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

AI Thread Summary
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.
  • #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.
 
  • #61
russ_watters said:
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.

Thank you russ, this has explained it.

Also, thank you for confirming the 0.75N value. (I'm just glad laser and myself got there eventually)
 
  • #62
russ_watters said:
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.

well that's good, it means there was no dependence on knowledge of some funky aspect of fans; it came down to properly using Bernoulli's equation.

i used to be afraid of being wrong, but the only way to improve and master things is by trying and learning from mistakes. thanks for the correction/information.
 
  • #63
last night i was too lazy to look at the NASA page (general thrust equation) i had previously linked to see why it didn't including the 1/2. it does include but not in a "thrust for idiots" type way. yet i looked at their page on propeller thrust and that one is totally a (propeller) "thrust for idiots" page, which really helped me understand. i trusted russ_watters answer, but i wanted an extended version. i figure there is a few others who may find it useful too. here is the link:

http://wright.nasa.gov/airplane/propth.html

btw, it does say that you use the blade length to calculate the area, so the fan in this topic will have a higher thrust than calculated. looking at typical ducted fans, the thrust will be 25% greater when taking the revised area into consideration. minus the 5% losses, it'll be about 20% greater thrust than calculated, so 0.9N per fan.
 
  • #65
Lazer57 said:
Keep it under 3 pounds if you want it to lift off the ground with two of those.

Yeah, I know. I can't imagine everything else in the aircraft weighing more than 3 pounds with those fans.
 
  • #66
What power source are you going to use? Weight considerations from batteries etc?
 

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