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

  1. Sep 15, 2010 #1
    Hey guys, I'm a college student looking to build a miniature RC plane based off this picture here: [PLAIN]http://www.kruxor.com/wp-content/uploads/2010/07/SC2_Banshee_Art.jpg [Broken]

    As you can see, the thrust needed to lift and mobilize the plane is provided by the two fans on the sides (does this make it a helicopter?)

    The model I want to build will be aproximately 3-4 feet long and 2 feet wide. I'm assuming that it will weigh approximately 15 pounds total.

    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'm only a freshman but I've taken a calculus based mechanics physics course so I should be able to understand any formulas you throw at me so don't hesitate to do so.

    Thank you all in advance,
    Leon
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 15, 2010 #2
    Thank you!

    That would be a total of roughly 1.7 ft lb / s from two fans. I understand that that means that if you had 1.7 pounds in the aircraft, it would take a second to rise one foot, but how would you know whether or not it would be enough to lift the aircraft off the ground at all? I'm sure that there's a cut-off thrust under which the plane wouldn't even budge from the ground.

    And a question about the math - how does the size of the fan (120 mm) affect the calculation? Like i see that you divided 4.2 by it but I don't understand why you did that.

    Thanks again guys
     
  4. Sep 15, 2010 #3
    Convert cubic ft per minute into cubic feet per second:
    252 CFM = 4.2 ft^3 / s

    Convert mm to ft:
    120mm = 0.3937 ft

    Which gives a rotor radius of:
    r = 0.1968 ft

    Convert volume airflow per second into flow velocity:
    4.2 / (0.1968^2 x Pi) = 34.5182 ft / s

    International standard for air:
    1 ft^3 of air (standard temp and pressure) = 0.0807lbs

    F = ((m1 x v1) - (m2 x v2)) / (t1 - t2)

    as initial velocity and time are zero, this gives:
    F = (- (m2 x v2)) / (- t1)

    Thrust:
    34.5182 x 0.0807 = 2.7856 ft lb / s

    I think, someone should probably check this just to be sure, but it seems right.

    THIS IS INCORRECT I MISSED DENSITY!
     
    Last edited: Sep 15, 2010
  5. Sep 15, 2010 #4
    Now THATTT is more like it!

    Thanks so much man, I appreciate it.

    So how much would the plane be able to lift off the ground? even if it's at a slow rate, it doesn't matter
     
  6. Sep 15, 2010 #5
    If thrust = weight then the helicopter will hover (or remain on the ground). The thrust has to be greater than the weight to allow it to rise.

    Thrust relies on the change in velocity of the air from 0m/s above the fan to 35 ft / s below the fan. Can you confirm the fan is passing 252 cubic ft per minute through it? That seems awfully high for such a small fan and implies a flow velocity of 24 mph.
     
  7. Sep 15, 2010 #6
    I still don't like my calculation, it seems wrong.
     
  8. Sep 15, 2010 #7
    Correct me if I'm wrong but doesn't 10 ft lb / s mean that if an object weight 10 pounds, it can rise at a rate of one foot per second? And if it is 20 pounds then it'll rise at the rate of one foot every 2 seconds..?

    Here's the link to the fan: http://www.ayagroup.com/product.php?productid=17263 [Broken]

    The calculations seem right, why do you think they're wrong?
     
    Last edited by a moderator: May 4, 2017
  9. Sep 15, 2010 #8
    If you aim the fan downwards and it supplies 10 ft lb / s then the reaction will be 10 ft lb / s. This equals the weight of the vehicle and so it will neither rise nor fall (in the air the helicopter would hover). You would have to supply 20 ft lb / in order to get the helicopter to rise at 1 ft per second.

    I still think there's something I'm missing in the calcs, something about the units ain't right.

    I'll have a look at the fan now.
     
    Last edited by a moderator: May 4, 2017
  10. Sep 15, 2010 #9
    Nono, the units match up perfectly! Have some confidence in your math ;)

    I just don't seem to have any belief in the concept that my physics professor taught us - any acceleration will provide lift or motion when ignoring friction.
     
  11. Sep 15, 2010 #10
    It will, but in this case, the friction would be the weight of the helicopter. You need to overcome that weight in order to lift off.

    If weight = lift you hover (or remain on the ground)

    If weight < lift you ascend

    If weight > lift you descend
     
  12. Sep 15, 2010 #11
    But the units you provided (Ft lb / s) imply a momentum, not a force. If there was a way to calculate the force of the fan, we'd be able to determine lift capacity.

    I'm in class right now so I can't concentrate on the math too hard but everything you've said so far has helped quite a bit - thank you for that; I think I have enough information now to do the calculations on my own.
     
  13. Sep 15, 2010 #12
    That's the problem, I gave momentum which I don't think is correct.

    I don't think that fan is suitable. That is a case fan designed for maximum airflow, not lift production.
     
  14. Sep 15, 2010 #13
    Airflow is what produces lift, though. Low airflow = weak lift. High airflow, however will almost always provide strong lift.

    When you say the fan is designed for max airflow and not lift production, can you define the difference between the two and the factors involved for each one? It's not like RPM vs. Torque here. We know how much air the fan can push at STP and I think that's all that's really important in these calculations.
     
  15. Sep 15, 2010 #14
    A coaxial fan with a large CSA turning at a low RPM can have a high airflow and produce no thrust.

    A small coaxial fan with a small CSA turning at a high RPM can have a high airflow and produce massive thrust.

    The amount of air passing through a fan is irrelevant to the lift it produces unless you include factors such as cross-sectional area, angle of attack, aerofoil design and RPM.

    The fan you are trying to use is specifically designed to draw as much air into / out of a computer case as possible. It has a large number of blades with incredibly high angles of attack and limited aerofoil. It has a low CSA.

    Ideally, you need a fan with a large CSA and a high RPM, with a relatively shallow angle of attack. There should be around 3 blades, with a good aerofoil designed for maximum lift production.
     
    Last edited: Sep 15, 2010
  16. Sep 15, 2010 #15
    new to forum and in a hurry so i'm gunna skip using the tools

    Area = 0.155 sq.ft
    4.2 cu.ft per second of flow
    velocity of air = 4.2cu.ft/s divided by 0.122sq.ft = 34.43ft/s
    air density = 0.0749lbm/cu.ft = 0.0024slugs/cu.ft
    mass flow rate = 4.2cu.ft/s times 0.0024slugs/cu.ft = 0.01 slugs/s
    Thrust = F = mass flow rate x velocity = 0.01slugs/s times 34.43ft/s = 0.3443lbf

    therefore with two fans you can hover a vehicle with a weight of 0.6886 lbs

    i'm rusty so if i'm wrong don't hesitate to point it out.
     
  17. Sep 15, 2010 #16
    Your last equation is wrong

    how could F = mass flow rate x velocity?

    F = kg * m / s^2

    according to your equation, F = kg * m / s * m / s which would be kg * m^2 / s^2
     
  18. Sep 15, 2010 #17
  19. Sep 15, 2010 #18
    thehacker3, if you lay a computer fan face down and apply the driving voltage of 12v, it doesn't hover. Therefore it cannot even provide enough thrust to lift it's own weight. You are looking to use the wrong type of fan for your design. I will have a look now for a more suitable fan type for you.

    I'm going to convert your initial values to metric so I can work with them and then give you a rough idea what thrust you need for the helicopter.
     
  20. Sep 15, 2010 #19
    Yes, but you have to keep in mind that most computer fans only push around 30-40 CFM each.

    Thank you, but you need not do research for me - just point me in the right direction!

    Needless to say, I appreciate you looking for a more suitable fan for me.

    After doing some more calculations, I figure that the helicopter shouldn't weigh more than 8 pounds.
     
  21. Sep 15, 2010 #20
    Thrust = Mass Flow Rate = r * V * A

    where r = 1.2kg/m^3

    Now with your original fan which is 0.12m diameter and 0.1189 m^3/s

    A = (0.06^2 * Pi) = 0.0113m^2

    V = (0.1189 / 0.0113) = 10.5130 m/s

    Mass Flow Rate = (1.2 * 0.0113 * 10.5130) = 0.1425 kg/s

    So with two fans you could hover 0.2850kgs.

    Now you have to bare in mind that this assumes you are achieving thrust from the full 120mm of fan blade, in reality, the central motor is taking a large portion of the centre of the fan so this figure is slightly less.

    Again, your airflow is not the only factor in lift production. The calculation above shows that you need a relatively large CSA and flow velocity. At the moment, the problem with these fans is that they don't have a large enough area.
     
  22. Sep 15, 2010 #21
    1 lbf = 1 slug x 1 ft/s^2 and that is what the units i used are; its a fluid dynamics equation.

    jarednjames latest calculation comes out the same as mine (just convert mine and you will see). i.e. with two fans, mine comes out to 0.3 kg
     
  23. Sep 15, 2010 #22
    Let's assume your helicopter weighs 10lb = 4.54kg.

    The minimum force from each fan is 2.27kg/s to get the helicopter to hover.

    If you insist on using such a small rotor, you would require a flow velocity of 12.94m/s or a volume flow rate of 0.14m^3/s (which converts to 309.6 cubic ft per minute) per fan.

    I somehow doubt any computer fans can push that mush air through them! That's why helicopters have such a large CSA for the rotors, you have to remember that the CSA increase as a square function, whereas the flow velocity is linear. A small change in rotor diameter can have a large effect on thrust, but it would take a much larger change in flow velocity to achieve the same effect.
     
    Last edited: Sep 15, 2010
  24. Sep 15, 2010 #23
    Wait, I just looked at this and noticed that you are saying that a fan has .1425 kg/s of thrust. Thrust is force, so it would be Newtons or pounds-force. I'll have to look over what you did but somethings not right. Maybe you got the same answer as me by accident.

    Also, you say that a smaller fan area (due to the motor) will cause less thrust. Actually it should be more because the CFM is the same and I get the velocity by dividing by area; so dividing by a smaller area will get a larger velocity and thus more thrust since we're multiplying by velocity to get the thrust.

     
  25. Sep 15, 2010 #24
    Why would it be wrong? mass flow units are kg/s and velocity is m/s that would give
    F=(Kg*m)/s2
     
  26. Sep 15, 2010 #25
    Thrust is the rate of change of mass with respect to time (mass flow rate of exhaust).

    The CSA is a square function, so a small change in that has to be countered by a much larger change in flow velocity. Remember, at the final stage where you calculate mass flow rate you multiply the values out so you need enough change in velocity to compensate for the area difference.

    mass flow rate = mass / time = ( (mass/length^3) * (mass / time) * (mass^2) ) = mass / time or kg/s

    I said something wasn't right with mine.

    If I multiply my mass flow rate value from the first calcs by velocity I get:

    1.5 kgm/s^2 or N of force

    So those two fans could supply 3N of lifting force.
     
    Last edited: Sep 15, 2010
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