How much lift thrust would a 252 CFM 120mm fan provide?

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The discussion revolves around calculating the lift thrust provided by a 120mm fan capable of 252 CFM for a miniature RC plane weighing approximately 15 pounds. Participants clarify that thrust must exceed the aircraft's weight for it to ascend, and they explore the relationship between airflow, fan size, and lift production. Calculations indicate that the fan's thrust is insufficient for the desired weight, emphasizing the need for a fan designed specifically for lift rather than airflow. The importance of factors like cross-sectional area, velocity, and blade design in generating effective thrust is highlighted. Ultimately, the consensus suggests that a more suitable fan type is necessary for the project to achieve flight.
  • #31
jarednjames said:
Yes but I'm currently 3 years into an aerospace engineering degree which is slightly concerning. Hence the embarrassment.

I have amended the above calcs to show require thrust for each fan to be 309.6 CFM.

Well at least you're on here doing calcs for random people. Its good practice, you'll learn from experienced people and it'll likely come in handy down the road. I just played video games all of my college days; didn't learn much except how to shoot people really well in fps games.
 
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  • #32
thehacker3 said:
309.6 and 252 seem to be comparable, especially when considering that the aircraft will weigh well under 10 pounds and will not require anywhere near as much thrust to sustain horizontal flight.

Also, CSA stands for cross-sectional area, right? If so, then wouldn't more blades = more CSA?

No, the CSA I'm referring to is the circular area the blades travel through. More blades = more drag plus more weight. Although in the case of this fan, it require such a large number to maintain airflow.

309 and 252 are not so comparable in that, given you require two of them, the 'missing' portion will amount to nearly 120 CFM. Which is the difference between flight or sink.
 
  • #33
Lazer57 said:
Well at least you're on here doing calcs for random people. Its good practice, you'll learn from experienced people and it'll likely come in handy down the road. I just played video games all of my college days; didn't learn much except how to shoot people really well in fps games.

I'll challenge you to a game of Crysis when I'm done building this thing :p
 
  • #34
jarednjames said:
No, the CSA I'm referring to is the circular area the blades travel through. More blades = more drag plus more weight. Although in the case of this fan, it require such a large number to maintain airflow.

309 and 252 are not so comparable in that, given you require two of them, the 'missing' portion will amount to nearly 120 CFM. Which is the difference between flight or sink.

That's about 1/6 of the airflow. If I get it under 8 pounds, wouldn't it hover?
 
  • #35
thehacker3 said:
309.6 and 252 seem to be comparable, especially when considering that the aircraft will weigh well under 10 pounds and will not require anywhere near as much thrust to sustain horizontal flight.

Also, CSA stands for cross-sectional area, right? If so, then wouldn't more blades = more CSA?

Well it better be well under 0.7 pounds if you want it to fly at all (if we're right about the thrust). And a difference of 57.6 CFM isn't comparable.
 
  • #36
thehacker3 said:
I'll challenge you to a game of Crysis when I'm done building this thing :p

I play that!

You want to start looking at model shops, they have ducted fans such as those you want (and look much closer to the picture you gave) and the prices are similar.

http://bestofferbuy.com/Ducted-2570...ource=gbase&utm_medium=cse&utm_campaign=gbase

They seem to quote it in grams of pull whatever that is, I'm assuming it means it will lift this many grams.

So this particular one will lift 1150g, not quite what you want, but certainly a good start.

Am I reading that page right? 41A of current?

It takes 550W. As you can see, it takes slightly more power than the standard PC fan (your original only required 48W).

And the RPM is only half of the computer fan.

Both of these fans are designed for different jobs. You can't expect the computer fan to be able to fly a helicopter. The ducted fan you are after needs to be designed for the job required, which is producing enough thrust for lift.
 
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  • #37
jarednjames said:
I play that!

You want to start looking at model shops, they have ducted fans such as those you want (and look much closer to the picture you gave) and the prices are similar.

http://bestofferbuy.com/Ducted-2570...ource=gbase&utm_medium=cse&utm_campaign=gbase

They seem to quote it in grams of pull whatever that is, I'm assuming it means it will lift this many grams.

So this particular one will lift 1150g, not quite what you want, but certainly a good start.

That one draws 41A and is less than 90mm in diameter.

http://www.mh-aerotools.de/airfoils/propuls4.htm

The first formula on that site is for Thrust in Newtons. How do I find out what delta V is? They say it's the additional velocity or acceleration by propellor but I don't know what that means.

We should seriously play some time btw, I've been looking for good competition. Anyone play SC2 by any chance too?
 
  • #39
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  • #40
v is the velocity of the air hitting the propeller, in your case, 0m/s as it is stationary.

delta v is the acceleration caused by the propeller, in your case it would be the flow velocity which is 10.something as we showed above.
 
  • #41
thehacker3 said:
I'll challenge you to a game of Crysis when I'm done building this thing :p

I haven't played much FPS games in the last 6 years, I have new vices that take precedence. Otherwise I would take you up on that offer.
 
  • #42
Interesting Lazer. According to the equation dragged up here (http://www.mh-aerotools.de/airfoils/propuls4.htm), the thrust of the original fan is only:

(3.1415 x 0.06^2) × (0 + (10.5130 / 2)) × 1.2 × 10.5130 = 0.75N

Simplified theirs shows:

Thrust = (A * (V / 2) * r) * V

Any ideas? I can't understand why they divide out delta v by 2. If it wasn't for that, our calculations match this one. Is it some sort of allowance for the internal area of the rotor not generating the same life as the outside?
 
  • #43
jarednjames said:
Interesting Lazer. According to the equation dragged up here (http://www.mh-aerotools.de/airfoils/propuls4.htm), the thrust of the original fan is only:

(3.1415 x 0.06^2) × (0 + (10.5130 / 2)) × 1.2 × 10.5130 = 0.75N

Simplified theirs shows:

Thrust = (A * (V / 2) * r) * V

Any ideas? I can't understand why they divide out delta v by 2. If it wasn't for that, our calculations match this one. Is it some sort of allowance for the internal area of the rotor not generating the same life as the outside?

I just did the math and got .75N

That's suspiciously low...
 
  • #44
It's not suspiciously low at all, it's realistic (assuming the /2 is correct).

Even assuming our calcs are best estimates, you wouldn't get more than 1.5N out of them (double the above 0.75N figure). Now until I know why they are dividing the V figure by 2 (I'd still go with it being margin of some sort), I can't tell you which is the correct figure. But either way, it isn't going to improve from 1.5N regardless of which calcs you do.
 
  • #45
jarednjames said:
It's not suspiciously low at all, it's realistic (assuming the /2 is correct).

Even assuming our calcs are best estimates, you wouldn't get more than 1.5N out of them (double the above 0.75N figure). Now until I know why they are dividing the V figure by 2 (I'd still go with it being margin of some sort), I can't tell you which is the correct figure. But either way, it isn't going to improve from 1.5N regardless of which calcs you do.

So far, to have a feasible lift / weight ratio, assuming our calcs correct at 1.5N per fan, your craft could weigh no more than around 1 - 1.4kg. Otherwise it won't lift off.

wouldn't it have to be from .1 - .15kg? Because g on Earth is around 10 so 1.5N / g = .15kg

That's why I'm saying it's suspiciously low. How does a 309 CFM achieve lift with a 10lb craft when a 252 can't even lift .15kg
 
  • #46
That 309 figure is wrong.

Basic maths now, to lift 10lbs = 4.54kg, it would take 44.5N. You need a fan which can supply half of that.

Which gives you 22.25N per fan.

If you want a 120mm fan, you would need a CFM of:

22.25 = (1.2 * 0.0113 * V) * V

(22.25 / (1.2 * 0.0113) = V^2 = 1640

V = 40.51m/s

Volume flow rate = 0.46 m^3/s = 969.88 ft^3/min

Which means you need 969.88 cubic feet per minute per fan to hover.
 
  • #47
I apologise for the mistakes, it is 3:13am where I am at the moment.
 
  • #48
Dude, don't apologize.. I can't begin to explain how much I'm grateful for what both of you are doing for me. Mistakes are just part of the process and I understand that.

At the moment, I'm working on a graph of CFM vs diameter to find the optimal values for both.
 
  • #49
Well the graph is based solely on the thrust equation but with a 120mm diameter, i would need a 1,100 CFM fan.

I decided to just go with two of these babies http://www.esky-heli.com/tower-pro-edf-fan-unit-74mm-257kv-860g-thrust-w-motor-p-6270.html

I'll build a small version of the plane first and work on creating my own custom fans in the meantime because none exist that match the specs I'm looking for (low profile, large diameter, high CFM)

Thanks again to everyone who helped me with all the calculations and guidance
 
  • #50
So, slightly different question, rather interested now, how do you plan to control it? Both fans are going to have rotation in the same direction. As I understand it, the only similar design to this is the Boeing Osprey which has counter-rotating props to eradicate yaw effects due to motor torque.

Note: Inverting a fan will not work as it will severely drop the thrust output of the fan and so destabilise the aircraft.

Also, what about motion control? Getting it to go forward, backward, left, right etc?
 
  • #51
pretty much like this



except, if possible, i want the wings to move on 2 axis (the other axis being like birds' wings)
 
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  • #53
I wonder if there's a real life, actually built RC version to see flying?

Still, they had a twin rotor design so that they canceled each others torque out. How will you achieve such a feat with only single prop engines?

If you get this thing operational I'd be really interested in seeing it fly (get a video on youtube).

Gonna be a pain getting those rotors to tilt.
 
  • #54
  • #55
jarednjames said:
Interesting Lazer. According to the equation dragged up here (http://www.mh-aerotools.de/airfoils/propuls4.htm), the thrust of the original fan is only:

(3.1415 x 0.06^2) × (0 + (10.5130 / 2)) × 1.2 × 10.5130 = 0.75N

Simplified theirs shows:

Thrust = (A * (V / 2) * r) * V

Any ideas? I can't understand why they divide out delta v by 2. If it wasn't for that, our calculations match this one. Is it some sort of allowance for the internal area of the rotor not generating the same life as the outside?

I read another forum and the answer is that the exit air velocity is divided by 2 because it is only twice of what the air velocity is going into the fan. As the fan pulls air in, it accelerates it and thus the velocity of the air going into the fan is half of the air velocity coming out of the fan. So the air velocity difference is NOT Vout - 0 even though the aircraft is hovering, it IS Vout - Vin and Vin is half of Vout due to the acceleration of the air as it pulls it in (that air is in front of the fan so you only count the velocity difference between the front and the back of the fan when considering thrust). Then you also get like 5% losses due to swirl as the site you link says, plus there are likely other losses. Yet, as I said before, the fan area doesn't include the center mounted motor cone, so that will give you a smaller area which will give you a larger air velocity, therefore:

T = 0.75N * 0.95 * [(Atotal)/(Atotal-Amotor)]

The 0.95 adjusts for the 5% losses due to swirl and the last portion adjusts for the actual area that air flows through which affects the air velocity.

i.e. if the motor takes half the area, then you will get double the air velocity because there is half the area for the air to flow through (which increases the velocity) and therefore double the thrust because the CFM remains the same since it is a factory measured/calculated value

To get the last part, look at the picture of the fan you're buying, measure it on screen size, get the total area and the motor area and plug them in (if you're doing it via screen size you'll have to measure the diameter and not use 120mm to get an accurate ratio).
 
  • #56
Ok, I can go with that, so we're looking at 0.75N per fan of thrust (minus losses).

As previously, a 10lb craft requires 44.5N of thrust simply to hover. So these fans are not able to do the job.
 
  • #57
Well you would still need to calculate the areas and include those in the final calculation. If the motor takes half the area, then it would be more like 1.4N per fan including losses. I'm guessing its probably close to 1N per fan without measuring anything.
 
  • #58
I've just done a spreadsheet so I don't have to keep doing the calcs by hand. I agree, as long as the CFM stays constant, the thrust increases with smaller diameter size.
 
  • #59
thehacker3 said:
I found a 120mm fan on the internet that is capable of pushing 252 CFM. How do I calculate how much weight a fan like that can lift? And at what rate?
I know I'm late and this thread moved fast, but the method I would use here is to use Bernoulli's equation to calculate the velocity pressure...

252 CFM is .1189 m^3/s and velocity through that fan is 10.5m/s

The density of air is 1.2 kg/m^3 and Bernoulli's equation is P=1/2 rho V^2 so P=66.4 N/sq m. So the thrust is 0.75N, as indicated in post #42. The equation in the link in that post combines Bernoulli's equation with the area of the fan.
 
  • #60
Lazer57 said:
I read another forum and the answer is that the exit air velocity is divided by 2 because it is only twice of what the air velocity is going into the fan. As the fan pulls air in, it accelerates it and thus the velocity of the air going into the fan is half of the air velocity coming out of the fan.
No. It comes from Bernoulli's equation and Bernoulli's equation is actually about kinetic energy. Note the similarity to the kinetic energy equation: 1/2 M V^2.
 

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