
Here is the answer:


This may be compensated at some helis but I cannot give you an example because the ones that I have seen. Bell, Robinson and others look like they have no twist on the blades.
But you have asymmetric lift when you go forward with an heli because the advancing blade has more airspeed than the receding blade. This is compensated by the cyclic pitch. And it is an example of relative motions of both "the air" and the object. The same effect is if you are stationary in a wind.
 
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so i guess to calculate the lift of a rotor blade,which will see increasing airspeed from rotor as you go towards the tip

the blade has to be divided into ten parts and the Lift equation should be applied to each part corresponding to the airspeed it sees,and integrates all the calculated lift ,and that will be close estimation of the total lift produced by the blade...

do you think this would be a good way to calculate the lift of a rotor blade?
 

I guess that you are right. An integral will give you a better calculation. And if the blade has not twisting or other variations over the length you can easily integrate to obtain the total lift.
Besides airspeed the lift will depend on various factors like air density, airfoil profile. angle of attack, reynolds number, mach number, etc. Some of them enclosed in the lift coefficient for a particular airfoil.
 


yes,i do have the lift co-efficient of the blade,which is 1.318,i got it from the FoilSim applet

and i divided the blade in 10 parts along using 1/10 of the total area of the blade in the equation

here i have made a rough calculation using the above discussed technique to calculate a rotor blade lift

for a flat plate with chord = 0.1014 meter and blade span = 0.5821meter and AOA set to 15 degree and Life co-efficient 1.318, air density = 1.0286 kg/m3 and with a RPM of 300, the blade can lift 1.716 Kilogram force..

here is my calculation,can anyone verify if it is a good approximation?

 

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